In the reaction, $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH} \stackrel{\mathrm{NaOH}}{\longrightarrow}(\mathrm{A})$ $\underset{140^{\circ} \mathrm{C},(4-7) \mathrm{atm}}{\mathrm{CO}_{2}}(\mathrm{~B}) \stackrel{\mathrm{HCl}}{\longrightarrow}(\mathrm{C})$, the compound (C) is

• A

  Benzoic acid
• B

  Salicylaldehyde
• C

  Chlorobenzene
• Correct answer
  D

  Salicylic acid

Solution

Treatment of sodium salt of phenol with $\mathrm{CO}_{2}$ under pressure brings about substitution of the-COOH group for the hydrogen of the ring. This is called as Kolbe's reaction

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