RI 2025 Intro to Organic tutorial answers qn6-12

1 Raffles Institution Year 5 H2 Chemistry 2025 Tutorial 9 – Introduction to Organic Chemistry 6. C5H10 can either be an alkene (CnH2n) or a cycloalkane (CnH2n). Steps to draw isomers of alkenes Structures (1) Draw positional isomers of pentene. II exhibits cis-trans isomerism. Draw its cis-isomer III. I II III (2) Draw positional isomers of the branched chain IV V VI Steps to draw isomers of cycloalkanes Structures (1) Draw the cyclopentane (with the biggest number of carbon in the ring) and methylcyclobutane . VII VIII (2) Draw cyclopropane with one ethyl group and two methyl groups on the same C. IX X (3) Draw stereoisomers of 1,2-dimethylcyclopropane. This compound has 2 chiral carbons and will have a maximum of 22 = 4 stereoisomers. However, cis-1,2-dimethylcyclopropane (XIII) has a plane of symmetry and is a meso compound. Thus, there are only 3 stereoisomers for 1,2-dimethylcyclopropane. XI XII XIII Type of isomerism Examples of isomers which exhibit the isomerism Constitutional isomerism I, IV, V, VI, VII, VIII, IX, X, (one of XI, XII or XIII) and (one of II or III) Cis-trans isomerism II and III ; XI (or XII) and XIII Enantiomerism XI and XII 7a b Suggested Solutions to Tutorial 9: Introduction to Organic Chemistry * * C O C C N H C H C H H H H H H O H C H H C C O H C H H H H H H

2 7c C C C C C C H H H H H H C H H H H C H H H H 8 (a) (b) Cl (c) 9(a) (b) O O CH3 CO OH CH2OH CH3 sp2 trigonal planar sp3 tetrahedral * * * * * * (c) OH CH3 C N sp2 trigonal planar sp linear 10. (a) substitution (b) reduction (c) addition (d) elimination (e) substitution (f) elimination (g) hydrolysis (h) step 1: addition step 2: substitution O O OH sp3 tetrahedral sp2 trigonal planar sp2 trigonal planar * * * Nootkatone (Grapefruit oil) C CH2 CH3 O CH3 H3C

3 11 cis,cis-isomer trans,cis-isomer cis,trans-isomer trans,trans-isomer Note: • As each C atom in the C=C bonds has two different groups attached to it, the maximum number of cis-trans isomers is 22 = 4.] • When drawing cis-trans isomers, the trigonal planar geometry around each sp2 C in the C=C bonds must be shown. 12(a) Cis-trans isomerism refers to the existence of two compounds with the same molecular and structural formulae but with different spatial arrangements of atoms due to restricted rotation about a double bond (e.g. C=C, C=N or N=N) or a bond in a ring structure. To exhibit cis - trans isomerism, the compound either has a double bond with two different groups attached to each of the two atoms in the double bond or a ring structure with two different groups attached to two atoms in the ring structure. Enantiomerism refers to the existence of compounds with the same molecular and structural formulae but with different spatial arrangements of atoms and are non-superimposable mirror images of each other. Molecules do not have a plane of symmetry and usually contain one or more chiral centres. Enantiomers rotate plane-polarised light in opposite directions. (b)(i) Hex-3-ene, CH 3CH2CH=CHCH2CH3 Type of stereoisomerism: cis-trans isomerism No of stereoisomers: 2 Hex-3-ene exhibits cis-trans isomerism since 2 different groups are attached to each of the carbon atoms in the C=C bond. Due to restricted rotation about the C=C bond, hex -3-ene has 2 stereoisomers: cis-hex-3-ene trans-hex-3-ene C C C C A H H H C H B H H C C C C H H A H C H B H H C C C C A H H H C B H H H C C C C H H A H C B H H H C CH3CH2 C H CH2CH3 H C H C CH3CH2 CH2CH3 H

4 (ii) 3-methylcyclohexene Type of stereoisomerism: Enantiomerism No of stereoisomers: 2 3-methylcyclohexene has one chiral carbon and does not have a plane of symmetry. It forms mirror images that are non-superimposable and hence exhibits enantiomerism. It exists as 2 enantiomers: CH3 CH3 can also be accepted. Note: There is restricted rotation about the C=C bond in the ring and there are also two different groups attached to each C atom of the C=C bond. However, the trans-isomer does not exist in the case of cyclic alkene due to the presence of ring strain. (iii) Type of stereoisomerism: cis-trans isomerism and enantiomerism No of stereoisomers: 4 The compound has one C=C double bond with 2 different groups attached to each of the C in C=C bond ⇒ it exhibits cis-trans isomerism. The compound has one chiral carbon and lacks a plane of symmetry. Thus it has mirror images that are non-superimposable ⇒ it exhibits enantiomerism. Total number of stereoisomers = 22 = 4: Enantiomers: P and Q; R and S Cis-trans isomers: P and R; Q and S Self-practice: Try drawing the skeletal formulae of the isomers. (iv) Type of stereoisomerism: none No of stereoisomers: not applicable (CH3)2C=CHCH2CH(OH)CH2CH=C(CH3)2 does not exhibit enantiomerism as it has a plane of symmetry (or no chiral centre) and it does not exhibit cis-trans isomerism as one carbon in each of the C=C bonds has the same groups attached. P Q C C CH3 H H CH2 C HO CH3 H CC CH3 H H CH2 C OHCH3 H SR C C H CH3 H CH2 C HO CH3 H CC H CH3 H CH2 C OHCH3 H CH3 H CH3 H mirror CH3 * mirror mirror * * * *

5 (v) Type of stereoisomerism: cis-trans isomerism and enantiomerism No of stereoisomers: 3 The compound has 2 chiral carbons and so, in theory, can have a maximum number of 22 = 4 stereoisomers. However, this compound has only 3 stereoisomers. X Y Z Z’ mirror Due to restricted rotation about the bonds in the ring and the ring having two carbon atoms each attached to 2 different groups (H and COOH) , the compound exhibits cis- trans isomerism. X (or Y) has no plane of symmetry . Therefore, it is not superimposable onto its mirror image, Y (or X). X and Y are non-superimposable mirror images of each other. X and Y are enantiomers. However, one of the isomers, cis-cyclobutane-1,2-dicarboxylic acid ( Z) has a plane of symmetry. Z is thus superimposable onto its mirror image (Z’). Z and Z’ are thus identical. Z is a meso compound. Enantiomers: X and Y Cis-trans isomers: X (or Y) and Z COOH COOH HOOC HOOC COOH COOH HOOC HOOC CO2H CO2H * *