Carboxylic Acids Question 62
Question: When propanamide reacts with $ Br_2 $ and $ NaOH $ then which of the following compound is formed
[Manipal 2001]
Options:
A) Ethyl alcohol
B) Propyl alcohol
C) Propyl amine
D) Ethylamine
Show Answer
Answer:
Correct Answer: D
Solution:
$ \underset{Propionamide}{\mathop{CH_3CH_2CONH_2}}\overset{Br_2/KOH}{\mathop{\xrightarrow[\begin{smallmatrix} Hofmannbromamide \\ reaction \end{smallmatrix}]{}}}\underset{Ethylamine}{\mathop{CH_3CH_2NH_2}} $
BETA
