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JEE Main 2019 · Chemistry · Alcohols, Phenols, Ethers and Haloalkanes · hard

2-Bromo-2,3-dimethylpentane, CH3C(Br)(CH3)CH(CH3)CH2CH3, is heated separately with (i) sodium ethoxide in ethanol and (ii) potassium tert-butoxide in…

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2-Bromo-2,3-dimethylpentane, CH3C(Br)(CH3)CH(CH3)CH2CH3\mathrm{CH_3C(Br)(CH_3)CH(CH_3)CH_2CH_3}, is heated separately with (i) sodium ethoxide in ethanol and (ii) potassium tert-butoxide in tert-butanol. The major alkenes (i), (ii) respectively are
  1. A. 2,3-dimethylpent-1-ene; 2,3-dimethylpent-2-ene
  2. B. 2,3-dimethylpent-2-ene; 2,3-dimethylpent-2-ene
  3. C. 2,3-dimethylpent-2-ene; 2,3-dimethylpent-1-ene
  4. D. 2,3-dimethylpent-1-ene; 2,3-dimethylpent-1-ene
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Answer: (C) 2,3-dimethylpent-2-ene; 2,3-dimethylpent-1-ene

The β\beta-hydrogens are on the two methyl groups at C-2 (giving the terminal alkene, 2,3-dimethylpent-1-ene) and on C-3 (giving the tetrasubstituted 2,3-dimethylpent-2-ene). The small base ethoxide gives the more substituted, more stable Zaitsev product. The bulky tert-butoxide cannot easily reach the crowded tertiary C-3 hydrogen and removes a less hindered methyl hydrogen instead, so the Hofmann product dominates.
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