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(D) The reaction involves the electrophilic addition of $HBr$ to the two double bonds present in the molecule,which is allyl acrylate $(CH_2=CH-COO-CH

Vedclass · Accepted Answer

$1-$bromopropan$-2-$yl $3-$bromopropanoate

Vedclass · Answer

(D) The reaction involves the electrophilic addition of $HBr$ to the two double bonds present in the molecule,which is allyl acrylate $(CH_2=CH-COO-CH_2-CH=CH_2)$.$1$. The first $HBr$ molecule adds to the more reactive double bond. The double bond in the allyl group (alkene part) is more electron-rich than the double bond conjugated with the carbonyl group (acrylate part). Thus,$HBr$ adds to the allyl double bond following Markovnikov's rule to form $CH_2=CH-COO-CH_2-CH(Br)-CH_3$.$2$. The second $HBr$ molecule adds to the remaining double bond in the acrylate part. This addition follows anti-Markovnikov's rule (if peroxide is present,though usually,in the absence of peroxide,it follows Markovnikov's rule,but here the electron-withdrawing carbonyl group directs the addition to the terminal carbon to form the more stable carbocation or via conjugate addition). The addition to the $CH_2=CH-$ group results in the formation of $Br-CH_2-CH_2-COO-CH_2-CH(Br)-CH_3$.Therefore,the final product is $1-$bromopropan$-2-$yl $3-$bromopropanoate.

The major product of the following reaction is:

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The major product of this reaction is:

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