Carboxylic Acids and their Derivatives 1 Question 3
4. An organic compound $A$ upon reacting with $\mathrm{NH}{3}$ gives $B$. On heating, $B$ gives $C$. $C$ in the presence of $\mathrm{KOH}$ reacts with $\mathrm{Br}{2}$ to give $\mathrm{CH}{3} \mathrm{CH}{2} \mathrm{NH}_{2}$. $A$ is
(2013 Main)
(a) $\mathrm{CH}_{3} \mathrm{COOH}$
(b) $\mathrm{CH}{3} \mathrm{CH}{2} \mathrm{CH}_{2} \mathrm{COOH}$
(c) $\mathrm{CH}{3}-\underset{\mathrm{CH}{3}}{\mathrm{CH}}-\mathrm{COOH}$
(d) $\mathrm{CH}{3} \mathrm{CH}{2} \mathrm{COOH}$
Show Answer
Answer:
Correct Answer: 4. (d)
Solution:
- $\mathrm{CH}{3} \mathrm{CH}{2} \underset{(A)}{\stackrel{\mathrm{O}}{\mathrm{C}}-\mathrm{OH} \xrightarrow{\mathrm{NH}_{3}}}$
(B)
(C)
$$ \underset{\begin{array}{l} \text { Hoffmann’s bromamide } \ \text { reaction } \end{array}}{\mathrm{Br}{2}, \mathrm{KOH}} \mathrm{CH}{3}-\mathrm{CH}{2}-\mathrm{NH}{2} $$
