Question

Acetic acid reacts with hydrazoic acid in the presence of conc. H$_2$SO$_4$ at 0$^\circ$C to form –

• A. Methane
• B. Methyl amine
• C. Methyl cyanide
• D. Ethyl amine

Hint:

The Schmidt reaction converts carboxylic acids into primary amines with the elimination of CO$_2$. It is a good method for the preparation of amines.

Answer 1

The Correct Option is B

Step 1: Recognize the reaction.
This reaction is the Schmidt reaction. In this reaction, a carboxylic acid (R–COOH) reacts with hydrazoic acid (HN$_3$) in the presence of conc. H$_2$SO$_4$ at low temperature to give a primary amine (R–NH$_2$) with one carbon atom less.

Step 2: Apply to given case.
Acetic acid (CH$_3$COOH) reacts as follows: \[ CH_3COOH + HN_3 \xrightarrow{conc. \, H_2SO_4, \, 0^\circ C} CH_3NH_2 + CO_2 \]

Step 3: Identify product.
The product is methyl amine (CH$_3$NH$_2$).
Step 4: Final Answer.
Thus, the reaction forms methyl amine.
\[ \boxed{\text{Methyl amine (CH$_3$NH$_2$)}} \]