2025 VJC JC2 H2 Organic Revision Lecture (students)

1 VICTORIA JUNIOR COLLEGE CHEMISTRY DEPARTMENT JC2 H2 CHEMISTRY ORGANIC CHEMISTRY REVISION LECTURE OUTLINE S/N TOPIC PAGE 1 Reaction Mechanisms 2 – 4 2 Hybridisation 4 – 5 3 Isomerism 5 – 7 4 Acidity and Basicity of Organic Compounds 7 – 10 5 Hydrolysis Reactions (relative rate) 11 6 Environmental Concerns of Organic Compounds 11 – 12 7 Amino Acids 12 – 14 8 Reactions of Organic Compounds by Homologous Series 15 – 21 9 Reactions of Organic Compounds by Reagents and Conditions 22 – 29 10 Structural Elucidation 29 – 31 11 Practice Questions (Assigned on SLS) -

2 1. REACTION MECHANISMS (a) Mechanism: Free radical substitution Stoichiometric equation: CH4 + Cl2 → CH3Cl + HCl Mechanism: Initiation: Generation of free radicals. Cl Cl 2Cluv Propagation: Free radicals generate other free radicals. Termination: Free radicals combine to form molecules. (b) Mechanism: Electrophilic addition Stoichiometric equation: CH3CH=CH2 + Br2 → CH3CHBrCH2Br Mechanism: CH3 C C H H H + Br Br + - CH3 C H C H H Br + Br N.B.: The most stable carbocation intermediate will be formed (Markovnikov’s rule). In the case, Br is attached to C with more hydrogen atoms to form the more stable carbocation. Relative stability of carbocations: tertiary > secondary > primary, because alkyl groups exert electron – donating inductive effect which disperse the positive charge on the carbocation, stabilising the carbocation. CH3 CH CH2 Br+ + Br- CH3 CH CH2 Br Br (c) Mechanism: Electrophilic substitution Stoichiometric equation: + HNO3 → NO2 + H2O Mechanism: Generation of strong electrophile: 2H2SO4 + HNO3 → 2HSO4− + NO2+ + H3O+ nitronium ion + NO2 + H NO2 + H NO2 + + HSO4 - NO2 + H2SO4 Brønsted- Lowry acid catalyst H2SO4 regenerated. For chlorination, bromination and Friedel – Crafts alkylation, Lewis acid catalysts (e.g anhydrous AlCl3 or FeBr3) are used to form strong electrophiles. (d) Mechanism: Nucleophilic addition Stoichiometric equation: H H R–C + HCN → R–C–OH O CN Mechanism: Catalyst CN‒ regenerated. CN‒ slow fast uv Cl HCl Cl Cl Cl CH3Cl Cl Cl Cl Cl CH3Cl

3 (e)(i) Mechanism: Bimolecular nucleophilic substitution, SN2 (for primary RX) Stoichiometric equation: CH3CH2Cl + NaOH(aq) → CH3CH2OH +NaCl Mechanism: Factors favouring the mechanism Low steric hindrance [due to less alkyl groups attached to electron deficient carbon atom] – usually favoured by primary halogenoalkanes Stereochemistry (applicable if the molecule is chiral) Nucleophile attacks electron deficient carbon atom from the side opposite to the halogen atom, resulting in inversion of the spatial arrangement of the organic molecule. (ii) Mechanism: Unimolecular nucleophilic substitution, SN1 (for tertiary RX) Stoichiometric equation: CH3CH(CH2CH3)Cl + NaOH(aq) → CH3CH(CH2CH3)OH +NaCl Mechanism: Factors favouring the mechanism High carbocation stability [tertiary > secondary > primary], also benzylic (C6H5CH2+) Stereochemistry Nucleophile attacks trigonal planar carbocation from top and bottom with equal chance, resulting in equal proportions of both enantiomers (racemic mixture). Examples 1 Describe the mechanism for the following conversion. , where R is an alkyl group, Solution

4 2 CJC Prelim 11/III/5 (modified) A general method for the preparation of -amino acids is the amidomalonate synthesis. The reaction below forms part of the synthesis of aspartic acid using this method: In Step I, the reaction begins with a treatment of compound A with CH3CH2ONa which acts as the base, followed by the reaction with BrCH2CO2CH2CH3 to form B. (a) Suggest the type of reaction undergone in Step I to form B. (b) Hence, construct a balanced equation for the acid -base reaction between A and CH3CH2ONa. Solution (a) (b) 2. HYBRIDISATION Type of hybridization Bonds formed by carbon Shape and bond angle Example sp3 4 sigma bonds Tetrahedral, 109.5o sp2 3 sigma and 1 pi bonds Trigonal planar, 120o sp 2 sigma and 2 pi bonds Linear, 180o

5 Example 3 N2008/I/19 The bond lengths in buta-1,3-diene differ from those which might be expected. The carbon-carbon bond length in ethane is 0.154 nm and in ethene 0.134 nm. The central single bond in buta-1,3,-diene (C2–C3), however, is shorter than the single bond in ethane: it is 0.147 nm. What helps to explain this C2–C3 bond length? A It is an sp2–sp2 overlap. B It is an sp2–sp3 overlap. C The electrons in the filled p orbitals on C2 and C3 repel each other. D The sp3–sp3 bonding is pulled shorter by a p–p (π-bond) overlap. Solution C1 to C4 are hybridised. Hence the sigma bond between C2 and C3 must be formed from the overlap of sp2 hybrid orbitals (options B and D incorrect). Both C atoms in ethane are hybridised. Hence, the C –C bond in ethane is formed from the overlap of sp3 hybrid orbitals. s-orbitals are closer to the nucleus. sp2 orbital has degree of s character, hence resulting in more effective overlap and a shorter and stronger bond. Answer: 3. ISOMERISM ISOMERISM (compounds with the same molecular formula) N.B. (a) To determine all the isomers: (1) If molecular formula is given, determine all the structural isomers first followed by stereoisomers. (2) If structural formula is given, determine only the stereoisomers. (b) Maximum n o. of stereoisomers = 2 n, where n = chiral centres + C=C that demonstrate cis-trans isomerism Structural (Constitutional) isomerism (same molecular formula, different structural formula) Stereoisomerism (same molecular formula, same structural formula, different arrangement in space) Enantiomerism - presence of chiral centre - ability to rotate plane -polarised light - no plane of symmetry in the molecule Cis–trans isomerism - arises due to restricted rotation, for example, about C=C, whereby 2 different groups attached to each alkene C

6 Examples 4 HCI Prelim 10/I/19 A pheromone is a secreted or excreted chemical factor that triggers a social response in members of the same species. Compound A below is an example of pheromones. Compound A How many possible isomers, including stereoisomers, can be obtained when A is reacted with excess concentrated sulfuric acid under heat? A 26 B 27 C 28 D 29 Solution Consider structural isomers first: For each structure, identify chiral centre and C=C that show cis-trans isomerism: Answer:

7 5 The complete hydrolysis of 1.76 g of an ester with molecular formula CnH2nO2 required 2.0 x 10−2 mol of sodium hydroxide. (a) Deduce the value of n. Hence, state the molecular formula of the ester. (b) Hence, give the structural formulae of all the esters with this molecular formula. Solution (a) (b) 4. ACIDITY AND BASICITY OF ORGANIC COMPOUNDS