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Consider the following reaction:
Phenol $\xrightarrow{Zn\ dust}$ X $\xrightarrow[Anhydrous\ AlCl_3]{CH_3Cl}$ Y $\xrightarrow{Alkaline\ KMnO_4}$ Z
The product Z is
Phenol $\xrightarrow{Zn\ dust}$ X $\xrightarrow[Anhydrous\ AlCl_3]{CH_3Cl}$ Y $\xrightarrow{Alkaline\ KMnO_4}$ Z
The product Z is
A
Benzene
B
Toluene
C
Benzaldehyde
D
Benzoic acid
Detailed Solution
Step 1: phenol heated with zinc dust is reduced to benzene: $C_6H_5OH + Zn \rightarrow C_6H_6 + ZnO$. X is benzene.
Step 2: benzene undergoes Friedel–Crafts alkylation with $CH_3Cl$ and anhydrous $AlCl_3$: $C_6H_6 + CH_3Cl \rightarrow C_6H_5CH_3 + HCl$. Y is toluene.
Step 3: alkaline $KMnO_4$ oxidises the alkyl side chain of an alkylbenzene to a carboxyl group: $C_6H_5CH_3 \xrightarrow{alk.\ KMnO_4} C_6H_5COOK \xrightarrow{H_3O^+} C_6H_5COOH$
Hence Z is benzoic acid.
Step 2: benzene undergoes Friedel–Crafts alkylation with $CH_3Cl$ and anhydrous $AlCl_3$: $C_6H_6 + CH_3Cl \rightarrow C_6H_5CH_3 + HCl$. Y is toluene.
Step 3: alkaline $KMnO_4$ oxidises the alkyl side chain of an alkylbenzene to a carboxyl group: $C_6H_5CH_3 \xrightarrow{alk.\ KMnO_4} C_6H_5COOK \xrightarrow{H_3O^+} C_6H_5COOH$
Hence Z is benzoic acid.