Carboxylic Acids and their Derivatives - Result Question 18
####4. An organic compound $A$ upon reacting with $NH _3$ gives $B$. On heating, $B$ gives $C$. $C$ in the presence of $KOH$ reacts with $Br _2$ to give $CH _3 CH _2 NH _2$. $A$ is
(2013 Main)
(a) $CH _3 COOH$
(b) $CH _3 CH _2 CH _2 COOH$
(c) $CH _3-\underset{CH _3}{CH}-COOH$
(d) $CH _3 CH _2 COOH$
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Answer:
Correct Answer: 4. (d)
Solution:
- $CH _3 CH _2 \underset{(A)}{\stackrel{O}{C}-OH \xrightarrow{NH _3}}$
(B)
(C)
$$ \underset{\begin{array}{l} \text { Hoffmann’s bromamide } \ \text { reaction } \end{array}}{Br _2, KOH} CH _3-CH _2-NH _2 $$
