Question

What is the end product when ${\mathrm{CH}}_3{\mathrm{CONH}}_2$reacts with ${\mathrm{Br}}_2+\mathrm{KOH}$?

• A. ${\mathrm{CH}}_3{\mathrm{CH}}_2-{\mathrm{NH}}_2$
• B. ${\mathrm{CH}}_3\mathrm{Br}$
• C. ${\mathrm{CH}}_3\mathrm{CONHBr}$
• D. ${\mathrm{CH}}_3{\mathrm{NH}}_2$

Answer 1

The correct option is D

${\mathrm{CH}}_3{\mathrm{NH}}_2$

Explanation for the correct option:

(D)${\mathrm{CH}}_3{\mathrm{NH}}_2$

• When amide is made to react with bromine and an aqueous or alcoholic solution of potassium hydroxide or sodium hydroxide it gives primary amine. This reaction is known as Hoffman's bromamide degradation of amide.
• It is a step-down reaction where the amine formed contains one carbon atom less than the corresponding amide used in the reactant.
• When Ethanamide or acetamide$\left({\mathrm{CH}}_3{\mathrm{CONH}}_2\right)$ reacts with ${\mathrm{Br}}_2$and $\mathrm{KOH}$ it forms Methanamine $\left({\mathrm{CH}}_3{\mathrm{NH}}_2\right)$ along with Potassium carbonate, potassium bromide, and water.

$$\begin{gathered} {\mathrm{CH}}_3{\mathrm{CONH}}_2\left(\mathrm{s}\right)+{\mathrm{Br}}_2\left(\mathrm{l}\right)+4\mathrm{KOH}\left(\mathrm{alc}\right)\to {\mathrm{CH}}_3{\mathrm{NH}}_2\left(\mathrm{g}\right)+{\mathrm{K}}_2{\mathrm{CO}}_3\left(\mathrm{aq}\right)+2\mathrm{KBr}\left(\mathrm{aq}\right)+2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right) \\ \left(\mathrm{Acetamide}\right)\left(\mathrm{Bromine}\right)(\mathrm{Potassium}\mathrm{hydroxide})(\mathrm{Methyl}\mathrm{amine})(\mathrm{Potassium}\mathrm{carbonate})(\mathrm{Potassium}\mathrm{bromide})\left(\mathrm{water}\right) \end{gathered}$$

• Thus, option (D) is the correct option: The end product when ${\mathrm{CH}}_3{\mathrm{CONH}}_2$reacted with ${\mathrm{Br}}_2+\mathrm{KOH}$ is the Methyl amine$\left({\mathrm{CH}}_3{\mathrm{NH}}_2\right)$.

Hence, the correct option is (D) Methyl amine$\left({\mathrm{CH}}_3{\mathrm{NH}}_2\right)$