The reaction of CH_3CD_2CHBr-CH_2CD_3 with al | vedclass

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Question

(A) The reaction of an alkyl halide with alcoholic $KOH$ is a dehydrohalogenation reaction ($E2$ mechanism). In $CH_3CD_2CHBr-CH_2CD_3$,the $\alpha$-c

Vedclass · Accepted Answer

$CH_3CD_2CH=CH-CD_3$

Vedclass · Answer

(A) The reaction of an alkyl halide with alcoholic $KOH$ is a dehydrohalogenation reaction ($E2$ mechanism). In $CH_3CD_2CHBr-CH_2CD_3$,the $\alpha$-carbon is the one attached to $Br$. The $\beta$-hydrogens are available on the adjacent carbons: $C_2$ (which has $D$ atoms) and $C_4$ (which has $H$ atoms). However,the $H$ atoms on $C_4$ are more acidic and easier to remove than $D$ atoms on $C_2$ due to the kinetic isotope effect. Thus,the elimination occurs at the $C_4$ position to form the alkene $CH_3CD_2CH=CH-CD_3$.

The reaction of $CH_3CD_2CHBr-CH_2CD_3$ with alc. $KOH$ yields:

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