In the reaction C_6H_5CH_3 xrightarrow{text{O | vedclass

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(A) Step $1$: Oxidation of toluene $(C_6H_5CH_3)$ gives benzoic acid $(C_6H_5COOH)$ as product $A$.Step $2$: Reaction of benzoic acid with $NaOH$ give

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$C_6H_6$

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(A) Step $1$: Oxidation of toluene $(C_6H_5CH_3)$ gives benzoic acid $(C_6H_5COOH)$ as product $A$.Step $2$: Reaction of benzoic acid with $NaOH$ gives sodium benzoate $(C_6H_5COONa)$ as product $B$.Step $3$: Decarboxylation of sodium benzoate with sodalime $(NaOH/CaO)$ gives benzene $(C_6H_6)$ as product $C$.Therefore,the final product $C$ is $C_6H_6$.

In the reaction $C_6H_5CH_3$ $\xrightarrow{\text{Oxidation}} A$ $\xrightarrow{NaOH} B$ $\xrightarrow{\text{Sodalime}} C$. Then $C$ is

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