Consider the following reaction, Phenol $\stackrel{\mathrm{Zn} \text { dust }}{\longrightarrow} \mathrm{X} \underset{\text { Anhydrous } \mathrm{AlCl}_3}{\stackrel{\mathrm{CH}_3 \mathrm{Cl}}{\longrightarrow}} \mathrm{Y} \stackrel{\begin{array}{l}\text { Alkaline } \\ \mathrm{KMnO}\end{array}}{\longrightarrow} \mathrm{Z}$, the product, $Z$, is

• A

  toluene
• B

  benzaldehyde
• Correct answer
  C

  benzoic acid
• D

  benzene

Solution

Key Idea (i) $Z n$ dust converts $-\mathrm{OH}$ group into- $\mathrm{H}$. (ii) Reaction with $\mathrm{CH}_3 \mathrm{Cl}$ in presence of anthy. $\mathrm{AlCl}_3$ is called Friedel-Craft's alkylation. (iii) Alkaline $\mathrm{KMnO}_4$ converts complete carbon chain which is directly attached with benzene ring, into - $\mathrm{COOH}$ group.

Hence, the correct answer is option: 'C'.