In the E2 dehydrohalogenation of 2-bromo-2-methylbutane, the proportion of the less substituted alkene 2-methylbut-1-ene (Hofmann product) increases when (select all that apply)

Athe base is changed from $NaOEt$ in ethanol to $(CH X_{3}) X_{3} COK$ in $(CH X_{3}) X_{3} COH$
Bthe leaving group is changed from $Br$ to $F$ (2-fluoro-2-methylbutane, same base)
Cthe leaving group is changed to $- N X^{+} (CH X_{3}) X_{3}$ (hydroxide, heat)
Dthe base is changed from $NaOEt$ in ethanol to $NaOMe$ in methanol

Answer: A, B, C

A bulky base removes the more accessible methyl hydrogens, so tert-butoxide raises the Hofmann fraction. Poor leaving groups such as fluoride and the trimethylammonium group give a carbanion-like transition state and favour the least hindered, most acidic hydrogens, again giving the Hofmann product. Methoxide is smaller than ethoxide, so it keeps Saytzeff orientation and does not raise the Hofmann fraction.

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In the E2 dehydrohalogenation of 2-bromo-2-methylbutane, the proportion of the less substituted alkene 2-methylbut-1-ene (Hofmann product) increases when (select all that apply)

Answer: A, B, C

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