Q.Which of the following compounds will give a racemic mixture on nucleophilic substitution by OH⁻ ion? i) CH₃-CH(C₂H₅)-CH₂Br ii) (CH₃)₂C(Br)(C₂H₅), i.e. H₃C-C(CH₃)(Br)-C₂H₅ iii) CH₃-CH(Cl)-C₂H₅, i.e. a secondary carbon bearing H, CH₃, C₂H₅ and Cl a)

Two mechanisms carry out nucleophilic substitution, distinguished by kinetics, stereochemistry and which substrates favour each. SN2 (bimolecular): Rate = k2[RX][Nu-], second order, ONE step. The nucleophile attacks from directly opposite the leaving halogen (backside attack) through a single transition state, inverting the carbon's configuration exactly as an umbrella flips in a strong wind (Walden inversion) -- so an optically active substrate with the leaving group ON the stereocentre gives a single inverted, still optically active product. SN2 favours LESS hindered carbons: methyl > primary > secondary >> tertiary (tertiary essentially never goes SN2, since three alkyl groups block the required backside approach). SN1 (unimolecular): Rate = k[RX] only, independent of nucleophile concentration, first order, TWO steps -- (1) slow, rate-determining heterolysis of C-X to a planar carbocation (sp2, empty p-orbital with two equal lobes), then (2) fast attack by the nucleophile from EITHER face. Because both faces are attacked with roughly equal probability, a chiral SN1 substrate (halogen on the stereocentre) gives a racemic (optically inactive) mixture. SN1 favours MORE substituted carbons: tertiary > secondary > primary > CH3X, since more alkyl groups (inductive donation plus hyperconjugation) stabilise the in …