Question

An organic compound (A) on treatment with acetic acid in the presence of sulphuric acid produces an ester (B), (A) on mild oxidation gives (C), (C) with $50\%$ potassium hydroxide followed by acidification with dilute hydrochloric acid generates (A) and (D), (D) with phosphorus pentachloride followed by reaction with ammonia gives (E), (E) on dehydration produces hydrocyanic acid. E is an:

• A. Amide
• B. Ester
• C. Alcohol
• D. Amine

Answer 1

The correct option is A Amide

$\underset{\left(A\right)}{\underset{}{C{H}_3-OH}}+C{H}_3-COOH⟶\underset{\left(B\right)}{\underset{}{C{H}_{3}COOC{H}_3}}$

$\underset{\left(A\right)}{\underset{}{C{H}_{3}OH}}+\left[O\right]⟶\underset{\left(C\right)}{\underset{}{HCHO}}$

$\underset{\left(C\right)}{\underset{}{HCHO}}\stackrel{50\%NaOH}{⟶}\underset{\left(A\right)}{\underset{}{C{H}_{3}OH}}+\underset{\left(D\right)}{\underset{}{HCOOH}}$

$HCOOH+PC{l}_5⟶HCOCl$

$HCOCl+N{H}_3⟶HCON{H}_2$

$\underset{\left(E\right)}{\underset{}{HCON{H}_2}}⟶HCN+H_{2}O$

So compound E is amide as reaction shown

Option A is correct.