Question:medium

Predict the % product formation in the given reaction :
$CH_3CH_2CBr(CH_3)_2 \xrightarrow{KOH(alc.)} CH_3CH=C(CH_3)_2 \ (X) + CH_3CH_2C(CH_3)=CH_2 \ (Y)$}

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Small bases (e.g., $NaOH$, $KOH$, $NaOEt$, $NaOMe$) favor the Zaitsev product (more substituted alkene). Bulky bases (e.g., potassium tert-butoxide, LDA) suffer from steric hindrance and thus favor the Hofmann product (less substituted alkene, formed by removing the most accessible proton).
Updated On: Jul 31, 2026
  • X = 29%, Y = 71%
  • X = 71%, Y = 29%
  • X = 50%, Y = 50%
  • X = 0%, Y = 100%
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The Correct Option is B

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