Q44 / 150 · MHT CET 2026 (11 April Shift 1) · Chemistry
Identify the product obtained when RCN is treated with sodium and alcohol.
- A$\text{RCONH}_2$
- B$\text{RCOONH}_4$
- Correct answerC$\text{RCH}_2\text{NH}_2$
- D$\text{R}(\text{CH}_2)_3\text{NH}_2$
Solution
The reduction of alkyl cyanides (nitriles) with sodium and alcohol is known as the Mendius reduction.
In this reaction, the cyano group ($-\text{C}\equiv\text{N}$) is reduced to a primary amine group ($-\text{CH}_2\text{NH}_2$).
$\text{R-C}\equiv\text{N} + 4[\text{H}] \xrightarrow{\text{Na/C}_2\text{H}_5\text{OH}} \text{R-CH}_2\text{-NH}_2$
The product obtained is a primary amine, $\text{RCH}_2\text{NH}_2$.
Answer: $\text{RCH}_2\text{NH}_2$
In this reaction, the cyano group ($-\text{C}\equiv\text{N}$) is reduced to a primary amine group ($-\text{CH}_2\text{NH}_2$).
$\text{R-C}\equiv\text{N} + 4[\text{H}] \xrightarrow{\text{Na/C}_2\text{H}_5\text{OH}} \text{R-CH}_2\text{-NH}_2$
The product obtained is a primary amine, $\text{RCH}_2\text{NH}_2$.
Answer: $\text{RCH}_2\text{NH}_2$
Hence, the correct answer is option: 'C'.