question_answer

In the reaction, \[{{C}_{6}}{{H}_{5}}OH\xrightarrow{NaOH}(A)\] \[\underset{140{}^\circ C,\ (4-7\,atm)}{\mathop{\xrightarrow{\ C{{O}_{2}}\ \ \ \ \ \ \ \ \ \ \ \ }}}\,(B)\] \[\xrightarrow{HCl}(C),\]the compound  is                          [Pb. CET 2001]

A)                 Benzoic acid      

B)                 Salicylaldehyde

C)                 Chlorobenzene                

D)                 Salicylic acid

Correct Answer: D

Solution :

           Treatment of sodium salt of phenol with \[C{{O}_{2}}\] under pressure bring about substitution of the carbonyl group \[-COOH,\]  for the hydrogen of the ring. This is called as Kolbe?s reaction