Convert, ethanoic acid to ethanol.(ii) Explain the mechanism of alkaline hydrolysis of bromomethane.

Answer:

(i) Ethanoic acid is reduced by lithium aluminium hydride in dry ether, followed by acid hydrolysis:

CH₃COOH + 4[H] —(i) LiAlH₄, dry ether; (ii) H₃O⁺→ CH₃CH₂OH (ethanol) + H₂O

(ii) Mechanism (S_N2): CH₃Br + OH⁻ (aq. NaOH/KOH) → CH₃OH + Br⁻

1. The C–Br bond is polar (C^δ+–Br^δ−). The nucleophile OH⁻ attacks the carbon from the side opposite to the bromine (back-side attack, 180° to the leaving group), because the front side is blocked by the large Br atom.
2. As the C–O bond starts forming, the C–Br bond starts breaking in the same single step. In the transition state HO^δ−···CH₃···Br^δ−, the carbon is bonded partly to both OH and Br, and its three H atoms lie in one plane (perpendicular to the HO–C–Br axis).
3. Br leaves as Br⁻, and methanol is formed. The three H atoms flip to the other side like an umbrella turned inside out, so the configuration is inverted (Walden inversion).

The rate depends on both reactants: rate = k[CH₃Br][OH⁻], so the reaction is second order and bimolecular, i.e. S_N2 (substitution, nucleophilic, bimolecular). Bromomethane reacts fastest by this path because it has the least steric hindrance.