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(B) The reaction of $2	ext{-bromobutane}$ with alcoholic $KOH$ is a dehydrohalogenation reaction (elimination reaction).
According to $Saytzeff'

Vedclass · Accepted Answer

But$-2-$ene

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(B) The reaction of $2	ext{-bromobutane}$ with alcoholic $KOH$ is a dehydrohalogenation reaction (elimination reaction).
According to $Saytzeff's$ rule,the more substituted alkene is the major product.
$CH_3-CH(Br)-CH_2-CH_3 + KOH (	ext{alc.}) 	o CH_3-CH=CH-CH_3 (	ext{But-2-ene}) + KBr + H_2O$.
Since $But-2-ene$ is more stable than $But-1-ene$,it is the major product.

The major product of the following reaction is $CH_3-CH(Br)-CH_2-CH_3 + \text{alc. } KOH \to $

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