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(B) Step $1$: Oxidation of toluene with $KMnO_4/H^+$ gives benzoic acid $(A = C_6H_5COOH)$.Step $2$: Reaction of benzoic acid with $SOCl_2$ gives benz

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Aniline

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(B) Step $1$: Oxidation of toluene with $KMnO_4/H^+$ gives benzoic acid $(A = C_6H_5COOH)$.Step $2$: Reaction of benzoic acid with $SOCl_2$ gives benzoyl chloride $(B = C_6H_5COCl)$.Step $3$: Reaction of benzoyl chloride with $NH_3$ gives benzamide $(C = C_6H_5CONH_2)$.Step $4$: Reaction of benzamide with $KOH/Br_2$ is the Hofmann bromamide degradation reaction,which converts an amide to a primary amine with one carbon atom less. Thus,benzamide is converted to aniline $(D = C_6H_5NH_2)$.

Consider the following reaction sequence. The final product $D$ will be:
$C_6H_5CH_3$ $\xrightarrow[\Delta]{KMnO_4/H^+} A$ $\xrightarrow{SOCl_2} B$ $\xrightarrow[\Delta]{NH_3} C$ $\xrightarrow{KOH/Br_2} D$

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Consider the following reaction sequence. The final product $D$ will be:$C_6H_5CH_3$ $\xrightarrow[\Delta]{KMnO_4/H^+} A$ $\xrightarrow{SOCl_2} B$ $\xrightarrow[\Delta]{NH_3} C$ $\xrightarrow{KOH/Br_2} D$