Author name: contact@devinevisiontech.com

NTA Level 5 • Semester 1 • PST05106 Alcohols of Pharmaceutical Importance Pharmaceutical Organic Chemistry • Source Session/Topic 11 Full source-text version: all educational wording from the extracted learning source is retained; only presenter/tutor metadata and web-layout noise are removed, while formatting is improved for readability. 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

NTA Level 5 • Semester 1 • PST05106 Alkynes of Pharmaceutical Importance Pharmaceutical Organic Chemistry • Source Session/Topic 10 Full source-text version: all educational wording from the extracted learning source is retained; only presenter/tutor metadata and web-layout noise are removed, while formatting is improved for readability. 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

NTA Level 5 • Semester 1 • PST05106 Alkenes of Pharmaceutical Importance Pharmaceutical Organic Chemistry • Source Session/Topic 9 Full source-text version: all educational wording from the extracted learning source is retained; only presenter/tutor metadata and web-layout noise are removed, while formatting is improved for readability. 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

NTA Level 5 • Semester 1 • PST05106 Alkanes of Pharmaceutical Importance Pharmaceutical Organic Chemistry • Source Session/Topic 8 Full source-text version: all educational wording from the extracted learning source is retained; only presenter/tutor metadata and web-layout noise are removed, while formatting is improved for readability. 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

NTA Level 5 • Semester 1 • PST05106 Isomerism Pharmaceutical Organic Chemistry • Source Session/Topic 7 Full source-text version: all educational wording from the extracted learning source is retained; only presenter/tutor metadata and web-layout noise are removed, while formatting is improved for readability. 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

NTA Level 5 • Semester 1 • PST05106 Chemical Reaction in Organic Compounds Pharmaceutical Organic Chemistry • Source Session/Topic 6 Full source-text version: all educational wording from the extracted learning source is retained; only presenter/tutor metadata and web-layout noise are removed, while formatting is improved for readability. 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

NTA Level 5 • Semester 1 • PST05106 General Properties of Organic Compounds Pharmaceutical Organic Chemistry • Source Session/Topic 5 Full source-text version: all educational wording from the extracted learning source is retained; only presenter/tutor metadata and web-layout noise are removed, while formatting is improved for readability. 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

NTA Level 5 • Semester 1 • PST05106 Nomenclature of Organic Compounds Pharmaceutical Organic Chemistry • Source Session/Topic 4 Full source-text version: all educational wording from the extracted learning source is retained; only presenter/tutor metadata and web-layout noise are removed, while formatting is improved for readability. 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

NTA Level 5 • Semester 1 • PST05106 Classification of Drugs According to Their Chemical Nature Pharmaceutical Organic Chemistry • Source Session/Topic 3 Full source-text version: all educational wording from the extracted learning source is retained; only presenter/tutor metadata and web-layout noise are removed, while formatting is improved for readability. 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 ← Previous TopicNext Topic →View all Pharmaceutical Organic Chemistry topicsOpen Complete Full Notes PDF / OFFLINE NOTES Unataka kutumiwa notes hizi kupitia WhatsApp?Kwa notes zilizopangiliwa vizuri kwa kusoma offline au PDF, bonyeza kitufe hapa chini. Ujumbe wenye Level, Semester, Module na Topic utaandaliwa moja kwa moja.TUMIWA NOTES WHATSAPP WhatsApp: 255620339260

NTA Level 5 • Semester 1 • PST05106 Classification of Organic Compounds Pharmaceutical Organic Chemistry • Source Session/Topic 2 Full source-text version: all educational wording from the extracted learning source is retained; only presenter/tutor metadata and web-layout noise are removed, while formatting is improved for readability. 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 ← Previous TopicNext Topic →View all Pharmaceutical Organic Chemistry topicsOpen Complete Full Notes PDF / OFFLINE NOTES Unataka kutumiwa notes hizi kupitia WhatsApp?Kwa notes zilizopangiliwa vizuri kwa kusoma offline au PDF, bonyeza kitufe hapa chini. Ujumbe wenye Level, Semester, Module na Topic utaandaliwa moja kwa moja.TUMIWA NOTES WHATSAPP WhatsApp: 255620339260