Consider the following reaction:Phenol xright | vedclass

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(B) Step $1$: Phenol reacts with $Zn$ dust to form benzene $(X)$.$C_6H_5OH + Zn \rightarrow C_6H_6 + ZnO$Step $2$: Benzene undergoes Friedel-Crafts al

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benzoic acid

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(B) Step $1$: Phenol reacts with $Zn$ dust to form benzene $(X)$.$C_6H_5OH + Zn \rightarrow C_6H_6 + ZnO$Step $2$: Benzene undergoes Friedel-Crafts alkylation with $CH_3Cl$ in the presence of anhydrous $AlCl_3$ to form toluene $(Y)$.$C_6H_6 + CH_3Cl \xrightarrow{Anhyd.\,AlCl_3} C_6H_5CH_3 + HCl$Step $3$: Toluene $(Y)$ is oxidized by alkaline $KMnO_4$ to form benzoic acid $(Z)$.$C_6H_5CH_3 \xrightarrow{Alkaline\,KMnO_4} C_6H_5COOH$Therefore,the product $Z$ is benzoic acid.

Consider the following reaction:
$Phenol$ $\xrightarrow{Zn\,dust} X$ $\xrightarrow[{Anhyd.\,AlCl_3}]{CH_3Cl} Y$ $\xrightarrow{Alkaline\,KMnO_4} Z$
The product $Z$ is:

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Consider the following reaction:$Phenol$ $\xrightarrow{Zn\,dust} X$ $\xrightarrow[{Anhyd.\,AlCl_3}]{CH_3Cl} Y$ $\xrightarrow{Alkaline\,KMnO_4} Z$The product $Z$ is: