2-Bromobutane on heating with alcoholic KOH f | vedclass

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(D) The reaction of $2-$bromobutane with alcoholic $KOH$ is a dehydrohalogenation reaction (elimination reaction).According to $Saytzeff's$ rule,the m

Vedclass · Accepted Answer

$80\% \beta-$Butylene $+ 20\% \alpha-$Butylene

Vedclass · Answer

(D) The reaction of $2-$bromobutane with alcoholic $KOH$ is a dehydrohalogenation reaction (elimination reaction).According to $Saytzeff's$ rule,the major product is the more substituted alkene.$2-$Bromobutane $(CH_3-CHBr-CH_2-CH_3)$ undergoes elimination to form $2-$butene ($\beta-$butylene) and $1-$butene ($\alpha-$butylene).$2-$Butene is more stable due to hyperconjugation and steric factors,making it the major product $(80\%)$.$1-$Butene is the minor product $(20\%)$.Therefore,the correct composition is $80\% \beta-$butylene and $20\% \alpha-$butylene.

$2-$Bromobutane on heating with alcoholic $KOH$ forms

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