question_answer

Phenol \[\xrightarrow[(ii)\,C{{O}_{2}}/140{{\,}^{o}}C]{(i)\,NaOH}A\xrightarrow[{}]{{{H}^{+}}/{{H}_{2}}O}B\xrightarrow{{{A}_{{{c}_{2}}}}O}C\] In this reaction, the end product C is

A)  salicylaldehyde      

B)  salicylic acid

C)  phenyl acetate       

D)  aspirin

Correct Answer: D

Solution :