An optically active alkyl halide C_4H_9Br [A] | vedclass

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(C) $1$. The optically active alkyl halide $[A]$ is $sec$-butyl bromide $(CH_3CH_2CH(Br)CH_3)$.$2$. Reaction with hot alcoholic $KOH$ (dehydrohalogena

Vedclass · Accepted Answer

Butan$-2-$one

Vedclass · Answer

(C) $1$. The optically active alkyl halide $[A]$ is $sec$-butyl bromide $(CH_3CH_2CH(Br)CH_3)$.$2$. Reaction with hot alcoholic $KOH$ (dehydrohalogenation) gives but$-2-$ene as the major product $[B]$ $(CH_3CH=CHCH_3)$.$3$. Reaction of $[B]$ with $Br_2$ gives $2,3-$dibromobutane $[C]$ $(CH_3CH(Br)CH(Br)CH_3)$.$4$. Reaction of $[C]$ with alcoholic $NaNH_2$ (dehydrohalogenation) gives but$-1-$yne $[D]$ $(CH_3CH_2C \equiv CH)$.$5$. Hydration of but$-1-$yne $[D]$ with $H_2O$ in the presence of $HgSO_4$ and $H^+$ (Kucherov reaction) follows Markovnikov's rule to form an enol intermediate,which tautomerizes to form butan$-2-$one $[E]$ $(CH_3CH_2COCH_3)$.$6$. The $IUPAC$ name of $[E]$ is butan$-2-$one.

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