Question

A compound 'X' on treatment with $Br_{2}/NaOH$ , provided $C_{3}H_{9}N$ (Y) which gives positive carbylamine test. Compound 'X' is

• A. $CH_{3}COCH_{2}NHCH_{3}$
• B. $CH_{3}CH_{2}COCH_{2}NH_{2}$
• C. $CH_{3}CH_{2}CH_{2}CONH_{2}$
• D. $CH_{3}CON\left(\right.CH_{3}\left(\left.\right)_{2}$

Answer 1

Answer: (C)

$\text{X}\overset{\left(\text{Br}\right)_{2} / \text{NaOH}}{ \rightarrow }\left(\text{C}\right)_{3}\left(\text{H}\right)_{9}\text{N}\left(Y\right)$
As Y gives +ve carbylamine test so it is a primary amine and X is an amide which on Hoffmann's bromamide degradation gives amine Y.
$\left(\text{CH}\right)_{\text{3}} \left(\text{CH}\right)_{\text{2}} \left(\text{CH}\right)_{\text{2}} \left(\text{CONH}\right)_{\text{2}} \text{=} \left(\text{X}\right) \overset{\left(\text{Br}\right)_{2} \text{+NaOH}}{ \rightarrow } \left(\left(\text{CH}\right)_{\text{3}} \left(\text{CH}\right)_{\text{2}} \left(\text{CH}\right)_{\text{2}} \left(\text{NH}\right)_{\text{2}}\right)$
Option $\left(\right.2\left.\right)$ is not possible as it is not an amide. $\left(\right.1,4\left.\right)$ are not possible as they will not give 1° amine on degradation.