Question

$Phenol\underset{\left(ii\right)C{O}_2/{140}^{\circ }C}{\stackrel{\left(i\right)NaOH}{}}A\stackrel{H^{+}/H_{2}O}{\to }B\stackrel{A{c}_{2}O}{\to }C$.

In the above reaction, end product ${}^{\prime }{C}^{\prime }$ is:

• A. salicylaldehyde
• B. salicylic acid
• C. phenyl aetate
• D. aspirin

Answer 1

The correct option is D aspirin

Option D is correct.