Toluene on reaction with the reagent X gives | vedclass

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(D) $1$. Toluene reacts with $KMnO_4 / OH^-, \Delta$ followed by $H_3O^+$ (reagent $X$) to form benzoic acid $(Y)$.$2$. Benzoic acid dissolves in $NaH

Vedclass · Accepted Answer

$(i) \ KMnO_4 / OH^-, \Delta \ (ii) \ H_3O^+; 	ext{m-bromobenzoic acid}$

Vedclass · Answer

(D) $1$. Toluene reacts with $KMnO_4 / OH^-, \Delta$ followed by $H_3O^+$ (reagent $X$) to form benzoic acid $(Y)$.$2$. Benzoic acid dissolves in $NaHCO_3$ because it is acidic.$3$. The $-COOH$ group is a meta-directing group in electrophilic aromatic substitution reactions.$4$. Therefore,reaction of benzoic acid with $Br_2 / Fe$ (electrophilic bromination) yields m-bromobenzoic acid $(Z)$.$5$. Thus,$X$ is $(i) \ KMnO_4 / OH^-, \Delta \ (ii) \ H_3O^+$ and $Z$ is m-bromobenzoic acid.

Toluene on reaction with the reagent $X$ gives $Y$,which dissolves in $NaHCO_3$ and when reacted with $Br_2 / Fe$ gives $Z$. What are $X$ and $Z$?

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