Q1(b). E2 of 2-bromo-2-methylpentane: Zaitsev and Hofmann

Introduction. A tertiary bromide with a strong base reacts by E2: the base removes a β-H anti-periplanar to Br while Br⁻ leaves. Two kinds of β-H are present: the six CH₃ hydrogens and the two hydrogens on C-3 (CH₂).

Scheme: Loss of H from C-3 gives the trisubstituted alkene A; loss from a methyl gives the disubstituted alkene B.

1. (i) NaOEt in ethanol. Ethoxide is small. The transition state has partial C=C character, so the route to the more substituted, more stable alkene has the lower barrier: A, 2-methylpent-2-ene, is major (Zaitsev product); B is minor.
2. (ii) ᵗBuOK in t-butanol. tert-Butoxide is bulky. The CH₂ hydrogens on C-3 are flanked by the CMe₂Br and ethyl groups and are hard to reach; the six primary methyl hydrogens are exposed (and statistically favoured). B, 2-methylpent-1-ene, becomes major (Hofmann product); A is minor.
3. The reaction type (E2, one step, second order) is the same in both; only the steric size of the base changes which β-H is removed.

Conclusion. (i) NaOEt: major A = 2-methylpent-2-ene (Zaitsev), minor B = 2-methylpent-1-ene. (ii) ᵗBuOK: major B = 2-methylpent-1-ene (Hofmann), minor A.

Application. Bulky bases (ᵗBuOK, LDA) are chosen in synthesis to make terminal alkenes from tertiary halides; with 2-bromo-2-methylbutane the same switch gives mainly 2-methylbut-1-ene.