PST05106 Pharmaceutical Organic Chemistry – Complete Full Notes

NTA Level 5 • Semester 1 • PST05106

Pharmaceutical Organic Chemistry – Complete Full Notes

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UNITED REPUBLIC OF TANZANIA

[pic]

Ministry of Health, Community Development, Gender, Elderly and Children

Facilitator Guide

Copyright © Ministry of Health, Community Development, Gender, Elderly and

Children – 2018

Table of Contents

Table of Contents iii

Background iv

Acknowledgment v

Introduction vii

Abbreviations/Acronym xi

Session 1: Introduction to Pharmaceutical Organic Chemistry. 1

Session 2: Classification of Organic Compounds. 8

Session 3: Classification of Drugs According to Their Chemical Nature.

14

Session 4: Nomenclature of Organic Compounds. 17

Session 5: General Properties of Organic Compounds. 23

Session 6: Chemical Reaction in Organic Compounds. 32

Session 7: Isomerism. 38

Session 8: Alkanes of Pharmaceutical Importance. 48

Session 9: Alkenes of Pharmaceutical Importance. 57

Session 10: Alkynes of Pharmaceutical Importance. 73

Session 11: Alcohols of Pharmaceutical Importance. 84

Session 12: Carboxylic Acids of Pharmaceutical Importance. 101

Session 13: Esters of Pharmaceutical Importance. 114

Session 14: Acyl Chlorides of Pharmaceutical Importance. 122

Session 15: Ethers of Pharmaceutical importance. 128

Session 16: Aldehydes of Pharmaceutical Importance. 138

Session 17: Ketones of Pharmaceutical Importance. 148

Session 18: Aromatic Organic Compounds of Pharmaceutical Importance .

161

Session 19: Phenols of Pharmaceutical Importance. 175

Session 20: Aryl Halides of Pharmaceutical Importance. 188

Session 21: Amines of Pharmaceutical Importance. 200

Session 22: Amides of Pharmaceutical Importance. 216

Session 23: Introduction to Heterocyclic Compounds. 224

Session 24: Chemical Reactions of Heterocyclic Compounds. 234

Session 25: Introduction to Structure – Activity Relationship of Drugs.

243

Session 26: Structure – Activity Relationship of Penicillins. 247

Session 27: Structure – Activity Relationship of Cephalosporins. 257

Session 28: Structure – Activity Relationship of Quinolones. 264

Session 29: Structure – Activity Relationship of Sulphonamides. 271

Session 30: Structure – Activity Relationship of Aspirin. 279

Session 31: Structure – Activity Relationship of Paracetamol. 285

Session 32: Biotransformation of Medicinal Products. 290

Background

There is currently an ever-increasing demand for pharmaceutical personnel

in Tanzania. This is due to expanding investment in public and private

pharmaceutical sector. Shortage of trained pharmaceutical human resource

contributes to poor quality of pharmaceutical services and low access to

medicines in the country (GIZ, 2012).

Through Public-Private-Partnership (PPP) the Pharmacy Council (PC) together

with Development Partners (DPs) in Germany and Pharmaceutical Training

Institutions (PTIs) worked together to address the shortage of human

resource for pharmacy by designing a project named “Supporting Training

Institutions for Improved Pharmaceutical Services in Tanzania” in order to

improve quality and capacity of PTIs in training, particularly of lower

cadre pharmaceutical personnel.

The Pharmacy Council formed a Steering committee that conducted a

stakeholder’s workshop from18th – 22ndAugust 2014 in Morogoro to initiate

the implementation of the project.

Key activities in the implementation of this project included carrying out

situational analysis, curriculum review and harmonization, development of

training manual/facilitators guide, development of assessment plan,

training of trainers and supportive supervision.

After the curricula were reviewed and harmonized, the process of developing

standardised training materials started through Writer’s Workshop approach.

The approach included a number of workshops for developing, reviewing,

editing and formatting the sessions of the modules.

The goals of writer’s workshops were to build capacity of tutors in the

development of training materials and to develop high-quality, standardized

teaching materials.

The training package for pharmacy cadres includes a facilitator guide,

assessment plan and practicum. There are 11 modules for NTA level 5 making

11 facilitator guides including one practicum guide.

Acknowledgment

The development of standardized training materials of a competence-based

curriculum for pharmaceutical sciences has been accomplished through

involvement of different stakeholders.

Special thanks go to the Pharmacy Council for spearheading the

harmonization of training materials in the pharmacy after noticing that

training institutions in Tanzania were using different curricula and train

their students differently.

I would also like to extend my gratitude to Christian Social Service

Commission (CSSC) for their tireless efforts to mobilize funds from

development partners. Special thanks to Multi Actors Partnership (MAP)

for the financial and technical support.

Particular thanks are due to those who led this important process to its

completion, Centre for Education Development in Health Arusha (CEDHA), Ms.

Diana Gamuya, Mr. Dickson Mtalitinya and Members from the secretariat of

National Council for Technical Education (NACTE) for facilitating the

process.

Finally, I very much appreciate the contributions of the tutors and content

experts representing PTIs, hospitals, and other health training

institutions. Their participation in meetings and workshops, and their

input in the development of this training manual/facilitators guide have

been invaluable.

These participants are listed with our gratitude below:

Ms. Elizabeth Shekalaghe Registrar, Pharmacy Council of Tanzania

Dr. Catherine Jincen Principal, CEDHA

Dr. Saitore Laizer Deputy Principal, CEDHA

Dr. Sungwa N. Kabissi Project Manager – MAP, CSSC

Ms. Diana Gamuya CEDHA

Ms. Emily Mwakibolwa Pharmacy Council

Ms. Tumaini H. Lyombe MUHAS

Ms. Dilisi J. Makawia KSP

Director of Human Resources Development

Ministry of Health, Community Development, Gender, Elderly and Children

Introduction

Module Overview

This module content is a guide for tutors of Pharmaceutical schools for

training of students. The session contents are based on sub-enabling

outcomes and their related tasks of the curriculum for Basic Technician

Course in Pharmaceutical Sciences. The module sub-enabling outcomes and

their related tasks are as indicated in the Ordinary Technician Certificate

in Pharmaceutical Sciences (NTA Level 5) Curriculum.

Target Audience

This module is intended for use primarily by tutors of pharmaceutical

schools. The module’s sessions give guidance on the time, activities and

provide information on how to teach the session. The sessions include

different activities which focus on increasing students’ knowledge, skills

and attitudes.

Organization of the Module

The module consists of thirty-two (32) sessions; each session is divided

into several parts as indicated below:

• Session Title: The name of the session
• Total Session Time: The estimated time for teaching the session,

indicated in minutes

• Pre-requisites: A module or session which needs to be covered before

teaching the session.

• Learning Tasks:Statements which indicate what the student is expected to

learn by the end of the session.

• Resources Needed: All resources needed for the session are listed

including handouts and worksheets.

• Session Overview: The session overview box lists the steps, time for each

step, the activity or method used in each step and the step title.

• Session Content: All the session contents are divided into steps. Each

step has a heading and an estimated time to teach that step as shown in

the overview box. Also, this section includes instructions for the tutor

and activities with their instructions to be done during teaching of the

contents.

• Key Points: Key messages for concluding the session contents at the end

of a session. This step summarizes the main points and ideas from the

session, based on the learning tasks of the session.

• Evaluation: The last section of the session consists of short questions

based on the learning tasks to check the understanding of students.

• Handouts: Additional information which can be used in the classroom while

teaching or later for students’ further learning. Handouts are used to

provide extra information related to the session topic that cannot fit

into the session time. Handouts can be used by the students to study

material on their own and to refer to them after the session. Sometimes,

a handout will have questions or an exercise for the participants

including the answers to the questions.

• Instructions for Use and Facilitators Preparation.

o Tutors are expected to use the module as a guide to train students in

the classroom and skills laboratory.

o The contents of the modules are the basis for teaching and learning

Pharmaceutical Organic Chemistry.

o Use the session contents as a guide.

o The tutors are therefore advised to read each session and the relevant

hand-outs and worksheets as preparation before facilitating the

session.

o Tutors need to prepare all the resources, as indicated in the resource

section or any other item, for an effective teaching and learning

process.

o Plan a schedule (timetable) of the training activities.

o Facilitators are expected to be innovative to make the teaching and

learning process effective.

o Read the sessions before facilitation; make sure you understand the

contents in order to clarify points during facilitation.

o Time allocated is estimated, but you are advised to follow the time as

much as possible and adjust as needed.

o Use session activities and exercises suggested in the sessions as a

guide.

o Always involve students in their own learning. When students are

involved, they learn more effectively.

o Facilitators are encouraged to use real life examples to make learning

more realistic.

o Make use of appropriate reference materials and teaching resources

available locally.

3 Preparation with Hand-outs and Worksheets

o Go through the session and identify hand-outs and worksheets needed for

the session.

o Reproduce pages of these hand-outs and worksheets for student use while

teaching the session. This will enable students to refer to hand-outs and

worksheets during the session in the class. You can reproduce enough

copies for students or for sharing.

o Give clear instructions to students on the student activity in order for

the students to follow the instructions of the activity.

o Refer students to the specific page in the student manual as instructed

in the facilitator guide.

4 Using Students Manual When Teaching

o The student manual is a document which has the same content as the

facilitator guide, which excludes facilitator instructions and answers

for exercises.

o The student manual is for assisting students to learn effectively and

acts as a reference document during and after teaching the session.

o Some of the activities included in facilitator guide are in the student

manual without facilitator instructions.

Abbreviations/Acronym

CEDHA Centre for Educational Development in Health Arusha

COX Cyclooxygenase

CUHAS Catholic University of Health and Allied Sciences

ELCT Evangelical Lutheran Church in Tanzania

Giz Deutsche GesellschftFürInternationaleZusammenarbeit

HKMU Hurbert Kairuki Memorial University

JSI John Snow Inc

KSP Kilimanjaro School of Pharmacy

LZHRC Lake Zone Health Resource Centre

MoHCDGEC Ministry of Health, Community Development, Gender, Elderly and

Children

MUHAS Muhimbili University of Health and Allied Sciences

NACTE National Council for Technical Education

NTA National Technical Award

PTI Pharmaceutical Training Institution

RuCU Ruaha Catholic University

SAR Structure-Activity Relationship

SIBS Spring Institute of Business and Science

SLF Saint Luke Foundation

USP United States Pharmacopeia

Session 1: Introduction to Pharmaceutical Organic Chemistry.

Total Session Time: 120 minutes

Prerequisites: PST 04210 Inorganic Chemistry

Learning Tasks

By the end of this session students are expected to be able to:

• Define pharmaceutical organic chemistry
• List characteristics of organic compounds
• Explain the importance of organic chemistry in pharmacy

Resources Needed:

• Flip charts, marker pens, and masking tape.
• Black/white board and chalk/whiteboard markers.

SESSION OVERVIEW.

|Step |Time |Activity/ |Content |

| | |Method | |

|1 |05 minutes |Presentation |Introduction, Learning Tasks |

|2 |15 minutes |Buzzing. |Definition of Pharmaceutical Organic|

| | |Presentation. |Chemistry. |

|3 | |Brainstorming.| |

| |35 minutes |Presentation. |Characteristics of Organic |

| | | |Compounds. |

|4 |45 minutes |Group |Importance of Organic Chemistry in |

| | |discussion. |Pharmacy. |

| | |Presentation. | |

|5 |10minutes |Presentation |Key Points |

| 6 |10 minutes |Presentation |Evaluation |

SESSION CONTENT.

STEP 1: Presentation of Session Title and Learning Tasks (5 minutes)

READ or ASK students to read the learning tasks and clarify.

ASK students if they have any questions before continuing.

STEP 2: Definition of Pharmaceutical Organic Chemistry (15 minutes)

|Activity: Buzzing (5 minutes) |

| |

|Ask students to brainstorm on the following question: |

| |

|What is a pharmaceutical Organic Chemistry? |

| |

|ALLOW few students to respond. |

| |

|WRITE their responses on the flip chart/ board. |

| |

|CLARIFY and SUMMARISE by using the content below. |

• Organic Chemistry is the branch of Chemistry which deals with the study

of carbon and its compounds.

o However, carbon monoxide, carbon dioxide, carbonates, hydrogen

carbonates, carbides and cyanides are excluded.

• Pharmaceutical Organic Chemistry is the study of all substances

containing carbon which are involved with design, chemical synthesis and

development of bioactive molecules (drugs).

• Two aspects are involved

o Medicinal Chemistry – which is devoted to discovery and

development of new agents for treating diseases.

o Biochemistry – study of the cell & analysis of its structure to

identify factors necessary for life of every cell.

• Occurrence of Carbon and Sources of Organic Compounds.

o Carbon is the principle element in organic compounds; most also

contain hydrogen and others contain the halogens, nitrogen, oxygen,

phosphorus, sulphur etc.

o Occurs widely in carbohydrates, proteins, fats, vitamins, nucleic

acids, hormones and synthetic materials such as drugs, ink and

dyes.

o Major sources are petroleum, coal and natural gases.

o Other sources include wood, oils and agricultural waste products

o Biggest source of organic compounds – plants & animals

• The Unique Nature of Carbon

o Ability to catenate

▪ Carbon atoms link together to form chains of varying length,

branched chains and rings of different sizes

o Ability to form four strong single covalent bonds (tetravalency)

▪ Each carbon atom has four unpaired electrons when excited, as

a 4A element in the periodic table, it can share 4 valence

electrons – tetravalent, Tend to form four strong covalent bonds

o Ability to Form Multiple Bonds

▪ Carbon atom has the ability to form the multiple bonds links

with itself.

▪ There two categories of multiple bond (i) double bonds eg C=C

(ii) triple bonds eg C≡C.

▪ The compounds formed in this way have different properties from

compounds of the same elements but single bond.

o Carbon atom can vary in oxidation state from -4 to +4

CH4 Carbon oxidation number -4

CCl4 Carbon oxidation number +4

STEP 3: Characteristics of Organic Compounds (35 minutes).

|Activity: Brainstorming (10minutes) |

| |

|ASK students to pair up and buzz on the following question for 5 |

|minutes |

| |

|What are the characteristics of organic compounds? |

| |

|ALLOW pairs to respond on the question |

| |

|WRITE their response on the flip chart/board |

| |

|CLARIFY and SUMMARIZE by using the content in the table 1 below |

The following are the characteristics of organic compounds

• Covalent nature

o Carbon atom from strong covalent bonds with one another, most

organic compounds are stable because of the strong carbon -carbon

bonds, since they have a covalent nature they do not ionizes in

solution and are non-conductors of electricity.

• Polarity and solubility of non-polar compounds

o Carbon-hydrogen bonds are non-polar like carbon-carbon bonds; this

is because of the almost equal electro negativities of the two

elements.

o Most of the organic compounds are non-polar unless the compounds

consist of very electronegative elements like chlorine or groups

like the hydroxyl group, they cannot therefore form bonds with

water molecules (insoluble in water).

o If an organic compound contain polar groups hydrogen bond can form

between polar group in the molecule of organic compound and the

water molecule.

o For the example ethanol molecule contains a hydroxyl group which is

polar, so it is soluble in water.

• Low melting and boiling points

o Organic compounds generally have low melting and boiling points

than inorganic compounds; this is because these compounds possess

relatively weak intermolecular bonds which can easily broke by heat

energy.

• Thermal instability

o Many organic compounds are; thermally unstable, decomposing into

simpler molecules when heated to temperature above 5000C.

o However, this property is sometimes of commercial importance as in

the cracking of petroleum.

o This property is very useful in the fractional distillation of

crude oils.

• Flammability

o Most organic compounds are flammable (catch fire easily) and burn

exothermically in a plentiful supply of air to yield carbon (IV)

oxide and water.

o Thus, most fuels such as woods, coal, oil, petrol and natural gas

are organic, and their combustion provides our main source of heat

energy.

• Reactivity

o Reactions involving organic compounds tend to be much slower than

the ionic reactions commonly encountered in inorganic Chemistry.

o They usually require heating, thoroughly mixing and catalyst to

speed up the reaction.

STEP 4: Importance of Pharmaceutical Organic Chemistry (45 minutes).

|Activity: Small Group Discussion (20 minutes) |

| |

|DIVIDE students into small manageable groups. |

| |

|ASK students to discuss on the following question |

|What is the importance of Pharmaceutical Organic Chemistry? |

| |

|ALLOW students to discuss for 15 minutes. |

| |

|ALLOW few groups to present and the rest to add points not mentioned. |

| |

|CLARIFY and SUMMARIZE by using the contents below. |

The following are the importance of pharmaceutical organic chemistry

• Medicine discovery

o Many medicines (most of which are organic compounds) come from

natural source in which the right extraction will benefit human

health.

o Moreover, medicines from organic compounds tend to be safer

compared to medicines from laboratory experiments.

o Example cambogia from mangosteen that helps to treat cancer and

prevents cell damage.

• Support study of diseases

o Knowledge of Pharmaceutical Organic Chemistry can also help in

discovering the cause of disease since many diseases originate from

natural source.

o For example, in order to understand G.I.T problems scientist can

learn from the organic chemicals inside the organ such as acids and

amino acids as well as strange intrusion of the chemicals.

• Diagnosis of Diseases

o This can be done in various ways, for example the study of

chemicals within brain will help to identify any disturbance inside

the brain and hence help to diagnose the disease occurring within

the brain.

• Diet regulation

o It is important that the right medicine or treatment will help

human to have better health through the attachment of proper

patterns of diet, example the amount of carbohydrate, protein,

mineral, and the level of acid in the body as well as the reaction

between two chemicals in the body helps to determine the food

intake for daily consumption and regulation of the diet based on

the balance of chemical in the body.

• Preparations of cleansing agent

o In industries and laboratories organic solvents are widely used to

clear impurities. For example, in the drug extraction from plants,

the fatty matter from the pulp is removed using petroleum ether.

o Thus, organic chemistry through its knowledge of polarity,

solubility, partition factors uses solvents to separate components

for better use.

• Sterilizing agents

o Most of the sterilizing agents and disinfectants like phenol,

formaldehyde etc. are carbon compound.

o Due to their properties like solubility, pH, they can kill microbes

and even human body cells.

• Analysis of substances

o Most substances we use like drugs, pesticides, etc. are analysed

qualitatively and quantitatively using different types of

titrations, chromatograph techniques, and spectrophotometry.

o Here the reagents used like acids or bases or reductive oxidative

species is organic in nature, further, the end point indicators in

titration are developed by organic chemistry.

STEP 5: Key Points (10 minutes)

• Organic Chemistry is the branch of Chemistry which deals with the

study of carbon and its compounds

• Organic compounds form covalent bonds and most of them are non-polar,

having low boiling points and melting points compared to inorganic

compounds.

• Organic Chemistry is important as it helps in discovery of drugs and

other agents used in medical field, diagnosis and study of diseases.

STEP 6: Evaluation (10 minutes)

• What is a pharmaceutical organic chemistry?
• What are the characteristics of organic compounds?
• What is the importance of organic chemistry in pharmacy?

References.

Ternay, A.L (1976). Contemporary Organic Chemistry. Philadelphia, United

States: W.B. Saunders Co.

Morrison R.T and Boyd R N (1997). Organic Chemistry (6th Ed.). New Delhi,

India: Prentice Hall of India

Graham Solomon et al (2014). Organic Chemistry (11th Ed.). New Jeysey,

United States: John Willey and Sons.

Rama R. N. (2005). Principles of Pharmaceutical Organic Chemistry. New

Delhi, India: MacMillan Publishers

Bruice Y (2013). Organic Chemistry (7th ed.). New York, United States:

Prentice Hall Pearson.

Delgado J. N. Et al (1998). Wilson and Gisvold's Textbook of Organic

Medicinal and Pharmaceutical Chemistry (10th Ed.). California, United

States: Lippincott Williams

Bhassin S.K, Gupta R.(2013). Pharmaceutical organic chemistry (E-book

Kindle edition). New Delhi, India: Elsevier Publishing Services

Session 2: Classification of Organic Compounds.

Total Session Time: 60 minutes

Prerequisites

• None

Learning Tasks

By the end of this session students are expected to be able to:

• Classify organic compounds

Resources Needed:

• Flip charts, marker pens, and masking tape.
• Black/white board and chalk/whiteboard markers.

SESSION OVERVIEW

|Step |Time |Activity/ |Content |

| | |Method | |

|1 |05 minutes |Presentation |Introduction, Learning Tasks |

|2 |40 minutes |Presentation |Classification of Organic compounds |

| | |Group | |

| | |Discussion | |

|3 |05 minutes |Presentation |Key Points |

| 4 |10 minutes |Presentation |Evaluation |

SESSION CONTENTS

STEP 1: Presentation of Session Title and Learning Tasks (5 minutes)

READ or ASK students to read the learning tasks and clarify.

ASK students if they have any questions before continuing.

STEP 2: Classification of Organic Compounds (40 minutes)

| |

|Activity: Small Group Discussion (20 minutes). |

| |

|DIVIDE students into small manageable groups. |

| |

|ASK students to discuss on the following question. |

|How do we classify Organic Compounds? |

| |

|ALLOW students to discuss for 15 minutes. |

| |

|ALLOW few groups to present and the rest to add points not mentioned. |

| |

|CLARIFY and SUMMARIZE by using the contents below |

• Organic compounds are classified as follows:

o Acyclic or open chain compounds,

o Alicyclic or closed chain or ring compounds,

o Aromatic compounds and

o Heterocyclic aromatic compounds

Acyclic or open chain compounds:

o These compounds are also known as aliphatic compounds, they have

branched or straight chains. Following are the examples in this

category.

[pic]

• Alicyclic or closed chain or ring compounds:

o These are cyclic compounds which contain carbon atoms connected to

each other in a ring (homocyclic).

o When atoms other than carbon are also present then it is called as

heterocyclic. Examples of this type are as follows:

[pic]

• Aromatic compounds

o They are a special type of compounds which contain benzene and other

ring related compounds.

o Similar to alicyclic, they can also have heteroatoms in the ring.

o Such compounds are called as heterocyclic aromatic compounds.

o Some of the examples are as follows:

▪ Benzenoid aromatic compounds

[pic]

▪ Non-benzenoid aromatic compounds

[pic]

Eg Tropolone

• Heterocyclic aromatic compounds

[pic]

• Organic compounds are also classified as saturated and unsaturated

hydrocarbons.

o Hydrocarbons are compounds containing carbon and hydrogen

o Saturated hydrocarbons are those in which adjacent carbon atoms are

joined by a single covalent bond and all other bonds are satisfied

by hydrogen.

o Unsaturated hydrocarbons have at least two carbon atoms that are

joined by more than one covalent bond and all remaining bonds are

satisfied by hydrogen.

[pic] [pic]

Fig.1. Diagrammatic Classification of Organic Compounds

[pic]

STEP 3: Key Points (05 minutes).

• Organic compounds are classified as Acyclic or open chain compounds,

Alicyclic or closed chain or ring compounds, Aromatic compounds and

Heterocyclic aromatic compounds.

• Organic compounds are also classified as saturated and unsaturated

hydrocarbons.

STEP 4: Evaluation (10 minutes).

• What are Acyclic or open chain compounds?
• What are Alicyclic or closed chain or ring compounds?
• What are Aromatic compounds?
• What are Heterocyclic aromatic compounds?
• What are saturated and unsaturated hydrocarbons?

References

Ternay, A.L (1976). Contemporary Organic Chemistry. Philadelphia, United

States: W.B. Saunders Co.

Morrison R.T and Boyd R N (1997). Organic Chemistry (6th Ed.). New Delhi,

India: Prentice Hall of India

Graham Solomon et al (2014). Organic Chemistry (11th Ed.). New Jeysey,

United States: John Willey and Sons.

Rama R. N. (2005). Principles of Pharmaceutical Organic Chemistry. New

Delhi, India: MacMillan Publishers

Bruice Y (2013). Organic Chemistry (7th ed.). New York, United States:

Prentice Hall Pearson.

Delgado J. N. Et al (1998). Wilson and Gisvold's Textbook of Organic

Medicinal and Pharmaceutical Chemistry (10th Ed.). California, United

States: Lippincott Williams

Bhassin S.K, Gupta R.(2013). Pharmaceutical organic chemistry (E-book

Kindle edition). New Delhi, India: Elsevier Publishing Services

Session 3: Classification of Drugs According to Their Chemical Nature.

Total Session Time: 60 minutes

Prerequisites

• None

Learning Tasks

By the end of this session students are expected to be able to:

• Classify drugs according to their chemical nature

Resources Needed:

• Flip charts, marker pens, and masking tape.
• Black/white board and chalk/whiteboard markers.

SESSION OVERVIEW

|Step |Time |Activity/ |Content |

| | |Method | |

|1 |05 minutes |Presentation |Introduction, Learning Tasks |

|2 | |Presentation |Classification of Drugs According to|

| |45 minutes |Small group |Their Chemical Nature |

| | |discussion | |

|3 |05 minutes |Presentation |Key Points |

| 4 | |Presentation |Evaluation |

| |05 minutes | | |

SESSION CONTENTS

STEP 1: Presentation of Session Title and Learning Tasks (5 minutes).

READ or ASK students to read the learning tasks and clarify

ASK students if they have any questions before continuing.

STEP 2: Clasification of Drugs According to Their Chemical Nature (45

minutes).

|Activity: Small Group Discussion (20 minutes) |

| |

|DIVIDE students into small manageable groups |

| |

|ASK students to discuss on the following question |

|How do you classify drugs according to their chemical nature? |

| |

|ALLOW students to discuss for 15 minutes |

| |

|ALLOW few groups to present and the rest to add points not mentioned |

| |

|CLARIFY and SUMMARIZE by using the contents below |

Chemically, drugs are classified as follows:

• Inorganic drugs

This includes:

o Metals and their salts (ferrous sulphate, zinc sulphate and magnesium

sulphate)

o Non-metals such as sulphur

• Organic drugs

They include the following:

o Alkaloids; examples are atropine, strychnine and morphine

o Glycosides; examples are digitoxin and digoxin

o Proteins; examples are oxytocin and insulin

o Esters, amides, alcohols, glycerides, carboxylic acids, phenols

STEP 4: Key Points (5 minutes).

• Drugs are classified chemically as Inorganic and organic drugs.
• Inorganic drugs include metal and their salts and non-metals.
• Organic drugs include alkaloids, glycosides, proteins, esters and amides

STEP 5: Evaluation (5 minutes).

• How are drugs classified according to their chemical nature?

References

Ternay, A.L (1976). Contemporary Organic Chemistry. Philadelphia, United

States: W.B. Saunders Co.

Morrison R.T and Boyd R N (1997). Organic Chemistry (6th Ed.). New Delhi,

India: Prentice Hall of India

Graham Solomon et al (2014). Organic Chemistry (11th Ed.). New Jeysey,

United States: John Willey and Sons.

Rama R. N. (2005). Principles of Pharmaceutical Organic Chemistry. New

Delhi, India: MacMillan Publishers

Bruice Y (2013). Organic Chemistry (7th ed.). New York, United States:

Prentice Hall Pearson.

Delgado J. N. Et al (1998). Wilson and Gisvold's Textbook of Organic

Medicinal and Pharmaceutical Chemistry (10th Ed.). Calfornia, United

States: Lippincott Williams

Bhassin S.K, Gupta R.(2013). Pharmaceutical organic chemistry (E-book

Kindle edition). New Delhi, India: Elsevier Publishing Services

Session 4: Nomenclature of Organic Compounds.

Total Session Time: 120 minutes

Prerequisites

• None

Learning Tasks

By the end of this session students are expected to be able to:

• Define nomenclature of organic compounds
• Explain nomenclature of organic compounds

Resources Needed:

• Flip charts, marker pens, and masking tape.
• Black/white board and chalk/whiteboard markers.

SESSION OVERVIEW

|Step |Time |Activity/ |Content |

| | |Method | |

|1 |05 minutes |Presentation |Introduction, Learning Tasks |

|2 |10 minutes |Buzzing |Definition of Nomenclature of |

| | |Presentation |Organic Compounds |

|3 |85 minutes |Group |Nomenclature of organic compounds |

| | |discussion | |

| | |Presentation | |

|4 |10 minutes |Presentation |Key Points |

| 5 |10 minutes |Presentation |Evaluation |

SESSION CONTENTS

STEP 1: Presentation of Session Title and Learning Tasks (5 minutes)

READ or ASK students to read the learning tasks and clarify.

ASK students if they have any questions before continuing.

STEP 2: Definition of Nomenclature of Organic Compounds (10 Minutes).

|Activity: Buzzing (5 minutes) |

| |

|Ask students to brainstorm on the following question: |

| |

|What is nomenclature of organic compounds? |

| |

|ALLOW few students to respond |

| |

|WRITE their responses on the flip chart/ board |

| |

|CLARIFY and SUMMARISE by using the content below |

• Nomenclature is the act or a system of naming.
• IUPAC nomenclature of organic chemistry is a systematic method of

naming organic chemical compounds as recommended by the International

Union of Pure and Applied Chemistry (IUPAC).

STEP 3: Nomenclature of organic compounds (85 minutes).

|Activity: Small Group Discussion (20 minutes). |

| |

|DIVIDE students into small manageable groups. |

| |

|ASK students to discuss on the following question |

|What are the rules for naming organic compounds? |

| |

|ALLOW students to discuss for 15 minutes. |

| |

|ALLOW few groups to present and the rest to add points not mentioned. |

| |

|CLARIFY and SUMMARIZE by using the contents below. |

The following are the rules that are followed in naming organic

compounds

1. Identification of the parent chain. This chain must obey the following

rules, in order of precedence:

i. It should have the maximum number of substituents of the suffix

functional group. By suffix, it is meant that the parent functional

group should have a suffix, unlike halogen substituents. If more

than one functional group is present, the one with highest

precedence should be used.

ii. It should have the maximum number of multiple bonds.
iii. It should have the maximum number of single bonds.
iv. It should have the maximum length.

2. Identification of the parent functional group, if any, with the

highest order of precedence.

3. Identification of the side-chains. Side chains are the carbon chains

that are not in the parent chain but are branched off from it.

4. Identification of the remaining functional groups, if any, and naming

them by their ionic prefixes (such as hydroxy for -OH, oxy for =O,

oxyalkane for O-R, etc.).

Different side-chains and functional groups will be grouped together

in alphabetical order. (The prefixes di-, tri-, etc. are not taken

into consideration for grouping alphabetically. For example, ethyl

comes before dihydroxy or dimethyl, as the "e" in "ethyl" precedes the

"h" in "dihydroxy" and the "m" in "dimethyl" alphabetically. The "di"

is not considered in either case). When both side chains and secondary

functional groups are present, they should be written mixed together

in one group rather than in two separate groups.

5. Identification of double/triple bonds.

6. Numbering of the chain. This is done by first numbering the chain in

both directions (left to right and right to left), and then choosing

the numbering which follows these rules, in order of precedence

i. Has the lowest-numbered locant (or locants) for the suffix

functional group. Locants are the numbers on the carbons to which

the substituent is directly attached.

ii. Has the lowest-numbered locants for multiple bonds (The locant of a

multiple bond is the number of the adjacent carbon with a lower

number).

iii. Has the lowest-numbered locants for prefixes.

7. Numbering of the various substituents and bonds with their locants. If

there is more than one of the same type of substituent/double bond, a

prefix is added showing how many there are ( di – 2 tri – 3 tetra – 4

then as for the number of carbons below with 'a' added)

• The numbers for that type of side chain will be grouped in ascending

order and written before the name of the side-chain. If there are two

side-chains with the same alpha carbon, the number will be written

twice. Example: 2,2,3-trimethyl- . If there are both double bonds and

triple bonds, "en" (double bond) is written before "yne" (triple

bond).

• When the main functional group is a terminal functional group (a group

which can exist only at the end of a chain, like formyl and carboxyl

groups), there is no need to number it.

1. Arrangement in this form: Group of side chains and secondary

functional groups with numbers made in step 3 + prefix of parent

hydrocarbon chain (eth, meth) + double/triple bonds with numbers (or

"ane") + primary functional group suffix with numbers.

Wherever it says "with numbers", it is understood that between the

word and the numbers, the prefix(di-, tri-) is used.

2. Adding of punctuation:

i. Commas are put between numbers (2 5 5 becomes 2,5,5)
ii. Hyphens are put between a number and a letter (2 5 5

trimethylheptane becomes 2,5,5-trimethylheptane)

iii. Successive words are merged into one word (trimethyl heptane

becomes trimethylheptane)

Note: IUPAC uses one-word names throughout. This is why all

parts are connected.

Example

Here is a sample molecule with the parent carbons numbered:

[pic]

For simplicity, here is an image of the same molecule, where the hydrogens

in the parent chain are removed and the carbons are shown by their numbers:

[pic]

Now, following the above steps:

1. The parent hydrocarbon chain has 23 carbons. It is called tricosa-.

2. The functional groups with the highest precedence are the two ketone

groups.

i. The groups are on carbon atoms 3 and 9. As there are two, we

write 3,9-dione.

ii. The numbering of the molecule is based on the ketone groups.

When numbering from left to right, the ketone groups are

numbered 3 and 9. When numbering from right to left, the ketone

groups are numbered 15 and 21. 3 is less than 15, therefore the

ketones are numbered 3 and 9. The smaller number is always used,

not the sum of the constituent’s numbers.

3. The side chains are: an ethyl- at carbon 4, an ethyl- at carbon 8, and

a butyl- at carbon 12.

Note:The -O-CH3 at carbon atom 15 is not a side chain, but it is a

methoxy functional group.

I. There are two ethyl- groups. They are combined to create, 4,8-

diethyl.

II. The side chains are grouped like this: 12-butyl-4,8-diethyl.

(But this is not necessarily the final grouping, as functional

groups may be added in between to ensure all groups are listed

alphabetically.)

1. The secondary functional groups are: a hydroxy- at carbon 5, a chloro-

at carbon 11, a methoxy- at carbon 15, and a bromo- at carbon 18.

Grouped with the side chains, this gives 18-bromo-12-butyl-11-chloro-

4,8-diethyl-5-hydroxy-15-methoxy

2. There are two double bonds: one between carbons 6 and 7, and one

between carbons 13 and 14. They would be called "6,13-diene", but the

presence of alkynes switches it to 6,13-dien. There is one triple bond

between carbon atoms 19 and 20. It will be called 19-yne.

3. The arrangement (with punctuation) is: 18-bromo-12-butyl-11-chloro-4,8-

diethyl-5-hydroxy-15-methoxytricosa-6,13-dien-19-yne-3,9-dione

4. Finally, due to Cis-trans isomerism, we have to specify the relative

orientation of functional groups around each double bond. For this

example, we have (6E,13E)

The final name is (6E,13E)-18-bromo-12-butyl-11-chloro-4,8-diethyl-5-

hydroxy-15-methoxytricosa-6,13-dien-19-yne-3,9-dione.

STEP 3: Key Points (5minutes)

• The system of naming drugs is called Nomenclature.
• Drugs are named according to IUPAC nomenclature of organic chemistry

which is a systematic method of naming organic chemical compounds as

recommended by the International Union of Pure and Applied Chemistry

(IUPAC).

STEP 4: Evaluation (5 minutes)

• What is nomenclature?
• What is IUPAC nomenclature of organic chemistry?
• What are the rules that are followed in naming organic compounds?

References

Ternay, A.L (1976). Contemporary Organic Chemistry. Philadelphia, United

States: W.B. Saunders Co.

Morrison R.T and Boyd R N (1997). Organic Chemistry (6th Ed.). New Delhi,

India: Prentice Hall of India

Graham Solomon et al (2014). Organic Chemistry (11th Ed.). New Jeysey,

United States: John Willey and Sons.

Rama R. N. (2005). Principles of Pharmaceutical Organic Chemistry. New

Delhi, India: MacMillan Publishers

Bruice Y (2013). Organic Chemistry (7th ed.). New York, United States:

Prentice Hall Pearson.

Delgado J. N. Et al (1998). Wilson and Gisvold's Textbook of Organic

Medicinal and Pharmaceutical Chemistry (10th Ed.). Calfornia, United

States: Lippincott Williams

Bhassin S.K, Gupta R.(2013). Pharmaceutical organic chemistry (E-book

Kindle edition). New Delhi, India: Elsevier Publishing Services

Session 5: General Properties of Organic Compounds.

Total Session Time: 120 minutes + 10 minutes home assignment.

Prerequisites

None

Learning Tasks

By the end of this session students are expected to be able to:

• List physical properties of organic compounds
• Explain the general properties of organic compounds

Resources Needed:

• Flip charts, marker pens, and masking tape.
• Black/white board and chalk/whiteboard markers.

SESSION OVERVIEW

|Step |Time |Activity/ |Content |

| | |Method | |

|1 |05 minutes |Presentation |Introduction, Learning Tasks |

|2 |25 minutes |Buzzing |Physical Properties of Organic |

| | |Presentation |Compounds |

|3 |60 minutes |Small group |General properties of Organic |

| | |discussion |Compounds |

| | |Presentation | |

|4 |10 minutes |Presentation |Key Points |

|5 | |Presentation |Evaluation |

| |10 minutes | | |

|6 |10 minutes |Presentation |Take Home Assignment |

SESSION CONTENTS.

STEP 1: Presentation of Session Title and Learning Tasks (5 minutes).

READ or ASK students to read the learning tasks and clarify.

ASK students if they have any questions before continuing.

STEP 2: Physical Properties of Organic Compounds (25 minutes).

|Activity: Buzzing (10minutes) |

| |

|ASK students to pair up and buzz on the following question for 5 |

|minutes. |

| |

|What are the physical properties of organic compounds? |

| |

|ALLOW pairs to respond on the question. |

| |

|WRITE their response on the flip chart/board. |

| |

|CLARIFY and SUMMARIZE by using the content in the table 1 below. |

The following are the physical properties of organic compounds;

Melting Point.

• It usually indicates the temperature in which a state of a compound

changes from solid to liquid state.

• There are few factors that affect the melting point such as:

o Size of a molecule:

▪ Melting Point identifies the characteristics of an organic

compound.

▪ Two different compounds consisting of a variant structural

arrangement of atoms or possess different configurations will have

difference of melting point.

▪ Two samples possessing same melting point will have same

configurations.

o Force of attraction between the molecules:

▪ Melting point of a compound is usually affected by the force of

attraction between the molecules.

▪ The existence of hydrogen bonds in organic compounds will result

to a higher melting point.

Boiling Point:

• Boiling Point varies depending on the surrounding environment.
• A boiling point of a liquid is high at high pressure and has a lower

boiling point when atmospheric pressure is low.

• Factors that affect boiling point and they are stated below.

o Polarity: Greater the polarity the higher the boiling point, that is,

polarity determines the force of attraction between the molecules.

Molecules are attracted by opposite charges in a polar compound.

o Carbon-carbon chain: Boiling point decreases with the increase in the

length of a carbon-carbon chain.

o Strength of Intermolecular forces: Various effects such as Vander

Waals dispersion hydrogen – bonding. Ionic bonding will affect the

strength of intermolecular forces.

Solubility

• Organic compounds may dissolve in solvents like mixture, ethyl

alcohol or white spirits.

Flammability and vapour pressure

• Flammability is a measure of how easy it would be for a substance to

catch alight and burn.

• When a substance is in the liquid or solid state there will be some

molecules in the gas state. The weaker the intermolecular forces within a

substance the higher the vapour pressure will be.

• Compounds with higher vapour pressures have lower flash points and are

therefore more flammable.

STEP 3: General properties of organic compounds (60 minutes).

|Activity: Small Group Discussion (20 minutes) |

| |

|DIVIDE students into small manageable groups. |

| |

|ASK students to discuss on the following question; |

|What are the general properties of organic compounds? |

| |

|ALLOW students to discuss for 15 minutes. |

| |

|ALLOW few groups to present and the rest to add points have not been |

|mentioned. |

| |

|CLARIFY and SUMMARIZE by using the contents below |

• An understanding of the various types of noncovalent forces allows us to

explain, on a molecular level, many observable physical properties of

organic compounds.

• Factors that influence physical properties of organic compound are:

o Intermolecular forces

o Type of function group

o Chain length

o Shape of the molecule

• Intermolecular forces are forces that exist between molecules. They

include;

o Van der waals forces –dipole-dipole forces

▪ -induced dipole-induced dipole forces (London dispersion

forces)

o Hydrogen bonding.

Flammability

• Flammability is a measure of how easy it would be for a substance to

catch alight and burn. The flash point of a substance is the lowest

temperature that is likely to form a gaseous mixture you could set

alight.

• If a liquid has a low enough flash point it is considered flammable (able

to be ignited easily) while those with higher flash points are considered

nonflammable.

• A substance that is classified as nonflammable can still be forced to

burn, but it will not ignite easily.

Vapor pressure

• When a substance is in the liquid or solid state there will be some

molecules in the gas state. These molecules have enough energy to

overcome the intermolecular forces holding the majority of the substance

in the liquid or solid phase

• These gas molecules exert a pressure on the liquid or solid (and the

container) and that pressure is the vapour pressure of that compound

• The weaker the intermolecular forces within a substance the higher the

vapour pressure will be

• Compounds with higher vapour pressures have lower flash points and are

therefore more flammable

Solubility

Solubility is a chemical property referring to the ability for a given

substance, the solute, to dissolve in a solvent.

• It is measured in terms of the maximum amount of solute dissolved in a

solvent at equilibrium. The resulting solution is called a saturated

solution.

Solubility of polar compounds in water

• When considering the solubility of an organic compound in a given

solvent, the most important issue to consider is how strong the

noncovalent interactions between the compound and the solvent molecules

are.

• If the solvent is polar, like water, then a smaller hydrocarbon component

and/or more charged, hydrogen bonding, and other polar groups will tend

to increase the solubility.

• If the solvent is non-polar, like hexane, then the exact opposite is

true.

• For example, table salt, or sodium chloride will dissolve almost

immediately in water, because water, as a very polar molecule, is able to

form many ion-dipole interactions with both the sodium cation and the

chloride anion, the energy from which is more than enough to make up for

energy required to break up the ion-ion interactions in the salt

crystal.

• The end result, then, is that in place of sodium chloride crystals, we

have individual sodium cations and chloride anions surrounded by water

molecules – the salt is now in solution.

• Charged species as a rule dissolve readily in water: in other words, they

are very hydrophilic (water-loving).

Solubility of non-polar compounds

• A compound called biphenyl, which in non-polar, will not dissolve in

water.

• This is because it is a very non-polar molecule, with only carbon-carbon

and carbon-hydrogen bonds.

o It is able to bond to itself very well through nonpolar van der Waals

interactions, but it is not able to form significant attractive

interactions with very polar solvent molecules like water.

o Thus, the energetic cost of breaking up the biphenyl-to-biphenyl

interactions in the solid is high, and very little is gained in terms

of new biphenyl-water interactions.

o Therefore, water is a terrible solvent for nonpolar hydrocarbon

molecules: they are very hydrophobic (water-fearing).

Solubility of alcohol in water

• Another example is alcohol compounds, starting with methanol (1 carbon)

and ending with octanol (8 carbons).

• Smaller alcohols – methanol, ethanol, and propanol – dissolve easily in

water, at any water/alcohol ratio that you try.

• This is because the water is able to form hydrogen bonds with the

hydroxyl group in these molecules, and the combined energy of formation

of these water-alcohol hydrogen bonds is more than enough to make up for

the energy that is lost when the alcohol-alcohol (and water-water)

hydrogen bonds are broken up.

• However, butanol is only sparingly soluble in water
• The longer-chain alcohols – pentanol, hexanol, heptanol, and octanol –

are increasingly non-soluble in water.

• This is because the larger alcohols have larger nonpolar, hydrophobic

regions in addition to their hydrophilic hydroxyl group.

• At about four or five carbons, the influence of the hydrophobic part of

the molecule begins to overcome that of the hydrophilic part, and water

solubility is lost.

Boiling point and melting point

• The observable melting and boiling points of different organic molecules

provides an additional illustration of the effects of noncovalent

interactions.

• Melting and boiling are processes in which noncovalent interactions

between identical molecules in a pure sample are disrupted.

• The stronger the noncovalent interactions, the more energy that is

required, in the form of heat, to break them apart

• As a rule, larger molecules have higher boiling (and melting) points.

Butane versus Octane

• Consider the boiling points of increasingly small versus larger

hydrocarbons.

• More carbons and hydrogens mean a greater surface area possible for van

der Waals interaction, and thus higher boiling points.

• Below zero degrees centigrade (and at atmospheric pressure) butane is a

liquid, because the butane molecules are held together by Van der Waals

forces.

• Above zero degrees, however, the molecules gain enough thermal energy to

break apart and enter the gas phase.

• Octane, in contrast, remains in the liquid phase all the way up to 128oC,

due to the increased van der Waals interactions made possible by the

larger surface area of the individual molecules.

• The strength of intermolecular hydrogen bonding and dipole-dipole

interactions is reflected in higher boiling points.

• Look at the trend for hexane (van der Waals interactions only), 3-

hexanone (dipole-dipole interactions), and 3-hexanol (hydrogen bonding).

• In all three molecules, van der Waals interactions are significant.
• The polar ketone group allows 3-hexanone to form intermolecular dipole-

dipole interactions, in addition to the weaker van der Waals

interactions. 3-hexanol, because of its hydroxyl group, is able to form

intermolecular hydrogen bonds, which are stronger yet.

• The effect of hydrogen bonding in water of particular interest.
• Because it is able to form tight networks of intermolecular hydrogen

bonds, water remains in the liquid phase at temperatures up to 100 OC

despite its small size.

STEP 4: Key Points (10 minutes).

• The physical properties of organic compounds include melting point,

boiling points, Solubility, flammability and vapor pressure

• When considering the solubility of an organic compound in a given

solvent, the most important issue to consider is how strong are the

noncovalent interactions between the compound and the solvent molecules

• Melting and boiling are processes in which noncovalent interactions

between identical molecules in a pure sample are disrupted.

STEP 5: Evaluation (10 minutes).

• What are physical properties of organic compounds?
• What is the solubility of polar compounds in water?
• What is the effect of larger molecules on melting and boiling points?

STEP 6: Take Home Assignment (10 minutes)

|Activity: Take home Assignment (10 minutes) |

| |

|DIVIDE students in groups or individual. |

| |

|ASK the students to work on the following assignment |

| |

|Write short notes on the solubility of glucose, benzoic acid and |

|acetic acid in water. |

| |

|ALLOCATE time for students to do the assignment and submit. |

| |

|REFER students to recommended references |

References

Ternay, A.L (1976). Contemporary Organic Chemistry. Philadelphia, United

States: W.B. Saunders Co.

Morrison R.T and Boyd R N (1997). Organic Chemistry (6th Ed.). New Delhi,

India: Prentice Hall of India

Graham Solomon et al (2014). Organic Chemistry (11th Ed.). New Jeysey,

United States: John Willey and Sons.

Rama R. N. (2005). Principles of Pharmaceutical Organic Chemistry. New

Delhi, India: MacMillan Publishers

Bruice Y (2013). Organic Chemistry (7th ed.). New York, United States:

Prentice Hall Pearson.

Delgado J. N. Et al (1998). Wilson and Gisvold's Textbook of Organic

Medicinal and Pharmaceutical Chemistry (10th Ed.). California, United

States: Lippincott Williams.

Bhassin S.K, Gupta R.(2013). Pharmaceutical organic chemistry (E-book

Kindle edition). New Delhi, India: Elsevier Publishing Services.

Session 6: Chemical Reaction in Organic Compounds.

Total Session Time: 60 minutes

Prerequisites

• None

Students Learning Tasks

By the end of this session students are expected to be able to:

• Define term chemical reaction
• List types of chemical reactions in organic compounds
• Explain chemical reactions in organic compounds

Resources Needed:

• Flip charts, marker pens, and masking tape.
• Black/white board and chalk/whiteboard markers.

SESSION OVERVIEW

|Step |Time |Activity/ |Content |

| | |Method | |

|1 |05 minutes |Presentation |Introduction, Learning Tasks |

|2 |15 minutes |Presentation |Definition of the term chemical |

| | |Buzzing |reaction |

|3 | |Group |Types of chemical reactions |

| |30 minutes |discussion |involving organic compounds |

| | |Presentation | |

|4 |05 minutes |Presentation |Key Points |

|5 | | Presentation |Evaluation |

| |05 minutes | | |

CONTENTS

STEP 1: Presentation of Session Title and Learning Tasks (5 minutes)

READ or ASK students to read the learning tasks and clarify.

ASK students if they have any questions before continuing.

STEP 2: Definition of Chemical Reaction (15 minutes)

|Activity: Buzzing (10minutes) |

| |

|ASK students to pair up and buzz on the following question for 5 |

|minutes |

| |

|What is chemical reaction? |

| |

|ALLOW pairs to respond on the question |

| |

|WRITE their response on the flip chart/board |

| |

|CLARIFY and SUMMARIZE by using the content in the table 1 below |

Definition

• A chemical reaction is process in which a substance is transformed into a

new substance through chemical change, OR

• Chemical reactions, a process in which one or more substances, the

reactants, are converted to one or more different substances, the

products, substances are either chemical elements or compounds.

• A chemical reaction rearranges the constituent atoms of the reactants to

create different substances as products.

STEP 3: Types of Chemical Reactions Involving Organic Compounds (30

minutes).

|Activity: Small Group Discussion (20 minutes) |

| |

|DIVIDE students into small manageable groups. |

|ASK students to discuss on the following question. |

| |

|What are the types of chemical reactions involving organic compounds? |

| |

|ALLOW students to discuss for 15 minutes |

|ALLOW few groups to present for 5 minutes and the rest to add points |

|not mentioned. |

| |

|CLARIFY and SUMMARIZE by using the contents below |

There are five main types of organic reactions that can take place. They

are as follows:

• Substitution reactions
• Elimination reactions
• Addition reactions
• Radical reactions
• Oxidation-Reduction Reactions

Let us study each of these reactions in detail, to understand more about

them.

[pic]

• Substitution Reactions

o In a substitution reaction, one atom or a group of atoms take place of

another atom or a group of atoms which leads to the formation of an

altogether new substance.

o We can take an example of C – Cl bond, in which the carbon atom

usually has a partial positive charge due to the presence of highly

electronegative chlorine atoms.

o In a nucleophilic substitution reaction, it is important that the

nucleophile must have a pair of electrons and it also should have a

high affinity for the electropositive species in comparison to the

substituent which was originally present in the element.

o In order for the substitution reaction to occur, there are certain

conditions that have to be present such as maintaining low

temperatures same as room temperature.

• Elimination Reactions

o These are reactions which involve the elimination and removal of the

adjacent atoms.

o After these multiple bonds are simultaneously formed and there is a

release of small molecules as product.

o One of the examples of a typical elimination reaction is the

conversion of ethyl chloride to ethylene.

o CH3CH2Cl → CH2= CH2 + HCl

o In the above reaction, the eliminated molecule is HCl, which can form

out of the combination of H+ from the carbon atom which is on the left

side and Cl– from the carbon atom which is on the right side.

• Addition Reactions

o An addition reaction is simply just the opposite of an elimination

reaction.

o In an addition reaction, the components or molecules of A and B are

added to the carbon-carbon multiple bonds and this is called an

addition reaction.

o In the reaction given below when HCl is added to ethylene, it will

give us ethylene chloride.

HCl + CH2 = CH2 → CH3CH2Cl
• Radical Reactions

o Most of the organic reactions involve radicals and their movement.

o Addition of a halogen to a typically saturated hydrocarbon involves

free radical mechanism.

o There are usually three stages involved in a radical reaction which

are;

▪ initiation
▪ propagation
▪ termination

o Initially when the weak bond is broken initiation of the reaction

takes place with the formation of free radicals.

o After that when the halogen is added to the hydrocarbon a radical is

produced and finally, it gives alkyl halide.

• Oxidation Reduction reactions (REDOX)

o Electrons in an organic redox reaction often are transferred in the

form of a hydride ion – a proton and two electrons.

o Because they occur in conjunction with the transfer of a proton, these

are commonly referred to

as hydrogenation and dehydrogenation reactions: a hydride plus a

proton adds up to a hydrogen (H2) molecule.

o When a carbon atom in an organic compound loses a bond to hydrogen and

gains a new bond to a heteroatom (or to another carbon), this means

the compound has been dehydrogenated, or oxidized.

o A very common biochemical example is the oxidation of an alcohol to a

ketone or aldehyde:

[pic]

o Conversely, when a carbon atom in an organic compound gains a bond to

hydrogen and loses a bond to a heteroatom (or to another carbon atom),

it means the compound has been hydrogenated, or reduced.

o The hydrogenation of a ketone to an alcohol, for example, is overall

the reverse of the alcohol dehydrogenation shown above.

o Illustrated below is another common possibility, the hydrogenation

(reduction) of an alkene to an alkane

[pic]

STEP 4: Key Points (5 minutes)

• Chemical reactions is a process in which one or more substances, the

reactants, are converted to one or more different substances, the

products, substances are either chemical elements or compounds.

• Types of organic reactions include substitution reactions, elimination

reactions, addition reactions, radical reactions and Oxidation-Reduction

Reactions.

STEP 5: Evaluation (10 minutes).

• What is a chemical reaction?
• What is a substitution reaction?
• What is an addition reaction?
• What is a REDOX reaction?

References

Ternay, A.L (1976). Contemporary Organic Chemistry. Philadelphia, United

States: W.B. Saunders Co.

Morrison R.T and Boyd R N (1997). Organic Chemistry (6th Ed.). New Delhi,

India: Prentice Hall of India

Graham Solomon et al (2014). Organic Chemistry (11th Ed.). New Jeysey,

United States: John Willey and Sons.

Rama R. N. (2005). Principles of Pharmaceutical Organic Chemistry. New

Delhi, India: MacMillan Publishers

Bruice Y (2013). Organic Chemistry (7th ed.). New York, United States:

Prentice Hall Pearson.

Delgado J. N. Et al (1998). Wilson and Gisvold's Textbook of Organic

Medicinal and Pharmaceutical Chemistry (10th Ed.). Calfornia, United

States: Lippincott Williams

Bhassin S.K, Gupta R.(2013). Pharmaceutical organic chemistry (E-book

Kindle edition). New Delhi, India: Elsevier Publishing Services

Session 7: Isomerism.

Total Session Time: 60 minutes

Prerequisites

• None

Learning Tasks

By the end of this session students are expected to be able to:

• Define isomer and isomerism
• List types of isomers
• Explain types of isomers
• Explain the importance of isomerism in pharmacy

Resources Needed:

• Flip charts, marker pens, and masking tape
• Black/white board and chalk/whiteboard markers

SESSION OVERVIEW

|Step |Time |Activity/ |Content |

| | |Method | |

|1 |05 minutes |Presentation |Introduction, Learning Tasks |

|2 |15 minutes |Buzzing |Definition of Isomer and Isomerism |

| | |Presentation | |

|3 |40 minutes |Group |Types of Isomers |

| | |discussion | |

| | |Presentation | |

|4 |40 minutes |Presentation |Importance of Isomerism in Pharmacy |

| | |Brainstorming | |

|5 |10 minutes |Presentation |Key Points |

|6 | | Presentation |Evaluation |

| |10 minutes | | |

SESSION CONTENTS

STEP 1: Presentation of Session Title and Learning Tasks (5 minutes)

READ or ASK students to read the learning tasks and clarify.

ASK students if they have any questions before continuing.

STEP 2: Definition of Isomer and Isomerism (15 minutes).

|Activity: Buzzing (10 minutes) |

| |

|ASK students to pair up and buzz on the following question for 5 |

|minutes |

| |

|What is an isomer? |

|What is isomerism? |

|ALLOW pairs to respond on the question. |

| |

|WRITE their response on the flip chart/board. |

| |

|CLARIFY and SUMMARIZE by using the content in the table 1 below |

• Isomers are compound having same molecular formula but different

structural formula

• Isomerism is the existence of a compound with the same molecular

formula but different structural formula

STEP 3: Types of Isomerism (40 minutes).

|Activity: Small Group Discussion (20 minutes) |

| |

|DIVIDE students into small groups |

| |

|ASK students to discuss in groups on the following questions |

|What are the types of isomerism |

| |

|[pic]REFER Students to Book |

| |

|ALLOW students to discuss for 15 minutes |

| |

|ALLOW each group to present for 5 minutes |

| |

|CLARIFY and SUMMARIZE by using the contents below |

The following are the descriptions of types of isomerism;

• Geometric isomerism
• Structural isomerism
• Constitutional isomerism
• Sterioisomerism
• Geometric isomerism.

o Arises due to restricted rotation across the C-C double bonds.

o Occurs only when two atoms/groups attached to each carbon of the

double bond are different from one another.

o This type of isomerism is also known as cis-trans isomerism.

o The cis-isomer has like groups on the same side of the double bond,

whereas the trans-isomer has like group on opposite sides of the

double bond.

Example,

[pic]

o Cis Butene (the methyl groups are on the same side)

[pic]

o Trans Butene (the methyl groups are on the opposite side)

o Another way to name the cis-trans isomers is to use the Z and E

nomenclature

[pic] E

E = Entgegen in German, Z =

Zusammen, means together

which means on opposite sides.

• Structural isomerism.

o These are Isomers which have the atoms of their molecules linked in a

different order

o The structural isomerism is further subdivided in the following

categories;

▪ Chain Isomerism
▪ Positional Isomerism
▪ Functional Group Isomerism

Chain Isomerism

• Chain isomers are the compounds having the same molecular formula but

different arrangement of carbon chain within the molecule.

• Chain isomers are also known as skeletal isomers or nuclear isomers.
• Chain isomers of the same compound are very similar.
• There may be small difference in physical properties such as melting or

boiling point due to different strengths of intermolecular bonding.

• Their chemistry is likely to be identical.

[pic]

Positional Isomerism

• Position isomers are the compounds which have the same molecular formula

and same carbon skeleton but differ in the position of attached atoms or

groups or in position of multiple bonds.

• Positional isomers are also usually similar.
• There are slight physical differences, but the chemical properties are

usually very similar.

• However, occasionally, positional isomers can have quite different

properties.

[pic]

• A simple example of isomerism is given by propanol:
• it has the formula C3H8O (or C3H7OH) and two isomers propan-1-ol (n-

propyl alcohol; I) and propan-2-ol (isopropyl alcohol; II)

• Note that the position of the oxygen atom differs between the two: it is

attached to an end carbon in the first isomer and to the center carbon in

the second.

• The number of possible isomers increases rapidly as the number of atoms

increases; for example, the next largest alcohol, named butanol (C4H10O),

has four different structural isomers.

[pic]

Functional Group Isomers

• Functional group isomers are the compounds having the same molecular

formula but different functional groups.

• Functional group isomers are likely to be both physically and

chemically dissimilar.

[pic]

• Constitutional Isomerism.

o Isomers that differ in connectivity are called constitutional

(sometimes structural) isomers.

o They have the same parts, but those parts are attached to each other

differently.

o The bracelets of red and green beads mentioned above are analogous to

constitutional isomers.

o The simplest hydrocarbons—methane (CH4), ethane (CH3CH3), and propane

(CH3CH2CH3)—have no constitutional isomers, as there is no other way

to connect the carbons and hydrogens of these molecules consistent

with the tetravalency of carbon and the univalency of hydrogen.

[pic]

o However, there are two different butanes, C4H10, and these two

molecules, called butane and isobutane, are constitutional isomers.

o They are different molecules with different chemical and physical

properties.

o Butane has its four carbon atoms bonded in a continuous chain.

Isobutane has a branched structure.

o [pic]

o The number of possible constitutional isomers increases greatly with

the number of available atoms.

o There are only two butanes, but there are three pentanes (C5H12), 18

octanes (C8H18), and no fewer than 366,319 constitutional isomers of

the hydrocarbon containing 20 carbon atoms and 42 hydrogens.

• Stereoisomers.

o Stereoisomers are isomers that have the same composition (that is, the

same parts) but that differ in the orientation of those parts in space.

o There are two kinds of stereoisomers:

▪ enantiomers
▪ Enantiomers are stereoisomers which are related to each as

an object and its non-superimposible mirror image, like one’s

hands.

▪ Enantiomers are also called optical isomers.
▪ Diastereomers.
▪ Diastereomers (sometimes called diastereoisomers) are a

type of a stereoisomer.

▪ Diastereomerism occurs when two or more stereoisomers of a

compound have different configurations at one or more (but

not all) of the equivalent (related) stereocenters and are

not mirror images of each other.

▪ When two diastereoisomers differ from each other at only

one stereocenter they are epimers. Each stereocenter gives

rise to two different configurations and thus increases the

number of stereoisomers by a factor of two.

▪ Diastereomers differ from enantiomers in that the latter

are pairs of stereoisomers that differ in all stereocenters

and are therefore mirror images of one another.

o Enantiomers of a compound with more than one stereocenter are also

diastereomers of the other stereoisomers of that compound that are not

their mirror image.

o Diastereomers have different physical properties (unlike enantiomers) and

different chemical reactivity.

[pic]

Fig. Isomerism

[pic]

STEP 4: The importance of Isomerism in Pharmacy (40 minutes).

|Activity: Brainstorming (5 minutes) |

| |

|Ask students to brainstorm on the following question: |

| |

|What is the importance of isomerism in pharmacy? |

| |

|ALLOW few students to respond |

| |

|WRITE their responses on the flip chart/ board |

| |

|CLARIFY and SUMMARISE by using the content below |

• Drug isomerism has opened a new era of drug development.
• Currently, knowledge of isomerism has helped us in introducing safer

and more effective drug alternatives of the newer as well as existing

drugs.

• Many existing drugs have gone chiral switch i.e., switching from

racemic mixture to one of its isomers.

• Cetrizine to levocetrizine is one of such examples, where effective

and safer drug has been made available.

• Isomerism can lead to different therapeutic uses and adverse drug

reactions like Quinine has antimalarial activity while quinidine has

an antiarrythmic property.

• Other examples include the following;
o L-sotalol is alpha-blocker while d-sotalol is antiarrythmic;

Levomethorphan is a potent opiod analgesic while dextromethorphan

is a cough suppressant

o R-Thalidomide is sedative while S-Thalidomide has been shown

teratogenic effects;

o R-Naproxen is used for arthralgic pain while S-Naproxen is

teratogenic

o D-Ethambutol is antituberculosis drug while L-ethambutol has been

found to cause blindness

o L-dopa used in treatment for Parkinson's disease has an isomer D-

dopa which has never been used because it causes deficiency of

white blood cells and thus susceptibility to infections.

STEP 5: Key Points (10 minutes).

• Isomers are compound having same molecular formula but different

structural formula.

• Types of isomerism include Geometric isomerism, Structural isomerism,

Constitutional isomerism and Stereoisomerism.

• Isomerism finds its importance in pharmacy, as isomers differ in their

pharmacokinetic and pharmacodyanmic properties which has helped in

introducing safer and more effective drug alternatives of the newer as

well as existing drugs.

STEP 6: Evaluation (10 minutes)

• What is isomerism?
• What are the types of isomerism?
• What is the importance of isomerism in Pharmacy?

References

Ternay, A.L (1976). Contemporary Organic Chemistry. Philadelphia, United

States: W.B. Saunders Co.

Morrison R.T and Boyd R N (1997). Organic Chemistry (6th Ed.). New Delhi,

India: Prentice Hall of India

Graham Solomon et al (2014). Organic Chemistry (11th Ed.). New Jeysey,

United States: John Willey and Sons.

Rama Rao Nadendla (2005). Principles of Pharmaceutical Organic Chemistry.

New Delhi, India: MacMillan Publishers

Bruice Y (2013). Organic Chemistry (7th ed.). New York, United States:

Prentice Hall Pearson.

Delgado J. N. Et al (1998). Wilson and Gisvold's Textbook of Organic

Medicinal and Pharmaceutical Chemistry (10th Ed.). California, United

States: Lippincott Williams

Bhassin S.K, Gupta R. (2013). Pharmaceutical organic chemistry (E-book

Kindle edition). New Delhi, India: Elsevier Publishing Services

Session 8: Alkanes of Pharmaceutical Importance.

Total Session Time: 120 minutes

Prerequisites

• None

Learning Tasks

By the end of this session students are expected to be able to:

• Define alkanes
• List alkanes and their isomers
• Explain nomenclature of alkanes
• Draw chemical structure of alkanes
• List chemical properties of alkanes
• Explain chemical reactions of alkanes

Resources Needed:

• Flip charts, marker pens, and masking tape.
• Black/white board and chalk/whiteboard markers.

SESSION OVERVIEW

|Step |Time |Activity/ |Content |

| | |Method | |

|1 |05 minutes |Presentation |Introduction, Learning Tasks |

|2 |10 minutes |Brainstorming |Definition of Alkanes |

| | |Presentation | |

|3 |15 minutes |Buzzing |Alkanes and their Isomers |

| | |Presentation | |

|4 |15 minutes |Presentation |Nomenclature of Alkanes |

|5 |15 minutes |Presentation |Chemical Structure of Alkanes |

|6 |10 minutes |Brainstorming |Chemical Properties of Alkanes |

| | |Presentation | |

|7 |30 minutes |Group |Chemical Reactions and Uses of |

| | |discussion |Alkanes |

| | |Presentation | |

|8 |10 minutes |Presentation |Key Points |

|9 |10 minutes |Presentation |Evaluation |

SESSION CONTENTS

STEP 1: Presentation of Session Title and Learning Tasks (5 minutes)

READ or ASK students to read the learning tasks and clarify

ASK students if they have any questions before continuing.

STEP 2: Definition of Alkanes (10 minutes).

|Activity: Brainstorming (5 minutes) |

| |

|Ask students to brainstorm on the following question: |

| |

|What is Alkane? |

| |

|ALLOW few students to respond. |

| |

|WRITE their responses on the flip chart/ board. |

| |

|CLARIFY and SUMMARISE by using the content below; |

• Alkane or Paraffin is an acyclic saturated hydrocarbon composed of only

carbons and hydrogen atoms and contain carbon-carbon single bonds.

• Compounds that contain only carbon and hydrogen are called hydrocarbons.

Homologous series (homo is Greek for “the same as”) is a family of

compounds in which each member differs from the next by one methylene

group (CH2).

• The general molecular formula for an alkane is CnH2n+2 where n is an

integer.

• Members (CnH2n+2)

o Methane CH4

o Ethane C2H6

o Propane C3H8

o Butane C4H10

o Pentane C5H12

o Hexane C6H14

o Heptane C7H16

o Octane C8H18

o Nonane C9H20

o Decane C10H22

o Undecane C11H24 etc

• So, if an alkane has one carbon atom, it must have four hydrogen atoms;

if it has two carbon atoms, it must have six hydrogen.

NOTE; Only one possible structure for an alkane with molecular formula CH4

(methane) and molecular formula C2H6 (ethane)

• There are two possible structures for an alkane straight-chain and a

branched structure.

• Both of these structures fulfill the requirement that each carbon forms

four bonds and each hydrogen forms only one bond.

STEP 3: Alkanes and their Isomers (15 minutes).

|Activity: Buzzing (5minutes) |

| |

|ASK students to pair up and buzz on the following question for 5 |

|minutes. |

| |

|What is an isomer? |

|What is isomerism? |

|ALLOW pairs to respond on the question. |

| |

|WRITE their response on the flip chart/board. |

| |

|CLARIFY and SUMMARIZE by using the content below |

• Isomers Refers to different compounds having the same molecular formula

but different structural formula (have different arrangements of atoms in

space).

• In n-alkanes, no carbon is bonded to more than two other carbons, give

rise to a linear chain.

• When carbon is bonded to more than two other carbons, a branched is

formed.

Example.1; C6H14 has 5 isomers as follows:

o Hexane

o 2-Methylpentane

o 3-Methylpentane

o 2,2-Dimethylbutane

o 2,3-Dimethylbutane

Example.2; C7H16 has 9 isomers as follows:

o Heptane

o 2-Methylhexane

o 3-Methylhexane

o 2,2-Dimethylpentane

o 2,3-Dimethylpentane

o 2,4-Dimethylpentane

o 3,3-Dimethylpentane

o 3-Ethylpentane

o 2,2,3-Trimethylbutane

Example 3; C5H12 has 3 isomers which are:

o Pentane

o 2-Methylbutane

o 2,2-Dimethylpropane

Example 4; C4H10 has 2 isomers which are;

o Butane

o 2-Methylpropane

STEP 4: Nomenclature of Alkanes (15 Minutes).

• Rule1. Determine the longest continuous carbon chain.

o This chain is called the parent hydrocarbon.

• Rule 2. In isomeric compounds (II and III), indicate by a number the

Carbon to which the alkyl group is attached.

• Rule 3. In numbering the parent chain, start at whichever end resulting

in the use of the lowest numbers; thus, II is called 2–‐ methylpentane

rather than 4–‐ methylpentane.

• Rule 4. If the same alkyl group occurs more than once as a side chain,

indicate this by the prefix di-, tri-, tetra- etc., to show how many of

these alkyl groups are there and indicate by various numbers the position

of each group, as in 2,2,4‐trimethyl-pentane.

[pic]

• Rule 5 If there are several different alkyl groups attached to the

parent chain, name them in alphabetical order, as in 3,3‐diethyl-5-

isopropyl-4-methyloctane

[pic]

STEP 5: Chemical Structure of Alkanes (15 minutes).

• All acyclic alkanes (unbranded and branched) have the characteristic

molecular formula CnH2n+2, where n is the number of carbon atoms in

the chain.

• Gives the molecular formulas and Lewis structure for the unbranched

and n-alkanes (n stands for normal)

[pic]

STEP 6: Chemical Properties of Alkanes (10 minutes).

|Activity: Brainstorming (5minutes) |

| |

|ASK students to pair up and brainstorm on the following question for 5 |

|minutes. |

| |

|What are the chemical properties of Alkanes? |

| |

|ALLOW pairs to respond on the question |

| |

|WRITE their response on the flip chart/board |

| |

|CLARIFY and SUMMARIZE by using the content in the table 1 below |

• Combustion

o Complete combustion (Under sufficient amount of oxygen (Air)) any

hydrocarbon produces carbon dioxide and water.

o Also, can react with very strong oxidizing agents like Potassium

permanganate, potassium dichromate, Manganese oxide, to produce carbon

dioxide and water.

• Halogenation of alkanes (Substitution of alkanes)

o Halogenation is the replacement of one or more hydrogen atoms in an

organic compound by a halogen (Fluorine, Chlorine Bromine or Iodine).

o Under ultra violet light or temperature that makes up the free radical

chain for halogenation reaction to proceed and that is the first step.

• Cracking

o Cracking is the breakdown of large alkanes under high temperature and

pressure into smaller, more useful alkenes.

o Cracking of alkanes produce a mixture of alkenes and alkanes, cracking

is an example of thermal decomposition reaction.

o Conditions are Temperature of about 500°C and moderately low

pressures.

STEP 7: Chemical Reactions and Uses of Alkanes (30 minutes).

|Activity: Small Group Discussion (15 minutes) |

| |

|DIVIDE students into small groups |

| |

|ASK students to discuss in groups on the following questions |

|What are the reactions involving Alkanes? |

| |

|[pic]REFER Students to Book |

| |

|ALLOW students to discuss for 15 minutes |

| |

|ALLOW each group to present for 5 minutes |

| |

|CLARIFY and SUMMARIZE by using the contents below |

• Combustion

o Complete combustion (Under sufficient amount of oxygen (Air)) any

hydrocarbon produces carbon dioxide and water.

o Example; [pic]

• Halogenation of alkanes (Substitution of alkanes)

o Halogenation is the replacement of one or more hydrogen atoms in an

organic compound by a halogen (Fluorine, Chlorine Bromine or

Iodine).

o Example;

[pic]

• Cracking

o Cracking is the breakdown of a large alkanes under high temperature

and pressure into smaller, more useful alkenes.

o Product of cracking of alkanes are alkenes and alkanes.

o Cracking is an example of thermal decomposition reaction at

temperature of about 500°C and moderately low pressures.

o Example;

[pic]

Uses of alkanes

• The pharmaceutical use of alkane is often used in general anesthesia.

o Alkane is a compound of halothane which is a general anesthetic agent.

• Petroleum and natural gas are largely mixtures of different alkanes.
• On refining, they give liquefied petroleum gas (LPG), gasoline, kerosene,

diesel, furnace oil and wax which are used as fuels.

• Some higher alkanes are used as lubricating oils and as vaseline.
• Alkanes are used as starting materials for the preparation of many other

useful organic compounds.

o For example, methane on chlorinate on gives chloromethane,

dichloromethane, trichloromethane (chloroform) and tetrachloromethane.

STEP 8: Key Points (10 minutes).

• Alkanes or Paraffin is an acyclic saturated hydrocarbon composed of only

carbons and hydrogen atoms and contain carbon-carbon single bonds.

• The alkanes general formula is CnH(2n+2) which is stable for all the
alkanes members, where n= the number of carbon atoms in the molecular

structure.

• The first 3 members of the alkane series, the methane, ethane & propane

have no isomers because there is only one way in arranging the molecule

in 3D structures but the rest of the alkane’s series might have an

isomer, which is called structure isomers.

• There are two basic systems for the naming of alkanes which are common

names also called general or primary names or the un-systematic name and

The IUPAC name, or the systematic name.

• Chemical properties of alkanes involve combustion, halogenation and

cracking.

• Alkanes can be used in preparation of pharmaceuticals such as halothane

which is a general anaesthetic.

STEP 9: Evaluation (10 minutes)

• What are alkanes?
• What are the rules of naming alkanes?
• What are the chemical properties of alkanes?
• What are the uses of alkanes in pharmacy?

References

Ternay, A.L (1976). Contemporary Organic Chemistry. Philadelphia, United

States: W.B. Saunders Co.

Morrison R.T and Boyd R N (1997). Organic Chemistry (6th Ed.). New Delhi,

India: Prentice Hall of India

Graham Solomon et al (2014). Organic Chemistry (11th Ed.). New Jeysey,

United States: John Willey and Sons.

Rama Rao Nadendla (2005). Principles of Pharmaceutical Organic Chemistry.

New Delhi, India: MacMillan Publishers

Bruice Y (2013). Organic Chemistry (7th ed.). New York, United States:

Prentice Hall Pearson.

Delgado J. N. Et al (1998). Wilson and Gisvold's Textbook of Organic

Medicinal and Pharmaceutical Chemistry (10th Ed.). California, United

States: Lippincott Williams

Bhassin S.K, Gupta R. (2013). Pharmaceutical organic chemistry (E-book

Kindle edition). New Delhi, India: Elsevier Publishing Services

Session 9: Alkenes of Pharmaceutical Importance.

Total Session Time: 120 minutes

Prerequisites

None

Learning Tasks

By the end of this session students are expected to be able to:

• Define alkenes
• List alkenes and their isomers
• Explain nomenclature of alkenes
• Draw chemical structure of alkenes
• List chemical properties of alkenes
• Explain chemical reactions of alkenes

Resources Needed:

• Flip charts, marker pens, and masking tape.
• Black/white board and chalk/whiteboard markers.

SESSION OVERVIEW

|Step |Time |Activity/ |Content |

| | |Method | |

|1 |05 minutes |Presentation |Introduction, Learning Tasks |

|2 |10 minutes |Brainstorming |Definition of Alkenes |

| | |Presentation | |

|3 |15 minutes |Buzzing |Alkenes and their Isomers |

| | |Presentation | |

|4 |15 minutes |Presentation |Nomenclature of Alkenes |

|5 |15 minutes |Presentation |Chemical Structure of Alkenes |

|6 |10 minutes |Presentation |Chemical Properties of Alkenes |

| | |Brainstorming | |

|7 |40 minutes |Group |Chemical Reactions and Uses of |

| | |discussion |Alkenes |

| | |Presentation | |

|8 |05 minutes |Presentation |Key Points |

|9 |05 minutes |Presentation |Evaluation |

SESSION CONTENTS.

STEP 1: Presentation of Session Title and Learning Tasks (5 minutes).

READ or ASK students to read the learning tasks and clarify.

ASK students if they have any questions before continuing.

STEP 2: Definition of Alkenes (5 minutes).

|Activity: Brainstorming (5 minutes) |

| |

|Ask students to brainstorm on the following question: |

| |

|What are Alkenes? |

| |

|ALLOW few students to respond. |

| |

|WRITE their responses on the flip chart/ board. |

| |

|CLARIFY and SUMMARISE by using the content below |

• These are unsaturated hydrocarbons, they contain two less hydrogen atoms

as compared to corresponding alkanes (sp2 hybrid), also known as OLEFINS

or ALKYLENES, general formula: (CnH2n).

• They contain carbon-carbon double bond, this is the distinguishing

feature of the alkenes.

• The simplest member of the alkene family is ethylene C2H2.

STEP 3: Alkenes and their Isomers (15 minutes).

|Activity: Buzzing (10minutes) |

| |

|ASK students to pair up and buzz on the following question for 5 |

|minutes. |

| |

|What is the isomer of Alkenes? |

| |

|ALLOW pairs to respond on the question. |

| |

|WRITE their response on the flip chart/board. |

| |

|CLARIFY and SUMMARIZE by using the content in the table 1 below |

Geometric isomerism

• Arises due to restricted rotation at the C-C double bonds
• Occurs only when two atoms/groups attached to each carbon of the double

bond are different from one another

• The cis-isomer has like groups on the same side of the double bond,

whereas the trans-isomer has like group on opposite sides of the double

bond

Example,

[pic]

Cis Butene (the methyl groups are on the same side)

[pic]

Trans Butene (the methyl groups are on the opposite side)

Structural isomerism

• Isomers which have the atoms of their molecules linked in a different

order

• This can come about in one of three ways:

o Chain Isomerism

[pic] [pic][pic]

▪ Chain isomers of the same compound are very similar.
▪ There may be small difference in physical properties such as

melting or boiling point due to different strengths of

intermolecular bonding.

▪ Their chemistry is likely to be identical.

o Positional Isomers

[pic] [pic]

▪ Positional isomers are also usually similar.
▪ There are slight physical differences, but the chemical

properties are usually very similar.

▪ However, occasionally, positional isomers can have quite

different properties

Example of isomerism is given by propanol

• It has the formula C3H8O (or C3H7OH) and two isomers propan-1-ol

(n-propyl alcohol; I) and propan-2-ol (isopropyl alcohol; II)

• Note that the position of the oxygen atom differs between the

two: it is attached to an end carbon in the first isomer, and to

the center carbon in the second.

• The number of possible isomers increases rapidly as the number

of atoms increases;

For example; the next largest alcohol, named butanol

(C4H10O), has four different structural isomers.

[pic] [pic]

STEP 4: Nomenclature of Alkenes (15 minutes).

• The IUPAC Rules are similar to those of alkanes, but few new rules must

be added to name and locate the double bond.

o Rule 1: Select as the parent structure the longest continuous chain

that contains the C-C double bond:

▪ C-C double bonds are designated by the ending -ene, if more than

one double bond is present, the ending is diene, triene, tetraene,

etc.

o Rule 2: Indicate by a number the position of the double bond in the

chain. Number it so that the C-atoms in the double bond have the

lowest possible numbers.

o Rule 3: The position of the double bond(s) is indicated by the

number(s) of the lower numbered carbon atom of each double bond. These

numbers are placed in front of the name of the compound.

Example,

[pic]

o Rule 4: In cyclic hydrocarbons, start numbering around the ring with

the carbons of the double bond indicates by numbers the positions of

alkyl groups attached to the parent chain.

Example,

[pic]

3-Methylcyclopenten

Table 1. Nomenclature of simple alkenes

|COMPOUND |COMMON NAME |IUPAC NAME |

|CH2=CH2 |Ethylene |Ethene |
|CH3CH=CH2 |Propylene |1-Propene |
|CH3CH2CH=CH2 |α-Butylene |1-Butene |
|CH3C(CH3)=CH2 |Isobutylene |2-Methylpropene |
|CH2=C(C2H5)CH2CH3 |- |2-Ethyl-1-butene |
|CH2=CHCl |Vinyl chloride |Chloroethene |
|CH2=CHCH2Cl |Allyl chloride |3-Chloropropene |
|CH3=CHCH=CH2 | |1,3-Butadiene |

STEP 5: Chemical Structure of Alkenes (15 minutes).

Definition

• The arrangement of chemical bonds between atoms in a molecule (or in an

iron or radical with multiple atoms) especially which atoms are

chemically bonded to what other atoms with what kind of chemical bonds,

together with any information on the geometric shape of the molecule

needed to uniquely identify the type of molecule.

OR

• It is the spatial arrangement of atoms in a molecule and the chemical

bonds that holds the atoms together. Example diatomic oxygen or nitrogen

molecules or DNA molecules.

Table 2: Chemical stuctures of alkenes CnH2n

| IUPAC Name | Molecular Formula |Condensed Structural |

| | |Formula |

|Ethane |C2H4 |CH2=CH2 |
|Propene |C3H6 |CH2=CHCH3 |
|1-butene |C4H8 |CH2=CHCH2CH3 |
|1-pentene |C5H10 |CH2=CH(CH2)2CH3 |
|1-hexene |C6H12 |CH2=CH(CH2)3CH3 |
|1-heptene |C7H14 |CH2=CH(CH2)4CH3 |
|1-octene |C8H16 |CH2=CH(CH2)5CH3 |

STEP 6: Chemical Properties of Alkenes (10 minutes).

|Activity: Brainstorming (5 minutes) |

| |

|Ask students to brainstorm on the following question: |

| |

|What are chemical properties of Alkenes? |

| |

|ALLOW few students to respond. |

| |

|WRITE their responses on the flip chart/ board. |

| |

|CLARIFY and SUMMARISE by using the content below |

• Alkenes are chemically more reactive than alkanes, this because alkenes
are unsaturated hydrocarbons that have a double bond, C=C, between two

carbon atoms, almost all of the chemical reactions of alkene occur at the

double bond.

Isomerization

• Alkenes when heated alone at high temperatures (500-700°C) or at lower

temperatures (200-300°C) isomerizes in the presence of catalyst, such as

Al2(SO4)3.

• Alkenes isomerism due to;

o The shifting of the double bond which tends to move towards the center

of chain, e.g., pentene-1 isomerizes to pentene-2.
CH3-CH2-CH2-CH=CH2[pic]CH3-CH2-CH=CH-CH3

pentene-1 pentene-2

o The migration of a methyl group, e.g., butene-1 isomerizes to 2-

methylpropene (iso-butene).

[pic]

• Alkenes undergoes electrophilic addition with halogens, water, hydrogen

and halogen acids to produce di halo alkanes, alcohol, alkane and halo

alkanes, also with oxygen to form epoxides.

• At elevated temperatures (500°C), higher alkenes give substitution
products with chlorine. For example, CH3-CH=CH2 + Cl2 [pic] ClCH2-CH=CH2

+HCl

• Alkenes has an ability to dimarize and polymerizes.
• Alkenes undergoes combustion to produce carbondioxide and water
CH2=CH2 + 3O2 → 2CO2 + 2H2O

STEP 7: Chemical Reactions and Uses of Alkenes (40 minutes).

|Activity: Small Group Discussion (20 minutes) |

| |

|DIVIDE students into small groups. |

| |

|ASK students to discuss in groups on the following questions |

|What are the chemical properties of Alkanes? |

| |

|[pic]REFER Students to Book |

| |

|ALLOW students to discuss for 15 minutes. |

| |

|ALLOW each group to present for 5 minutes. |

| |

|CLARIFY and SUMMARIZE by using the contents below |

• All alkenes have a common feature: a carbon-carbon double bond.
• The reactions of alkenes arise from the reactivity of the carbon-carbon

double bond.

• Because single bonds (sigma bonds) are more stable than pi bonds, the

most common reactions of double bonds transform the pi bond into a sigma

bond.

• For example, catalytic hydrogenation converts the C = C pi bond and the H-

H sigma bond into two C – H sigma bonds

• The double bond serves as an electron source

o These are because atoms are added & the double bond becomes a

single bond

[pic]

• Alkenes can undergo the following reactions:

o Electrophilic addition

o Dimerazation

o Polymerisation

o Combustion

• Electrophilic addition to alkenes

o Addition is the most common reaction of alkenes

o Most addition reactions involve a second step in which a nucleophile

attacks the carbocation (as in the second step of the SN1 reaction),

forming a stable addition product.

o In the product, both the electrophile and the nucleophile are bonded

to the carbon atoms that were connected by the double bond.

o Example 1;

[pic]

Step 1: Attack of the pi bond on the electrophile forms a carbocation.

[pic]

Step 2: Attack by a nucleophile gives the addition product.

[pic]

o Example 2: Ionic addition of HBr to 2-butene

o When gaseous HBr adds to 2-butene the proton in HBr is electrophilic;

it reacts with the alkene to form a carbocation.

o Bromide ion reacts rapidly with the carbocation to give a stable

product in which the elements of HBr have added to the ends of the

double bond.

o Step 1: Protonation of the double bond forms a carbocation.

[pic]

Step 2: Bromide ion attacks the carbocation.

[pic]

The electrophilic addition reactions to alkenes include the following

reactions

1. Catalytic hydrogenation =Addition of Hydrogen
• A solution of alkenes is shaken under slight pressure with H2(g) in the

presence of a small amount of Ni / pt catalyst.

• The hydrogen reduces the double bond to a single bond
• One molecule of H2 is absorbed for each double bond in the unsaturated

compound (heterogeneous)

[pic]

2. Addition of Halogens to Alkenes.

• Halogens add to alkenes to form vicinal dihalides.
• Alkenes are readily converted by Cl2 & Br2 into saturated compounds.
• The halogens will be attached to consecutive positions on carbon chain/

adjacent carbon atoms.

[pic]

EXAMPLE: Addition of Br2 to propene.

Step 1: Electrophilic attack forms a bromonium ion.

[pic]

Step 2: Bromide ion opens the bromonium ion

[pic]

• Chlorine and bromine commonly add to alkenes by the halonium ion

mechanism.

• Iodination is used less frequently because diiodide products decompose

easily.

• Any solvents used must be inert to the halogens; methylene chloride

(CH2CI2), chlorofonn (CHCI3 ), and carbon tetrachloride (CCI4) are the

most frequent choices.

3. Addition of Hydrogen Halides to Alkenes

• Alkenes are converted by HCl, HBr & HI into corresponding alkyl halide

[pic]

• Addition occurs at the double bond; following ‘Markovnikov’s rule’
• ‘Markovnikov’s rule’ The addition of a proton acid to the double bond of

an alkene results in a product with the acid proton bonded to the carbon

atom that already holds the greater number of hydrogen atoms’.

• Reactions that follow this rule are said to follow Markovnikov

orientation and give the Markovnikov product.

• We are often interested in adding electrophiles other than proton acids

to the double bonds of alkenes.

• Markovnikov's rule can be extended to include a wide variety of other

additions, based on the addition of the electrophile in such a way as to

produce the most stable carbocation

‘In an electrophilic addition to an alkene, the electrophile adds in such

a way as to generate the most stable intermediate’.

▪ Mechanism summarized

Step 1: Protonation of the pi bond forms a carbocation.

[pic]

Step 2: Attack by the halide ion gives the addition product .

[pic]

Example.

[pic]

• Where the structure permits, electrophilic addition is accompanied by

rearrangement

[pic]

[pic]

• The more stable carbocation is preferred because the positive charge is

stabilized by the alkyl groups which release electrons

[pic]

Order of stability of carbocations: 3o>2o>1o

4. Hydration of Alkenes: Addition of Water

• An alkene may react with water in the presence of a strongly acidic

catalyst to form an alcohol.

• Formally, this reaction is a hydration (the addition of water), with a

hydrogen atom adding to one carbon and a hydroxyl group adding to the

other.

• Hydration of an alkene is the reverse of the dehydration of alcohols

[pic]

• The reaction follows ‘Markorvonikov’s rule’
• Hydration of an alkene is accomplished by adding excess water to drive

the equilibrium toward the alcohol.

EXAMPLE: Acid-catalyzed hydration of propene.

Step 1: Protonation of the double bond forms a secondary carbocation.

[pic]

Step 2: Nucleophilic attack by water gives a protonated alcohol

[pic]

Step 3: Deprotonation gives the alcohol

[pic]

• Dimerization of Alkenes

o Under proper conditions isobutene is converted by sulfuric acid/

phosphoric acid into a mixture of two (2) alkenes – C8H16

o Hydrogenation of these alkenes produce the same alkane 2,2,4-

trimethylpentane.

[pic]

o The alkenes produced contain exactly twice the number of carbon and

hydrogen atoms as the original isobutylene, they are known as

dimers of isobutylene

[pic]

[pic]

[pic]

• Polymerization of Alkenes

o A polymer is a large molecule composed of many smaller repeating

units (the monomers) bonded together.

o Alkenes serve as monomers for some of the most common polymers,

such as polyethylene, polypropylene, polystyrene, poly (vinyl

chloride), and many others.

o Alkenes generally undergo addition polymerization, the rapid

addition of one molecule at a time to a growing polymer chain.

o There is generally a reactive intermediate (cation, anion, or

radical) at the growing end of the chain; for that reason, addition

polymers are also called chain-growth polymers

o Polymerization is an important industrial process to produce

synthetic polymers such as poly(ethene)=polyethene, Teflon,

plastic, nylon & Bakelite

o In nature there are some natural polymers as well such as; starch,

cellulose, proteins & rubber

• Combustion

o Alkenes, like alkanes, are highly combustible.

o Alkenes burn with a luminous flame to give carbon dioxide and

water; the flame becomes luminous because of the higher carbon

content of alkenes than alkanes.

o Their combustion reactions are exothermic.

[pic]

o Due to the luminosity of the flame, the lower alkenes may be used

as illuminants.

Uses of Alkenes

• Ethene is used in the manufacture of polyethylene or polythene-a

plastic material.

• Propene is used in the manufacture of polypropene.

o These polymers are used in making plastic bags, pipes electrical

insulation.

• Ethene is used in the preparation of solvents like ethylene glycol,

dioxane.

o Ethylene glycol is also used as an antifreeze in automobile

radiators.

• Ethene is a plant hormone which controls growth, seed germination and

fruit development.

o Therefore, ethene is used for artificial ripening of fruits,

flower maturation.

• Alkenes are used as a starting material for the manufacture of many

compounds such as alkyl halides, ethylene oxide, ethanol and other

alkanols.

STEP 8: Key Points (05 minutes).

• Alkenes are organic compounds made up of carbon and hydrogen atoms with

one or more carbon- carbon double bonds.

• The IUPAC Rules is used for naming of alkenes.
• Alkenes undergo Electrophilic addition reactions, Dimerization,

Polymerization and Combustion reactions.

• The most important alkenes for the chemical industry are ethene, propene

and 1,3-butadiene which are used as starting materials in the syntheses

of alcohols, plastics, detergents, and fuels

STEP 9: Evaluation (05 minutes)

• What are alkenes?
• What are rules for naming alkenes?
• What are chemical reactions involving alkenes?
• What are the uses of alkenes?

References

Ternay, A.L (1976). Contemporary Organic Chemistry. Philadelphia, United

States: W.B. Saunders Co.

Morrison R.T and Boyd R N (1997). Organic Chemistry (6th Ed.). New Delhi,

India: Prentice Hall of India

Graham Solomon et al (2014). Organic Chemistry (11th Ed.). New Jeysey,

United States: John Willey and Sons.

Nadendla R. R. (2005). Principles of Pharmaceutical Organic Chemistry. New

Delhi, India: MacMillan Publishers

Bruice Y (2013). Organic Chemistry (7th ed.). New York, United States:

Prentice Hall Pearson.

Delgado J. N. Et al (1998). Wilson and Gisvold's Textbook of Organic

Medicinal and Pharmaceutical Chemistry (10th Ed.). Calfornia, United

States: Lippincott Williams

Bhassin S.K, Gupta R.(2013). Pharmaceutical organic chemistry (E-book

Kindle edition). New Delhi, India: Elsevier Publishing Services

Session 10: Alkynes of Pharmaceutical Importance.

Total Session Time: 120 minutes

Prerequisites

None

Learning Tasks

By the end of this session students are expected to be able to:

• Define alkynes
• List alkynes and their isomers
• Explain nomenclature of alkynes
• Draw chemical structure of alkynes
• List physical properties of alkynes
• Explain chemical reactions of alkynes

Resources Needed:

• Flip charts, marker pens, and masking tape.
• Black/white board and chalk/whiteboard markers.

SESSION OVERVIEW

|Step |Time |Activity/ |Content |

| | |Method | |

|1 |05 minutes |Presentation |Introduction, Learning Tasks |

|2 |10 minutes |Brainstorming |Definition of Alkynes |

| | |Presentation | |

|3 |15 minutes |Buzzing |Alkynes and their Isomers |

| | |Presentation | |

|4 |15 minutes |Presentation |Nomenclature of Alkynes |

|5 |15 minutes |Presentation |Chemical Structure of Alkynes |

|6 |10 minutes |Brainstorming |Physical Properties of Alkynes |

| | |Presentation | |

|7 |40 minutes |Group |Chemical Reactions involving Alkynes|

| | |discussion | |

| | |Presentation | |

|8 |05 minutes |Presentation |Key Points |

|9 |05 minutes |Presentation |Evaluation |

SESSION CONTENTS.

STEP 1: Presentation of Session Title and Learning Tasks (5 minutes).

READ or ASK students to read the learning tasks and clarify.

ASK students if they have any questions before continuing.

STEP 2: Definition of Alkynes (10 minutes).

|Activity: Brainstorming (5 minutes) |

| |

|Ask students to brainstorm on the following question: |

| |

|What are Alkynes? |

| |

|ALLOW few students to respond |

| |

|WRITE their responses on the flip chart/ board |

| |

|CLARIFY and SUMMARISE by using the content below |

• These are unsaturated hydrocarbons, they contain four less hydrogen atoms

as compared to corresponding alkanes, also known as Acetylenes, general

formula: (CnH2n-2).

• They contain carbon-carbon triple bond, this is the distinguishing

feature of the alkynes.

• The simplest member of the alkyne family is ethylene C2H2.

STEP 3: Alkynes and their Isomers (15 minutes).

|Activity: Buzzing (10minutes) |

| |

|ASK students to pair up and buzz on the following question for 5 |

|minutes. |

| |

|What are the isomers of Alkynes? |

| |

|ALLOW pairs to respond on the question. |

| |

|WRITE their response on the flip chart/board. |

| |

|CLARIFY and SUMMARIZE by using the content in the table 1 below |

Structural isomerism

• Isomers which have the atoms of their molecules linked in a different

order

• This can come about in one of three ways:

o Chain Isomerism.

▪ Chain isomers of the same compound are very similar.
▪ There may be small difference in physical properties such as

melting or boiling point due to different strengths of

intermolecular bonding.

▪ Their chemistry is likely to be identical.

o Positional Isomer

▪ Positional isomers are also usually similar.
▪ There are slight physical differences, but the chemical

properties are usually very similar.

▪ However, occasionally, positional isomers can have quite

different properties

STEP 4: Nomenclature of Alkynes (15 minutes).

• The IUPAC Rules are similar to those of alkanes, but few new rules

must be added to name and locate the triple bond.

o Rule 1: Select as the parent structure the longest continuous chain

that contains the C-C triple bond:

▪ C-C triple bonds are designated by the ending -yne, if more than one

triple bond is present, the ending is diyne, triyne, tetrayne, etc.

o Rule 2: Indicate by a number the position of the triple bond in the

chain. Number it so that the C-atoms in the triple bond have the

lowest possible numbers.

o Rule 3: The position of the triple bond(s) is indicated by the

number(s) of the lower numbered carbon atom of each triple bond. These

numbers are placed in front of the name of the compound.

Examples

o [pic] [pic]

o [pic][pic]

o [pic][pic]

o Rule 4: In cyclic hydrocarbons, start numbering around the ring with

the carbons of the double bond indicates by numbers the positions of

alkyl groups attached to the parent chain.

STEP 5: Chemical Structure of Alkynes (15 minutes).

• The sigma bond is sp-sp overlap

[pic]

• The two pi bonds are unhybridized p overlaps at 90(, which blend into

a

cylindrical shape.

[pic]

• Bond Lengths
• More s character, so shorter length than alkenes or alkanes.Three bonding

overlaps, so shorter

[pic]

[pic]

• Acidity Table

[pic]

Table 2: Chemical stuctures of alkynes CnH2n-2

| IUPAC Name | Molecular Formula |Condensed Structural |

| | |Formula |

|Ethyne |C2H2 |CHCH |

|Propyne |C3H4 |CHCCH3 |

|1-butyne |C4H6 |CHCCH2CH3 |

|1-pentyne |C5H8 |CHC(CH2)2CH3 |

|1-hexyne |C6H10 |CHC(CH2)3CH3 |

|1-heptyne |C7H12 |CHC(CH2)4CH3 |

|1-octyne |C8H14 |CHC(CH2)5CH3 |

STEP 6: Physical Properties of Alkynes (10 minutes).

|Activity: Brainstorming (5 minutes) |

| |

|Ask students to brainstorm on the following question: |

| |

|What are physical properties of Alkynes? |

| |

|ALLOW few students to respond. |

| |

|WRITE their responses on the flip chart/ board. |

| |

|CLARIFY and SUMMARISE by using the content below |

• Nonpolar, insoluble in water
• Soluble in most organic solvent
• Boiling points similar to alkane of same size
• Less dense than water
• Up to 4 carbons, gas at room temperature

STEP 7: Chemical Reactions involving Alkynes (40 minutes).

|Activity: Small Group Discussion (20 minutes) |

| |

|DIVIDE students into small groups. |

| |

|ASK students to discuss in groups on the following questions |

|What are the chemical properties of Alkanes? |

| |

|[pic]REFER Students to Book |

| |

|ALLOW students to discuss for 15 minutes. |

| |

|ALLOW each group to present for 5 minutes. |

| |

|CLARIFY and SUMMARIZE by using the contents below |

| |

• Reduction Reactions

o Addition of Hdrogen; Alkynes can be reduced to form alkenes and/or

alkanes depending on upon the reaction condition.

Examples;

o Reduction with Metal – Amonia: Alkynes on treatment with sodium or

lithium in liquid ammonia produce trans- alkenes as almost exclusive

product.

[pic]

o Catalytic reduction: Catalytic reduction of alkynes to alkanes

[pic]

• Electrophilic Addition

o Addition of halogens:Addition of halogens to alkyenes takes place in

two stages, first forming a trans-1, 2-dihaloalkene and then

1,1,2,2,-tetrahaloalkane

[pic]

o Addition of halogen acids: On treatment with halogens acids, alkynes

first form vinly halides followed by alkylidene halides. The second

molecule of halogens acid adds according to Markowkwnikoff’s rule

[pic]

o Hydration (addition of water): React with water in presence of

sulphuric acid and mercuric sulphate to form a vinyl alcohol which

readily tautomerizes to the corresponding carbonyl derivative.

[pic]

o Ozonization: Ozon adds to carbon-carbon triple bond forming ozonide.

The ozonides are hydrolyzed by water and the products – the 1,2-

diketones undergo oxidative cleavage to acids by hydrogen peroxide

formed in the reaction.

[pic]

o Hydroboration: Addition of borane to carbon –carbon triple bond

produces vinlyboranes which is oxidized by hydrogen peroxide to

given enol. Tautomerazation then gives either ketone or aldehyde

depending on the structure of alkyne reactant.

[pic]

[pic]

STEP 8: Key Points (05 minutes)

• Alkynes or acetylenes are unsaturated hydrocarbons containing carbon

–carbon triple bond and having general formula (CnH2n-2).

• The IUPAC Rules is used for naming of alkynes
• Alkynes undergo reduction and electrophilic addition reactions.
• Alkynes occur in some pharmaceuticals including the contraceptives

noretynodrel, antiretroviral Efavirenz and antifungal terbinafine.

STEP 9: Evaluation (05 minutes)

• What are alkynes?
• What are rules for naming alkynes?
• What are chemical reactions involving alkynes?
• What are the uses of alkynes?

References

Ternay, A.L (1976). Contemporary Organic Chemistry. Philadelphia, United

States: W.B. Saunders Co.

Morrison R.T and Boyd R N (1997). Organic Chemistry (6th Ed.). New Delhi,

India: Prentice Hall of India

Graham Solomon et al (2014). Organic Chemistry (11th Ed.). New Jeysey,

United States: John Willey and Sons.

Nadendla R. R. (2005). Principles of Pharmaceutical Organic Chemistry. New

Delhi, India: MacMillan Publishers

Bruice Y (2013). Organic Chemistry (7th ed.). New York, United States:

Prentice Hall Pearson.

Delgado J. N. Et al (1998). Wilson and Gisvold's Textbook of Organic

Medicinal and Pharmaceutical Chemistry (10th Ed.). Calfornia, United

States: Lippincott Williams

Bhassin S.K, Gupta R.(2013). Pharmaceutical organic chemistry (E-book

Kindle edition). New Delhi, India: Elsevier Publishing Services

Session 11: Alcohols of Pharmaceutical Importance.

Total Session Time: 120 minutes

Prerequisites

None

Learning Tasks

By the end of this session students are expected to be able to:

• Define alcohols
• List alcohols and their isomers
• Explain nomenclature of alcohols
• Draw chemical structure of alcohols
• List chemical properties of alcohols
• Explain chemical reactions of alcohols

Resources Needed:

• Flip charts, marker pens, and masking tape.
• Black/white board and chalk/whiteboard markers.

SESSION OVERVIEW

|Step |Time |Activity/ |Content |

| | |Method | |

|1 |05 minutes |Presentation |Introduction, Learning Tasks |

|2 |10 minutes |Brainstorming |Definition of Alcohols |

| | |Presentation | |

|3 |15 minutes |Presentation |Alcohols and their Isomers |

|4 |15 minutes |Presentation |Nomenclature of Alcohols |

|5 |15 minutes |Presentation |Chemical Structure of Alcohols |

|6 |15 minutes |Presentation |Chemical Properties of Alcohols |

| | |Buzzing | |

|7 |35 minutes |Group |Chemical Reactions involving |

| | |discussion |Alcohols |

| | |Presentation | |

|8 |05 minutes |Presentation |Key Points |

|9 |05 minutes |Presentation |Evaluation |

SESSION CONTENTS.

STEP 1: Presentation of Session Title and Learning Tasks (5 minutes)

READ or ASK students to read the learning tasks and clarify

ASK students if they have any questions before continuing.

STEP 2: Definition of Alcohols (10 minutes).

|Activity: Brainstorming (5 minutes) |

| |

|Ask students to brainstorm on the following question: |

| |

|What are Alcohols? |

| |

|ALLOW few students to respond. |

| |

|WRITE their responses on the flip chart/ board. |

| |

|CLARIFY and SUMMARISE by using the content below |

• Alcohols are a family of organic compounds containing a hydroxyl (OH)

group bonded to an sp3 hybridized carbon atom.

• The general formula is CnH(2n+1)OH or R-OH.
• Alcohols are named in similar manner as in alkenes except that the suffix

–e from alkanes is replace by –ol.

• Alcohols are classified into three groups: primary, secondary, and

tertiary, depending on the carbon atom bonded to the – OH group.

• If the carbon atom is primary (bonded to one other carbon atom), the

compound is a primary alcohol.

• If the OH group is attached to a carbon atom that is joined to two other

carbon atoms, it is a secondary alcohol, and the carbon atom to which it

is attached is a secondary carbon atom.

• If the OH group is attached to a carbon atom that is joined to three

other carbon atoms, it is a tertiary alcohol, and the carbon atom to

which it is attached is a tertiary carbon atom.

Primary alcohols

[pic]

[pic]

[pic]

STEP 3: Alcohols and their Isomers (15 Minutes).

• Alcohols exhibit following three types of isomerism.

o Chain isomerism.

o Position isomerism.

o Functional isomerism.

Chain isomerism

• Alcohols containing at least 4-carbon atoms form chain isomerism due

to the different structure of C-skeleton in the longest chain.

[pic]

Position isomerism

• Alcohols containing at least 3 C atoms form position isomerism due to

a different position of a hydroxyl group (OH).

Example

[pic]

Functional isomer

• Alcohols containing at least 2 carbon atoms give functional isomers.

The functional isomer of an alcohol is ether.

Example

[pic]

STEP 4: Nomenclature of Alcohols (15 minutes).

• The IUPAC system provides unique names for alcohols, based on rules that

are similar to those for other classes of compounds.

• In general, the name carries the -ol suffix, together with a number to

give the location of the hydroxyl group

Rules

1. Select the longest continuous carbon atom chain containing the carbinol

(hydroxyl) group(s).

2. Number the chain, giving the hydroxyl (alcohol) substituent(s) the

lowest number possible.

3. Name the longest chain as an alkane, but drop the terminal -e and add

-ol. Ethane would become ethanol.

4. For monohydric alcohols the letter ‘-e’ at the end of the root name is

replaced by the ending ‘-ol’ with a number, when necessary, to show the

position of the –OH group on the carbon skeleton.

[pic]

5. If more than one hydroxyl group is present (for polyols i.e.

dihydric,trihydric etc.) the name becomes -diol, -triol, etc. and the

terminal -e in the parent name is not dropped from the alkane name.

But the letters ‘diol’ ‘triol’ etc & numbers 1,2,3 etc are added to the

ending to show how many –OH groups & their position

• For example ethane with two hydroxyl groups would become ethanediol.
• Since the hydroxyl groups could be on or different carbon atoms, one

needs to specify where the hydroxyl groups are attached.

• The name would be 1,2-ethanediol if the hydroxyl groups are on adjacent

(vicinal) carbon atoms.

• Updated nomenclature rules suggest that the position of substituent
attachment should precede the functional group name, e.g., ethane-1,2-

diol rather than 1,2-ethanediol.

6. Indicate by numbers the positions of other groups attached to the

parent chain

• OH group takes priority (even over -ene or -yne)

[pic]

• Considering the example below:

[pic]

• The complete IUPAC name is 1-bromo-3,3-dimethyl-2-butanol.
• The new IUPAC positioning of numbers would place the 2 next to the

group it locates (-ol), giving the name 1-bromo-3,3-dimethylbutan-2-

ol.

7. Cyclic alcohols are named using the prefix cyclo-; the hydroxyl

group is assumed to be on carbon number 1, C1.

[pic]

[pic]

STEP 5: Chemical Structure of Alcohols (15 minutes).

• Alcohols fall into different classes depending on how the -OH group is

positioned on the chain of carbon atoms. There are some chemical

differences between the various types.

Primary alcohols

• In a primary (1°) alcohol, the carbon atom that carries the -OH group

is only attached to one alkyl group. Some examples of primary alcohols

are shown below:

[pic]

• Notice that the complexity of the attached alkyl group is irrelevant.
• In each case there is only one linkage to an alkyl group from the CH2

group holding the -OH group.

• There is an exception to this. Methanol, CH3OH, is counted as a primary

alcohol even though there are no alkyl groups attached to the the -OH

carbon atom.

• Secondary alcohols

In a secondary (2°) alcohol, the carbon atom with the -OH group attached is

joined directly to two alkyl groups, which may be the same or different.

Examples include the following:

[pic]

• Tertiary alcohols
• In a tertiary (3°) alcohol, the carbon atom holding the -OH group is

attached directly to three alkyl groups, which may be any combination of

the same or different groups.

• Examples of tertiary alcohols are given below:

[pic]

STEP 6: Chemical Properties of Alcohols (15 minutes).

|Activity: Buzzing (5minutes) |

| |

|ASK students to pair up and buzz on the following question for 5 |

|minutes |

| |

|What are the chemical properties of Alcohols? |

| |

|ALLOW pairs to respond on the question |

| |

|WRITE their response on the flip chart/board |

| |

|CLARIFY and SUMMARIZE by using the content below |

The following are some chemical properties of alcohols

• Combustion

o Alcohols burns in oxygen to produce carbon dioxide and water. An

alcohol burns cleanly and easily and does not produce soot.

o It becomes increasingly more difficult to burn alcohols as the

molecules get bigger.

o The general molecular equation for the reaction is:

CnH2n+1OH + (1.5n)O2 → (n+1)H2O + nCO2

• Dehydration of Alcohol to alkene

o Dehydration of alcohols is done by heating with concentrated sulfuric

acid, which acts as the dehydrating agent, at 180°C.

o This reaction uses alcohols to produce corresponding alkenes and water

as byproduct.

[pic]

• Oxidation

o Primary alcohols (R-CH2-OH) can be oxidized either to aldehydes (R-

CHO) or to carboxylic acids (R-CO2H), while the oxidation of secondary

alcohols (R1R2CH-OH) normally terminates at the ketone (R1R2C=O)

stage.

o Tertiary alcohols (R1R2R3C-OH) are resistant to oxidation.

o e.g. oxidation of ethanol:

C2H5OH + [O] → CH3COOH + H2O

o Oxidation can be done by using oxidising agents such as acidified

potassium dichromate (VI), acidified potassium manganate (VII) etc

• Esterification

o Alcohols can be reacted with carboxylic acid to form esters.

[pic]

STEP 7: Chemical Reactions involving Alcohols (35 minutes).

|Activity: Small Group Discussion (15 minutes) |

| |

|DIVIDE students into small groups. |

| |

|ASK students to discuss in groups on the following questions. |

|What are the chemical reactions involving Alcohols? |

| |

|[pic]REFER Students to Book |

| |

|ALLOW students to discuss for 10 minutes. |

| |

|ALLOW each group to present for 5 minutes. |

| |

|CLARIFY and SUMMARIZE by using the contents below |

• Oxidation

o Oxidation products depend on whether the alcohol is 1o, 2o or 3o.

o In oxidation one or more hydrogen atoms are lost from the carbon

having OH group

o Primary and secondary alcohols are easily oxidized by a variety of

reagents, including chromium reagents, permanganate, nitric acid, and

even household bleach (NaOCl, sodium hypochlorite).

• Oxidation of Primary Alcohols

o Oxidation of a primary alcohol initially forms an aldehyde.

o Unlike a ketone, however, an aldehyde is easily oxidized further to

give a carboxylic acid.

[pic]

o Chromic acid generally oxidizes a primary alcohol all the way to the

carboxylic acid.

[pic]

o A better reagent for the limited oxidation of primary alcohols to

aldehydes is pyridinium chlorochromate (PCC), a complex of chromium

trioxide with pyridine and HCI.

[pic]

• Oxidation of Secondary Alcohols

o Secondary alcohols are easily oxidized to give excellent yields of

ketones.

o The chromic acid reagent is often best for laboratory oxidations of

secondary alcohols.

[pic]

[pic]

Example:

[pic]

• Resistance of Tertiary Alcohols to Oxidation

o Oxidation of tertiary alcohols is not an important reaction in organic

chemistry.

o Tertiary alcohols have no hydrogen atoms on the carbinol carbon atom,

so oxidation must take place by breaking carbon-carbon bonds.

o These oxidations require severe conditions and result in mixtures of

products.

[pic]

Summary of alcohol oxidations

[pic]

o Two other strong oxidants are potassium permanganate and nitric

acid.

o Both of these reagents are less expensive than the chromium

reagents, and both of them give byproducts that are less

environmentally hazardous than spent chromium reagents.

o Both permanganate and nitric acid oxidize secondary alcohols to

ketones and primary alcohols to carboxylic acids.

• Dehydration

o Dehydration requires an acidic catalyst to protonate the hydroxyl

group of the alcohol and convert it to a good leaving group.

o Loss of water, followed by loss of a proton, gives the alkene

o Dehydration results from E1 elimination of the protonated alcohol

o Example:

[pic]

o Alcohol dehydrations generally take place through the E1 mechanism.

o Protonation of the hydroxyl group converts it to a good leaving

group.

o Water leaves, forming a carbocation.

o Loss of a proton gives the alkene.

[pic]

[pic]

o Because the rate limiting step is formation of a carbocation, the

ease of dehydration follows from the ease of formation of

carbocations: 3° > 2° > 1

o As in other carbocation reactions, rearrangements are common.

• Esterification

o Alcohols can combine with many kinds of acids to form esters.

o When no type of acid is specified, the word ester is assumed to

mean a carboxylic ester, the ester of an alcohol and a carboxylic

acid.

o The reaction, called Fischer esterification, is characterized by

the combining of an alcohol and an acid (with acid catalysis) to

yield an ester plus water.

[pic]

o Under appropriate conditions, inorganic acids also react with

alcohols to form esters.

▪ To form these esters, a wide variety of specialized

reagents and conditions can be used.

[pic]

Acidity of alcohols: formation of alkoxides

• Alcohols are weak acids.
• The most acidic simple alcohols (methanol and ethanol) are about as

acidic as water, and most other alcohols are somewhat less acidic.

• A strong base can deprotonate an alcohol to yield an alkoxide ion (R−O−).
• For example, sodamide (NaNH2), a very strong base, abstracts the hydrogen

atom of an alcohol.

• Metallic sodium (Na) or potassium (K) is often used to form an alkoxide

by reducing the proton to hydrogen gas.

[pic]

• Alkoxides can be useful reagents.
• For example, the most common synthesis of ethers involves the attack of

an alkoxide ion on an alkyl halide.

• This method is called Williamson ether synthesis.

Summary of common reactions of alcohols

[pic]

STEP 8: Key Points (10 minutes)

• Alcohols are a family of organic compounds containing a hydroxyl (OH)

group bonded to an sp3 hybridized carbon atom.

• The general formula is CnH(2n+1)OH or R-OH.
• Alcohols are classified into three groups: primary, secondary, and

tertiary, depending on the carbon atom bonded to the – OH group.

• Alcohols shows chain, position and functional isomerism.
• Primary alcohols (R-CH2-OH) can be oxidized either to aldehydes (R-CHO)

or to carboxylic acids (R-CO2H), while the oxidation of secondary

alcohols (R1R2CH-OH) normally terminates at the ketone (R1R2C=O) stage
• Alcohols can combine with many kinds of acids to form esters.

STEP 9: Evaluation (10 minutes)

• What are alcohols?
• List three alcohols and their isomers
• How are alcohols classified?
• Draw chemical structure of tertiary alcohol
• What is Fischer esterification?

References

Ternay, A.L (1976). Contemporary Organic Chemistry. Philadelphia, United

States: W.B. Saunders Co.

Morrison R.T and Boyd R N (1997). Organic Chemistry (6th Ed.). New Delhi,

India: Prentice Hall of India

Graham Solomon et al (2014). Organic Chemistry (11th Ed.). New Jeysey,

United States: John Willey and Sons.

Nadendla R. R. (2005). Principles of Pharmaceutical Organic Chemistry. New

Delhi, India: MacMillan Publishers

Bruice Y (2013). Organic Chemistry (7th ed.). New York, United States:

Prentice Hall Pearson.

Delgado J. N. Et al (1998). Wilson and Gisvold's Textbook of Organic

Medicinal and Pharmaceutical Chemistry (10th Ed.). California, United

States: Lippincott Williams

Bhassin S.K, Gupta R.(2013). Pharmaceutical organic chemistry (E-book

Kindle edition). New Delhi, India: Elsevier Publishing Services

Session 12: Carboxylic Acids of Pharmaceutical Importance.

Total Session Time: 120 minutes

Prerequisites

• None

Learning Tasks.

By the end of this session students are expected to be able to:

• Define carboxylic acids
• Explain nomenclature of carboxylic acids
• Draw chemical structure of carboxylic acids
• List chemical properties of carboxylic acids
• Explain chemical reactions of carboxylic acids

Resources Needed:

• Flip charts, marker pens, and masking tape.
• Black/white board and chalk/whiteboard markers.

SESSION OVERVIEW

|Step |Time |Activity/ |Content |

| | |Method | |

|1 |05 minutes |Presentation |Introduction, Learning Tasks |

|2 |10 minutes |Brainstorming |Definition of Carboxylic Acids |

| | |Presentation | |

|3 |15 minutes |Presentation |Nomenclature of Carboxylic Acids |

|4 |15 minutes |Presentation |Chemical Structure of Carboxylic |

| | | |Acids |

|5 |15 minutes |Buzzing |Chemical Properties of Carboxylic |

| | |Presentation |Acids |

|6 |40 minutes |Group |Chemical Reactions involving |

| | |discussion |Carboxylic Acids |

| | |Presentation | |

|7 |10 minutes |Presentation |Key Points |

|8 |10 minutes |Presentation |Evaluation |

SESSION CONTENTS

STEP 1: Presentation of Session Title and Learning Tasks (5 minutes)

READ or ASK students to read the learning tasks and clarify

ASK students if they have any questions before continuing.

STEP 2: Definition of Carboxylic Acids (10 minutes).

|Activity: Brainstorming (5 minutes) |

| |

|Ask students to brainstorm on the following question: |

| |

|What is carboxylic acid? |

| |

|ALLOW few students to respond |

| |

|WRITE their responses on the flip chart/ board |

| |

|CLARIFY and SUMMARISE by using the content below |

• The combination of a carbonyl group and a hydroxyl on the same carbon

atom is called a carboxyl group.

• Compounds containing the carboxyl group are distinctly acidic and are

called carboxylic acids.

[pic]

Condensed structures

[pic]

• Therefore, carboxylic acids are acidic organic compounds containing the

carboxyl group as a functional group, attached to hydrogen HCOOH or an

alkyl group as RCOOH or an aryl group as ArCOOH

• The general formula would be CnH(2n+1)COOH or R-CO2H.

STEP 3: Nomenclature of Carboxylic Acids (15 minutes).

IUPAC Names

• The IUPAC nomenclature for carboxylic acids uses the name of the alkane

that corresponds to the longest continuous chain of carbon atoms.

• The final -e in the alkane name is replaced by the suffix -oic acid.
• The chain is numbered, starting with the carboxyl carbon atom, to give

positions of substituents along the chain.

In naming, the carboxyl group takes priority over any of the functional

groups discussed previously

=Examples

[pic]

[pic]

[pic]

[pic]

[pic]

[pic]

[pic]

[pic]

Some more examples of traditional names most widely used are:

• Formic acid- HCOOH
• Acetic acid – CH3COOH
• Propionic acid – CH3CH2COOH
• Butyric acid – CH3(CH2)2COOH
• Valeric acid – CH3(CH2)3COOH
• Caproic acid – CH3(CH2)4COOH
• Capyrylic acid – CH3(CH2)6COOH
• Capric acid – CH3(CH2)8COOH

STEP 4: Chemical Structure of Carboxylic Acids (15 minutes)

• The CO2H unit is planar and consistent with sp2 hybridization and a

resonance interaction of the lone pairs of the hydroxyl oxygen with the π

system of the carbonyl.

| Carboxylic Acid | | |

| |Structure | |

|Ethanoic acid |CH3CO2H | |

|Propanoic acid |CH3CH2CO2H | |

|Fluoroethanoic acid |CH2FCO2H | |

|Chloroethanoic acid |CH2ClCO2H | |

|Dichloroethanoic acid |CHCl2CO2H | |

|Trichloroethanoic acid |CCl3CO2H | |

|Nitroethanoic acid |O2NCH2CO2H | |

STEP 5: Chemical Properties of Carboxylic Acids (15 minutes).

|Activity: Buzzing (10minutes) |

| |

|ASK students to pair up and buzz on the following question for 5 |

|minutes |

| |

|What are the chemical properties of carboxylic acid? |

| |

|ALLOW pairs to respond on the question |

| |

|WRITE their response on the flip chart/board |

| |

|CLARIFY and SUMMARIZE by using the content in the table 1 below |

The following are chemical properties of carboxylic acids;

Acidity of Carboxylic Acids

Carboxylic acids are weak acids and their carboxylic anions are strong

conjugate bases are slightly alkaline due to the hydrolysis of carboxylate

anion compared to other species, the order of acidity and basicity or

corresponding conjugate bases are as follows:

Acidity RCOOH > HOH > ROH > HC[pic] CH > NH3 > RH

Basicity RCOO– < HO– < RO– < HCC– < NH2-< R–

Reaction of Carboxylic Acids with Metals

• The carboxylic acids react with metals to liberate hydrogen and are

soluble in both NaOH and NaHCO3 solutions.

For example;

▪ 2CH3COOH + 2Na → 2CH3COO–Na+ + H2
▪ CH3COOH + NaOH → CH3COO–Na+ + H2O
▪ CH3COOH + NaHCO3 → CH3COO–Na+ + H2O + CO2
• Carboxylic acids dissociate in water to give resonance stabilised

carboxylate anions and hydronium ion.

[pic]

Effect of substituents on the acidity of Carboxylic Acids

• Any factor that stabilizes the anion more than it stabilizes the acid

would increase the acidity of carboxylic acids.

• While any factor that decreases the stability of anion would decrease the

acidity of carboxylic acids.

• Electron withdrawing groups disperse the negative charge and thus

stabilize the anion which results in increase in acidity of the

carboxylic acids.

• Electron donating groups intensify the negative charge and destabilize

the anion which results in decrease in acidity of carboxylic acid.

[pic]

Conversion of Carboxylic Acids into functional derivatives

• Carboxylic acids can be converted into number of other compounds (known

as derivatives of carboxylic acids or simply acid derivatives)

by replacement of its –OH group by a Cl, OR or NH2 .

o Replacement of -OH by -Cl forms acid chlorides.

o Replacement of -OH by -OR forms ester.

o Replacement of -OH by -NH2 forms amide.

• Carboxylic acid can be recovered simply by hydrolysis of the acid

derivatives or one can say that the functional derivatives are all

readily reconverted into the acid by simple hydrolysis.

Conversion of Carboxylic Acids into Amide

• Reaction of carboxylic acids with ammonia results in formation of give

ammonium salt which on further heating at high temperature give

amides.

[pic]

Conversion of Carboxylic Acids into Acid Anhydrides

Lower monocarboxylic acid on heating with dehydrating agent (say P2O5)

forms anhydrides

[pic]

Note: Anhydride of formic acid is not known, it gives CO and H2O on

heating with conc. H2SO4.

HCOOH +Conc. H2SO → H2O + CO

Reduction of Carboxylic Acids to Alcohols

Lithium aluminium hydride can reduce an acid to an alcohol; the

initial product is an alkoxide from which the alcohol is liberated by

hydrolysis.

4 R–COOH + 3LiAlH4 → 4H2 + 2LiAlO2 + (RCH2O)4 AlLi + H2

Hydrolysis of (RCH2O)4 → alcohols

STEP 6: Chemical Reactions involving Carboxylic Acids (40 minutes).

|Activity: Small Group Discussion (20 minutes) |

| |

|DIVIDE students into small groups. |

| |

|ASK students to discuss in groups on the following questions |

|What are the chemical reactions involving carboxylic acids? |

| |

|[pic]REFER Students to Book |

| |

|ALLOW students to discuss for 15 minutes |

| |

|ALLOW each group to present for 5 minutes |

| |

|CLARIFY and SUMMARIZE by using the contents below |

An overview of the chemical reactions involving carboxylic acids

[pic]

Acid chloride (ROCl)

• Acid chlorides are formed when carboxylic acids react with thionyl

chloride (SOCl2).

• They are the most reactive derivatives of carboxylic acid.

[pic]

Ester (RCOOR’)

• Esters are derived when a carboxylic acid reacts with an alcohol.
• Esters containing long alkyl chains (R) are main constituents of animal

and vegetable fats and oils.

• Many esters containing small alkyl chains are fruity in smell and are

commonly used in fragrances.

[pic]

The acid-catalyzed esterification of carboxylic acids with alcohols to give

esters is termed Fischer esterification.

Thioester (RCOSR’)

• Thioesterification: A thioester is formed when a carboxylic acid

reacts with a thiol (RSH) in the presence of an acid.

[pic]

• Thioesters are commonly found in biochemistry, the best-known example

being acetyl CoA.

Acid anhydride

• An acid anhydride is a compound that has two acyl groups (R-C=O) bonded

to the same oxygen atom.

• Anhydrides are commonly formed when a carboxylic acid reacts with an acid

chloride in the presence of a base.

[pic]

• The mechanism by which a carboxylic acid anhydride is synthesized is as

follows;

[pic]

• This reaction follows an addition-elimination mechanism in which the

chloride anion (Cl-) is the leaving group.

o In the first step, the base abstracts a proton (H+) from the

carboxylic acid to form the corresponding carboxylate anion (1).

o The carboxylate anion's negatively charged oxygen attacks the

considerably electrophilic acyl chloride's carbonyl carbon.

o As a result, a tetrahedral intermediate (2) is formed.

o In the final step, chloride – a good leaving group – is eliminated

from the tetrahedral intermediate to yield the acid anhydride.

Amide

• The direct conversion of a carboxylic acid to an amide is difficult

because amines are very basic and tend to convert carboxylic acids to

their highly unreactive carboxylate ions.

o Therefore, DCC (Dicyclohexylcarbodiimide) is used to drive this

reaction.

[pic]

[pic]

• A carboxylic acid first adds to the DCC molecule to form a good leaving

group, which can then be displaced by an amine during nucleophilic

substitution to form the corresponding amide.

Relative reactivity of the carboxylic acid derivatives towards a

nucleophilic substitution reaction

[pic]

• Consider the carboxylic acid derivatives as an acyl group, R-C=O,

attached to a substituent (X).

o These derivatives also undergo a nucleophilic substitution reaction

with a nucleophile (Nu) as shown above.

o The reactivity of these derivatives towards nucleophilic substitution

is governed by the nature of the substituent X present in the acid

derivative.

o If the substituent (X) is electron donating, it reduces the

electrophilic nature of the carbonyl group by neutralizing the partial

positive charge developed on the carbonyl carbon, and thus makes the

derivative less reactive to nucleophilic substitution.

o If the substituent (X) is electron withdrawing, then it increases the

electrophilic nature of carbonyl group by pulling the electron density

of the carbonyl bond towards itself, making the carbonyl carbon more

reactive to nucleophilic substitution.

|Derivative |

|Substituent (X) |

|Electronic effect of X |

|Relative reactivity |

| |

|Acid chloride |

|-Cl |

|electron withdrawing |

|1 (most reactive) |

| |

|Acid anhydride |

|-OC=OR |

|electron withdrawing |

|2 (almost as reactive as 1) |

| |

|Thioester |

|-SR |

|weakly electron donating |

|3 |

| |

|Ester |

|-OR |

|alkoxy (-OR) group is weakly electron donating |

|4 |

| |

|Amide |

|-NH2 |

|very strongly donating |

|5 |

| |

|Carboxylate ion |

|-O |

|Carboxylate ions are not reactive because their negative charge |

|repels the approach of other nucleophiles |

|6 (least reactive) |

| |

• Thus, on a reactivity scale, the order of reactivity of various

carboxylic acid derivatives towards nucleophilic substitution is as

follows:

Acid halide > acid anhydride > thioester > ester > amide

STEP 7: Key Points (10 minutes)

• A carboxylic acid is an organic compound that contains a carboxyl
group (C(=O)OH).
• The general formula of a carboxylic acid is R–COOH, with R referring

to the rest of the molecule.

• Carboxylic acids are generally more acidic than other

organic compounds containing hydroxyl groups but are generally weaker

than the familiar mineral acids (e.g., hydrochloric acid,

HCl, sulfuric acid, H2SO4, etc.).

• The chemical reactions involving carboxylic acids include conversion

of carboxylic acids into esters, amides, carboxylate salts, acid

chlorides, and alcohols.

• Carboxylic acids react with bases to form carboxylate salts, in which

the hydrogen of the hydroxyl (–OH) group is replaced with a

metal cation.

STEP 8: Evaluation (10 minutes)

• What are carboxylic acids?
• Write chemical five structures of carboxylic acids
• What are the chemical properties of carboxylic acids?
• Mention three chemical reactions involving carboxylic acids.

References

Ternay, A.L (1976). Contemporary Organic Chemistry. Philadelphia, United

States: W.B. Saunders Co.

Morrison R.T and Boyd R N (1997). Organic Chemistry (6th Ed.). New Delhi,

India: Prentice Hall of India

Graham Solomon et al (2014). Organic Chemistry (11th Ed.). New Jeysey,

United States: John Willey and Sons.

Nadendla R. R. (2005). Principles of Pharmaceutical Organic Chemistry. New

Delhi, India: MacMillan Publishers

Bruice Y (2013). Organic Chemistry (7th ed.). New York, United States:

Prentice Hall Pearson.

Delgado J. N. Et al (1998). Wilson and Gisvold's Textbook of Organic

Medicinal and Pharmaceutical Chemistry (10th Ed.). California, United

States: Lippincott Williams

Bhassin S.K, Gupta R.(2013). Pharmaceutical organic chemistry (E-book

Kindle edition). New Delhi, India: Elsevier Publishing Services

Session 13: Esters of Pharmaceutical Importance.

Total Session Time: 120 minutes

Prerequisites

• None

Learning Tasks

By the end of this session students are expected to be able to:

• Define esters
• Explain nomenclature of esters
• Draw chemical structure of esters
• List chemical properties of esters
• Explain chemical reactions of esters

Resources Needed:

• Flip charts, marker pens, and masking tape
• Black/white board and chalk/whiteboard markers

SESSION OVERVIEW

|Step |Time |Activity/ |Content |

| | |Method | |

|1 |05 minutes |Presentation |Introduction, Learning Tasks |

|2 |10 minutes |Brainstorming |Definition of Esters |

| | |Presentation | |

|3 |15 minutes |Presentation |Nomenclature of Esters |

|4 |15 minutes |Presentation |Chemical Structure of Esters |

|5 |20 minutes |Buzzing |Chemical Properties of Esters |

| | |Presentation | |

|6 |40 minutes |Group |Chemical Reactions involving Esters |

| | |discussion | |

| | |Presentation | |

|7 |05 minutes |Presentation |Key Points |

|8 |10 minutes |Presentation |Evaluation |

SESSION CONTENTS

STEP 1: Presentation of Session Title and Learning Tasks (5 minutes)

READ or ASK students to read the learning tasks and clarify.

ASK students if they have any questions before continuing.

STEP 2: Definition of Esters (10 minutes)

|Activity: Brainstorming (5 minutes) |

| |

|Ask students to brainstorm on the following question: |

| |

|What is Ester? |

| |

|ALLOW few students to respond. |

| |

|WRITE their responses on the flip chart/ board. |

| |

|CLARIFY and SUMMARISE by using the content below |

• An ester (“carboxylic ester” in the textbook) is a derivative of a

carboxylic acid in which there is a carbon group connected to the single-

bonded oxygen:

• In an ester, The H in the carboxyl group is replaced with an alkyl group.

[pic] [pic]

Some common esters are as follows;

[pic]

STEP 3: Nomenclature of Esters (15 minutes).

• Name the alkyl or aromatic portion contributed by the “alcohol part”

first.

[pic]

• The “acid part” is named as a carboxylic acid, with the -ic acid suffix

changed to -ate

[pic]

[pic]

STEP 4: Chemical Structure of Esters (15 minutes).

• Esters contain a carbonyl center, which gives rise to 120-degree C-C-O

and O-C-O bond angles due to sp2 hybridization.

• Unlike amides, esters are structurally flexible functional groups because

rotation about the C-O-C bonds has a lower energy barrier.

• Their flexibility and low polarity affect their physical properties on a

macroscopic scale.

• They tend to be less rigid, leading to a lower melting point, and more

volatile, leading to a lower boiling point, than the corresponding

amides.

• The pKa of the alpha-hydrogens, or the hydrogens attached to the carbon

adjacent to the carbonyl, on esters is around 25, making them essentially

non-acidic except in the presence of very strong bases.

[pic]

• An ester is characterized by the orientation and bonding of the atoms

shown, where R and R’ are both carbon-initiated chains of varying length,

also known as alkyl groups.

• As usual, R and R’ are either alkyl groups or groups initiating with

carbon.

• Esters are derivative of carboxylic acids where the hydroxyl (OH) group

has been replaced by an alkoxy (O-R) group.

• They are commonly synthesized from the condensation of a carboxylic acid

with an alcohol.

STEP 5: Chemical Properties of Esters (20 minutes).

|Activity: Buzzing (5minutes) |

| |

|ASK students to pair up and buzz on the following question for 5 |

|minutes |

| |

|What are the chemical properties of Esters? |

| |

|ALLOW pairs to respond on the question |

| |

|WRITE their response on the flip chart/board |

| |

|CLARIFY and SUMMARIZE by using the content in the table 1 below |

In acid hydrolysis

• An ester reacts with water to produce a carboxylic acid and an alcohol.
• An acid catalyst is required.

[pic]

Base hydrolysis

Base hydrolysis is the reaction of an ester with a strong base. Produces

the salt of the carboxylic acid and an alcohol.

[pic]

STEP 6: Chemical Reactions involving Esters (40 minutes).

|Activity: Small Group Discussion (15 minutes) |

| |

|DIVIDE students into small groups |

| |

|ASK students to discuss in groups on the following questions |

|What are the chemical reactions involving Esters? |

| |

|[pic]REFER Students to Book |

| |

|ALLOW students to discuss for 10 minutes |

| |

|ALLOW each group to present for 5 minutes |

| |

|CLARIFY and SUMMARIZE by using the contents below |

Esterification Reaction

• The simplest way to synthesize an ester is to heat a carboxylic acid with

an alcohol or phenol (plus an acid catalyst).

o The oxygen of the alcohol adds to the carboxyl group, splitting out

a molecule of water in the process (an esterification reaction).

[pic]

• Since this reaction is a reversible reaction, it often reaches an

equilibrium with a large amount of unreacted starting material still

present.

• Better yields are obtained using either acid chlorides or acid anhydrides

as starting materials.

o These reactions are nonreversible

[pic]

Examples

[pic]

Ester Hydrolysis

• Esters may be broken apart under acidic conditions by water (a hydrolysis

reaction) to form a carboxylic acid and an alcohol.

[pic]

• This is essentially the reverse reaction of the synthesis of esters from

carboxylic acids and alcohols.

Base hydrolysis (Saponification)

Esters may be broken apart under basic conditions by sodium hydroxide (lye)

or potassium hydroxide to form carboxylate salts and alcohols.

[pic]

This reaction is important in the production of soaps

STEP 8: Key Points (05 minutes)

• An ester is a chemical compound derived from carboxylic acid in which

at least one –OH (hydroxyl) group is replaced by an –O–alkyl (alkoxy)

group.

• An ester reacts with water to produce a carboxylic acid and an alcohol.
• Esters may be broken apart under acidic conditions by water (a hydrolysis

reaction) to form a carboxylic acid and an alcohol.

STEP 9: Evaluation (10 minutes)

• What are esters?
• How are the esters named?
• Write general chemical structure of esters.
• What is saponification?

References

Ternay, A.L (1976). Contemporary Organic Chemistry. Philadelphia, United

States: W.B. Saunders Co.

Morrison R.T and Boyd R N (1997). Organic Chemistry (6th Ed.). New Delhi,

India: Prentice Hall of India

Graham Solomon et al (2014). Organic Chemistry (11th Ed.). New Jeysey,

United States: John Willey and Sons.

Nadendla R. R (2005). Principles of Pharmaceutical Organic Chemistry. New

Delhi, India: MacMillan Publishers

Bruice Y (2013). Organic Chemistry (7th ed.). New York, United States:

Prentice Hall Pearson.

Delgado J. N. Et al (1998). Wilson and Gisvold's Textbook of Organic

Medicinal and Pharmaceutical Chemistry (10th Ed.). Calfornia, United

States: Lippincott Williams

Bhassin S.K, Gupta R. (2013). Pharmaceutical organic chemistry (E-book

Kindle edition). New Delhi, India: Elsevier Publishing Services

Session 14: Acyl Chlorides of Pharmaceutical Importance.

Total Session Time: 120 minutes

Prerequisites

• None

Learning Tasks

By the end of this session students are expected to be able to:

• Define acyl chlorides
• Explain nomenclature of acyl chlorides
• Explain physical properties of acyl chlorides
• Describe the preparation of acyl chlorides
• Explain chemical reactions of acyl chlorides

Resources Needed:

• Flip charts, marker pens, and masking tape
• Black/white board and chalk/whiteboard markers

SESSION OVERVIEW

|Step |Time |Activity/ |Content |

| | |Method | |

|1 |05 minutes |Presentation |Introduction, Learning Tasks |

|2 |10 minutes |Brainstorming |Definition of acyl chlorides |

| | |Presentation | |

|3 |15 minutes |Presentation |Nomenclature of acyl chlorides |

|4 |15 minutes |Presentation |Physical Properties of Acyl |

| | | |Chlorides |

|5 |20 minutes |Buzzing |Preparation of Acyl Chlorides |

| | |Presentation | |

|6 |40 minutes |Group |Chemical Reactions involving acyl |

| | |discussion |chlorides |

| | |Presentation | |

|7 |05 minutes |Presentation |Key Points |

|8 |10 minutes |Presentation |Evaluation |

SESSION CONTENTS

STEP 1: Presentation of Session Title and Learning Tasks (5 minutes).

READ or ASK students to read the learning tasks and clarify.

ASK students if they have any questions before continuing.

STEP 2: Definition of Acyl Chlorides (10 minutes).

|Activity: Brainstorming (5 minutes) |

| |

|Ask students to brainstorm on the following question: |

|What are acyl chlorides? |

| |

|ALLOW few students to respond. |

| |

|WRITE their responses on the flip chart/ board. |

| |

|CLARIFY and SUMMARISE by using the content below |

• Acyl chlorides as "acid derivatives".
• A carboxylic acid such as ethanoic acid has the structure:

[pic]

• There are a number of related compounds in which the -OH group in the

acid is replaced by something else.

• Compounds like this are described as acid derivatives.
• Acyl chlorides (also known as acid chlorides) are one example of an

acid derivative.

• In this case, the -OH group has been replaced by a chlorine atom.

[pic]

STEP 3: Nomenclature of Acyl Chlorides (15 minutes).

• The easiest way of thinking about the names is to see the relationship

with the corresponding.

|carboxylic acid |acyl chloride |acyl chloride |

|name |name |formula |

|ethanoic acid |ethanoyl |CH3COCl |

| |chloride | |

|propanoic acid |propanoyl |CH3CH2COCl |

| |chloride | |

|butanoic acid |butanoyl |CH3CH2CH2COCl |

| |chloride | |

• The acyl group name is derived from the carboxylic acid name by

replacing -oic acid by -ly.

• If you have something substituted into the hydrocarbon chain, the carbon

in the -COCl group counts as the number 1 carbon.

• For example, 2-methylbutanoyl chloride is:

[pic]

• Note: Hardly anyone ever mentions methanoyl chloride, HCOCl – derived

from methanoic acid.

• That is because methanoyl chloride is very unstable, decomposing to give

carbon monoxide and HCl.

STEP 4: Physical properties of acyl chlorides (15 minutes).

• Appearance

o An acyl chloride like ethanoyl chloride is a colourless fuming liquid.

o The strong smell of ethanoyl chloride is a mixture of the smell of

vinegar (ethanoic acid) and the acrid smell of hydrogen chloride gas.

o The smell and the fumes are because ethanoyl chloride reacts with

water vapour in the air.

• Solubility in water

o Acyl chlorides can't be said to dissolve in water because they react

(often violently) with it.

o The strong reaction means that it is impossible to get a simple

aqueous solution of an acyl chloride.

• Boiling points

o Taking ethanoyl chloride as typical:

o Ethanoyl chloride boils at 51°C.

o It is a polar molecule, and so has dipole-dipole attractions between

its molecules as well as van der Waals dispersion forces.

o However, it doesn't form hydrogen bonds.

o Its boiling point is therefore higher than, say, an alkane of similar

size (which has no permanent dipoles), but not as high as a similarly

sized alcohol (which forms hydrogen bonds in addition to everything

else.)

STEP 5: Preparation of Acyl Chlorides (20 minutes).

|Activity: Buzzing (5minutes) |

| |

|ASK students to pair up and buzz on the following question for 5 |

|minutes. |

|How are acyl chlorides prepared? |

| |

|ALLOW pairs to respond on the question. |

| |

|WRITE their response on the flip chart/board. |

| |

|CLARIFY and SUMMARIZE by using the content in the table 1 below |

• Acyl chlorides are prepared by treatment of carboxylic acids with

thionyl (SOCl2).

[pic]

• Example,

[pic]

STEP 6: Chemical Reactions involving acyl chlorides (40 minutes).

|Activity: Small Group Discussion (15 minutes) |

| |

|DIVIDE students into small groups |

| |

|ASK students to discuss in groups on the following questions |

|What are the chemical reactions involving Esters? |

|[pic]REFER Students to Book |

|ALLOW students to discuss for 10 minutes |

| |

|ALLOW each group to present for 5 minutes |

| |

|CLARIFY and SUMMARIZE by using the contents below |

• Substitution of the chlorine atom by other groups

o Acyl chlorides are extremely reactive, and in their reactions the

chlorine atom is replaced by other groups.

o In each case, in the first instance, hydrogen chloride gas is produced

as steamy acidic fumes.

o However, in some cases the hydrogen chloride goes on to react with one

of the substances in the reaction mixture.

o Taking ethanoyl chloride as typical, the initial reaction is of this

kind:

[pic][pic]

o The reactions involve compounds like water, alcohols and phenols, or

ammonia and amines.

o All of these particular cases contain a very electronegative element

with an active lone pair of electrons – either oxygen or nitrogen.

STEP 8: Key Points (05 minutes).

• An acyl chloride a chemical compound derived from carboxylic acid in

which at least one –OH (hydroxyl) group is replaced by a halogen

chlorine.

• Acyl chlorides are prepared by treatment of carboxylic acids with

thionyl.

• Acyl chlorides are extremely reactive, and in their reactions the

chlorine atom is replaced by other groups

STEP 9: Evaluation (10 minutes).

• What are acyl chlorides?
• How are acyl chlorides named?
• How are acyl chlorides prepared?
• What is solubility of acyl chlorides in water?

References.

Ternay, A.L (1976). Contemporary Organic Chemistry. Philadelphia, United

States: W.B. Saunders Co.

Morrison R.T and Boyd R N (1997). Organic Chemistry (6th Ed.). New Delhi,

India: Prentice Hall of India

Graham Solomon et al (2014). Organic Chemistry (11th Ed.). New Jeysey,

United States: John Willey and Sons.

Rama R. N. (2005). Principles of Pharmaceutical Organic Chemistry. New

Delhi, India: MacMillan Publishers

Bruice Y (2013). Organic Chemistry (7th ed.). New York, United States:

Prentice Hall Pearson.

Delgado J. N. Et al (1998). Wilson and Gisvold's Textbook of Organic

Medicinal and Pharmaceutical Chemistry (10th Ed.). California, United

States: Lippincott Williams.

Bhassin S.K, Gupta R.(2013). Pharmaceutical organic chemistry (E-book

Kindle edition). New Delhi, India: Elsevier Publishing Services.

Session 15: Ethers of Pharmaceutical importance.

Total Session Time: 120 minutes

Prerequisites

• None

Learning Tasks

By the end of this session students are expected to be able to:

• Define ethers
• List ethers and their isomers
• Explain nomenclature of ethers
• Draw chemical structure of ethers
• List chemical properties of ethers
• Explain chemical reactions of ethers

Resources Needed:

• Flip charts, marker pens, and masking tape.
• Black/white board and chalk/whiteboard markers.

SESSION OVERVIEW

|Step |Time |Activity/ |Content |

| | |Method | |

|1 |05 minutes |Presentation |Introduction, Learning Tasks |

|2 |10 minutes |Brainstorming |Definition of Ethers |

| | |Presentation | |

|3 |10 minutes |Presentation |Ethers and their Isomers |

|4 |15 minutes |Presentation |Nomenclature of Ethers |

|5 |15 minutes |Presentation |Chemical Structure of Ethers |

|6 |15 minutes |Buzzing |Chemical Properties of Ethers |

| | |Presentation | |

|7 |35 minutes |Group |Chemical Reactions involving Ethers |

| | |discussion | |

| | |Presentation | |

|8 |05 minutes |Presentation |Key Points |

|9 |05 minutes |Presentation |Evaluation |

SESSION CONTENTS.

STEP 1: Presentation of Session Title and Learning Tasks (5 minutes)

READ or ASK students to read the learning tasks and clarify

ASK students if they have any questions before continuing.

STEP 2: Definition of Ethers (10 minutes).

|Activity: Brainstorming (5 minutes) |

| |

|Ask students to brainstorm on the following question: |

| |

|What are Ethers? |

| |

|ALLOW few students to respond |

| |

|WRITE their responses on the flip chart/ board |

| |

|CLARIFY and SUMMARISE by using the content below |

• Ethers are compounds of formula R – 0 – R ', where R and R' may be alkyl

groups or aryl (benzene ring) groups.

• Ethers can be considered derivatives of water (HOH) or alcohols (ROH) by

replacing an H with an R group.

• In an alcohol, one hydrogen atom of water is replaced by an alkyl group.
• In an ether, both hydrogens are replaced by alkyl groups.

[pic]

• The R groups in ethers can be alkyl, aryl, or alkenyl.
• The R groups can be the same or different.
• If the R groups are the same, the ethers are called symmetrical ethers.
• If the R groups are different, the ethers are called unsymmetrical

ethers.

• Ethers can be cyclic or acyclic.
• Rings that contain an atom other than carbon are called heterocyclic

compounds.

• Oxygen is the heteroatom in cyclic ethers.

Examples of ethers

[pic]

• The most important commercial ether is diethyl ether, often called "ethyl

ether’’, or simply "ether."

STEP 3: Ethers and their Isomers (10 minutes).

Aliphatic Ethers can give two different types of isomers.

• Chain isomerism

o Ethers with the same formula and having different carbon chain

skeletons are called chain isomers.

o Examples:

[pic]

• Functional isomers.

o Ethers are isomeric with alcohols.

o Example:

[pic] is isomeric with ethyl alcohol C2H5OH

• Metamerism

o Isomers with the same molecular formula but different alkyl groups

(around the functional group) are called metamers. An ether with

formula C4H10O has 3 metamers.

[pic]

STEP 4: Nomenclature of Ethers (15 minutes).

• The IUPAC system, generally used with more complicated ethers, is

sometimes called the alkoxy alkane system.

• The common nomenclature of ethers, which is sometimes called the alkyl

alkyl ether system has also been widely used

• IUPAC names use the more complex alkyl group as the root name, and the

rest of the ether as an alkoxy group.

• For example, cyclohexyl methyl ether is named methoxycyclohexane.
• This systematic nomenclature is often the only clear way to name complex

ethers.

Example,

[pic]

[pic]

|Table 1: Common Alkyl and Alkoxy Groups |

|Alkyl Group |Name | |Alkoxy Group |Name |

|CH3– |Methyl | |CH3O– |Methoxy |

|CH3CH2– |Ethyl | |CH3CH2O– |Ethoxy |

|(CH3)2CH– |Isopropyl | |(CH3)2CHO– |Isopropoxy |

|(CH3)3C– |tert-Butyl | |(CH3)3CO– |tert-Butoxy |

|C6H5– |Phenyl | |C6H5O– |Phenoxy |

Common Names (Alkyl Alkyl Ether Names)

• Common names of ethers are formed by naming the two alkyl groups on

oxygen and adding the word ether.

• Under the current system, the alkyl groups should be named in

alphabetical order, but many people still use the old system, which named

the groups in order of increasing complexity.

• For example, if one of the alkyl groups is methyl and the other is t-

butyl, the current common name should be "t-butyl methyl ether,’’

• But most chemists use the older common name, "methyl t-butyl ether"
• If both groups are methyl, the name is "dimethyl ether.’’
• If just one alkyl group is described in the name, it implies the ether is

symmetrical, as in "ethyl ether."

STEP 5: Chemical Structure of Ethers (15 minutes).

• Ethers are a class of organic compounds that contain an ether group.
• An ether group is an oxygen atom connected to two alkyl or aryl groups.
• They follow the general formula R-O-R’. The C-O-C linkage is

characterized by bond angles of 104.5 degrees, with the C-O distances

being about 140 pm.

• The oxygen of the ether is more electronegative than the carbons.
• Thus, the alpha hydrogens are more acidic than in regular hydrocarbon

chains.

[pic]

STEP 6: Chemical Properties of Ethers (15 minutes).

|Activity: Buzzing (5minutes) |

| |

|ASK students to pair up and buzz on the following question for 5 |

|minutes |

| |

|What are the chemical properties of Ethers? |

| |

|ALLOW pairs to respond on the question |

| |

|WRITE their response on the flip chart/board |

| |

|CLARIFY and SUMMARIZE by using the content in the table 1 below |

• Ether are like alkanes, in that they undergo halogenation reactions.
• The oxygen of the ether linkage makes ethers basic, react with proton

donors to form oxonium salts.

• Ethers resist attack by nucleophiles and by bases.

STEP 7: Chemical Reactions involving Ethers (35 minutes).

|Activity: Small Group Discussion (15 minutes) |

| |

|DIVIDE students into small groups |

| |

|ASK students to discuss in groups on the following questions |

|What are the chemical reactions involving Ethers? |

| |

|[pic]REFER Students to Book |

| |

|ALLOW students to discuss for 10 minutes |

| |

|ALLOW each group to present for 5 minutes |

| |

|CLARIFY and SUMMARIZE by using the contents below |

Reactions of Ether Due to an Alkyl Group

• Combustion: Ethers are highly inflammable, and they form extremely

explosive mixtures with air giving CO2 and water.

C2H5O C2H5 + 6O2 → 4CO2 + 5H2O

• Halogenation: The alkyl group undergoes substitution reaction with

chlorine or bromine. The resultant product is halogenated ether in

absence of sunlight. However, in presence of sunlight, it substitutes all

the hydrogen atoms of ethers.

CH3CH2OCH2CH3 [pic] CH3CHCIOCHCICH3 (α α’-dichloro diethyl ether)

CH3CH2OCH2CH3 [pic] C2CI2OC2CI5 (Perchloro diethyl ether)

Reaction of Ether Due to Ethereal Oxygen

• Ethers behave as Lewis bases because of the presence of two lone pairs of

electrons on the oxygen atom.

• Therefore, they form salts with strong acids.
• The oxonium salts are soluble in acid solution. We can facilitate the

regeneration of ether by hydrolysis of these salts.

• [pic]

Ethers also form coordination complexes with Lewis acids like BF3, AICI3,

RMgX etc.

• Therefore, we can derive the fact that ethers are very good solvents for

Grignard reagents.

Formation of Peroxides

• Ethers form peroxide linkage with oxygen when we expose them to air or

ozonized oxygen in presence of sunlight or ultraviolet light.

• These peroxides are highly poisonous in nature.
• They are oily liquids and decompose violently even at low concentrations.
• Therefore, we must ensure never to evaporate esters to dryness because it

might lead to explosive reactions.

• Besides this, we must also check the purity of ether before its use as an

anaesthetic agent.

• An impure ether (having peroxide linkage) gives red colour when shaken

with ferrous ammonium sulphate and potassium thiocyanate.

• This could prove to be lethal for the patients on whom we try

anaesthesia.

• [pic]

On mixing with KI solution, it liberates I2 which turns starch paper

blue.

[pic]

• We can make these ethers free from peroxide linkages by distilling

them with highly concentrated sulphuric acid, H2SO4.

• Also, we can check for the peroxide formation by adding a little

amount of Cu2O to the ether.

Reactions of Ether Involving Cleavage of Carbon-Oxygen Bond

• Action of dil. H2SO4: Ethers, on heating with dilute H2SO4, under high

pressure, hydrolyse to corresponding alcohols.

• Action of Conc. H2SO4: Ethers, on warming with conc. H2SO4, give alkyl

hydrogen sulphate.

R-OR + conc. H2SO4 → 2R HSO4

R-OR’ + conc. H2SO4 → RHSO4 + R’HSO4

• Action of HI:
• The products that we get during the action of HI on ethers depend

mainly upon the temperature in which we carry out the reaction.

• R-OR + HI [pic] R-OH + RI
• R-OR’ + HI [pic] R’-OH + RI
• Note: In case of a mixed ether, halogen atom attaches itself to the

simpler alkyl group.

• CH3OC2H5 + HI → CH3I + C2H5OH

• R-R + HI [pic] 2RI + H2O

We would observe similar reactions with HCI, HBr & the reactivity order is

HI > HBr > HCI.

• Action of PCI5: In the presence of heat, we get the following reaction:

R-O-R + PCI5 [pic] 2RCI + POCI3

• Action of Acetyl chloride or Acetic anhydride:

[pic]

• Dehydration of Ethers:
C2H5OC2H5 [pic] 2CH2=CH2 + H2

Action of Carbon Monoxide:

C2H5OC2H5, + CO [pic] C2H5COOC2H5

ROR + CO → RCOOR

STEP 8: Key Points (05 minutes).

• Ethers are a class of organic compounds that contain an ether group—an

oxygen atom connected to two alkyl or aryl groups.

• Ethers have the general formula R–O–R′, where R and R′ represent the

alkyl or aryl groups.

• Ethers can be named either by identifying the alkyl groups on either side

of the oxygen atom in alphabetical order, then write “ether” or by

using the formal, IUPAC method.

• Ethers have relatively low chemical reactivity, but they are still more

reactive than alkanes.

• Although they resist undergoing hydrolysis, ethers are often cleaved by

acids, which results in the formation of an alkyl halide and an alcohol.

• Ethers tend to form peroxides in the presence of oxygen or air.

STEP 9: Evaluation (05 minutes).

• What are ethers?
• List two ethers and their isomers?
• Draw chemical structure of ethers.
• What are chemical properties of ethers?
• List chemical reactions of ethers.

References

Ternay, A.L (1976). Contemporary Organic Chemistry. Philadelphia, United

States: W.B. Saunders Co.

Morrison R.T and Boyd R N (1997). Organic Chemistry (6th Ed.). New Delhi,

India: Prentice Hall of India

Graham Solomon et al (2014). Organic Chemistry (11th Ed.). New Jeysey,

United States: John Willey and Sons.

Nadendla R. R. (2005). Principles of Pharmaceutical Organic Chemistry. New

Delhi, India: MacMillan Publishers

Bruice Y (2013). Organic Chemistry (7th ed.). New York, United States:

Prentice Hall Pearson.

Delgado J. N. Et al (1998). Wilson and Gisvold's Textbook of Organic

Medicinal and Pharmaceutical Chemistry (10th Ed.). California, United

States: Lippincott Williams

Bhassin S.K, Gupta R.(2013). Pharmaceutical organic chemistry (E-book

Kindle edition). New Delhi, India: Elsevier Publishing Services

Session 16: Aldehydes of Pharmaceutical Importance.

Total Session Time: 120 minutes

Prerequisites

• None

Learning Tasks

By the end of this session students are expected to be able to:

• Define aldehydes
• List aldehydes and their isomers
• Explain nomenclature of aldehydes
• Draw chemical structure of aldehydes
• List chemical properties of aldehydes
• Explain chemical reactions of aldehydes

Resources Needed:

• Flip charts, marker pens, and masking tape.
• Black/white board and chalk/whiteboard markers.

SESSION OVERVIEW

|Step |Time |Activity/ |Content |

| | |Method | |

|1 |05 minutes |Presentation |Introduction, Learning Tasks |

|2 |10 minutes |Brainstorming |Definition of Aldehydes |

| | |Presentation | |

|3 |10 minutes |Presentation |Aldehydes and their Isomers |

|4 |15 minutes |Presentation |Nomenclature of Aldehydes |

|5 |15 minutes |Presentation |Chemical Structure of Aldehydes |

|6 |15 minutes |Buzzing |Chemical Properties of Aldehydes |

| | |Presentation | |

|7 |35 minutes |Group |Chemical Reactions involving |

| | |discussion |Aldehydes |

| | |Presentation | |

|8 |05 minutes |Presentation |Key Points |

|9 |05 minutes |Presentation |Evaluation |

SESSION CONTENTS.

STEP 1: Presentation of Session Title and Learning Tasks (5 minutes).

READ or ASK students to read the learning tasks and clarify

ASK students if they have any questions before continuing.

STEP 2: Definition of Aldehydes (10 minutes).

|Activity: Brainstorming (5 minutes) |

| |

|Ask students to brainstorm on the following question: |

| |

|What are Aldehydes? |

| |

|ALLOW few students to respond |

| |

|WRITE their responses on the flip chart/ board |

| |

|CLARIFY and SUMMARISE by using the content below |

Aldehydes are compounds of the general formula RCHO (R may be aliphatic or

aromatic group)

Examples of aldehydes

In aldehydes, the carbonyl group has a hydrogen atom attached to it

together with either

• a second hydrogen atom Or
• More commonly, a hydrocarbon group which might be an alkyl group or one

containing a benzene ring.

[pic]

ALDEHYDE

STEP 3: Aldehydes and their Isomers (10 minutes).

Aldehydes exhibit the following type of isomerism:

• Chain (nuclear) Isomerism.

o Aldehydes with 4 or more carbon atoms show chain isomerism. For

example:

[pic]

• Position isomerism.

o Aromatic aldehydes and higher ketones give position isomers. For

example:

[pic]

• Functional Isomerism.

o The general formula of aldehydes, ketones, unsaturated alcohols

oxiranes and oxolanes is CnH2nO

o C3H6O has isomers as:

[pic]

STEP 4: Nomenclature of Aldehydes (15 minutes).

I. The longest chain carrying the –CHO group is considered as the parent

structure. Aldehydes are named by replacing the ‘-e’ of the

corresponding alkane by the ending ‘-al’ [pic]

II. When R is aromatic just the word aldehyde is added to the aromatic name

III.The common name of simple aldehydes end with ‘aldehyde’

Examples:

Formaldehyde (=methanal)
Acetaldehyde (=ethanal)
Propionoaldehyde (=propanal)
Butyraldehyde (=butanal)
IV. If the aldehyde group is attached to a large unit (ring), the suffix

carbaldehyde is used.[pic]

[pic]

STEP 5: Chemical Structure of Aldehydes (15 minutes).

In an aldehyde, at least one of the attached groups must be a hydrogen

atom. The following compounds are aldehydes:

[pic]

• In condensed formulas, we use CHO to identify an aldehyde rather than

COH, which might be confused with an alcohol.

• This follows the general rule that in condensed structural formulas H

comes after the atom it is attached to (usually C, N, or O).

[pic]

Some structures of simple aldehydes

[pic]

STEP 6: Chemical Properties of Aldehydes (15 minutes).

|Activity: Buzzing (5minutes) |

| |

|ASK students to pair up and buzz on the following question for 5 |

|minutes. |

| |

|What are the chemical properties of Aldehydes? |

| |

|ALLOW pairs to respond on the question. |

| |

|WRITE their response on the flip chart/board. |

| |

|CLARIFY and SUMMARIZE by using the content in the table 1 below |

• Aldehydes can be reduced to a variety of compounds under different

conditions with different reducing agents.

• Aldehydes are reduced to the corresponding alcohols by;

o Addition of hydrogen in the presence of catalysts like finely divided

platinum, palladium, nickel and ruthenium.

o Treatment with chemical reagents such as sodium borohydride (NaBH4) or

Lithium aluminium hydride (LiAlH4).

• Aldehydes are easily oxdised to carboxylic acids on treatment with common

oxidising agents like nitric acid, potassium permanganate, potassium

dichromate etc.

• Aldehydes respond to the Fehlings' test.

o Fehlings' solution is an alkaline solution of copper sulphate

containing sodium potassium tartrate (Rochelle Salt) as a complexing

agent.

o Aldehydes on warming with solution, give a red precipitate of cuprous

oxide as a result of the redox reaction.

o Aromatic aldehydes give very poor results in this test.

STEP 7: Chemical Reactions involving Aldehydes (35 Minutes).

|Activity: Small Group Discussion (15 minutes) |

| |

|DIVIDE students into small groups. |

| |

|ASK students to discuss in groups on the following questions |

|What are the chemical reactions involving Aldehydes? |

| |

|[pic]REFER Students to Book |

| |

|ALLOW students to discuss for 10 minutes. |

| |

|ALLOW each group to present for 5 minutes. |

| |

|CLARIFY and SUMMARIZE by using the contents below |

• Aldehydes undergo many reactions to give a wide variety of useful

derivatives.

• Their most common reaction is nucleophilic addition, addition of a
nucleophile and a proton across the C = O double bond.
• The reactivity of the carbonyl group arises from the electronegativity of

the oxygen atom and the resulting polarization of the carbon-oxygen

double bond.

• The electrophilic carbonyl carbon atom is sp2 hybridized and flat,

leaving it relatively unhindered and open to attack from either face of

the double bond.

• As a nucleophile attacks the carbonyl group, the carbon atom changes

hybridization from sp2 to sp3

• The electrons of the pi bond are forced out to the oxygen atom to form an

alkoxide anion, which protonates to give the product of nucleophilic

addition.

[pic]

• Aldehydes are oxidized to carboxylic acids.

[pic]

Nucleophilic addition of Grignard Reagents

• A Grignard reagent (a strong nucleophile resembling a carbanion, R : -)

attacks the electrophilic carbonyl carbon atom to give an alkoxide

intermediate.

• Subsequent protonation gives an alcohol.

o [pic]

Hydride Reduction of Aldehydes

• Hydride reduction of an aldehyde is another example of nucleophilic

addition, with hydride ion (H : -) serving as the nucleophile.

• Attack by hydride gives an alkoxide that proto nates to form an alcohol.

[pic]

STEP 8: Key Points (05 minutes).

• An aldehyde is an organic compound containing a functional group with

the structure −CHO, consisting of a carbonyl center (a carbon double-

bonded to oxygen) with the carbon atom also bonded to hydrogen and to

an R group, which is any generic alkyl or side chain.

• In condensed structural formulas, the carbonyl group of an aldehyde is

commonly represented as −CHO.

• Aldehydes shows chain, position and functional isomerism
• Aldehydes can be easily oxidised to carboxylic acids due to the

presence of a hydrogen atom on carbonyl group which can be easily

converted to OH group.

• Reactions of aldehydes with alcohols produce either hemiacetals (a

functional group consisting of one —OH group and one —OR group bonded

to the same carbon) or acetals (a functional group consisting of two

—OR groups bonded to the same carbon), depending upon conditions

STEP 9: Evaluation (05 minutes).

• What are aldehydes?
• List two aldehydes and their isomers
• How are aldehydes named?
• Draw chemical structure of aldehydes
• What are chemical reactions of aldehydes?

References

Ternay, A.L (1976). Contemporary Organic Chemistry. Philadelphia, United

States: W.B. Saunders Co.

Morrison R.T and Boyd R N (1997). Organic Chemistry (6th Ed.). New Delhi,

India: Prentice Hall of India

Graham Solomon et al (2014). Organic Chemistry (11th Ed.). New Jeysey,

United States: John Willey and Sons.

Nadendla R. R. (2005). Principles of Pharmaceutical Organic Chemistry. New

Delhi, India: MacMillan Publishers

Bruice Y (2013). Organic Chemistry (7th ed.). New York, United States:

Prentice Hall Pearson.

Delgado J. N. Et al (1998). Wilson and Gisvold's Textbook of Organic

Medicinal and Pharmaceutical Chemistry (10th Ed.). California, United

States: Lippincott Williams

Bhassin S.K, Gupta R.(2013). Pharmaceutical organic chemistry (E-book

Kindle edition). New Delhi, India: Elsevier Publishing Services

Session 17: Ketones of Pharmaceutical Importance.

Total Session Time: 120 minutes

Prerequisites

• None

Learning Tasks

By the end of this session students are expected to be able to:

• Define ketones
• List ketones and their isomers
• Explain nomenclature of ketones
• Draw chemical structure of ketones
• List chemical properties of ketones
• Explain chemical reactions of ketones

Resources Needed:

• Flip charts, marker pens, and masking tape.
• Black/white board and chalk/whiteboard markers.

SESSION OVERVIEW

|Step |Time |Activity/ |Content |

| | |Method | |

|1 |05 minutes |Presentation |Introduction, Learning Tasks |

|2 |10 minutes |Brainstorming |Definition of Ketones |

| | |Presentation | |

|3 |15 minutes |Presentation |Ketones and their Isomers |

|4 |15 minutes |Presentation |Nomenclature of Ketones |

|5 |15 minutes |Presentation |Chemical Structure of Ketones |

|6 |15 minutes |Buzzing |Chemical Properties of Ketones |

| | |Presentation | |

|7 |35 minutes |Group |Chemical Reactions involving Ketones|

| | |discussion | |

| | |Presentation | |

|8 |05 minutes |Presentation |Key Points |

|9 |05 minutes |Presentation |Evaluation |

SESSION CONTENTS.

STEP 1: Presentation of Session Title and Learning Tasks (5 minutes)

READ or ASK students to read the learning tasks and clarify.

ASK students if they have any questions before continuing.

STEP 2: Definition of Ketones (10 minutes).

|Activity: Brainstorming (5 minutes) |

| |

|Ask students to brainstorm on the following question: |

| |

|What is Ketone? |

| |

|ALLOW few students to respond. |

| |

|WRITE their responses on the flip chart/ board. |

| |

|CLARIFY and SUMMARISE by using the content below |

• A ketone is a carbonyl compound which has two alkyl (or aryl) groups

bonded to the carbonyl carbon atom.

[pic]

• Aldehydes and Ketones are similar in structure, and they have similar

properties

• There are some differences, particularly in their reactions with

oxidizing agents and with nucleophiles

• In most cases Aldehydes are more reactive than Ketones

STEP 3: Ketones and their Isomers (10 minutes).

Ketones exhibit the following type of isomerism:

• Chain (nuclear) Isomerism.

o Aldehydes with 4 or more carbon atoms and ketones with five or more

carbon atoms show chain isomerism. For example:

[pic]

• Position lsomerism.

o Aromatic aldehydes and higher ketones give position isomers. For

example:

[pic]

• Functional Isomerism.

o The general formula of aldehydes, ketones, unsaturated alcohols

oxiranes and oxolanes is CnH2nO

o C3H6O has isomers as:

[pic]

STEP 4: Nomenclature of Ketones (15 minutes).

• The longest chain carrying the –CHO group is considered as the parent

structure.

o Ketones are named by replacing the ‘-e’ of the corresponding

alkane by the ending ‘-one’

o The alkane name becomes alkanone

[pic]

• The position of the attached group is indicated with number. The
carbonyl carbon= 1

o When R is aromatic just the word ketone is added to the

aromatic name

• In open-chain ketones, we number the longest chain that includes the

carbonyl carbon from the end closest to the carbonyl group, and we

indicate the position of the carbonyl group by a number

• In cyclic ketones, the carbonyl carbon atom is assigned the number 1

[pic]

[pic]

4-hydroxy-4-methyl-2-pentanone

4-hydroxy-4-methylpentan-2-one

• The common names of ketones are derived from the two alkyl groups

that are attached to the carbonyl carbon & followed by the word

‘ketone’

[pic]

• Some ketones have historical common names.

o Dimethyl ketone is always called acetone, and alkyl phenyl

ketones are usually named as the acyl group followed by the

suffix -phenone.

[pic]

[pic]

STEP 5: Chemical Structure of Ketones (15 minutes).

• In a ketone, two carbon groups are attached to the carbonyl carbon atom.
• The following general formulas, in which R represents an alkyl group and

Ar stands for an aryl group, represent ketones.

[pic]

• In condensed formulas, we use CHO to identify an aldehyde rather than

COH, which might be confused with an alcohol.

• This follows the general rule that in condensed structural formulas H

comes after the atom it is attached to (usually C, N, or O).

[pic]

Some simple structures of ketones

[pic]

STEP 6: Chemical Properties of Ketones (15 minutes).

|Activity: Buzzing (5minutes). |

| |

|ASK students to pair up and buzz on the following question for 5 |

|minutes. |

| |

|What are the chemical properties of Ketones? |

| |

|ALLOW pairs to respond on the question. |

| |

|WRITE their response on the flip chart/board. |

| |

|CLARIFY and SUMMARIZE by using the content in the table 1 below |

• Ketones undergoes addition reactions in presence of catalyst like finely

divided platinum, palladium and nickel.

• Ketones having at least one methyl group linked to the carbonyl carbon

atom (i.e., methyl ketones) are oxidised by sodium hypohalite to sodium

salts of carboxylic acids with one carbon atom less than that of the

ketones.

• Ketones can be reduced to a variety of compounds under different

conditions with different reducing agents.

• Ketones are oxidised only under vigorous conditions using powerful

oxidising agents such as conc. HNO3, KMnO4/H2SO4, K2Cr2O7/H2SO4 etc.

o Oxidation of ketones involves cleavage of bond between carbonyl

carbon and a-carbon on either side of keto group giving a mixture

of carboxylic acids.

STEP 7: Chemical Reactions involving Ketones (35 Minutes)

|Activity: Small Group Discussion (15 minutes) |

| |

|DIVIDE students into small groups |

| |

|ASK students to discuss in groups on the following questions. |

|What are the chemical reactions involving Ketones? |

| |

|[pic]REFER Students to Book |

| |

|ALLOW students to discuss for 10 minutes. |

| |

|ALLOW each group to present for 5 minutes. |

| |

|CLARIFY and SUMMARIZE by using the contents below |

Most common reaction is nucleophilic addition, addition of a nucleophile

and a proton across the C = O double bond.
• The reactivity of the carbonyl group arises from the electronegativity of

the oxygen atom and the resulting polarization of the carbon-oxygen

double bond.

• The electrophilic carbonyl carbon atom is sp2 hybridized and flat,

leaving it relatively unhindered and open to attack from either face of

the double bond.

• As a nucleophile attacks the carbonyl group, the carbon atom changes

hybridization from sp2 to sp3.

• The electrons of the pi bond are forced out to the oxygen atom to form an

alkoxide anion, which protonates to give the product of nucleophilic

addition.

[pic]

• The Witting Reaction

o The Witting reaction converts the carbonyl group of a ketone or an

aldehyde into a new C = C double bond where no bond existed before.

o A phosphorus-stabilized carbanion is added to a ketone or aldehyde.

o The product is not an alcohol, however, because the intermediate

undergoes elimination to an alkene.

[pic]

o The phosphorus-stabilized carbanion is an ylide (pronounced "ill -id")-

a molecule that bears no overall charge but has a negatively charged

carbon atom bonded to a positively charged heteroatom.

o Because of its carbanion character, the ylide carbon atom is strongly

nucleophilic.

o It attacks a carbonyl group to give a charge-separated intermediate

called a betaine

▪ Example 1.

[pic]

Example 2.

[pic]

• Hydration of Ketones and Aldehydes

o In an aqueous solution, a ketone or an aldehyde is in equilibrium with

its hydrate, a geminal diol.

o With most ketones, the equilibrium favors the unhydrated keto form

of the carbonyl

[pic]

[pic]

o Hydration occurs through the nucleophilic addition mechanism, with

water (in acid) or hydroxide ion (in base) serving as the

nucleophile.

o The electrophilic carbonyl group of a ketone is stabilized by its

two electron-donating alkyl groups

• Acid-catalyzed addition of Water and Alcohols

o Weak nucleophiles, such as water and alcohols, can add to activated

carbonyl groups under acidic conditions.

o A carbonyl group is a weak base, and it can become protonated in

an acidic solution.

o A carbonyl group that is protonated (or bonded to some other

electrophile) is strongly electrophilic, inviting attack by a weak

nucleophile.

[pic]

o The following reaction is the acid-catalyzed nucleophilic addition

of water across the carbonyl group of acetones.

[pic]

[pic]

o The base-catalyzed addition to a carbonyl group results from

nucleophilic attack of a strong nucleophile followed by

protonation.

• Base-catalyzed addition of water

[pic]

STEP 8: Key Points (05 minutes)

• A ketone is a carbonyl compound which has two alkyl (or aryl) groups

bonded to the carbonyl carbon atom.

• Ketones shoes chain, position and functional isomerism.
• Oxidation of ketones involves cleavage of bond between carbonyl carbon

and a-carbon on either side of keto group giving a mixture of carboxylic

acids.

• Ketones undergoes oxidation-reduction and nucleophilic addition

reactions.

STEP 9: Evaluation (05 minutes)

• What are ketones?
• List three isomers of ketones
• How are ketones named?
• Draw general chemical structure of ketones.
• What is the Witting Reaction?

References

Ternay, A.L (1976). Contemporary Organic Chemistry. Philadelphia, United

States: W.B. Saunders Co.

Morrison R.T and Boyd R N (1997). Organic Chemistry (6th Ed.). New Delhi,

India: Prentice Hall of India

Graham Solomon et al (2014). Organic Chemistry (11th Ed.). New Jeysey,

United States: John Willey and Sons.

Nadendla R. R, (2005). Principles of Pharmaceutical Organic Chemistry. New

Delhi, India: MacMillan Publishers

Bruice Y (2013). Organic Chemistry (7th ed.). New York, United States:

Prentice Hall Pearson.

Delgado J. N. Et al (1998). Wilson and Gisvold's Textbook of Organic

Medicinal and Pharmaceutical Chemistry (10th Ed.). Calfornia, United

States: Lippincott Williams

Bhassin S.K, Gupta R.(2013). Pharmaceutical organic chemistry (E-book

Kindle edition). New Delhi, India: Elsevier Publishing Services

Session 18: Aromatic Organic Compounds of Pharmaceutical Importance .

Total Session Time: 120 minutes

Prerequisites

• None

Learning Tasks

By the end of this session students are expected to be able to:

• Define aromatic organic compounds
• List aromatic compounds and their isomers
• Explain nomenclature of aromatic organic compounds
• Draw chemical structure of aromatic organic compounds
• List chemical properties of aromatic organic compounds
• Explain chemical reactions of aromatic organic compounds

Resources Needed:

• Flip charts, marker pens, and masking tape
• Black/white board and chalk/whiteboard markers

SESSION OVERVIEW

|Step |Time |Activity/ |Content |

| | |Method | |

|1 |05 minutes |Presentation |Introduction, Learning Tasks |

|2 |10 minutes |Brainstorming |Definition of Aromatic Organic |

| | |Presentation |Compounds |

|3 |15 minutes | |Aromatic organic compounds and their|

| | |Buzzing |Isomers |

| | |Presentation | |

|4 |15 minutes |Presentation |Nomenclature of Aromatic Organic |

| | | |Compounds |

|5 |15 minutes |Presentation |Chemical Structure of Aromatic |

| | | |Organic Compounds |

|6 |10 minutes |Presentation |Chemical Properties of Aromatic |

| | |Brainstorming |Organic Compounds |

|7 |40 minutes |Group |Chemical Reactions and Uses of |

| | |discussion |Aromatic Organic Compounds |

| | |Presentation | |

|8 |05 minutes |Presentation |Key Points |

|9 |05 minutes |Presentation |Evaluation |

SESSION CONTENTS.

STEP 1: Presentation of Session Title and Learning Tasks (5 minutes).

READ or ASK students to read the learning tasks and clarify

ASK students if they have any questions before continuing.

STEP 2: Definition of Aromatic Organic Compounds (10 minutes).

|Activity: Brainstorming (5 minutes) |

| |

|Ask students to brainstorm on the following question: |

| |

|What are Aromatic Organic Compounds? |

| |

|ALLOW few students to respond |

| |

|WRITE their responses on the flip chart/ board |

| |

|CLARIFY and SUMMARISE by using the content below |

Aromatic hydrocarbons are those compounds that have molecular structures

based on that of benzene C6H6 & resemble benzene in chemical behaviour.

[pic]

The Kekule Benzene structure

• It suggests the presence of alternating single & double bonds.
• Kekule suggested that 2 forms of benzene were in rapid equilibrium.

[pic]

STEP 3: Aromatic Organic Compounds and their Isomers (15 minutes).

|Activity: Buzzing (10minutes) |

| |

|ASK students to pair up and buzz on the following question for 5 |

|minutes |

| |

|What are the isomers of Aromatic organic compounds? |

| |

|ALLOW pairs to respond on the question. |

| |

|WRITE their response on the flip chart/board. |

| |

|CLARIFY and SUMMARIZE by using the content in the table 1 below |

• In a disubstituted benzene, three different position isomers are possible

depending upon the position of one substituent with respect to the other.

• Ortho (o−) is used to indicate that the relative position of the two

substituents is 1,2−. Similarly, meta (m−) and para (p−) are used to

indicate the relative positions 1,3− and 1,4− respectively.

ortho, meta and para isomers of dimethylbenzene (xylene)

[pic]

STEP 4: Nomenclature of Aromatic Organic Compounds (15 minutes)

• For many of the derivatives we simply prefix the name of the substituent

group to the word benzene.

• Other derivatives have special names, which show no resemblance to the

name of the attached substituent group.

[pic]

[pic]

[pic]

STEP 5: Chemical Structure of Aromatic Organic Compounds (15 minutes)

• Aromatic compounds are cyclic structures in which each ring atom is a

participant in a bond, resulting in delocalized electron density on both

sides of the ring.

• Due to this connected network of bonds, the rings are planar, unlike the

boat or table structures typical of cycloalkanes.

Structure of benzene: resonance theory

• “Whenever 2 or more structures can be written for a molecule and the only

difference between the structures is in the position of electrons.”

[pic]

[pic]

• If two groups are attached to the benzene ring their relative position

must be indicated. The three possible isomers of di-substituted benzene

are differentiated by use of the names;

o ortho-(o) at carbon 1 & 2,

o meta-(m) at carbon 1 & 3

o para-(p) at carbon 1 &4

[pic]

[pic]

[pic]

• If the two groups are different, and neither gives a common name, the two

groups are named successively, and the name is ended with –benzene:

• When benzene ring is a substituent, it is named as the prefix “phenyl”.

[pic]

[pic]

[pic]

STEP 6: Chemical Properties of Aromatic Organic Compounds (10 minutes).

|Activity: Brainstorming (5 minutes) |

| |

|Ask students to brainstorm on the following question: |

| |

|What are chemical properties of Aromatic Organic Compounds? |

| |

|ALLOW few students to respond |

| |

|WRITE their responses on the flip chart/ board |

| |

|CLARIFY and SUMMARISE by using the content below |

• Properties of Aromatic Hydrocarbons include that their major sources are

Petroleum and coal.

o Poly-aromatic hydrocarbons are defined as aromatic compounds with

more than one benzene.

o When they include in atmospheric pollution then it is known as

carcinogenic in nature.

• They go through electrophilic substitution reactions and nucleophile

aromatic substitution.

• Hydrocarbons which have multiple bonds are unsaturated in nature like

alkenes and alkynes.

o They tend to give addition reactions due to this unsaturation.

• Due to resonance and give characteristic electrophilic substitution

reactions aromatic hydrocarbons are stable.

o The carbon ring acts as a nucleophile in these reactions and to

form a substituted product an electrophile attack on benzene.

• With the coming electrophile, one of the H-atom of a ring is substituted

because of this the product also holds its stability and aromatic in

nature.

o On the opposite side in the addition reactions, aromatic compound

may lose their aromaticity, so they do not prefer to give such

reactions.

STEP 7: Chemical Reactions and Uses of Aromatic Organic Compounds (40

minutes).

|Activity: Small Group Discussion (20 minutes) |

| |

|DIVIDE students into small groups. |

| |

|ASK students to discuss in groups on the following questions. |

|What are the chemical reactions involving Aromatic Organic Compounds? |

| |

|[pic]REFER Students to Book |

| |

|ALLOW students to discuss for 15 minutes. |

| |

|ALLOW each group to present for 5 minutes. |

| |

|CLARIFY and SUMMARIZE by using the contents below |

Reactions with Substituent Effects

• Electrophilic Aromatic Substitutions

[pic]

• In an EAS reaction, an electrophile reacts with an aromatic ring and

substitutes it for one hydrogen.

General Mechanism

[pic]

There are five types of EAS reactions

o Halogenation

o Nitration

o Sulfonation

o Friedel-Crafts Alkylation

o Friedel-Crafts Acylation

o Nitration

▪ Reagents are HNO3 and H2SO4 with the electrophile being NO2+

[pic]

o Sulfonation

▪ Reagents are SO3 and H2SO4 (fuming sulfuric acid) with the

electrophile being HSO3+

[pic]

o Halogenation

▪ Reagents are X2 and FeX3 (catalyst) where X is a halogen

(generally Cl or Br)

[pic]

o Friedel-Crafts Alkylation

▪ Reagents vary, as there are three ways to produce the

electrophile—a carbocation; the overall reaction remains the same.

▪ Because this requires the formation of carbocation,

rearrangements are always possible.

R-X and AlX3 where X is either Br or Cl

[pic]

o Friedel-Crafts Acylation

▪ Reagents are an acyl halide (R-C=O-X) and AlX3 in which the
electrophile is R-C=O+

[pic]

▪ The acylation procedure can be used to produce alkyl benzenes

that otherwise cannot be prepared directly by alkylation through

Clemmensen Reduction.

▪ All that is required is the reduction of the acyl carbonyl group

to a CH2.

▪ The following is the Clemmensen Reduction reaction

[pic]

▪ The reagents used are a zinc/mercury amalgam and aqueous

hydrochloric acid.

▪ Therefore, to synthesize n-propyl benzene (which we could not do

via direct FC alkylation), acetalization using propanoyl chloride

can be done, and then reduce the phenyl ketone product which gives

our final product.

[pic]

The Benzene Elimination Addition Mechanism

• The previous addition elimination reaction mechanism required powerfully

electron withdrawing groups on the benzene ring.

• However, under forcing conditions, unactivated halobenzenes can react

with strong bases.

[pic]

• For example, phenol is produced commercially via the reaction of sodium

hydroxide with chlorobenzene.

• Analogously, aniline is produced via reaction of chlorobenzene with

sodium amide.

[pic]

• A clue to the mechanism of this type of reaction was provided by the

below reaction:

[pic]

• The products were found to be a 50:50 mixture of meta and para

substituted compounds.

• These two isomers can be explained as coming from the same intermediate,

a Benzyne.

Addition reactions of benzene

Although substitution is by far the most common reaction type of benzene

and its derivatives, addition reactions can occur if forcing conditions are

employed.

• Chlorination

o For example, if benzene is treated with an excess of chlorine under

conditions of heat and pressure, then 6 chlorine atoms will add,

generating 1,2,3,4,5,6-hexachlorocyclohexane.

[pic]

o This is believed to proceed through free radical intermediates, but

the mechanism is not relevant here.

• Catalytic Hydrogenation

o The addition of hydrogen to benzene occurs at elevated temperatures

and pressures, and requires a catalyst

[pic]

o Intermediate unsaturated compounds like cyclohexene or dienes

cannot be prepared because of the high pressures involved

Reactions of the Side Chains in Benzene Derivatives

• Permanganate Oxidation

o An aromatic ring imparts extra stability to the carbon atoms directly

bonded to it.

o Therefore, when an alkyl benzene is oxidized with permanganate, the

product is the carboxylate salt of di-benzoic acid.

[pic]

• Side Chain Halogenation

o Alkyl benzenes undergo free radical halogenation very easily at the

benzylic position, since the required intermediate radical is a

benzylic radical, and is therefore resonance stabilized

o For example, ethylbenzene reacts with bromine (or NBS) under UV

irradiation to give (1-bromoethyl)benzene and (1,1-

dibromoethyl)benzene.

[pic]

Uses of aromatic hydrocarbons

• In several industries, aromatic hydrocarbons have wide applications.

o For example, for model glues, toluene is used as solvent while

naphthalene is used as mothballs.

• For manufacturing of dyes, explosives, and drugs, Phenanthrene is an

intermediate product which has a different synthetic process.

o Trinitrotoluene (TNT) or 2, 4, 6 trinitrotoluene is an important

aromatic compound which is mainly used as explosive along with

the preparation of explosive.

• 1, 2 benzenediols or pyrocatechol is advertised as catechol which is

one of the most important components of a photographic developer.

STEP 8: Key Points (05 minutes).

• Aromatic compound, any of a large class of unsaturated chemical

compounds characterized by one or more planar rings of atoms joined

by covalent bonds of two different kinds.

• The unique stability of these compounds is referred to as aromaticity.
• Benzene (C6H6) is the best-known aromatic compound and the parent to

which numerous other aromatic compounds are related.

• The largest group of aromatic compounds are those in which one or more of

the hydrogens of benzene are replaced by some other atom or group, as

in toluene (C6H5CH3) and benzoic acid (C6H5CO2H).

• In the International Union of Pure and Applied Chemistry (IUPAC) system,

aromatic hydrocarbons are named as derivatives of benzene.

• The double bonds in aromatic compounds are less likely to participate in

addition reactions than those found in typical alkenes.

STEP 9: Evaluation (05 minutes).

• What are Aromatic organic compounds?
• Give three chemical structures of three different aromatic compounds
• What are the reactions involving aromatic organic compounds?
• What are the uses of aromatic organic compounds?

References

Ternay, A.L (1976). Contemporary Organic Chemistry. Philadelphia, United

States: W.B. Saunders Co.

Morrison R.T and Boyd R N (1997). Organic Chemistry (6th Ed.). New Delhi,

India: Prentice Hall of India

Graham Solomon et al (2014). Organic Chemistry (11th Ed.). New Jeysey,

United States: John Willey and Sons.

Rama Rao Nadendla (2005). Principles of Pharmaceutical Organic Chemistry.

New Delhi, India: MacMillan Publishers

Bruice Y (2013). Organic Chemistry (7th ed.). New York, United States:

Prentice Hall Pearson.

Delgado J. N. Et al (1998). Wilson and Gisvold's Textbook of Organic

Medicinal and Pharmaceutical Chemistry (10th Ed.). California, United

States: Lippincott Williams

Bhassin S.K, Gupta R.(2013). Pharmaceutical organic chemistry (E-book

Kindle edition). New Delhi, India: Elsevier Publishing Services

Session 19: Phenols of Pharmaceutical Importance.

Total Session Time: 120 minutes

Prerequisites

• None

Learning Tasks

By the end of this session students are expected to be able to:

• Define phenols
• List phenols and their isomers
• Explain nomenclature of phenols
• Draw chemical structure of phenols
• List chemical properties of phenols
• Explain chemical reactions of phenols

Resources Needed:

• Flip charts, marker pens, and masking tape.
• Black/white board and chalk/whiteboard markers.

SESSION OVERVIEW

|Step |Time |Activity/ |Content |

| | |Method | |

|1 |05 minutes |Presentation |Introduction, Learning Tasks |

|2 |10 minutes |Presentation |Definition of Phenols |

| | |Brainstorming | |

|3 |15 minutes |Presentation |Phenols and their Isomers |

|4 |15 minutes |Presentation |Nomenclature of Phenols |

|5 |15 minutes |Presentation |Chemical Structure of Phenols |

|6 |15 minutes |Presentation |Chemical Properties of Phenols |

| | |Buzzing | |

|7 |35 minutes |Presentation |Chemical Reactions involving Phenols|

| | |Group | |

| | |discussion | |

|8 |05 minutes |Presentation |Key Points |

|9 |05 minutes |Presentation |Evaluation |

SESSION CONTENTS.

STEP 1: Presentation of Session Title and Learning Tasks (5 minutes)

READ or ASK students to read the learning tasks and clarify

ASK students if they have any questions before continuing.

STEP 2: Definition of Phenols (10 minutes).

|Activity: Brainstorming (5 minutes) |

| |

|Ask students to brainstorm on the following question: |

| |

|What are Phenols? |

| |

|ALLOW few students to respond. |

| |

|WRITE their responses on the flip chart/ board. |

| |

|CLARIFY and SUMMARISE by using the content below |

• Phenols are organic compounds which contain a hydroxyl (—OH) group

attached to a carbon atom in a benzene ring.

• Their chemical behaviour is very distinct from that of alcohols, because

they are not capable of undergoing the same oxidation reactions that

alcohols participate it.

• Also, unlike alcohols, phenols are weak acids, since the phenoxide anion

generated by the loss of the hydroxyl proton is resonance-stabilized:

[pic]

[pic]

• Phenol is the active ingredient in some treatments for sore throats and

is found in several lozenges and throat sprays.

• It is used in the manufacture of many other compounds, such as aspirin,

and Bakelite.

STEP 3: Phenols and their Isomers (10 minutes).

• Cresol is a trivial name for the three isomeric methylphenols.

ortho-Cresol

[pic]

• The structure shown above is ortho-cresol, or 2-methylphenol; the other

isomers are shown below.

2,6 xylenol

[pic]

• The cresols are commonly used as solvents, in disinfectants and

deodorizers, and in the manufacture of other compounds (for example, the

BHA and BHT shown below).

• Xylenol is a trivial name given to the six isomeric dimethylphenols.
• The name is derived from xylene, which is the trivial name for the three

isomeric dimethylbenzenes.

• The structure shown above is 2,6-xylenol, or 2,6-dimethylphenol; the

other isomers are shown below.

[pic]

• The xylenols are found in a number of pesticides and are also used in the

manufacture of many other compounds.

STEP 4: Nomenclature of Phenols (15 minutes).

• Locate the position of hydroxyl group attached to the benzene ring.
• Benzene rings attached to more than one hydroxyl groups are labeled with

the Greek numerical prefixes such as di, tri, tetra to denote the number

of similar hydroxyl groups attached to the benzene ring.

• If two hydroxyl groups are attached to the adjacent carbon atoms of

benzene ring, it is named as benzene1, 2-diol

• In case of substituted phenols, we start locating the positions of the

other function groups with respect to the position where the hydroxyl

group is attached. For example, if a methyl group is attached at fourth

carbon atom with respect to hydroxy group; compound is named as, 4-Methyl

phenol.

[pic]

• Depending on the position of substituted functional group with respect to

the hydroxyl group, words like ortho (when the functional group is

attached to the adjacent carbon atom), para (when the functional group is

attached to the third carbon atom from the hydroxyl group), meta (when

the functional group is attached to the second carbon atom from the

hydroxyl group) are also used for the nomenclature of phenols.

• Compounds with two or more –OH groups have special names:

[pic]

• Certain groups, e.g., -COOH, -CHO, -SO3H, if present in the ring, take

priority; the –OH group is then used as a modifying prefix:

STEP 5: Chemical Structure of Phenols (15 minutes).

• The simplest way to draw the structure of phenol is

[pic]

• There is an interaction between the delocalized electrons in the benzene

ring and one of the lone pairs on the oxygen atom.

• This has an important effect on both the properties of the ring and of

the -OH group.

• One of the lone pairs on the oxygen overlaps with the delocalized ring

electron system

[pic]

Giving a structure rather like this

[pic]

• The donation of the oxygen's lone pair into the ring system increases the

electron density around the ring. That makes the ring much more reactive

than it is in benzene itself. That is explored in another page in this

phenol section.

• It also helps to make the -OH group's hydrogen a lot more acidic than it

is in alcohols. That will also be explored elsewhere in this section.

STEP 6: Chemical Properties of Phenols (15 minutes).

|Activity: Buzzing (5minutes) |

| |

|ASK students to pair up and buzz on the following question for 5 |

|minutes. |

| |

|What are the chemical properties of Phenols? |

| |

|ALLOW pairs to respond on the question. |

| |

|WRITE their response on the flip chart/board. |

| |

|CLARIFY and SUMMARIZE by using the content below |

Acidity:

• Phenols are more acidic (pKa»10) than alcohols (pKa»16 – 20), but less

acidic than carboxylic acids (pKa»5).

• The negative charge of the phenolate ion is stabilised by resonance due

to electron delocalisation onto the ring as shown below:

[pic]

• The acidity difference means that it is possible to separate phenols from

alcohols and/or carboxylic acids.

o Mixing an ether solution, of phenol and alcohol or phenol and

carboxylic acid, with dilute base (sodium hydroxide and sodium

bicarbonate, respectively), results in the stronger acid being

converted to its alkali salt, this is then extracted to the aqueous

phase and can be separated from the organic phase.

• Nucleophilic substitution reactions of phenols are generally carried out

under basic conditions as the phenolate ion is a better nucleophile.

Substituent Effects on Acidity

• Substituents, particularly those located ortho or para to the -OH group,

can dramatically influence the acidity of the phenol due to resonance and

/ or inductive effects.

• Electron withdrawing groups enhance the acidity; electron donating

substituents decrease the acidity. The resonance stabilisation of o-

nitrophenol is shown below:

[pic]

|Compound | | |Compound | |

| |pKa | | |pKa |

|Phenol | | | | |

| |10.0| | | |

|o-Methoxyphenol | | |p-Methoxyphenol | |

| |10.0| | |10.2|

|o-Methylphenol | | |p-Methylphenol | |

| |10.3| | |10.3|

|o-Chlorophenol | | |p-Chlorophenol | |

| |8.6 | | |9.4 |

|o-Nitrophenol | | |p-Nitrophenol | |

| |7.2 | | |7.2 |

|m-Nitrophenol | | | | |

| |8.4 | | | |

Reactivity

|[pic] |The image to the left shows the |

|[pic] |electrostatic potential for phenol. |

| |The redder an area is, the higher the |

| |electron density and the bluer an area is,|

| |the lower the electron density. |

| |The hydroxyl O atom is a region of high |

| |electron density (red) due to the lone |

| |pairs. |

| |The hydroxyl O atom can function as a |

| |nucleophile or Lewis base. |

| |There is low electron density (blue) on H |

| |atom of the hydroxyl group, i.e. H+ |

| |character, therefore phenols are acidic |

| |(pKa ~ 10) |

| |Due to conjugation with the ring, phenols |

| |are more acidic than alcohols (pKa ~ 16). |

| |Removal of the proton generates a |

| |phenolate ion. |

| |Note the increased electron density on the|

| |oxygen compared to the phenol |

STEP 7: Chemical Reactions involving Phenols (35 minutes).

|Activity: Small Group Discussion (15 minutes) |

| |

|DIVIDE students into small groups |

| |

|ASK students to discuss in groups on the following questions |

|What are the chemical reactions involving Phenols? |

| |

|[pic]REFER Students to Book |

| |

|ALLOW students to discuss for 10 minutes |

| |

|ALLOW each group to present for 5 minutes |

| |

|CLARIFY and SUMMARIZE by using the contents below |

Electrophilic Aromatic Substitution Reactions of Phenols

• Phenols are potentially very reactive towards electrophilic aromatic

substitution.

• This is because the hydroxy group, -OH, is a strongly activating, ortho-

/ para- directing substituent.

• Substitution typically occurs para to the hydroxyl group unless the para

position is blocked, then ortho substitution occurs.

• The strong activation often means that milder reaction conditions than

those used for benzene itself can be used (see table below for a

comparison).

• Phenols are so activated that polysubstitution can be a problem (similar

problems occur with anilines).

[pic]

Summary

|Reaction |Phenol |Benzene |

|Nitration |dil. HNO3 in H2O or |HNO3 / H2SO4 |

| |CH3CO2H | |

|Sulfonation |conc. H2SO4 |H2SO4 or SO3 / H2SO4 |

|Halogenation |X2 |X2 / Fe or FeX3 |

|Alkylation |ROH / H+ or RCl / AlCl3 |RCl / AlCl3 |

|Acylation |RCOCl / AlCl3 |RCOCl / AlCl3 |

|Nitrosation |aq. NaNO2 / H+ | |

• Acylation of Phenols

o Phenols are examples of bidentate nucleophiles, meaning that they can

react at two positions:

▪ on the aromatic ring giving an aryl ketone via C-acylation, a

Friedel-Crafts reaction or,

▪ on the phenolic oxygen giving an ester via O-acylation, an

esterification

o Reagents:

|C-acylation: acylating agent (acyl chloride or anhydride) and |

|AlCl3 |

|O-acylation: acylating agent (acyl chloride or anhydride) |

o The product of C-acylation is more stable and predominates under

conditions of thermodynamic control (i.e. when AlCl3 is present).

o The product of O-acylation forms faster and predominates under

conditions of kinetic control.

o O-acylation can be promoted by either:

▪ acid catalysis via protonation of the acylating agent,

increasing its' electrophilicity or

▪ Base catalysis via deprotonation of the phenol, increasing

its' nucleophilicity.

o It is also known that aryl esters readily rearrange to aryl ketones

in the presence of AlCl3, a reaction known as the Fries

rearrangement:

[pic]

[pic]

• Carboxylation of Phenols (Kolbe-Schmitt reaction)

o Heating the nucleophilic phenolate salt with carbon dioxide under high

pressure / temperature results in regioselective ortho-substitution.

o This process is also known as the Kolbe-Schmitt synthesis.

o O-hydroxybenzoic acid is more commonly known as salicyclic acid.

[pic]

|MECHANISM FOR CARBOXYLATION OF PHENOLS |

| |[pic] |

|The nucleophilic phenolate (reacting | |

|like an enolate) reacts with the | |

|electrophilic carbon of carbon dioxide | |

|in the ortho position (compare this | |

|with an Aldol reaction) | |

| | |

|The non-aromatic | |

|cyclohexadienonecarboxylate | |

|intermediate tautomerises to the more | |

|stable aromatic enol which is further | |

|stabilised by an intramolecular | |

|hydrogen bond. An acidic work-up will | |

|generate the carboxylic acid. | |

• Oxidation of Phenols

o In general, phenols are more easily oxidized than simple alcohols.

o Oxidation can be achieved by reaction with silver oxide (Ag2O) or

chromic acid (Na2Cr2O7), or other oxidizing agents.

o Particularly important are the oxidation of 1,2- and 1,4-benzenediol

(pyrocatechol and hydroquinone, respectively) and their derivatives

(see examples below):

[pic] [pic]

o These types of systems are important in biological redox-systems such

as coenzyme Q and vitamin K.

o Here's a closer look at the two one electron transfers that are

believed to take place when hydroquinone is oxidized to benzoquinone.

|[pic] |Loss of a proton and|

| |an electron |

| |generates a phenoxy |

| |radical |

|[pic] |Loss of a second |

| |proton and a second |

| |electron completes |

| |the oxidation. |

STEP 8: Key Points (05 minutes)

• Phenol, or hydroxybenzene, is the parent compound of the phenols,

consisting of an OH group directly connected to a benzene ring.

• Phenols are more acidic (pKa»10) than alcohols (pKa»16 – 20), but less

acidic than carboxylic acids (pKa»5).

• Phenols are potentially very reactive towards electrophilic aromatic

substitution.

• Phenols undergoes Carboxylation and acetylation chemical reactions.

STEP 9: Evaluation (05 minutes)

• What are phenols?
• List phenols and their isomers.
• Draw chemical structure of phenols.
• List chemical reactions of phenols.
• What is Kolbe-Schmitt reaction?

References

Ternay, A.L (1976). Contemporary Organic Chemistry. Philadelphia, United

States: W.B. Saunders Co.

Morrison R.T and Boyd R N (1997). Organic Chemistry (6th Ed.). New Delhi,

India: Prentice Hall of India

Graham Solomon et al (2014). Organic Chemistry (11th Ed.). New Jeysey,

United States: John Willey and Sons.

Nadendla R. R. (2005). Principles of Pharmaceutical Organic Chemistry. New

Delhi, India: MacMillan Publishers

Bruice Y (2013). Organic Chemistry (7th ed.). New York, United States:

Prentice Hall Pearson.

Delgado J. N. Et al (1998). Wilson and Gisvold's Textbook of Organic

Medicinal and Pharmaceutical Chemistry (10th Ed.). California, United

States: Lippincott Williams

Bhassin S.K, Gupta R.(2013). Pharmaceutical organic chemistry (E-book

Kindle edition). New Delhi, India: Elsevier Publishing Services

Session 20: Aryl Halides of Pharmaceutical Importance.

Total Session Time: 120 minutes

Prerequisites

None

Learning Tasks

By the end of this session students are expected to be able to:

• Define aryl halides
• Explain nomenclature of aryl halides
• Draw chemical structure of aryl halides
• List chemical properties of aryl halides
• Explain chemical reactions of aryl halides

Resources Needed:

• Flip charts, marker pens, and masking tape
• Black/white board and chalk/whiteboard markers

SESSION OVERVIEW

|Step |Time |Activity/ |Content |

| | |Method | |

|1 |05 minutes |Presentation |Introduction, Learning Tasks |

|2 |10 minutes |Brainstorming |Definition of Aryl Halides |

| | |Presentation | |

|3 |15 minutes |Presentation |Nomenclature of Aryl Halides |

|4 |15 minutes |Presentation |Chemical Structure of Aryl Halides |

|5 |20 minutes |Buzzing |Chemical Properties of Aryl Halides |

| | |Presentation | |

|6 |45 minutes |Group |Chemical Reactions involving Aryl |

| | |discussion |Halides |

| | |Presentation | |

|7 |05 minutes |Presentation |Key Points |

|8 |05 minutes |Presentation |Evaluation |

SESSION CONTENTS

STEP 1: Presentation of Session Title and Learning Tasks (5 minutes).

READ or ASK students to read the learning tasks and clarify

ASK students if they have any questions before continuing.

STEP 2: Definition of Aryl Halides (10 minutes).

|Activity: Brainstorming (5 minutes) |

| |

|Ask students to brainstorm on the following question: |

| |

|What are Phenols? |

| |

|ALLOW few students to respond. |

| |

|WRITE their responses on the flip chart/ board. |

| |

|CLARIFY and SUMMARISE by using the content below |

Aryl halides are the compounds that contain halogen atom directly attached

to the benzene ring. They have general formula ArX.

[pic]

Any halogen compound that contains a benzene ring is not classified as aryl

halide. e.g. Benzyl chloride is not an aryl halide but is a substituted

alkyl halide.

STEP 3: Nomenclature of Aryl Halides (15 minutes).

• Functional group suffix = -halobenzene
Functional group prefix = halo-

Numbering of the ring begins at the halogen-substituted carbon and

proceeds in the direction of the next substituted carbon that possesses

the lower number.

• Mono-substituted aryl halides are characterized using the prefix ortho (o-

), meta (m-) or para (p-) depending on the placement of the substituent

from the halogen or the halogen from a higher priority functional group:

1,2-, 1,3- or 1,4- respectively.

| | | |

|1-chloro-2-ethylbenze|1-chloro-3-ethylbenze|1-chloro-4-ethylbenze|

|ne |ne |ne |

|or |or |or |

|o-ethylchlorobenzene |m-ethylchlorobenzene |p-ethylchlorobenzene |

STEP 4: Chemical Structure of Aryl Halides (15 minutes).

• An aryl halide is classified by its distinct bonding of a halogen

directly to a benzene ring.

o From a structural standpoint, one of the more important things to

realize is the trend in bond lengths of the four aryl halides.

• Since the trend for atomic size goes F < Cl < Br < I, (meaning fluorine

is smaller than chlorine, which is smaller than bromine, etc.) it's

probably not surprising that in terms of bond length, the observation is

as follows:

|[pic] |

|Bond lengths (given in picometers) of the four |

|aryl halides |

• The bond lengths here are measured in picometers, which is a small unit

of measurement used because we are talking about chemical bonds on a

microscopic scale.

• Notice that as we go from fluorine, to chlorine, to bromine, to iodine,

the bond lengths get longer and longer.

• That is because as the size of the halogen gets bigger, the bond has to

elongate to make room for the larger atom that's bonded to the benzene

ring.

STEP 5: Chemical Properties of Aryl Halides (20 minutes).

|Activity: Buzzing (5minutes) |

| |

|ASK students to pair up and buzz on the following question for 5 |

|minutes. |

| |

|What are the chemical properties of Aryl Halides? |

| |

|ALLOW pairs to respond on the question |

| |

|WRITE their response on the flip chart/board |

| |

|CLARIFY and SUMMARIZE by using the content below |

Reactivity of Aryl halides

• In haloalkane, the carbon atom attached to halogen is sp3 hybridised

while in case of haloarene, the carbon atom attached to halogen is sp2-

hybridised.

[pic]

• The sp2 hybridized carbon with a greater s-character is

more electronegative and can hold the electron pair of C—X bond more

tightly than sp3 -hybridized carbon in haloalkane with less s-

character.

• Thus, C—X bond length in halo alkane is shorter than those present

in haloarene.

• Since it is difficult to break a shorter bond than a longer bond,

therefore, haloarenes are less reactive than haloalkanes towards

nucleophilic substitution reaction.

• Unlike alkyl halides, aryl halides are less reactive towards

Nucleophilic substitution reactions, this can be attributed to their

electron release via resonance

[pic]

• Structures III, IV and V stabilise chlorobenzene molecule and give a

double bond character to the carbon-chlorine bond.

• Now because of this the carbon-chlorine bond has more strength and hence

aryl halides are more stable towards Nucleophilic substitution reactions.

• In Alkyl halides the carbon atom attached to halogen is sp3 hybridized

and in aryl halides it is sp2, hybridized, as sp2 hybridized carbon is

more electronegative it does not permit the chlorine atom to get

displaced with the bonded pair of electrons.

Nucleophilic Substitution Reactions of Aryl Halides

• Aryl halides undergo Nucleophilic substitution reactions when a strong

Electron withdrawing group is present on the benzene ring.

• Electron withdrawing groups activate the benzene ring towards

nucleophilic substitution in aryl halides whereas Electron donating

groups deactivate the ring.

Elimination – Addition Mechanism

• In the absence of an electron withdrawing group, nucleophilic

substitution takes place in presence of very strong bases, but the

mechanism is entirely different from what we have seen in bimolecular

nucleophilic substitution reactions.

• This reaction proceeds by a mechanism called benzyne mechanism.

STEP 6: Chemical Reactions involving Aryl Halides (45 minutes).

|Activity: Small Group Discussion (20 minutes) |

| |

|DIVIDE students into small groups. |

| |

|ASK students to discuss in groups on the following questions. |

|What are the chemical reactions involving Aryl Halides? |

| |

|[pic]REFER Students to Book |

| |

|ALLOW students to discuss for 15 minutes. |

| |

|ALLOW each group to present for 5 minutes. |

| |

|CLARIFY and SUMMARIZE by using the contents below |

Preparations of Aryl Halides

|Halogenation of |[pic] |

|arenes | |

|Via Aryl Diazonium |[pic] |

|salts | |

Electrophilic Aromatic Substitution of Aryl Halides

• Aryl halides are themselves reactive towards electrophilic aromatic

substitution but they are less reactive than benzene.

• This is because halides are weak deactivators.
• Halides direct subsequent reactions ortho, para.
• This makes them a little unusual (activators are usually ortho, para-

directing, deactivators meta-directing).

• The weak deactivation is due to the electronegativity of the halogen

making the intermediate cations less stable than those produced when

benzene undergoes substitution:

[pic]

• The directing effect is due to the resonance stabilisation of the

cationic intermediates derived by ortho or para attack but not by meta

attack. For example, the stabilisation during ortho attack is shown

below:

[pic]

• However, aryl halides can undergo many of the same electrophilic aromatic

substitution reactions that benzene can (review) including nitration,

sulfonation, further halogenation and Friedel-Crafts alkylation or

acylation reactions.

• The following are Electrophilic Aromatic Substitution of aryl halides;

o Halogenation of Aryl halides

[pic]

o Nitration of Aryl halides

[pic]

o Sulphonation of Aryl halides

[pic]

o Friedel-Crafts reaction for Aryl halides

▪ Friedel–Crafts reaction involves the alkylation of

an aromatic ring with an alkyl halide using a strong Lewis

acid catalyst.

▪ With anhydrous ferric chloride as a catalyst, the alkyl

group attaches at the former site of the chloride ion.

[pic]

Addition-Elimination Mechanism

• The generally accepted mechanism for nucleophilic aromatic substitution

in nitro-substituted aryl halides is shown by example below:

[pic]

• Attack of the strong nucleophile on the halogen substituted aromatic

carbon forming an anionic intermediate.

• Loss of the leaving group, the halide ion restores the aromaticity.
• Kinetics of the reaction are observed to be second order.
• The addition step is the rate determining step (loss of aromaticity).
• Nucleophilic substitution, and therefore reaction rate, is facilitated by

the presence of a strong electron withdrawing group (esp. NO2) ortho or

para to the site of substitution, which stabilize the cyclohexadienyl

anion through resonance.

[pic]

• Aryl halide reactivity : -F > -Cl > -Br > -I (note the contrast to

simple nucleophilic substitution)

• The more electronegative the group the greater the ability to attract

electrons which increases the rate of formation of the cyclohexadienyl

anion.

[pic]

Elimination-Addition Mechanism:

• This pathway is followed when the nucleophile is an exceptionally strong
base (e.g. amide ion, NH2-) and the absence of the strong electron

withdrawing groups:

[pic]

[pic]

• Nucleophilic substitution can lead to substitution on either

o the same carbon that bore the leaving group (see addition mechanism

above) or on an adjacent carbon (see addition mechanism below)

[pic]

• This is most readily apparent when the benzyne is substituted:

[pic]

Aryl Grignards

Aryl Grignards are formed by the reaction of aryl halides (X= Cl, Br or I)

with magnesium metal

[pic]

• Typical solvents are normally anhydrous diethyl ether or tetrahydrofuran.
• Halide reactivity: I > Br > Cl
• Organolithium reagents can also be made.
• Aryl Grignard reactions allow for the introduction of C substituents

other than via Friedel-Crafts alkylation or acylation reactions.

Reaction of Haloarenes With Metals

• Wurtz-Fittig reaction

o The Wurtz–Fittig reaction is the chemical reaction of aryl halides

with alkyl halides and sodium metal in the presence of dry ether to

give substituted aromatic compounds

[pic]

• Fittig reaction.

o Fittig reaction is a chemical reaction of two Aryl halide and

sodium metal in presence of dry ether to give biphenyl as the

product.

[pic]

STEP 7: Key Points (05 minutes).

• Aryl halides are the compounds that contain halogen atom directly

attached to the benzene ring.

• Aryl halides undergo Nucleophilic substitution reactions when a strong

Electron withdrawing group is present on the benzene ring

• The Nucleophilic substitution of Aryl halides is facilitated by the

presence of a strong electron withdrawing group (esp. NO2) ortho or

para to the site of substitution, which stabilize the cyclohexadienyl

anion through resonance

STEP 8: Evaluation (05 minutes).

• What are aryl halides?
• Draw chemical structure of aryl halides
• List chemical properties of aryl halides
• List three chemical reactions under electrophilic aromatic substitution

of Aryl Halides

References

Ternay, A.L (1976). Contemporary Organic Chemistry. Philadelphia, United

States: W.B. Saunders Co.

Morrison R.T and Boyd R N (1997). Organic Chemistry (6th Ed.). New Delhi,

India: Prentice Hall of India

Graham Solomon et al (2014). Organic Chemistry (11th Ed.). New Jeysey,

United States: John Willey and Sons.

Nadendla R. R. (2005). Principles of Pharmaceutical Organic Chemistry. New

Delhi, India: MacMillan Publishers

Bruice Y (2013). Organic Chemistry (7th ed.). New York, United States:

Prentice Hall Pearson.

Delgado J. N. Et al (1998). Wilson and Gisvold's Textbook of Organic

Medicinal and Pharmaceutical Chemistry (10th Ed.). California, United

States: Lippincott Williams

Bhassin S.K, Gupta R.(2013). Pharmaceutical organic chemistry (E-book

Kindle edition). New Delhi, India: Elsevier Publishing Services

Session 21: Amines of Pharmaceutical Importance.

Total Session Time: 120 minutes

Prerequisites

None

Learning Tasks

By the end of this session students are expected to be able to:

• Define Amines
• Explain nomenclature of amines
• Draw chemical structure of amines
• List chemical properties of amines
• Explain chemical reactions of amines

Resources Needed:

• Flip charts, marker pens, and masking tape
• Black/white board and chalk/whiteboard markers

SESSION OVERVIEW

|Step |Time |Activity/ |Content |

| | |Method | |

|1 |05 minutes |Presentation |Introduction, Learning Tasks |

|2 |10 minutes |Brainstorming |Definition of Amines |

| | |Presentation | |

|3 |15 minutes |Presentation |Nomenclature of Amines |

|4 |15 minutes |Presentation |Chemical Structure of Amines |

|5 |20 minutes |Buzzing |Chemical Properties of Amines |

| | |Presentation | |

|6 |45 minutes |Group |Chemical Reactions involving Amines |

| | |discussion | |

| | |Presentation | |

|7 |05 minutes |Presentation |Key Points |

|8 |05 minutes |Presentation |Evaluation |

SESSION CONTENTS

STEP 1: Presentation of Session Title and Learning Tasks (5 minutes)

READ or ASK students to read the learning tasks and clarify

ASK students if they have any questions before continuing.

STEP 2: Definition of Amines (10 minutes).

|Activity: Brainstorming (5 minutes) |

| |

|Ask students to brainstorm on the following question: |

| |

|What are Amines? |

| |

|ALLOW few students to respond |

| |

|WRITE their responses on the flip chart/ board |

| |

|CLARIFY and SUMMARISE by using the content below |

• Amines are organic derivatives of ammonia NH3 with one or more alkyl or

aryl groups bonded to the nitrogen atom.

• Amines contain a nitrogen atom with a lone pair of electrons; basic &

nucleophilic.

• Amines are present widely throughout both plants & animals.
• As a class amine include some of the most important biological compounds.
• Amines serve many functions in living organisms such as bioregulation,

neurotransmission, and defense against predators.

Examples of some biologically active amines

[pic]

[pic]

[pic]

[pic]

[pic]

STEP 3: Nomenclature of Amines (15 minutes).

• The IUPAC nomenclature for amines is similar to that for alcohols.
• The longest continuous chain of the carbon atoms determines the root

name.

• The -e ending in the alkane name is changed to -amine, and a number shows

the position of the amino group along the chain.

• Other substituents on the carbon chain are given numbers, and the prefix

N- is used for each substituent on nitrogen.

Examples

[pic]

Common Names

• Common names of amines are formed from the names of the alkyl groups

bonded to nitrogen, followed by the suffix –amine

• The pefixes di-, tri and tetra- are used to describe two, three, or four

identical substituents

Examples

[pic]

[pic]

• In naming amines with more complicated structures, the -NH2 group is

called the amino group.

• The amino group is treated like any other substituent, with a number or

other symbol indicating its position on the ring or carbon chain.

Examples

[pic]

[pic]

• Aromatic and heterocyclic amines are generally known by historical names.
• For example, phenylamine is called aniline, and its derivatives are named

as derivatives of aniline.

[pic]

[pic]

Heterocyclic amines

• Compounds in which the nitrogen atom occurs as part of a ring.
• The heterocyclic nitrogen atom is always numbered as position 1.

Examples

[pic]

STEP 4: Chemical Structure of Amines (15 minutes).

• The basic chemical structure is that of ammonia (NH3) with the key atom

being the central nitrogen atom.

• The basic ammonia structure is changed when the hydrogen atoms are

replaced by alkyl groups to form amines.

• There are primary, secondary and tertiary amines.

|[pic] |

| |Secondary amine | |

| | | |

[pic]

Primary amine

[pic]

Tertiary amine

• The naming of amines is pretty straightforward. Primary amines are called

things like methylamine (CH3-NH2) and ethylamine (CH3-CH2-NH2).

• Simple secondary and tertiary amines are also easy to name. Dimethylamine

is CH3-NH-CH3 and trimethylamine is CH3-N(CH3)-CH3.

• Larger amines have names beginning with amino. For example, CH3-CH(NH)

-CH2-CH2-CH3 is called 2-aminopentane.

STEP 5: Chemical Properties of Amines (15 minutes).

|Activity: Buzzing (5minutes) |

| |

|ASK students to pair up and buzz on the following question for 5 |

|minutes. |

| |

|What are the chemical properties of Amines? |

| |

|ALLOW pairs to respond on the question. |

| |

|WRITE their response on the flip chart/board. |

| |

|CLARIFY and SUMMARIZE by using the content below |

• They react with acids to form acid-base salts
• They react with electrophiles in polar reactions
• Amines are stronger bases than alcohols, ethers or water
• Kb (basicity constant) – used to measure the base strength of an amine
• Simple methylated amines are prepared by reaction of NH3 with CH3OH in

the presence of alumina catalyst.

STEP 6: Chemical Reactions involving Amines (45 minutes)

|Activity: Small Group Discussion (20 minutes) |

| |

|DIVIDE students into small groups |

| |

|ASK students to discuss in groups on the following questions |

|What are the chemical reactions involving Amines? |

| |

|[pic]REFER Students to Book |

| |

|ALLOW students to discuss for 15 minutes |

| |

|ALLOW each group to present for 5 minutes |

| |

|CLARIFY and SUMMARIZE by using the contents below |

Due to the unshared electron pair, amines can act as both bases and

nucleophiles.

• Reaction with acids

When reacted with acids, amines donate electrons to form ammonium

salts.

[pic]

• Reaction with acid halides

o Acid halides react with amines to form substituted amides.

[pic]

o Aldehydes and ketones react with primary amines to give a reaction

product (a carbinolamine) that dehydrates to yield aldimines and

ketimines (Schiff bases).

[pic]

o If you react secondary amines with aldehydes or ketones, enamines

form.

[pic]

• Reaction with sulfonyl chlorides

o Amines react with sulfonyl chlorides to produce sulfonamides. A

typical example is the reaction of benzene sulfonyl chloride with

aniline.

[pic]

• The Hinsberg test

o The Hinsberg reaction is a lab test for the detection of primary,

secondary and tertiary amines.

o In this test, the amine is shaken well with Hinsberg reagent in the

presence of aqueous alkali (either KOH or NaOH).

o A reagent containing an aqueous sodium hydroxide solution

and benzenesulfonyl chloride is added to a substrate.

o A primary amine will form a soluble sulfonamide salt.

▪ Acidification of this salt then precipitates the sulfonamide of

the primary amine.

o A secondary amine in the same reaction will directly form an insoluble

sulfonamide.

o A tertiary amine will not react with the sulfonamide but is insoluble.

▪ After adding dilute acid this insoluble amine is converted to a

soluble Ammonium salt.

o In this way the reaction can distinguish between the three types of

amines.

[pic]

• Oxidation

o Although you can oxidize all amines, only tertiary amines give easily

isolated products.

o The oxidation of a tertiary amine leads to the formation of an amine

oxide.

[pic]

o Arylamines tend to be easily oxidized, with oxidation occurring on the

amine group as well as in the ring.

• Reaction with nitrous acid

o Nitrous acid is unstable and must be prepared in the reaction solution

by mixing sodium nitrite with acid.

o Primary amines react with nitrous acid to yield a diazonium salt,

which is highly unstable and degradates into a carbocation that is

capable of reaction with any nucleophile in solution.

o Therefore, reacting primary amines with nitrous acid leads to a

mixture of alcohol, alkenes, and alkyl halides

[pic]

o Primary aromatic amines form stable diazonium salts at zero degrees.

[pic]

o Secondary aliphatic and aromatic amines form nitrosoamine with nitrous

acid.

[pic]

o Tertiary amines react with nitrous acid to form N‐nitrosoammonium

compounds.

[pic]

Reactions of aromatic diazonium salts

• Diazonium salts of aromatic amines are very useful as intermediates to

other compounds.

• Because aromatic diazonium salts are only stable at very low temperatures

(zero degrees and below), warming these salts initiates decomposition

into highly reactive cations.

• These cations can react with any anion present in solution to form a

variety of compounds. Figure below illustrates the diversity of the

reactions.

[pic]

STEP 7: Key Points (05 minutes)

• Amines are organic derivatives of ammonia NH3 with one or more alkyl or

aryl groups bonded to the nitrogen atom

• The basic chemical structure is that of ammonia (NH3) with the key atom

being the central nitrogen atom

• Due to the unshared electron pair, amines can act as both bases and

nucleophiles

• The Hinsberg reaction is a lab test for the detection of primary,

secondary and tertiary amines

STEP 8: Evaluation (05 minutes)

• What are Amines?
• Classify amines
• Draw general chemical structure of tertiary amines
• List chemical properties of amines
• What is Hinsberg reaction?

References

Ternay, A.L (1976). Contemporary Organic Chemistry. Philadelphia, United

States: W.B. Saunders Co.

Morrison R.T and Boyd R N (1997). Organic Chemistry (6th Ed.). New Delhi,

India: Prentice Hall of India

Graham Solomon et al (2014). Organic Chemistry (11th Ed.). New Jeysey,

United States: John Willey and Sons.

Nadendla R. R. (2005). Principles of Pharmaceutical Organic Chemistry. New

Delhi, India: MacMillan Publishers

Bruice Y (2013). Organic Chemistry (7th ed.). New York, United States:

Prentice Hall Pearson.

Delgado J. N. Et al (1998). Wilson and Gisvold's Textbook of Organic

Medicinal and Pharmaceutical Chemistry (10th Ed.). California, United

States: Lippincott Williams

Bhassin S.K, Gupta R. (2013). Pharmaceutical organic chemistry (E-book

Kindle edition). New Delhi, India: Elsevier Publishing Services

Session 22: Amides of Pharmaceutical Importance.

Total Session Time: 120 minutes

Prerequisites

None

Learning Tasks

By the end of this session students are expected to be able to:

• Define Amides
• Explain nomenclature of amides
• Draw chemical structure of amides
• List chemical properties of amides
• Explain chemical reactions of amides

Resources Needed:

• Flip charts, marker pens, and masking tape.
• Black/white board and chalk/whiteboard markers.

SESSION OVERVIEW

|Step |Time |Activity/ |Content |

| | |Method | |

|1 |05 minutes |Presentation |Introduction, Learning Tasks |

|2 |10 minutes |Brainstorming |Definition of Amides |

| | |Presentation | |

|3 |15 minutes |Presentation |Nomenclature of Amides |

|4 |15 minutes |Presentation |Chemical Structure of Amides |

|5 |20 minutes |Buzzing |Chemical Properties of Amides |

| | |Presentation | |

|6 |45 minutes |Group |Chemical Reactions involving Amides |

| | |discussion | |

| | |Presentation | |

|7 |05 minutes |Presentation |Key Points |

|8 |05 minutes |Presentation |Evaluation |

SESSION CONTENTS

STEP 1: Presentation of Session Title and Learning Tasks (5 minutes)

READ or ASK students to read the learning tasks and clarify.

ASK students if they have any questions before continuing.

STEP 2: Definition of Amides (10 minutes).

|Activity: Brainstorming (5 minutes) |

| |

|Ask students to brainstorm on the following question: |

| |

|What are Amides? |

| |

|ALLOW few students to respond |

| |

|WRITE their responses on the flip chart/ board |

| |

|CLARIFY and SUMMARISE by using the content below |

• Amides are usually regarded as derivatives of carboxylic acids in which

the hydroxyl group has been replaced by an amine or ammonia.

• The lone pair of electrons on the nitrogen is delocalized into the

carbonyl, thus forming a partial double bond between N and the carbonyl

carbon.

• So, amides contain the -CONH2 group.

STEP 3: Nomenclature of Amides (15 minutes).

• Primary amides are named by changing the name of the acid by dropping the

-oic acid or -ic acid endings and adding -amide.

• The carbonyl carbon is given the #1 location number.
• It is not necessary to include the location number in the name because it

is assumed that the functional group will be on the end of the parent

chain.

[pic] [pic] [pic]

methanamide or formamide (left), ethanamide or acetamide (center) ,

benzamide (right).

[pic]

• Secondary amides are named by using an upper-case N to designate that the

alkyl group is on the nitrogen atom.

• Alkyl groups attached to the nitrogen are named as substituents.
• The letter N is used to indicate they are attached to the nitrogen.

[pic]

Tertiary amides are named in the same way as secondary amides, but with two

N's

[pic]

STEP 4: Chemical Structure of Amides (15 minutes).

• The amide functional group has a nitrogen atom attached to a carbonyl

carbon atom.

• If the two remaining bonds on the nitrogen atom are attached to hydrogen

atoms, the compound is a simple amide.

• If one or both of the two remaining bonds on the atom are attached to

alkyl or aryl groups, the compound is a substituted amide.

[pic]

• The carbonyl carbon-to-nitrogen bond is called an amide linkage.
• This bond is quite stable and is found in the repeating units of protein

molecules, where it is called a peptide linkage.

• Simple amides are named as derivatives of carboxylic acids.
• The -ic ending of the common name or the -oic ending of the International

Union of Pure and Applied Chemistry (IUPAC) name of the carboxylic acid

is replaced with the suffix –amide

[pic]

STEP 5: Chemical Properties of Amides (20 minutes).

|Activity: Buzzing (5minutes) |

| |

|ASK students to pair up and buzz on the following question for 5 |

|minutes. |

| |

|What are the chemical properties of Amides? |

| |

|ALLOW pairs to respond on the question. |

| |

|WRITE their response on the flip chart/board. |

| |

|CLARIFY and SUMMARIZE by using the content below |

• Amphoteric Character.

o Amides are very weak bases.

o This is due to the fact that the lone pair of electrons on nitrogen

atom is involved in resonance with carbonyl group.

o This is due to the contribution of resonating structure II as shown

below;

[pic]

o Thus, electron pair of nitrogen is not easily available for

protonation.

o Consequently, the basic character is considerably decreased.

o However, under suitable conditions amides can also exhibit a feeble

acidic character.

• Basic character.

o In accordance with resonating structure I already shown, it is evident

that nitrogen atom of amide molecule has a lone pair of electrons.

o Therefore, it can act as a base.

o For example, acetamide (as base) reacts with hydrochloric acid (an

acid) to form a salt.

o CH3 CONH2 + HCl à CH3 CONH2 HCl

• Acidic character.

o In accordance with resonating structure II shown earlier, it is clear

that the development of positive character on nitrogen atom

facilitates the release of proton.

o Thus, amide can act as acid.

o For example, acetamide (as acid) reacts with mercuric oxide (a base)

to form mercury salt and water.

o 2CH3 COHN2 + HgO → (CH3 CONH)2 Hg + H2O

• Hydrolysis.

o On boiling with dilute acid or alkali, amides rapidly undergo

hydrolysis.

o For example:

[pic]

STEP 6: Chemical Reactions involving Amides (45 minutes).

|Activity: Small Group Discussion (20 minutes). |

| |

|DIVIDE students into small groups. |

| |

|ASK students to discuss in groups on the following questions. |

|What are the chemical reactions involving Amides? |

| |

|[pic]REFER Students to Book |

| |

|ALLOW students to discuss for 15 minutes. |

| |

|ALLOW each group to present for 5 minutes. |

| |

|CLARIFY and SUMMARIZE by using the contents below |

• Dehydration of amides

o Amides are dehydrated by heating a solid mixture of the amide and

phosphorus(V) oxide, P4O10.

o Water is removed from the amide group to leave a nitrile group, -CN.

o The liquid nitrile is collected by simple distillation.

o For example, with ethanamide, the product is ethanenitrile.

[pic]

• The Hofmann Degradation

o The Hofmann degradation is a reaction between an amide and a mixture

of bromine and sodium hydroxide solution.

o The net effect of the reaction is a loss of the -CO- part of the amide

group.

o The product is primary amine with one less carbon atom than the

original amide.

o [pic]

o If ethanamide is used, the product will be methylamine.

▪ The full equation for the reaction is:

o CH3CONH2+Br2+4NaOH→CH3NH2+Na2CO3+2NaBr+2H2O

o The Hofmann degradation is used as a way of cutting a single carbon

atom out of a chain.

• The reduction of amides

o Amides can be reduced to primary amines by reaction with lithium

tetrahydridoaluminate, LiAlH4, in dry ether (ethoxyethane) at room

temperature.

o The initial reaction is followed by treatment with dilute acid, such

as dilute sulphuric or hydrochloric acid.

▪ For example, ethanamide can be reduced ethylamine.

o CH3CONH2+4[H]→CH3CH2NH2+H2O

• Reaction with Nitrous Acid.

o Amides react with nitrous acid to give carboxylic acids. and nitrogen

gas.

o Nitrous acid required is prepared in situ by reaction of NaNO2 and

HCI.

[pic]

STEP 7: Key Points (05 minutes).

• Amides are usually regarded as derivatives of carboxylic acids in which

the hydroxyl group has been replaced by an amine or ammonia

• Amides are classified into primary, secondary and tertiary amides
• Chemical properties of amides involve amphoteric character, reduction,

hydrolysis and dehydration reactions

• Amides (carrying primary nitrogen atom) react with bromine in the

presence of alkali to form a primary amine carrying one carbon atom less

than the parent amide.

STEP 8: Evaluation (05 minutes).

• What are Amides?
• How are amides named
• Draw chemical structure of secondary amides
• List chemical properties of amides
• What is Hoffman degradation?

References

Ternay, A.L (1976). Contemporary Organic Chemistry. Philadelphia, United

States: W.B. Saunders Co.

Morrison R.T and Boyd R N (1997). Organic Chemistry (6th Ed.). New Delhi,

India: Prentice Hall of India

Graham Solomon et al (2014). Organic Chemistry (11th Ed.). New Jeysey,

United States: John Willey and Sons.

Nadendla R. R. (2005). Principles of Pharmaceutical Organic Chemistry. New

Delhi, India: MacMillan Publishers

Bruice Y (2013). Organic Chemistry (7th ed.). New York, United States:

Prentice Hall Pearson.

Delgado J. N. Et al (1998). Wilson and Gisvold's Textbook of Organic

Medicinal and Pharmaceutical Chemistry (10th Ed.). California, United

States: Lippincott Williams

Bhassin S.K, Gupta R. (2013). Pharmaceutical organic chemistry (E-book

Kindle edition). New Delhi, India: Elsevier Publishing Services

Session 23: Introduction to Heterocyclic Compounds.

Total Session Time: 120 minutes

Prerequisites

None

Learning Tasks

By the end of this session students are expected to be able to:

• Define heterocyclic compounds
• Classify heterocyclic compounds
• Explain nomenclature of heterocyclic compounds
• Draw chemical structure of heterocyclic compounds

Resources Needed:

• Flip charts, marker pens, and masking tape.
• Black/white board and chalk/whiteboard markers.

SESSION OVERVIEW

|Step |Time |Activity/ |Content |

| | |Method | |

|1 |05 minutes |Presentation |Introduction, Learning Tasks |

|2 |10 minutes |Brainstorming |Definition of Heterocyclic Compounds|

| | |Presentation | |

|3 |25 minutes |Presentation |Classification of Heterocyclic |

| | | |Compounds |

|4 |30 minutes |Presentation |Nomenclature of Heterocyclic |

| | | |Compounds |

|5 |30 minutes |Presentation |Chemical Structure of Heterocyclic |

| | | |Compounds |

|6 |10 minutes |Presentation |Key Points |

|7 |10 minutes |Presentation |Evaluation |

SESSION CONTENTS.

STEP 1: Presentation of Session Title and Learning Tasks (5 minutes).

READ or ASK students to read the learning tasks and clarify.

ASK students if they have any questions before continuing.

STEP 2: Definition of Heterocyclic Compounds (10minutes).

|Activity: Brainstorming (5 minutes) |

| |

|Ask students to brainstorm on the following question: |

| |

|What are Heterocyclic compounds? |

| |

|ALLOW few students to respond. |

| |

|WRITE their responses on the flip chart/ board. |

| |

|CLARIFY and SUMMARISE by using the content below |

• Heterocyclic compound, also called heterocycle, any of a major class of

organic chemical compounds characterized by the fact that some or all of

the atoms in their molecules are joined in rings containing at least one

atom of an element other than carbon (C).

• The cyclic part (from Greek kyklos, meaning “circle”) of heterocyclic

indicates that at least one ring structure is present in such a compound,

while the prefix hetero- (from Greek heteros, meaning “other” or

“different”) refers to the noncarbon atoms, or heteroatoms, in the ring.

• Heterocyclic compounds include many of the biochemical material essential

to life. For example, nucleic acids, pigments, vitamins, and antibiotics.

STEP 3: Classification of Heterocyclic Compounds (25 minutes).

• Classification of heterocyclic compounds depends on ring size because

heterocyclic rings of a given size has many common features.

• Therefore, heterocyclic compounds can be classified as:

o Three-membered rings

o four-membered rings

o Five-membered rings

o six-membered rings

o Seven-membered rings

• Three-membered rings

o The three-membered ring heterocycles containing single atoms of

nitrogen, oxygen, and sulfur—aziridine, oxirane (or ethylene oxide),

and thiirane, respectively—and their derivatives can all be prepared

by nucleophilic reactions, of the type shown.

o Thus, aziridine is formed by heating β-aminoethyl hydrogen sulfate

with a base (in this case Y is −OSO3H).

[pic]

• Four-membered rings

o Azetidine, oxetane, and thietane—four-membered rings containing,

respectively, one nitrogen, oxygen, or sulfur atom—are prepared by

nucleophilic displacement reactions similar to those used to prepare

the corresponding three-membered rings.

[pic]

• Five-membered rings with one heteroatom

o The parent aromatic compounds of this family—pyrrole, furan, and

thiophene—have the structures shown.

[pic]

o The saturated derivatives are called pyrrolidine, tetrahydrofuran, and

thiophane, respectively.

o The bicyclic compounds made of a pyrrole, furan, or thiophene ring

fused to a benzene ring are called indole (or isoindole), benzofuran,

and benzothiophene, respectively.

• Six-membered rings with one heteroatom

o The nomenclature used for the various monocyclic nitrogen-containing

six-membered ring compounds is given below.

o Positions on the ring are shown for pyridine, Arabic numerals being

preferred to Greek letters, although both systems are used.

o The pyridones are aromatic compounds because of contributions to the

resonance hybrid from charged resonance forms such as that shown for 4-

pyridone.

[pic]

STEP 4: Nomenclature of Heterocyclic Compounds (30 minutes).

• Many heterocycles, especially amines, were identified early on, and

received trivial names which are still preferred.

• Some monocyclic compounds of this kind are shown in the following chart,

with the common (trivial) name in bold and a systematic name based on the

Hantzsch-Widman system given beneath it in blue.

[pic]

• An easy to remember, but limited, nomenclature system makes use of an

elemental prefix for the heteroatom followed by the appropriate

carbocyclic name.

• A short list of some common prefixes is given in the following table,

priority order increasing from right to left.

• Examples of this nomenclature are: ethylene oxide = oxacyclopropane,
furan = oxacyclopenta-2,4-diene, pyridine = azabenzene, and morpholine =

1-oxa-4-azacyclohexane.

|Element|oxygen|sulfu|seleniu|nitroge|phosphorou|silic|boron|

| | |r |m |n |s |on | |

|Valence|II |II |II |III |III |IV |III |

|Prefix |Oxa |Thia |Selena |Aza |Phospha |Sila |Bora |

| | | | | | | | |

• The Hantzsch-Widman system provides a more systematic method of naming

heterocyclic compounds that is not dependent on prior carbocyclic names.

• It makes use of the same hetero atom prefix defined above (dropping the

final "a"), followed by a suffix designating ring size and saturation.

• As outlined in the following table, each suffix consists of a ring size

root (blue) and an ending intended to designate the degree of

unsaturation in the ring.

• In this respect, it is important to recognize that the saturated suffix

applies only to completely saturated ring systems, and the unsaturated

suffix applies to rings incorporating the maximum number of non-cumulated

double bonds.

• Systems having a lesser degree of unsaturation require an appropriate

prefix, such as "dihydro"or "tetrahydro".

|Ring Size |3 |4 |5 |6 |7 |8 |9 |10 |

|Suffix | | | | | | | | |

| |iren|ete |ole |ine |epine|ocine|onine|ecine|

|Unsaturated|e |etan|olan|inane|epane|ocane|onane|ecane|

| Saturated|iran|e |e | | | | | |

| |e | | | | | | | |

• Despite the general systematic structure of the Hantzsch-Widman system,

several exceptions and modifications have been incorporated to

accommodate conflicts with prior usage.

• Some examples are:

o The terminal "e" in the suffix is optional though recommended.

o Saturated 3, 4 & 5-membered nitrogen heterocycles should use

respectively the traditional "iridine", "etidine" & "olidine"

suffix.

o Unsaturated nitrogen 3-membered heterocycles may use the

traditional "irine" suffix.

o Consistent use of "etine" and "oline" as a suffix for 4 & 5-

membered unsaturated heterocycles is prevented by their former use

for similar sized nitrogen heterocycles.

o Established use of oxine, azine and silane for other compounds or

functions prohibits their use for pyran, pyridine and

silacyclohexane respectively.

• Examples of these nomenclature rules are written in blue, both in the

previous diagram and that shown below.

• Note that when a maximally unsaturated ring includes a saturated atom,

its location may be designated by a "#H " prefix to avoid ambiguity, as

in pyran and pyrrole above and several examples below.

• When numbering a ring with more than one heteroatom, the highest priority

atom is #1 and continues in the direction that gives the next priority

atom the lowest number.

[pic]

• All the previous examples have been monocyclic compounds. Polycyclic

compounds incorporating one or more heterocyclic rings are well known.

• A few of these are shown in the following diagram.
• As before, common names are in black and systematic names in blue.
• The two quinolines illustrate another nuance of heterocyclic

nomenclature.

• Thus, the location of a fused ring may be indicated by a lowercase letter

which designates the edge of the heterocyclic ring involved in the

fusion, as shown by the pyridine ring in the green shaded box.

[pic]

• Heterocyclic rings are found in many naturally occurring compounds.
• Most notably, they compose the core structures of mono and

polysaccharides, and the four DNA bases that establish the genetic code.

STEP 5: Chemical Structure of Heterocyclic Compounds (30 minutes).

• Three- and four-membered rings, because of their small size, are

geometrically strained and thus readily opened; they are also readily

formed.

o Such heterocycles are well-known reactive intermediates.

• Five- and six-membered rings are readily formed and are very stable;

their sizes also allow the development of aromatic character.

• Seven-membered rings and larger are stable but less readily formed and

relatively less well investigated.

[pic]

[pic]

[pic]

[pic]

STEP 6: Key Points (10 minutes)

• A heterocyclic compound or ring structure is a cyclic compound that

has atoms of at least two different elements as members of its

ring(s).

• Heterocyclic compounds are classified based on ring size because

heterocyclic rings of a given size have many common features.

• The Hantzsch-Widman system provides a more systematic method of naming

heterocyclic compounds that is not dependent on prior carbocyclic

names.

• Three- and four-membered rings, are geometrically strained because of

their small size, and thus readily opened.

STEP 7: Evaluation (10 minutes)

• What are heterocyclic compounds?
• Classify heterocyclic compounds.
• Name three heterocyclic compounds using Hantzsch-Widman system.
• Draw three chemical structure of heterocyclic compounds.

References

Ternay, A.L (1976). Contemporary Organic Chemistry. Philadelphia, United

States: W.B. Saunders Co.

Morrison R.T and Boyd R N (1997). Organic Chemistry (6th Ed.). New Delhi,

India: Prentice Hall of India

Graham Solomon et al (2014). Organic Chemistry (11th Ed.). New Jeysey,

United States: John Willey and Sons.

Nadendla R. R. (2005). Principles of Pharmaceutical Organic Chemistry. New

Delhi, India: MacMillan Publishers

Bruice Y (2013). Organic Chemistry (7th ed.). New York, United States:

Prentice Hall Pearson.

Delgado J. N. Et al (1998). Wilson and Gisvold's Textbook of Organic

Medicinal and Pharmaceutical Chemistry (10th Ed.). California, United

States: Lippincott Williams

Bhassin S.K, Gupta R.(2013). Pharmaceutical organic chemistry (E-book

Kindle edition). New Delhi, India: Elsevier Publishing Services

Session 24: Chemical Reactions of Heterocyclic Compounds.

Total Session Time: 120 minutes

Prerequisites

None

Learning Tasks

By the end of this session students are expected to be able to:

• List chemical properties of heterocyclic compounds
• Explain chemical reactions of heterocyclic compounds

Resources Needed:

• Flip charts, marker pens, and masking tape
• Black/white board and chalk/whiteboard markers

SESSION OVERVIEW

|Step |Time |Activity/ |Content |

| | |Method | |

|1 |05 minutes |Presentation |Introduction, Learning Tasks |

|2 |35 minutes |Brainstorming |Chemical Properties of Heterocyclic |

| | |Presentation |Compounds |

|3 |60 minutes |Group |Chemical Reactions involving |

| | |discussion |Heterocyclic Compounds |

| | |Presentation | |

|4 |10 minutes |Presentation |Key Points |

|5 |10 minutes |Presentation |Evaluation |

SESSION CONTENTS

STEP 1: Presentation of Session Title and Learning Tasks (5 minutes)

READ or ASK students to read the learning tasks and clarify.

ASK students if they have any questions before continuing.

STEP 2: Chemical Properties of Heterocyclic Compounds (35 minutes).

|Activity: Brainstorming (5 minutes) |

| |

|Ask students to brainstorm on the following question: |

| |

|What are the chemical properties of heterocyclic compounds? |

| |

|ALLOW few students to respond. |

| |

|WRITE their responses on the flip chart/ board. |

| |

|CLARIFY and SUMMARISE by using the table below |

|Chemical Name |Chemical properties |

|2-AMINOPYRIDINE |The substance decomposes on burning |

| |producing toxic gases and vapours |

| |including nitrous oxides |

| |Reacts with strong oxidants causing |

| |fire and explosion hazard |

| |The substance is a strong base that |

| |is soluble in water |

|3,6-DICHLOROPICOLINIC ACID |The substance decomposes on burning |

| |producing toxic and corrosive gases |

| |Reacts with bases forming salts |

| |Solutions of them are corrosive to |

| |aluminum, iron, and tin |

|2-MERCAPTOBENZOTHIAZOLE |On combustion, forms toxic gases |

| |(carbon monoxide and sulphur |

| |compounds) |

| |The substance decomposes on heating |

| |and on burning producing toxic and |

| |irritating fumes (sulphur and |

| |nitrogen oxides) |

| |Reacts with acids with the formation |

| |of highly toxic fumes of sulphur |

| |compounds |

| |Reacts with acids or acid fumes |

| |producing toxic fumes (sulphur |

| |compounds) |

|2-MERCAPTOBENZOTHIAZOLE |On combustion, forms toxic gases: |

|DISULPHIDE |carbon, sulphur and nitrogen oxides |

| |Reacts with strong oxidants and acids|

| |The substance decomposes on heating |

|2-METHYLPYRIDINE |producing toxic fumes (nitrogen |

| |oxides) |

| |Reacts with oxidants and strong acids|

| | |

| |Attacks copper and its alloys |

|3-METHYLPYRIDINE |The substance decomposes on heating |

| |producing toxic fumes (nitrogen |

| |oxides) |

| |Reacts with oxidants and strong acids|

|4-METHYLPYRIDINE |The substance decomposes on heating |

| |producing toxic fumes (nitrogen |

| |oxides) |

| |Reacts with oxidants and strong acids|

|1-METHYL-2-PYRROLIDONE |The substance decomposes on heating |

| |above 315 °C producing toxic fumes |

| |Reacts with strong acids |

| |Attacks aluminium |

|MORPHOLINE |The substance decomposes on heating |

| |producing toxic fumes (nitrogen |

| |oxides) |

| |The substance is a weak base |

| |Reacts with strong oxidants causing |

| |fire hazard |

| |Attacks copper and its compounds |

|PHENOTHIAZINE |The substance decomposes on heating |

| |and on burning producing toxic and |

| |irritating fumes including nitrogen |

| |oxides and sulphur oxides |

|PHENYLENEPYRENE |Upon heating, toxic fumes are formed |

|PIPERIDINE |The substance decomposes on heating |

| |and on burning producing toxic gases |

| |such as nitrogen oxides |

| |The substance is a medium strong base|

| | |

| |Reacts violently with oxidants |

|PYRIDINE |On combustion, forms toxic fumes |

| |(amines) |

| |The substance decomposes on heating |

| |or on burning producing toxic fumes |

| |(nitrogen oxides and hydrogen |

| |cyanide) |

| |Reacts violently with strong oxidants|

| |and strong acids |

|2-PYRROLIDINONE |The substance decomposes on heating |

| |producing toxic fumes |

| |Reacts with strong acids cf |

| |• methylpyrrolidone Attacks aluminium|

|QUINOLINE |The substance decomposes on heating |

| |and on burning producing toxic fumes |

| |of nitrogen oxides |

| |Reacts with strong oxidants and |

| |maleine anhydride |

|TETRAHYDROTHIOPHENE |On combustion, forms toxic fumes |

| |• Reacts violently with strong |

| |oxidants and nitric acid • Attacks |

| |rubber |

|THIOPHENE |The substance decomposes on heating |

| |and on burning producing toxic and |

| |irritating fumes (sulphur oxides) |

| |• Reacts violently with oxidizing |

| |materials, including fuming nitric |

| |acid |

STEP 3: Chemical Reactions involving Heterocyclic Compounds (60 minutes).

|Activity: Small Group Discussion (20 minutes) |

| |

|DIVIDE students into small groups |

| |

|ASK students to discuss in groups on the following questions |

|What are the chemical reactions involving heterocyclic compounds? |

| |

|[pic]REFER Students to Book |

| |

|ALLOW students to discuss for 15 minutes |

| |

|ALLOW each group to present for 5 minutes |

| |

|CLARIFY and SUMMARIZE by using the contents below |

Three-Membered Rings

• Oxiranes (epoxides) are the most commonly encountered three-membered

heterocycles. Epoxides are easily prepared by reaction of alkenes with

peracids, usually with good stereospecificity.

• Because of the high angle strain of the three-membered ring, epoxides are

more reactive that unstrained ethers.

• Addition reactions proceeding by electrophilic or nucleophilic opening of

the ring constitute the most general reaction class.

• Example 1 in the following diagram shows one such transformation, which

is interesting due to subsequent conversion of the addition intermediate

into the corresponding thiirane.

o The initial ring opening is stereoelectronically directed in a

trans-diaxial fashion, the intermediate relaxing to the

diequatorial conformer before cyclizing to a 1,3-oxathiolane

intermediate.

o Other examples show similar addition reactions to thiiranes and

aziridines.

• The acid-catalyzed additions in examples 2 and 3, illustrate the

influence of substituents on the regioselectivity of addition.

• Example 2 reflects the SN2 character of nucleophile (chloride anion)

attack on the protonated aziridine (the less substituted carbon is the

site of addition).

o The phenyl substituent in example 3 serves to stabilize the

developing carbocation to such a degree that SN1 selectivity is

realized.

o The reduction of thiiranes to alkenes by reaction with phosphite

esters (example 6) is highly stereospecific and is believed to take

place by an initial bonding of phosphorous to sulfur.

[pic]

Four-Membered Rings

• Reactions of four-membered heterocycles also show the influence of ring

strain.

• Some examples are given in the following diagram.
• Acid-catalysis is a common feature of many ring-opening reactions, as

shown by examples 1, 2 & 3a.

• In the thietane reaction (2), the sulfur undergoes electrophilic

chlorination to form a chlorosulfonium intermediate followed by a ring-

opening chloride ion substitution.

• Strong nucleophiles will also open the strained ether, as shown by

reaction 3b.

• Cleavage reactions of β-lactones may take place either by acid-catalyzed

acyl exchange, as in 4a, or by alkyl-O rupture by nucleophiles, as in 4b.

• Example 5 is an interesting case of intramolecular rearrangement to an

ortho-ester.

• Finally, the β-lactam cleavage of penicillin G (reaction 6) testifies to

the enhanced acylating reactivity of this fused ring system.

• Most amides are extremely unreactive acylation reagents, thanks to

stabilization by p-π resonance.

• Such electron pair delocalization is diminished in the penicillins,

leaving the nitrogen with a pyramidal configuration and the carbonyl

function more reactive toward nucleophiles.

[pic]

Five-Membered Rings

The chemical reactivity of the saturated members of this class of

heterocycles: tetrahydrofuran, thiolane and pyrrolidine, resemble that of

acyclic ethers, sulfides, and 2º-amines, and will not be described here.

• 1,3-Dioxolanes and dithiolanes are cyclic acetals and thioacetals.
• These units are commonly used as protective groups for aldehydes and

ketones, and may be hydrolyzed by the action of aqueous acid.

It is the "aromatic" unsaturated compounds, furan, thiophene and pyrrole

that require our attention.

• In each case the heteroatom has at least one pair of non-bonding

electrons that may combine with the four π-electrons of the double bonds

to produce an annulene having an aromatic sextet of electrons.

• This is illustrated by the resonance description at the top of the

following diagram.

• The heteroatom Y becomes sp2-hybridized and acquires a positive charge as

its electron pair is delocalized around the ring.

• An easily observed consequence of this delocalization is a change in

dipole moment compared with the analogous saturated heterocycles, which

all have strong dipoles with the heteroatom at the negative end.

• As expected, the aromatic heterocycles have much smaller dipole moments,

or in the case of pyrrole a large dipole in the opposite direction.

• An important characteristic of aromaticity is enhanced thermodynamic

stability, and this is usually demonstrated by relative heats of

hydrogenation or heats of combustion measurements.

• By this standard, the three aromatic heterocycles under examination are

stabilized, but to a lesser degree than benzene.

Additional evidence for the aromatic character of pyrrole is found in its

exceptionally weak basicity (pKa ca. 0) and strong acidity (pKa = 15) for

a 2º-amine.

• The corresponding values for the saturated amine pyrrolidine are:

basicity 11.2 and acidity 32.

[pic]

Electrophilic Substitution of Pyridine

Pyridine is a modest base (pKa=5.2).
• Since the basic unshared electron pair is not part of the aromatic

sextet, as in pyrrole, pyridinium species produced by N-substitution

retain the aromaticity of pyridine.

• As shown below, N-alkylation and N-acylation products may be prepared as

stable crystalline solids in the absence of water or other reactive

nucleophiles.

• The N-acyl salts may serve as acyl transfer agents for the preparation of

esters and amides. Because of the stability of the pyridinium cation, it

has been used as a moderating component in complexes with a number of

reactive inorganic compounds.

• Several examples of these stable and easily handled reagents are shown at

the bottom of the diagram.

• The poly(hydrogen fluoride) salt is a convenient source of HF for

addition to alkenes and conversion of alcohols to alkyl fluorides,

pyridinium chlorochromate (PCC) and its related dichromate analog are

versatile oxidation agents and the tribromide salt is a convenient source

of bromine.

• Similarly, the reactive compounds sulfur trioxide and diborane are

conveniently and safely handled as pyridine complexes.

Amine oxide derivatives of 3º-amines and pyridine are readily prepared by

oxidation with peracids or peroxides, as shown by the upper right

equation.

• Reduction back to the amine can usually be achieved by treatment with

zinc (or other reactive metals) in dilute acid.

[pic]

Other Reactions of Pyridine

• Thanks to the nitrogen in the ring, pyridine compounds undergo

nucleophilic substitution reactions more easily than equivalent benzene

derivatives.

• In the following diagram, reaction 1 illustrates displacement of a 2-

chloro substituent by ethoxide anion.

• The addition-elimination mechanism shown for this reaction is helped by

nitrogen's ability to support a negative charge.

• A similar intermediate may be written for substitution of a 4-

halopyridine, but substitution at the 3-position is prohibited by the the

failure to create an intermediate of this kind.

• The two Chichibabin aminations in reactions 2 and 3 are remarkable in

that the leaving anion is hydride (or an equivalent). Hydrogen is often

evolved in the course of these reactions.

• In accord with this mechanism, quinoline is aminated at both C-2 and C-4.

Addition of strong nucleophiles to N-oxide derivatives of pyridine

proceed more rapidly than to pyridine itself, as demonstrated by

reactions 4 and 5.

• The dihydro-pyridine intermediate easily loses water or its equivalent by

elimination of the –OM substituent on nitrogen.

[pic]

STEP 4: Key Points (10 minutes)

• Chemical properties of heterocyclic compounds are variable depending

on the class of the compound.

• Chemical reactions of heterocyclic compounds depend on the class of

the heterocyclic compound.

STEP 5: Evaluation (10 minutes)

• List chemical properties of heterocyclic compounds.
• List chemical reactions of heterocyclic compounds.

References

Ternay, A.L (1976). Contemporary Organic Chemistry. Philadelphia, United

States: W.B. Saunders Co.

Morrison R.T and Boyd R N (1997). Organic Chemistry (6th Ed.). New Delhi,

India: Prentice Hall of India

Graham Solomon et al (2014). Organic Chemistry (11th Ed.). New Jeysey,

United States: John Willey and Sons.

Rama Rao Nadendla (2005). Principles of Pharmaceutical Organic Chemistry.

New Delhi, India: MacMillan Publishers

Bruice Y (2013). Organic Chemistry (7th ed.). New York, United States:

Prentice Hall Pearson.

Delgado J. N. Et al (1998). Wilson and Gisvold's Textbook of Organic

Medicinal and Pharmaceutical Chemistry (10th Ed.). California, United

States: Lippincott Williams

Bhassin S.K, Gupta R.(2013). Pharmaceutical organic chemistry (E-book

Kindle edition). New Delhi, India: Elsevier Publishing Services

Session 25: Introduction to Structure – Activity Relationship of Drugs.

Total Session Time: 60 minutes

Prerequisites

None

Learning Tasks

By the end of this session students are expected to be able to:

• Define structure-activity relationship
• Explain the importance of structure-activity relationship in pharmacy

Resources Needed:

• Flip charts, marker pens, and masking tape
• Black/white board and chalk/whiteboard markers

SESSION OVERVIEW

|Step |Time |Activity/ |Content |

| | |Method | |

|1 |05 minutes |Presentation |Introduction, Learning Tasks |

|2 |10 minutes |Brainstorming |Definition of structure – activity |

| | |Presentation |relationship |

|2 |30 minutes |Group |Importance of structure – activity |

| | |discussion |relationship in pharmacy |

| | |Presentation | |

|4 |05 minutes |Presentation |Key Points |

| 5 |10 minutes |Presentation |Evaluation |

SESSION CONTENTS

STEP 1: Presentation of Session Title and Learning Tasks (5 minutes)

READ or ASK students to read the learning tasks and clarify

ASK students if they have any questions before continuing.

STEP 2: Definition of Structure – Activity Relationship (10 minutes).

|Activity: Brainstorming (5 minutes) |

| |

|Ask students to brainstorm on the following question: |

| |

|What is structure- activity relationship? |

| |

|ALLOW few students to respond. |

| |

|WRITE their responses on the flip chart/ board. |

| |

|CLARIFY and SUMMARISE by using the table below |

Structure Activity Relationship (SAR):

• A structure activity relationship relates features of a chemical

structure to a property, effect, or biological activity associated

with that chemical.

• SAR is the subject which brings about the awareness of the

relationships between the chemistry of a particular compound or group

of compounds and their interaction with the body hence activity.

STEP 3: Importance of Structure – Activity Relationship in Pharmacy (30

minutes).

|Activity: Small Group Discussion (15 minutes) |

| |

|DIVIDE students into small groups. |

| |

|ASK students to discuss in groups on the following questions. |

|What is the importance of structure- activity relationship in Pharmacy?|

| |

|[pic]REFER Students to Book |

| |

|ALLOW students to discuss for 10 minutes. |

| |

|ALLOW each group to present for 5 minutes. |

| |

|CLARIFY and SUMMARIZE by using the contents below |

• SAR enables the determination of the chemical groups responsible for

evoking a target biological effect in the organism.

• It allows modification of the effect or the potency of a bioactive

compound (typically a drug) by changing its chemical structure

• Medicinal chemists use the techniques of chemical synthesis to insert new

chemical groups into the biomedical compound and test the modifications

for their biological effects.

STEP 4: Key Points (05 minutes).

• The structure–activity relationship (SAR) is the relationship between

the chemical or 3D structure of a molecule and its biological activity.

• SAR enables the determination of the chemical groups responsible for

evoking a target biological effect in the organism.

• It allows modification of the effect or the potency of a bioactive

compound (typically a drug) by changing its chemical structure.

• Medicinal chemists use the techniques of chemical synthesis to insert new

chemical groups into the biomedical compound and test the modifications

for their biological effects.

STEP 5: Evaluation (10 minutes)

• What is Structure-activity relationship?
• What is the importance of SAR in pharmacy?

References

Ternay, A.L (1976). Contemporary Organic Chemistry. Philadelphia, United

States: W.B. Saunders Co.

Morrison R.T and Boyd R N (1997). Organic Chemistry (6th Ed.). New Delhi,

India: Prentice Hall of India

Graham Solomon et al (2014). Organic Chemistry (11th Ed.). New Jeysey,

United States: John Willey and Sons.

Nadendla R. R. (2005). Principles of Pharmaceutical Organic Chemistry. New

Delhi, India: MacMillan Publishers

Bruice Y (2013). Organic Chemistry (7th ed.). New York, United States:

Prentice Hall Pearson.

Delgado J. N. Et al (1998). Wilson and Gisvold's Textbook of Organic

Medicinal and Pharmaceutical Chemistry (10th Ed.). Calfornia, United

States: Lippincott Williams

Bhassin S.K, Gupta R. (2013). Pharmaceutical organic chemistry (E-book

Kindle edition). New Delhi, India: Elsevier Publishing Services

Session 26: Structure – Activity Relationship of Penicillins.

Total Session Time: 120 minutes

Prerequisites

None

Learning Tasks

By the end of this session students are expected to be able to:

• Define Penicillins
• Explain Chemical structure of penicillins
• Explain the structure – activity relationship of penicillins

Resources Needed:

• Flip charts, marker pens, and masking tape
• Black/white board and chalk/whiteboard markers

SESSION OVERVIEW

|Step |Time |Activity/ |Content |

| | |Method | |

|1 |05 minutes |Presentation |Introduction, Learning Tasks |

|2 |10 minutes |Brainstorming |Definition of Penicillins |

| | |Presentation | |

|3 |40 minutes |Presentation |Chemical Structure of Penicilins |

|4 |45 minutes |Group |Structure – Activity Relationship of|

| | |discussion |Penicillins |

| | |Presentation | |

|5 |10 minutes |Presentation |Key Points |

| 6 |10 minutes |Presentation |Evaluation |

SESSION CONTENTS.

STEP 1: Presentation of Session Title and Learning Tasks (5 minutes).

READ or ASK students to read the learning tasks and clarify.

ASK students if they have any questions before continuing.

STEP 2: Definition of Penicillins (10 minutes).

|Activity: Brainstorming (5 minutes) |

| |

|Ask students to brainstorm on the following question: |

| |

|What are penicillins? |

| |

|ALLOW few students to respond |

| |

|WRITE their responses on the flip chart/ board |

| |

|CLARIFY and SUMMARISE by using the table below |

• There are two major classes of antibacterial agents which act by

inhibiting ell wall synthesis.

• Penicillins are a group of β-lactam antibiotics consisting of natural

penicillins and semisynthetic penicillins.

• Penicillin inhibit peptidoglycan synthesis in bacteria and the drugs

are mainly active against gram (+) ve bacteria.

STEP 3: Chemical Structure of Penicillins (40 minutes).

Fig. 24.1 General Chemical structure of penicillin

[pic]

G.L Patrick, an introduction to medical chemistry.

• Penicillin contains a highly unstable-looking bicyclic system

consisting of a four membered β-lactam ring fused to a five-membered

thiazolidine ring.

• The skeleton of the molecule suggests that it is derived from the

amino acids cysteine and valine. According to biogenesis, antibiotics

can be derived from various natural substances like β-lactam

antibiotics from cysteine and valine amino acids.

[pic]

Penicillin appears to be derived from

cysteine and valine

[pic]

Side chain varies according to carboxylic acids presents in

fermentation medium.

[pic]

• All penicillin has the same β-lactam-thiazolidine general structure

that contains three chiral centers.

• Therefore, theoretically this structure could present eight optically

active forms.

• However, the natural isomer, presumably the only one with biological

activity, has the stereochemistry of 3S:5R:6R. (According to BP & USP

2S:5R:6R).

[pic]

Penicillin analogues

[pic]

STEP 4: Structure – Activity relationship of Penicillins (45 minutes).

|Activity: Small Group Discussion (20 minutes) |

| |

|DIVIDE students into small manageable groups. |

| |

|ASK students to discuss on the following question. |

|What is the importance of structure activity relationship of |

|Penicillins? |

| |

|ALLOW students to discuss for 15 minutes. |

| |

|ALLOW few groups to present and the rest to add points not mentioned. |

| |

|CLARIFY and SUMMARIZE by using the contents below |

A large number of penicillin analogues have been synthesized and studied.

The results of these studies have demonstrated following features are

important for penicillins activity.

• The strained β-lactam ring is essential.
• The bicyclic system is important.
• The acidic functional group (free carboxylic acid) is essential.
• The acylamino (amide) side-chain is essential.
• The stereochemistry of the bicyclic ring with respect to the acylamino

side-chain is important.

• The acyl side-chain (R) can varies.

Very little variation is possible in penicillin nucleus.

Structure activity relationships of penicillin

[pic]

The acid sensitivity of penicillin.

There are three reasons for the acid sensitivity of penicillin.

• Ring strain

o The bicyclic system in penicillin consists of a four-membered ring and

a five membered ring.

o As a result, penicillins suffers large angle and torsional strains.

o Acid-catalyzed ring opening relieves these strains by breaking open

the more highlystrained four-memberedβ-lactam ring.

[pic]

[pic]

Ring opening

• A highly reactive β-lactam carbonyl group

o The carbonyl group in the β-lactam ring is highly susceptible to

nucleophiles and it does not behave like a normal tertiary amide which

is usually quite resistant to nucleophilic attack.

o A normal tertiary amide is far less susceptible to nucleophiles since

the resonance structures reduce the electrophilic character of the

carbonyl group.

o The β-lactam nitrogen is unable to show such effect.

o To show the similar effect like tertiary amide, penicillin had to

obtain astrained flat structure, which is highly unstable.

o As a result, the lone pair is localized on the nitrogen atom and the

carbonyl group is far more electrophilic than a tertiary amide.

[pic]

• Influence of the acyl side-chain

o Figure above demonstrates how the neighboring acyl group canactively

participate in a mechanism to open up the lactam ring.

o Thus,penicillin Ghas a self-destruct mechanism built into its

structure.

[pic]

Influence of the acyl side chain on acid sentivity

[pic]

STEP 5: Key Points (10 minutes).

• Penicillin refers to any of several antibiotics produced naturally by

molds of genus Penicillium and also semi-synthetically, having a

bactericidal action on many susceptible Gram positive and Gram-

negative bacteria.

negative cocciand bacilli, some also being effective against certain sp

irochetes.

• The term "penam" is used to describe the common core skeleton of a

member of the penicillins which has the molecular formula R-

C9H11N2O4S, where R is the variable side chain that differentiates the

penicillins from one another.

• SAR of penicillins has enabled development of acid resistant drugs

such as ampicillin, improving their spectrum of activity and

bioavailability (amoxicillin), development of beta-lactamase resistant

drugs, (oxacillin, cloxacillin), and broad spectrum penicillins

(Carbenicillin, ureidopenicillins)

STEP 6: Evaluation (10 minutes)

• What are Penicillins?
• Draw Chemical structure of penicillins
• What is the importance of structure – activity relationship of

penicillins?

References

Ternay, A.L (1976). Contemporary Organic Chemistry. Philadelphia, United

States: W.B. Saunders Co.

Morrison R.T and Boyd R N (1997). Organic Chemistry (6th Ed.). New Delhi,

India: Prentice Hall of India

Graham Solomon et al (2014). Organic Chemistry (11th Ed.). New Jeysey,

United States: John Willey and Sons.

Nadendla R. R. (2005). Principles of Pharmaceutical Organic Chemistry. New

Delhi, India: MacMillan Publishers

Bruice Y (2013). Organic Chemistry (7th ed.). New York, United States:

Prentice Hall Pearson.

Delgado J. N. Et al (1998). Wilson and Gisvold's Textbook of Organic

Medicinal and Pharmaceutical Chemistry (10th Ed.). California, United

States: Lippincott Williams

Bhassin S.K, Gupta R.(2013). Pharmaceutical organic chemistry (E-book

Kindle edition). New Delhi, India: Elsevier Publishing Services

Session 27: Structure – Activity Relationship of Cephalosporins.

Total Session Time: 120 minutes

Prerequisites

None

Learning Tasks

By the end of this session students are expected to be able to:

• Define Cephalosporins
• Explain Chemical structure of cephalosporins
• Explain the structure – activity relationship of cephalosporins

Resources Needed:

• Flip charts, marker pens, and masking tape.
• Black/white board and chalk/whiteboard markers.

SESSION OVERVIEW

|Step |Time |Activity/ |Content |

| | |Method | |

|1 |05 minutes |Presentation |Introduction, Learning Tasks |

|2 |10 minutes |Brainstorming |Definition of Cephalosporins |

| | |Presentation | |

|3 |40 minutes |Presentation |Chemical Structure of Cephalosporins|

|4 |45 minutes | Group |Structure – Activity Relationship of|

| | |discussion |Cephalosporins |

| | |Presentation | |

|5 |10 minutes |Presentation |Key Points |

| 6 |10 minutes |Presentation |Evaluation |

SESSION CONTENTS.

STEP 1: Presentation of Session Title and Learning Tasks (5 minutes).

READ or ASK students to read the learning tasks and clarify

ASK students if they have any questions before continuing.

STEP 2: Definition of Cephalosporins (10 minutes).

|Activity: Brainstorming (5 minutes) |

| |

|Ask students to brainstorm on the following question: |

| |

|What are Cephalosporins? |

| |

|ALLOW few students to respond |

| |

|WRITE their responses on the flip chart/ board |

| |

|CLARIFY and SUMMARISE by using the table below |

• Cephalosporins are Antibacterial agents which inhibit bacterial cell

wall synthesis.

• Cephalosporins are the second group of β-lactam antibiotic family to be

discovered after penicillin, which showed broad spectrum of action

against Staphylococcus aureus, Vibrio cholerae, and B. anthracis, but

moderate antibacterial activity was isolated from the fermentation broth

of a strain of A. chrysogenum

• Cephalosporins are bactericidal antibiotics, chemically closely related

to penicillins, and have the same mode of action, disrupting the

synthesis of the peptidoglycan layer of bacterial cell walls of bacteria,

causing their death.

STEP 3: Chemical Structure of Cephalosporins (40 minutes).

• Cephalosporin is a β-lactam antibiotic that inhibits bacterial cell

wall synthesis.

• In 1948 Dr. Abraham first isolated cephalosporin C from a fungus

Cephalosporium acremonium.

• Cephalosporins have broader gram –ve coverage than penicillin yet no

one of the cephalosporins is active against MRSA and enterococci.

Basic structure of cephalosporin is 7-aminocephalosporanic acid

[pic]

Structural classification of cephalosparins

[pic]

[pic]

STEP 4: Structure – Activity relationship of Cephalosporins (45 minutes).

|Activity: Small Group Discussion (20 minutes) |

| |

|DIVIDE students into small manageable groups. |

| |

|ASK students to discuss on the following question |

|What is the importance of SAR of Cephalosporins? |

| |

|ALLOW students to discuss for 15 minutes. |

| |

|ALLOW few groups to present and the rest to add points not mentioned. |

| |

|CLARIFY and SUMMARIZE by using the contents below |

Many analogues of Cephalosporin C have been made and the structure-activity

relationship (SAR) conclusions are as follows

• The β-lactam ring is essential
• A free carboxyl group is needed at position 4.
• The bicyclic system is essential
• The stereochemistry of the side-groups and the rings is important

These are very close to penicillin and there are only a limited number of

place where modifications can be made. Those places are:

• Variations of the 7-acylamino side chain
• Variations of the 3-acetoxymethyl side chain
• Extra substitution at carbon 7.

[pic]

Positions which can be varied

Beta-Lactam Ring:

• Required for PBP reactivity and antibacterial activity
• Reactivity reduced compared to the penicillins
• Compare mechanism of action, resistance, pharmacodynamics, etc to

penicillins

2-Carboxyl Group:

• Acidic: Salt formation, product formulation
• Prodrug formation
• Elimination profile: Renal

X-Substituent:

• Cephalosporins and cephamycins
• Determines, in part, resistance to beta-lactamase inactivation

3- Substituent (R3)

• Chemical/acid stability/instability
• Metabolic stability/instability
• Minimal impact on antibacterial activity
• Protein binding and half-life: Heterocycles
• Adverse Reaction and Drug Interaction
• Some role in cephalosporin classification (generation)

7-Substituent (R7)

• Incorporated by semi synthesis: Variable structures
• Impact on spectrum of activity (beta-lactamases, PBP affinity, etc.)
• Significant role in activity and classification by generation

Cephalosporin analogues

[pic]

• Different cephalosporins are developed by changing the moieties

attached at the 3 and/or 7 positions of the 7-ACA.

• Usually, substituents at C-3 (R2) modify the overall pharmacokinetic

properties, whereas those at C-7 (R1) alter the antibacterial

spectrum.

• R1 – the substituents at this position effects β-lactamase

resistance and its activity against Gram –ve and/or Gram +ve bacteria

(its spectrum).

• R2 –these substituents primarily affect the pharmacokinetics: the oral

activity, the extent of metabolism, and the duration of action.

• Electron withdrawing group at this position provide resonance

structure and thus increase stability of the structure.

• Carboxylic acid group –necessary for activity, this functional group

mimics the carboxylic acid group of alanine when binding the enzyme

active site.

[pic]

STEP 5: Key Points (10 minutes)

• The cephalosporins are a class of β-lactam antibiotics originally derived

from the fungus Acremonium, which was previously known as

"Cephalosporium".

• Cephalosporins have broder gram –ve coverage than penicillin yet no one

of the cephalosporins is active against MRSA and enterococci. Basic

structure of cephalosporin is 7-aminocephalosporanic acid.

• Important parameters for the SAR of Cephalosporins which are used to

modify its activity includes beta lactam ring, 2-Carboxyl Group, X-

Substituent,3- Substituent (R3) and 7- substituent (R7).

STEP 6: Evaluation (10 minutes)

• What are Cephalosporins?
• Draw general chemical structure of cephalosporins.
• What is the importance of structure – activity relationship of

cephalosporins?

References.

Ternay, A.L (1976). Contemporary Organic Chemistry. Philadelphia, United

States: W.B. Saunders Co.

Morrison R.T and Boyd R N (1997). Organic Chemistry (6th Ed.). New Delhi,

India: Prentice Hall of India

Graham Solomon et al (2014). Organic Chemistry (11th Ed.). New Jeysey,

United States: John Willey and Sons.

Nadendla R. R. (2005). Principles of Pharmaceutical Organic Chemistry. New

Delhi, India: MacMillan Publishers

Bruice Y (2013). Organic Chemistry (7th ed.). New York, United States:

Prentice Hall Pearson.

Delgado J. N. Et al (1998). Wilson and Gisvold's Textbook of Organic

Medicinal and Pharmaceutical Chemistry (10th Ed.). Calfornia, United

States: Lippincott Williams

Bhassin S.K, Gupta R.(2013). Pharmaceutical organic chemistry (E-book

Kindle edition). New Delhi, India: Elsevier Publishing Service

Session 28: Structure – Activity Relationship of Quinolones.

Total Session Time: 120 minutes

Prerequisites

None

Learning Tasks

By the end of this session students are expected to be able to:

• Define Quinolones
• Explain Chemical structure of Quinolones
• Explain the structure – activity relationship of Quinolones

Resources Needed:

• Flip charts, marker pens, and masking tape.
• Black/white board and chalk/whiteboard markers.

SESSION OVERVIEW

|Step |Time |Activity/ |Content |

| | |Method | |

|1 |05 minutes |Presentation |Introduction, Learning Tasks |

|2 |10 minutes |Brainstorming |Definition of Quinolones. |

| | |Presentation | |

|3 |40 minutes |Presentation |Chemical Structure of |

| | | |Quinolones. |

|4 |45 minutes | Group |Structure – Activity |

| | |discussion |Relationship of Quinolones. |

| | |Presentation | |

|5 |10 minutes |Presentation |Key Points |

| 6 |10 minutes |Presentation |Evaluation |

SESSION CONTENTS

STEP 1: Presentation of Session Title and Learning Tasks (5 minutes)

READ or ASK students to read the learning tasks and clarify

ASK students if they have any questions before continuing.

STEP 2: Definition of Quinolones (10 minutes).

|Activity: Brainstorming (5 minutes) |

| |

|Ask students to brainstorm on the following question: |

| |

|What are Quinolones? |

| |

|ALLOW few students to respond |

| |

|WRITE their responses on the flip chart/ board |

| |

|CLARIFY and SUMMARISE by using the table below |

• The Quinolones (fluoroquinolones) are broad-spectrum antibiotics with

particular activity against gram-negative organisms,

especially Pseudomonas aeruginosa.

• The fluoroquinolones are bactericidal antibiotics that act by

specifically targeting DNA gyrase.

o Fluoroquinolones can be classified as;

o First-generation drugs (e.g., nalidixic acid)
o Second-generation quinolones (e.g., ciprofloxacin)
o Third-generation drugs (e.g., levofloxacin)
o Fourth-generation quinolone drugs (e.g., trovafloxacin)

STEP 3: Chemical Structure of Quinolones (40 minutes).

The following are chemical structure of quinolones; [pic]

STEP 4: Structure – Activity relationship of Quinolones (45 minutes).

|Activity: Small Group Discussion (20 minutes) |

| |

|DIVIDE students into small manageable groups. |

| |

|ASK students to discuss on the following question |

|What is the importance of SAR of fluoroquinolones? |

| |

|ALLOW students to discuss for 15 minutes. |

| |

|ALLOW few groups to present and the rest to add points not mentioned. |

| |

|CLARIFY and SUMMARIZE by using the contents below |

[pic]

Structure of quinolone

• Structure of the quinolone molecule, using the accepted numbering scheme

for positions on the molecule.

• An R indicates possible sites for structural modification.
• Molecules at positions marked by a dashed box can also be changed;

however, the most commonly used structure is shown.

Position 1.

• This position is part of the enzyme-DNA binding complex and has a

hydrophobic interaction with the major grove of DNA.

• A cyclopropyl substituent is now considered the most potent modification

here, followed by addition of a 2,4-difluorophenyl.

• Most other substituents, including one with only the wrong stearic

position ((R)-ofloxacin) can presumably lower the number of molecules

capable of binding to the enzyme-DNA pocket, and therefore reduce

potency.

• Interestingly, ofloxacin has a tricyclic ring structure with a CH3

attached to the asymmetric C-3 position on the oxazine ring, thus

connecting positions 1 and 8 with a fused ring.

• Although this has been a useful alternative to the cyclopropyl

substituent, the S- isomer exhibits twice the order of magnitude of

activity as the R- isomer, which seems to be determined by the number of

molecules that can be assembled, or stacked, in the enzyme-DNA complex

binding pocket.

• Even the potency of the purified S- isomer fused ring is less than that

of the cyclopropyl substituent, suggesting the difficulty of improving

upon this latter modification.

Position 2.

• This location is very close to the site for DNA gyrase (or topoisomerase
IV) binding so it is believed that any added bulk inhibits access and

results in a lower level of microbiological activity.

• Only sulfur, incorporated into a small ring, has been able to replace

hydrogen at the R-2 position.

• To accomplish this, researchers reconfigured positions 3 and 4.

Positions 3 and 4.

• These two positions on the quinolone nucleus are considered critical for

binding to cleaved or perturbed DNA, and no useful substitutions have yet

been reported.

• Therefore, the 3-carboxylate and 4-carbonyl groups are considered

essential for antimicrobial activity.

Oxoquinolizines.

• This is a new addition to the quinolone class, in which nitrogen replaces

the carbon between ring carbons C-4 and C-5.

• Making this alteration renumbers the other positions so that 5- becomes 6-

, 6- becomes 7-, 7- becomes 8-, and 8- becomes position 9.

• This substitution enhances the in vitro and in vivo (mouse protection)

activity against gram-positive cocci, including methicillin-resistant S.

aureus (MRSA) that are resistant to ciprofloxacin.

Position 5.

• Substituents at this position of the basic quinolone nucleus appear to

have the capacity to alter overall stearic configuration (planar

structure) of the molecule, which is how changes here are thought to

affect activity.

• Modestly sized additions, such as an amino, hydroxyl, or methyl group can

markedly increase in vitro activity against gram-positive bacteria, as

well as enhance potency against Toxoplasma gondii.

• Also, it was found that the methyl group enhances action against gram-

positive but not against gram-negative bacteria.

Position 6.

The addition of a fluorine molecule here markedly improved antimicrobial

activity compared to the original quinolone agents and gave rise to the now

widely used and clinically successful fluoroquinolone compounds.

Position 7.

• This position is considered to be one that directly interacts with DNA

gyrase, or topoisomerase IV. The optimal substituents at this position

have been found to be groups that contain, at a minimum, a 5- or 6-

membered nitrogen heterocycle.

• The most common of these are aminopyrrolidines and piperazines.
• Placement of an aminopyrrolidine improves gram-positive activity, whereas

a piperazine generally enhances potency against gram-negative bacteria.

• Alkylation (-CH3) of the 5-membered or 6-membered heterocycle

(pyrrolidines and piperazines, respectively) also enhances activity

against gram-positive bacteria.

Position 8.

• This position is considered to affect overall molecular stearic

configuration, similar to position 5.

• Therefore, changes made here affect target affinity, probably by altering

drug access to the enzyme or DNA binding sites.

• A free halogen (F or Cl) here may improve activity against anaerobes.
• Halogen substituents, as well as a methyl or methoxy also increase the in

vitro activity against gram-positive cocci, even in those bacteria

resistant to older fluoroquinolones.

• Interestingly, the R-8–substituted quinolones also exhibit enhanced

bacteriostatic and lethal activities against GyrA mutants of both E. coli

and Mycobacterium species. Furthermore, in S. aureus a substitution here

created the most lethal agent for both wild type cells as well as those

strains with a preexisting topoisomerase IV mutation.

• Substituents such as ring nitrogen or a free C-8 methyl or methoxy in the

core quinolone molecule substantially reduce the likelihood of emergence

of resistant microbial strains that have no preexisting QRDR mutations.

• Recently, the optimal substituent placed here, combined with a bulky

addition at the C-7 position, and has also been shown to markedly reduce

the development of fluoroquinolone resistance in S. aureus.

STEP 5: Key Points (10 minutes).

• The Quinolones (fluoroquinolones) are broad-spectrum antibiotics with

particular activity against gram-negative organisms,

especially Pseudomonas aeruginosa.

• The fluoroquinolones are bactericidal antibiotics that act by

specifically targeting DNA gyrase.

• A number of substitutions at position 1 to 8 can be done in the

fluoroquinolones that may result in modification of their therapeutic

effect.

STEP 6: Evaluation (10 minutes).

• What are Quinolones?
• Draw a general chemical structure of Quinolones
• What is the importance of structure – activity relationship of

Quinolones?

References

Ternay, A.L (1976). Contemporary Organic Chemistry. Philadelphia, United

States: W.B. Saunders Co.

Morrison R.T and Boyd R N (1997). Organic Chemistry (6th Ed.). New Delhi,

India: Prentice Hall of India

Graham Solomon et al (2014). Organic Chemistry (11th Ed.). New Jeysey,

United States: John Willey and Sons.

Rama Rao Nadendla (2005). Principles of Pharmaceutical Organic Chemistry.

New Delhi, India: MacMillan Publishers

Bruice Y (2013). Organic Chemistry (7th ed.). New York, United States:

Prentice Hall Pearson.

Delgado J. N. Et al (1998). Wilson and Gisvold's Textbook of Organic

Medicinal and Pharmaceutical Chemistry (10th Ed.). Calfornia, United

States: Lippincott Williams

Bhassin S.K, Gupta R. (2013). Pharmaceutical organic chemistry (E-book

Kindle edition). New Delhi, India: Elsevier Publishing Services

Session 29: Structure – Activity Relationship of Sulphonamides.

Total Session Time: 120 minutes

Prerequisites

None

Learning Tasks

By the end of this session students are expected to be able to:

• Define Sulphonamides
• Explain chemical structure of sulphonamides
• Explain the structure – activity relationship of sulphonamides

Resources Needed:

• Flip charts, marker pens, and masking tape
• Black/white board and chalk/whiteboard markers

SESSION OVERVIEW

|Step |Time |Activity/ |Content |

| | |Method | |

|1 |05 minutes |Presentation |Introduction, Learning Tasks |

|2 |10 minutes |Brainstorming |Definition of Sulphonamides. |

| | |Presentation | |

|3 |40 minutes |Presentation |Chemical Structure of Sulphonamides.|

|4 |45 minutes | Group |Structure – Activity Relationship of|

| | |discussion |Sulphonamides. |

| | |Presentation | |

|5 |10 minutes |Presentation |Key Points |

| 6 |10 minutes |Presentation |Evaluation |

SESSION CONTENTS.

STEP 1: Presentation of Session Title and Learning Tasks (5 minutes)

READ or ASK students to read the learning tasks and clarify.

ASK students if they have any questions before continuing.

STEP 2: Definition of Sulphonamides (10 minutes).

|Activity: Brainstorming (5 minutes) |

| |

|Ask students to brainstorm on the following question: |

| |

|What are Sulphonamides? |

| |

|ALLOW few students to respond. |

| |

|WRITE their responses on the flip chart/ board. |

| |

|CLARIFY and SUMMARISE by using the table below |

Sulphonamides are antibacterial agents which acts against cell metabolism

(antimetabolites).

Sulfonamides:

• Once known as ‘wonder drug’.
• Once mainstays of antimicrobial chemotherapy.
• The relative cheapness of the sulphonamides is one of their most

attractive features and accounts for much of their persistence in the

market.

STEP 3: Chemical Structure of Sulphonamides (40 minutes).

• Sulphonamides are composed of a sulphur atom that has two sets of

double bonds to two oxygen atoms, a carbon-based side group, and a

nitrogen atom bonded to the sulphur itself.

• In organic chemistry, an amide contains a carbonyl group bonded to a

nitrogen atom.

• Sulphonamides are similar, but the carbonyl group is replaced with

sulfone sulphur with two oxygen atoms).

• That's why the term 'amide' appears in the name.

[pic]

General structure of amides and

sulphonamides.

• The 'R' groups in the figure simply represent any generic carbon-based

side chain and could be virtually anything.

• For example, R could be a methyl group, a benzene ring, an alkane

ring, or some other group.

• If the nitrogen atom contains two hydrogens, the sulphonamide is

classified primary, if there is one hydrogen it's secondary, and if no

hydrogens are present on the nitrogen, it's a tertiary sulphonamide

[pic]

Structures of primary, secondary, and tertiary

sulphonamides.

Important Sulfonamide Derivatives

• Sulfamethoxazole
• Sulfamethoxazole is another sulfonamide with antibacterial activity

and is commonly used in the treatment of urinary tract infections and

bronchitis.

• Sulfamethoxazole looks very similar to sulfanilamide in terms of its

structure but contains an extra ring system called an oxazole.

[pic]

STEP 4: Structure – Activity Relationship of Sulphonamides (45 minutes).

|Activity: Small Group Discussion (20 minutes). |

| |

|DIVIDE students into small manageable groups. |

| |

|ASK students to discuss on the following question. |

|What is the importance of SAR of sulphonamides? |

| |

|ALLOW students to discuss for 15 minutes. |

| |

|ALLOW few groups to present and the rest to add points not mentioned. |

| |

|CLARIFY and SUMMARIZE by using the contents below |

The synthesis of a large number of sulphonamide analogues led to the

following conclusions

[pic]

Sulfonamides analogues

• The para -amino group is essential for activity and must be unsubstituted
(i.e. R1=H). The only exception is when R1=acyl (i.e. amides).
• The amides themselves are inactive but can be metabolized in the body to

regenerate the active compound.

• Thus, amides can be used as sulfonamide prodrugs.
• Incorporation of other groups (halogen, alkyl, etc.) destroys the

activity.

[pic]

Metabolism of acyl group to regenerate active

compound

• The aromatic ring and the sulphonamide functional group are both

required.

• Total loss of antibacterial activity occurs if sulphonamide group is

replaced by other acidic groups (sulfonic, phosphoric etc.)

• The aromatic ring must be para -substituted only.
• Extra substitution eliminates activity for steric reasons.
• The sulfonamide nitrogen must be primary (sulfanilamide) or secondary

(acidic proton is essential for antibacterial activity).

• R2 is the only possible site that can be varied in sulfonamides.

Sulphonamide analogues

• R2 can be varied by incorporating a large range of heterocyclic or

aromatic structures, which affects the extent to which the drug binds to

plasma protein.

• This in turn controls the blood levels of the drug such that it can be

short acting or long acting.

• Thus, a drug which binds strongly to plasma protein will be slowly

released into the blood circulation and will be longer lasting.

• R2 affects pharmacokinetic properties but not the pharmacodynamics

properties.

Sulfonamide analogues with reduced toxicity

• Changing the nature of the group R2 has also helped to reduce the

toxicity of some sulfonamides.

• The primary amino groups of sulfonamides are acetylated in the body and

the resulting amides have reduced solubility which can lead to toxic

effects.

• For example, the metabolite formed from sulfathiazole is poorly soluble

and can prove fatal if it blocks the kidney tubules

[pic]

Insoluble

• It was discovered that the solubility problem could be overcome by

replacing the thiazole ring in sulfathiazole with a pyrimidine ring to

give sulfadiazine.

[pic]

• Its metabolites will also be water soluble
• The reason for the improved solubility lies in the acidity of the

sulphonamide NH proton.

• In sulfathiazole, this proton is not very acidic.
• Therefore, sulfathiazole and its metabolite are mostly un-ionized at

blood pH.

• Replacing the thiazole ring with amore electron withdrawing pyrimidine

ring increases the acidity of the NH proton by stabilizing the resulting

anion.

• Therefore, sulfadiazine and its metabolite are significantly ionized at

blood pH.

• As a consequence, they are more soluble and less toxic.

Treatment of intestinal infections

• Sulphonamides are useful against intestinal infections
• Prodrugs are used for this purpose
• For example, succinyl sulfathiazole is a prodrug of sulfathiazole

[pic]

• The succinyl moiety contains an acidic group which means that the prodrug

is ionized in the slightly alkaline conditions of the intestine.

• As a result, it is not absorbed into the blood stream and is retained in

the intestine.

• Slow enzymatic hydrolysis of the succinyl group then releases the active

sulfathiazole where it is needed.

[pic]

• Benzoyl substitution on the aniline nitrogen has also given useful

prodrugs, which are poorly absorbed through the gut wall since they

are too hydrophobic

STEP 5: Key Points (10 minutes).

• Sulphonamides are composed of a sulphur atom that has two sets of double

bonds to two oxygen atoms, a carbon-based side group, and a nitrogen atom

bonded to the sulphur itself.

• Modification at R2 group may alter the duration of action and toxicity of

sulfonamides.

STEP 6: Evaluation (10 minutes).

• What are Sulphonamides?
• Draw general chemical structure of sulphonamides
• What is the importance of the structure – activity relationship of

sulphonamides?

References

Ternay, A.L (1976). Contemporary Organic Chemistry. Philadelphia, United

States: W.B. Saunders Co.

Morrison R.T and Boyd R N (1997). Organic Chemistry (6th Ed.). New Delhi,

India: Prentice Hall of India

Graham Solomon et al (2014). Organic Chemistry (11th Ed.). New Jeysey,

United States: John Willey and Sons.

Rama Rao Nadendla (2005). Principles of Pharmaceutical Organic Chemistry.

New Delhi, India: MacMillan Publishers

Bruice Y (2013). Organic Chemistry (7th ed.). New York, United States:

Prentice Hall Pearson.

Delgado J. N. Et al (1998). Wilson and Gisvold's Textbook of Organic

Medicinal and Pharmaceutical Chemistry (10th Ed.). California, United

States: Lippincott Williams

Bhassin S.K, Gupta R.(2013). Pharmaceutical organic chemistry (E-book

Kindle edition). New Delhi, India: Elsevier Publishing Service

Session 30: Structure – Activity Relationship of Aspirin.

Total Session Time: 120 minutes

Prerequisites

None

Learning Tasks

By the end of this session students are expected to be able to:

• Define Aspirin
• Explain Chemical structure of aspirin
• Explain the structure – activity relationship of aspirin

Resources Needed:

• Flip charts, marker pens, and masking tape.
• Black/white board and chalk/whiteboard markers.

SESSION OVERVIEW

|Step |Time |Activity/ |Content |

| | |Method | |

|1 |05 minutes |Presentation |Introduction, Learning Tasks |

|2 |10 minutes |Brainstorming |Definition of Aspirin |

| | |Presentation | |

|3 |40 minutes |Presentation |Chemical Structure of Aspirin |

|4 |45 minutes |Group |Structure – Activity Relationship of|

| | |discussion |Aspirin |

| | |Presentation | |

|5 |10 minutes |Presentation |Key Points |

| 6 |10 minutes |Presentation |Evaluation |

SESSION CONTENTS

STEP 1: Presentation of Session Title and Learning Tasks (5 minutes)

READ or ASK students to read the learning tasks and clarify.

ASK students if they have any questions before continuing.

STEP 2: Definition of Aspirin (10 minutes).

|Activity: Brainstorming (5 minutes) |

| |

|Ask students to brainstorm on the following question: |

| |

|What is Aspirin? |

| |

|ALLOW few students to respond |

| |

|WRITE their responses on the flip chart/ board |

| |

|CLARIFY and SUMMARISE by using the table below |

• Aspirin is a nonsteroidal anti-inflammatory drug (NSAID) effective in

treating fever, pain, and inflammation in the body.

• It also prevents blood clots (i.e., is antithrombotic). As a group,

NSAIDs are non-narcotic relievers of mild to moderate pain of many

causes, including

o Headaches,

o Injury,

o Menstrual cramps,

o Arthritis, and other musculoskeletal conditions

• Other members of this class include

o ibuprofen (Motrin),

o indomethacin (Indocin),

o nabumetone (Relafen) and several others.

• They all work by reducing the levels of prostaglandins, chemicals that

are released when there is inflammation and that cause pain and fever.

• NSAIDs block the enzyme that makes prostaglandins (cyclooxygenase),

resulting in lower concentrations of prostaglandins.

• As a consequence, inflammation, pain, and fever are reduced.
• Inhibition of prostaglandins also reduces the function of platelets

and the ability of blood to clot.

STEP 3: Chemical Structure of Aspirin (40 minutes).

• Acetylsalicylic acid –or, as it is more commonly known, aspirin –has a

simple chemical structure.

• It consists of a small number of carbons, hydrogen and oxygen atoms that

form the chemical bonding patterns shown below.

• Aspirin Acetylsalicylic acid C9H8O4

[pic]

• One of the best-known aromatic acetates is acetylsalicylic acid, or

aspirin, which is prepared by the esterification of the phenolic

hydroxyl group of salicylic acid.

[pic]

• Aspirin possesses a number of properties that make it the most often

recommended drug which are;

o analgesia, leading to pain relief

o anti-inflammatory effect, providing some relief from the swelling

associated with arthritis and minor injuries.

o antipyretic effect, which means it reduces fever.

Synthesis

• The key compound in the synthesis of aspirin, salicylic acid, is

prepared from phenol by Kolbe synthesis (also known as the Kolbe-

Schmitt reaction) whereby sodium phenoxide is heated with CO2 under

pressure and the reaction mixture is subsequently acidified to yield

salicylic acid.

[pic]

STEP 4: Structure – Activity relationship of Aspirin (45 minutes).

|Activity: Small Group Discussion (20 minutes) |

| |

|DIVIDE students into small manageable groups. |

| |

|ASK students to discuss on the following question |

|What is the importance of SAR of aspirin? |

| |

|ALLOW students to discuss for 15 minutes. |

| |

|ALLOW few groups to present and the rest to add points not mentioned. |

| |

|CLARIFY and SUMMARIZE by using the contents below |

• Despite the vast effort that has been expended in the search to find a

“better” aspirin—that is, one possessing fewer GI side effects, but a

greater potency and a longer duration of action yet is inexpensive and an

antipyretic, analgetic, and anti-inflammatory agent that is overall

superior to aspirin—none has yet to be discovered.

The following structure–activity relationships have been established:

[pic]

• Possesses a free carboxylic acid (COOH) for an ionic interaction with

the positively charged arginine residue at the active site of the

cyclooxygenases (i.e., Arg-120 in COX-1 or Arg-106 in COX-2 isozymes).

• This acidic moiety is further linked to an aromatic (or

heteroaromatic) ring for binding to either the Δ5-double-bond or Δ8-

double-bond binding regions.

• The active moiety appears to the salicylate anion.
• The side effects of aspirin, particularly the GI effects, appear to be

associated with the carboxylic acid function.

• Reducing the acidity of this group (e.g., converting to an amide,

salicylamide) maintains the analgesic actions of salicylic acid

derivatives but eliminates the anti-inflammatory properties.

• Substitution on either the carboxyl or phenolic hydroxyl groups may

affect potency and toxicity.

• Benzoic acid itself has only weak anti-inflammatory activity.
• Placing the phenolic hydroxyl group meta or para to the carboxyl group

abolishes this activity.

• Substitution of halogen atoms on the aromatic ring enhances potency

and toxicity.

• Substitution of aromatic rings at the 5-position of salicylic acid
increases anti-inflammatory activity (e.g., diflunisal).

STEP 5: Key Points (10 minutes).

• Aspirin is a nonsteroidal anti-inflammatory drug (NSAID) effective in

treating fever, pain, and inflammation in the body.

• Aspirin possesses a free carboxylic acid (COOH) for an ionic

interaction with the positively charged arginine residue at the active

site of the cyclooxygenases.

• Modification of carboxylic acid function group of aspirin may help in

reducing its GI toxicity.

STEP 6: Evaluation (10 minutes).

• What is Aspirin?
• Draw chemical structure of aspirin.
• What is the importance of structure – activity relationship of

aspirin?

References

Ternay, A.L (1976). Contemporary Organic Chemistry. Philadelphia, United

States: W.B. Saunders Co.

Morrison R.T and Boyd R N (1997). Organic Chemistry (6th Ed.). New Delhi,

India: Prentice Hall of India

Graham Solomon et al (2014). Organic Chemistry (11th Ed.). New Jeysey,

United States: John Willey and Sons.

Rama Rao Nadendla (2005). Principles of Pharmaceutical Organic Chemistry.

New Delhi, India: MacMillan Publishers

Bruice Y (2013). Organic Chemistry (7th ed.). New York, United States:

Prentice Hall Pearson.

Delgado J. N. Et al (1998). Wilson and Gisvold's Textbook of Organic

Medicinal and Pharmaceutical Chemistry (10th Ed.). Calfornia, United

States: Lippincott Williams

Bhassin S.K, Gupta R.(2013). Pharmaceutical organic chemistry (E-book

Kindle edition). New Delhi, India: Elsevier Publishing Services

Session 31: Structure – Activity Relationship of Paracetamol.

Total Session Time: 120 minutes

Prerequisites

None

Learning Tasks

By the end of this session students are expected to be able to:

• Define Paracetamol
• Explain Chemical structure of paracetamol
• Explain the structure – activity relationship of paracetamol

Resources Needed:

• Flip charts, marker pens, and masking tape.
• Black/white board and chalk/whiteboard markers.

SESSION OVERVIEW

|Step |Time |Activity/ |Content |

| | |Method | |

|1 |05 minutes |Presentation |Introduction, Learning Tasks |

|2 |10 minutes |Brainstorming |Definition of Paracetamol |

| | |Presentation | |

|3 |40 minutes |Presentation |Chemical Structure of Paracetamol |

|4 |45 minutes |Group |Structure – Activity Relationship of|

| | |discussion |Paracetamol |

| | |Presentation | |

|5 |10 minutes |Presentation |Key Points |

| 6 |10 minutes |Presentation |Evaluation |

SESSION CONTENTS

STEP 1: Presentation of Session Title and Learning Tasks (5 minutes).

READ or ASK students to read the learning tasks and clarify

ASK students if they have any questions before continuing.

STEP 2: Definition of Paracetamol (10 minutes).

|Activity: Brainstorming (5 minutes) |

| |

|Ask students to brainstorm on the following question: |

| |

|What is paracetamol? |

| |

|ALLOW few students to respond. |

| |

|WRITE their responses on the flip chart/ board. |

| |

|CLARIFY and SUMMARISE by using the table below |

• Paracetamol also known as acetaminophen is an analgesic antipyretic

derivative of acetanilide.

• Acetaminophen has weak anti-inflammatory properties and is used as a

common analgesic, but may cause liver, blood cell, and kidney damage.

• Acetaminophen is a p-aminophenol derivative with analgesic and

antipyretic activities.

• Although the exact mechanism through which acetaminophen exert its

effects has yet to be fully determined, acetaminophen may inhibit the

nitric oxide (NO) pathway mediated by a variety of neurotransmitter

receptors including N-methyl-D-aspartate (NMDA) and substance P,

resulting in elevation of the pain threshold.

• The antipyretic activity may result from inhibition of prostaglandin

synthesis and release in the central nervous system (CNS) and

prostaglandin-mediated effects on the heat-regulating center in the

anterior hypothalamus.

• Acetaminophen is a widely used nonprescription analgesic and antipyretic

medication for mild-to-moderate pain and fever.

• Harmless at low doses, acetaminophen has direct hepatotoxic potential

when taken as an overdose and can cause acute liver injury and death from

acute liver failure.

• Even in therapeutic doses, acetaminophen can cause transient serum

aminotransferase elevations.

Summary

• Paracetamol (acetaminophen) is a pain reliever and a fever reducer.
• Paracetamol is used to treat many conditions such as headache, muscle

aches, arthritis, backache, toothaches, colds, and fevers. It relieves

pain in mild arthritis but has no effect on the underlying

inflammation and swelling of the joint.

STEP 3: Chemical Structure of Paracetamol (40 minutes).

• The acetaminophen has the IUPAC name N-(4-hydroxyphenyl) acetamide and

its chemical formula us C8H9NO2 and its extended formula is

HOC6H4NHCOCH3.

• Its molar mass is 151.165 g mol-1.
• The molecule is formed by an aromatic phenyl ring, which has two

substituents in position -para (1,4).

• The first substituent is an amide group (acetamide) and the second is

a hydroxy group (-OH).

• The molecule is planar with 7 carbon atoms with sp2 hybridization.
• Its chemical structure can be written as below, in the common

representations used for organic molecules.

[pic]

STEP 4: Structure – Activity relationship of Paracetamol (45 minutes)

|Activity: Small Group Discussion (20 minutes) |

| |

|DIVIDE students into small manageable groups |

| |

|ASK students to discuss on the following question. |

|What is the importance of SAR of paracetamol? |

| |

| |

|ALLOW students to discuss for 15 minutes. |

| |

|ALLOW few groups to present for 5 minutes and the rest to add points |

|not mentioned. |

| |

|CLARIFY and SUMMARIZE by using the contents below |

• Paracetamol consists of a benzene ring core, substituted by one

hydroxyl group and the nitrogen atom of an amide group in the para

(1,4) pattern.

• The amide group is acetamide (ethanamide).
• It is an extensively conjugated system, as the lone pair on the

hydroxyl oxygen, the benzene pi cloud, the nitrogen lone pair, the p

orbital on the carbonyl carbon, and the lone pair on the carbonyl

oxygen is all conjugated.

• The presence of two activating groups also makes the benzene ring

highly reactive toward electrophilic aromatic substitution.

• As the substituents are ortho,para-directing and para with respect to

each other, all positions on the ring are more or less equally

activated.

• The conjugation also greatly reduces the basicity of the oxygens and

the nitrogen, while making the hydroxyl acidic through delocalisation

of charge developed on the phenoxide anion.

• Structure and reactivity of acetaminophen accounts for its chemical

properties

• Based on the comparative toxicity of acetanilide and acetaminophen,

aminophenols are less toxic than the corresponding aniline derivatives,

although p-aminophenol itself is too toxic for therapeutic purposes.

• Etherification of the phenolic function with methyl or propyl groups

produces derivatives with greater side effects than with ethyl groups.

Substituent

• The nitrogen atoms that reduce basicity reduce activity unless that

substituent is metabolically labile (e.g., acetyl).

• Amides derived from aromatic acids (e.g., N-phenylbenzamide) are less

active or inactive.

• As indicated, both acetanilide and phenacetin are metabolized to

acetaminophen. Additionally, both undergo hydrolysis to yield Aniline

derivatives that produce directly, or through their conversion to

hydroxylamine derivatives, significant methemoglobinemia and hemolytic

anemia, which resulted in their removal from the U.S. market.

STEP 5: Key Points (10 minutes)

• Paracetamol also known as acetaminophen is an analgesic antipyretic

derivative of acetanilide

• The SAR of paracetamol may help in reducing its hepatotoxicity.

STEP 6: Evaluation (10 minutes)

• What is Paracetamol?
• Draw the Chemical structure of paracetamol.
• What is the importance of structure – activity relationship of

paracetamol?

References

Ternay, A.L (1976). Contemporary Organic Chemistry. Philadelphia, United

States: W.B. Saunders Co.

Morrison R.T and Boyd R N (1997). Organic Chemistry (6th Ed.). New Delhi,

India: Prentice Hall of India

Graham Solomon et al (2014). Organic Chemistry (11th Ed.). New Jeysey,

United States: John Willey and Sons.

Nadendla R. R. (2005). Principles of Pharmaceutical Organic Chemistry. New

Delhi, India: MacMillan Publishers

Bruice Y (2013). Organic Chemistry (7th ed.). New York, United States:

Prentice Hall Pearson.

Delgado J. N. Et al (1998). Wilson and Gisvold's Textbook of Organic

Medicinal and Pharmaceutical Chemistry (10th Ed.). California, United

States: Lippincott Williams

Bhassin S.K, Gupta R.(2013). Pharmaceutical organic chemistry (E-book

Kindle edition). New Delhi, India: Elsevier Publishing Services

Session 32: Biotransformation of Medicinal Products.

Total Session Time: 120 minutes

Prerequisites

None

Learning Tasks

By the end of this session students are expected to be able to:

• Define biotransformation
• Explain metabolism of different organic compounds
• Explain the importance of biotransformation

Resources Needed:

• Flip charts, marker pens, and masking tape.
• Black/white board and chalk/whiteboard markers.

SESSION OVERVIEW

|Step |Time |Activity/ |Content |

| | |Method | |

|1 |05 minutes |Presentation |Introduction, Learning Tasks |

|2 |10 minutes |Brainstorming |Definition of Biotransformation |

| | |Presentation | |

|3 |60 minutes |Presentation |Metabolism of Organic Compounds |

|4 |25 minutes |Group |Importance of Biotransformation |

| | |discussion | |

| | |Presentation | |

|5 |10 minutes |Presentation |Key Points |

| 6 |10 minutes |Presentation |Evaluation |

SESSION CONTENTS

STEP 1: Presentation of Session Title and Learning Tasks (5 minutes)

READ or ASK students to read the learning tasks and clarify

ASK students if they have any questions before continuing.

STEP 2: Definition of Biotransformation (10 minutes).

|Activity: Brainstorming (5 minutes) |

| |

|Ask students to brainstorm on the following question: |

| |

|What is Biotransformation? |

| |

|ALLOW few students to respond. |

| |

|WRITE their responses on the flip chart/ board. |

| |

|CLARIFY and SUMMARISE by using the table below |

• Biotransformation is Chemical alteration of the drug in body that

converts non-polar or lipid soluble compounds to polar or lipid

insoluble compounds.

OR

• Biochemical alteration of chemicals such as (but not limited to)

nutrients, amino acids, toxins, and drugs in the body.

• It is also needed to render non-polar compounds polar so that they are

not reabsorbed in renal tubules and are excreted.

• The body typically deals with a foreign compound (DRUGS) by making it

more water-soluble, to increase the rate of its excretion through the

urine.

STEP 3: Metabolism of Organic Compounds (60 minutes).

• Termination of drug effect is by the process of drug elimination which

involves mainly 2 processes: one of them is drug metabolism.

• Metabolism is predominantly done in the liver due to its richness in

enzymes

• The enzymes that are responsible for drug metabolism in the liver are;

o Microsomal mixed function oxidases (MFOs) system involved.

o Cytochrome P450 enzymes play important role.

• Other organs responsible for drug metabolism are;

o Lungs

o Kidney

o Intestine

o Placenta

o Skin

o Brain

o Testes

o Muscle

o Spleen

• Metabolism of drugs makes them:

o More polar

o Ionizable

o Water soluble to enhance renal excretion

o More active (for pro drugs)

Drug Metabolism in the Liver

• There are two phases of drug metabolism in the liver.
• Phase I reaction
• Phase II reaction

PHASE I REACTIONS

• A polar group is introduced/ unmasked to make the drug molecule more

water-soluble & less active to be excreted.

• Reactions are non-synthetic in nature.
• The majority of metabolites are generated by a common hydroxylating

enzyme system known as Cytochrome P450.

o Oxidation

o Reduction

o Hydrolytic cleavage

o Dealkylation

o Ring cyclization

o N-carboxylation

o Dimerization

o Transamidation

o Isomerization

o Decarboxylation

• Oxidation of aromatic carbon atoms (aromatic hydroxylation):

[pic]

• Oxidation of olefins (C=C bonds):
o Oxidation of non-aromatic C=C bonds is analogous to aromatic

hydroxylation. i.e. it proceeds via formation of epoxides to yield 1,2-

dihydrodiols

[pic]

• Oxidation of Benzylic Carbon Atoms

o Carbon atoms attached directly to the aromatic ring are hydroxylated.

[pic]

• Oxidation of Allylic carbon Atoms

o Carbon atoms adjacent to Olefinic double bonds (are allylic carbon

atoms) also undergo hydroxylation in a manner similar to Benzylic

Carbons.

[pic]

• Oxidation of Carbon Atoms Alpha to Carbonyls and Imines

o Several Benzodiazepines contain a carbon atom (C-3) alpha to both

Carbonyl (C=0) and imino (C=N) function which readily undergoes

Hydroxylation.

[pic]

• Oxidation of Aliphatic Carbon Atoms (Aliphatic Hydroxylation)

o Terminal hydroxylation of methyl group yields primary alcohols which

undergoes further oxidation to aldehydes and then to carboxylic acid.

[pic]

• Oxidation of Alicyclic Carbon Atoms (Alicyclic Hydroxylation)

o Cyclohexane (alicyclic) and piperidine (non-aromatic heterocyclic)

rings are commonly found in a number of molecules.

o Such rings are generally hydroxylated at C-3 or C-4 positions.

[pic]

• Oxidation Of Carbon-Heteroatom Systems

o Biotransformation of C-N, C-0 & C-S system proceed in one of the two

ways:

▪ Hydroxylation of carbon atom attached to the heteroatom and
subsequent cleavage at carbon-heteroatom bond. E.g. N-, O- & S-

dealkylation, oxidative deamination & desulfuration.

▪ Oxidation of the heteroatom itself. E.g. N- & S- oxidation.
• Oxidation of Carbon-Nitrogen System

o N-DEALKYLATION:

▪ Mechanism of N-dealkylation involve oxidation of α-carbon to

generate an intermediate carbinolamine which rearranges by cleavage of

C-N bond to yield the N dealkylated product and the corresponding

carbonyl of the alkyl group.

[pic]

▪ A tertiary nitrogen attached to different alkyl groups

undergoes dealkylation by removal of smaller alkyl group first.

Example:

o 2º aliphatic amine e.g. Methamphetamine.
o 3º aliphatic amine e.g. imipramine
o 3º alicyclic amine e.g. hexobarbital

o Amides e.g. Diazepam

o N-HYDROXYLATION:

▪ Converse to basic compounds that form N-oxide, N- hydroxy formation

is usually displayed by non-basic nitrogen atoms such as amide

Nitrogen.

[pic]

• Oxidation of Carbon-Sulfur Systems

o S-DEALKYLATION:

▪ The mechanism of S-Dealkylation of thioethers is analogous to N-

dealkylation .IT proceed via α-carbon hydroxylation.

▪ The C-S bond cleavage results in formation of a thiol and a carbonyl

product.

[pic]

• Desulfuration:
o This reaction also involves cleavage of carbon-sulfur bond (C=S).
o The product is the one with C=0 bond.

o Such a desulfuration reaction is commonly observed in thioamides such

as thiopental

[pic]

• S-Oxidation:

o Apart from S-dealkylation, thioethers can also undergo S-oxidation

reaction to yield sulfoxides which may be further oxidized to sulfones

several phenothiazines.

o E.g. Chlorpromazine undergo S-oxidation
• Oxidation of Carbon-Oxygen Systems:

o O-Dealkylation:

▪ This reaction is also similar to N-Dealkylation and proceeds by α-

carbon hydroxylation to form an unstable hemiacetal or hemiketal

intermediate.

▪ Which spontaneously undergoes C-0 bond cleavage to form alcohol and

a carbonyl moiety.

[pic]

• Oxidation of Alcohol, Carbonyl & Carboxylic Acid

o In case of ethanol, Oxidation to acetaldehyde is reversible and

further oxidation of the latter to acetic acid is very rapid since

Acetaldehyde is highly toxic and should not accumulate in body.

[pic]

• Miscellaneous Oxidative Reactions:

o Oxidative Aromatization /Dehydrogenation

o E.g. Metabolic aromatization of drugs is

[pic]

• Oxidative Dehalogenation

o This reaction is common with halogen containing drugs such as

chloroform.

o Dehalogenation of this drug yields phosgene which may results in

electrophiles capable of covalent binding to tissue.

[pic]

• Reductive Reaction Reductive Reaction [pic]
• Reductive Reaction

o Bioreductions are also capable of generating polar functional group

such as hydroxy and amino which can undergo further biotransformation

or conjugation.

o Reduction of carbonyls:

▪ Aliphatic aldehydes:

[pic]

▪ Aliphatic ketones:

[pic]

▪ Aromatic Ketone:

[pic]

• Reduction of Alcohols and C=C:

o These two reductions are considered together because the groups are

interconvertible by simple addition or loss of a water molecule.

Before an alcohol is reduced it is dehydrated to C=C bond.

[pic]

• Reduction of N-Compounds:

o Reduction of nitro groups proceeds via formation of nitro so and

hydroxyl amine intermediates to yield amines.

[pic]

o Reduction of azo compounds yield primary amines via formation of

hydrazo intermediate which undergo cleavage at N-N bond.

[pic]

o It is reduced to active Sulfanilamide.

• Miscellaneous Reductive Reactions

o REDUCTIVE DEHALOGENATION:

▪ This reaction involves replacement of halogen attached to the

carbon with the H-atom

[pic]

o REDUCTION OF SULFUR CONTAINING FUNCTIONAL GROUPS:

[pic]

• Hydrolytic Reactions:

o The reaction doesn’t involve change in the state of oxidation of

substrate.

o The reaction results in a large chemical chain in the substrate

brought about by loss of relatively large fragments of the molecule.

o HYDROLYSIS OF ESTERS AND ETHERS:

o Esters on hydrolyisis yield alcohol & carboxylic acid. The reaction is

catalyzed by esterases.

[pic]

• Hydrolysis of Amides:

o The reactions catalyzed by amides, involves C-N cleavage to yield

carboxylic acid and amine.

[pic]

PHASE II REACTIONS

• Involve covalent attachment of small polar endogenous molecule e.g

glucuronic acid, sulfate, or glycine to form water-soluble compounds

• They are known as conjugation reactions
• Conjugation reactions

They involve the following metabolic processes;

o Glucuronidation by UDP-Glucuronosyltransferase:

(on -OH, -COOH, -NH2, -SH groups)

o Sulfation by Sulfotransferase:

(on -NH2, -SO2NH2, -OH groups)

o Acetylation by acetyltransferase:

(on -NH2, -SO2NH2, -OH groups)

o Methylation

o Amino acid conjugation (on -COOH groups)

o Glutathione conjugation by Glutathione-S-transferase:

(to epoxides or organic halides)

o Fatty acid conjugation (on -OH groups)

o Glycine Conjugation

o Cyanide Conjugation

o Condensation reactions

• Glucuronide Conjugation

o Very important Synthetic reactions carried out by Uridine Di Phosphate

Glucuronosyl Transferase.

o Hydroxyl & Carboxylic acid groups are easily combined with Glucuronic

acid

Glucuronide formation occurs in 2 steps: –

1. Synthesis of an activated coenzyme uridine-5’- diphospho – ( – D-

Glucuronic acid (UDPGA) from UDP- glucose (UDPG).

[pic]

2. Transfer of the glucuronyl moiety from UDPGA to the substrate RXH in

presence of enzyme UDP- glucuronyl transferase to form the conjugate.

[pic]

FACTORS AFFECTING OF BIOTRANSFORMATION OF DRUGS:

• Physicochemical property of drug.

o Induction of drug metabolizing enzyme.

o Inhibition of drug metabolizing enzyme

o Environmental chemicals.

• Biological factors.

o Species differences.

o Strain differences.

o Sex differences.

o Pharmacogenetics – genetic differences in metabolic pathways

affecting individual responses to drugs effects.

• Age.
• Diet.
• Altered pharmacologic factors:

o Pregnancy.

o Hormonal imbalance.

o Disease state.

STEP 4: Importance of Biotransformation (25 minutes).

|Activity: Small Group Discussion (10 minutes). |

| |

|DIVIDE students into small manageable groups. |

| |

|ASK students to discuss on the following question. |

| |

|What is the importance of biotransformation? |

| |

|ALLOW students to discuss for 8 minutes. |

| |

|ALLOW few groups to present and the rest to add points not mentioned. |

| |

|CLARIFY and SUMMARIZE by using the contents below |

• Biotransformation is vital to survival because it transforms absorbed

nutrients (food, oxygen, etc.) into substances required for normal

body functions.

• Drugs are made to be more water soluble ready for renal excretion
• For some pharmaceuticals, it is a metabolite that is therapeutic and

not the absorbed drug.

o For Example, phenoxybenzamine, a drug given to relieve hypertension

caused by pheochromocytoma, a kind of tumor, is biotransformed into a

metabolite, which is the active agent.

• Biotransformation also serves as an important defence mechanism since

toxic xenobiotics and body wastes are converted into less harmful

substances and substances that can be excreted from the body.

STEP 5: Key Points (10 minutes).

• Biotransformation is Chemical alteration of the drug in body that

converts non-polar or lipid soluble compounds to polar or lipid insoluble

compounds.

• Metabolism of organic compounds involve Phase I and Phase II reactions.
• Biotransformation helps organic compounds to be water soluble in hence be

readily excreted renally.

STEP 6: Evaluation (10 minutes).

• What is biotransformation?
• List four reactions involved in phase I and phase II drug metabolism
• What is the importance of biotransformation?

References

Ternay, A.L (1976). Contemporary Organic Chemistry. Philadelphia, United

States: W.B. Saunders Co.

Morrison R.T and Boyd R N (1997). Organic Chemistry (6th Ed.). New Delhi,

India: Prentice Hall of India

Graham Solomon et al (2014). Organic Chemistry (11th Ed.). New Jeysey,

United States: John Willey and Sons.

Nadendla R. R. (2005). Principles of Pharmaceutical Organic Chemistry. New

Delhi, India: MacMillan Publishers

Bruice Y (2013). Organic Chemistry (7th ed.). New York, United States:

Prentice Hall Pearson.

Delgado J. N. Et al (1998). Wilson and Gisvold's Textbook of Organic

Medicinal and Pharmaceutical Chemistry (10th Ed.). California, United

States: Lippincott Williams

Bhassin S.K, Gupta R.(2013). Pharmaceutical organic chemistry (E-book

Kindle edition). New Delhi, India: Elsevier Publishing Services

———————–

| |

PST 05106 Pharmaceutical Organic Chemistry

NTA Level 5 Semester 1

December 2018

Z

N

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