What are X and Y respectively in the followin | vedclass

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(C) $1$. The reaction of styrene $(C_6H_5-CH=CH_2)$ with alkaline $KMnO_4$ followed by acidic workup $(H_3O^+)$ is an oxidative cleavage reaction that

Vedclass · Accepted Answer

$X$ = Benzoic acid,$Y$ = $m$-Bromobenzoic acid

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(C) $1$. The reaction of styrene $(C_6H_5-CH=CH_2)$ with alkaline $KMnO_4$ followed by acidic workup $(H_3O^+)$ is an oxidative cleavage reaction that converts the vinyl group into a carboxylic acid group,yielding benzoic acid $(C_6H_5COOH)$ as product $X$.$2$. Benzoic acid contains a $-COOH$ group attached to the benzene ring. The $-COOH$ group is a strongly deactivating group and is meta-directing for electrophilic aromatic substitution reactions.$3$. Therefore,the bromination of benzoic acid $(X)$ using $Br_2 / FeBr_3$ will result in the electrophilic substitution of the bromine atom at the meta-position,yielding $m$-bromobenzoic acid as product $Y$.

What are $X$ and $Y$ respectively in the following reaction sequence?
Styrene $\xrightarrow[(ii) H_3O^+]{(i) KMnO_4 / KOH} X$ $\xrightarrow{Br_2 / FeBr_3} Y$

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What are $X$ and $Y$ respectively in the following reaction sequence?Styrene $\xrightarrow[(ii) H_3O^+]{(i) KMnO_4 / KOH} X$ $\xrightarrow{Br_2 / FeBr_3} Y$