Question

When propanamide reacts with $B{{r}_{2}}$ and $NaOH$ then which of the following compounds is formed?
1) Ethyl alcohol
2) Propyl alcohol
3) Propylamine
4) Ethylamine

Answer 1

Hint: The answer is based on the balanced chemical reaction to be written when propanamide reacts with bromine and sodium hydroxide and when this reaction takes place the product obtained is amine.

Complete Solution :
We have come across the chapters in organic chemistry which deals with the basic concepts of reaction of halogens and bases like sodium hydroxide that give sodium halides and also about the named reactions.
Now, we shall see what happens when propanamide is treated with bromine and sodium hydroxide.
- The reaction when amide is treated with halogen in the presence of base is called the Hofmann degradation reaction.
- This reaction in general can be said as when the amide is treated with halogen in the presence of basic media then the product formed is primary amine with release of carbon dioxide.
Thus, the general reaction can be written as follows,
\[R-CON{{H}_{2}}\xrightarrow[NaOH]{{{X}_{2}}}R-N{{H}_{2}}+C{{O}_{2}}\]

Therefore, when propanamide reacts with bromine in the presence of sodium hydroxide, it undergoes Hofmann degradation reaction to yield ethanamine or ethylamine which is the primary amine. The reaction taking place can therefore be depicted as shown below:
\[C{{H}_{3}}C{{H}_{2}}CON{{H}_{2}}\xrightarrow[NaOH]{B{{r}_{2}}}C{{H}_{3}}C{{H}_{2}}N{{H}_{2}}+C{{O}_{2}}\]
So, the correct answer is “Option 4”.

Note: Note that in the Hofmann degradation reaction the product forms will have one carbon atom lesser as this carbon atom is released out in the form of carbon dioxide gas and therefore, while writing the product be careful about this fact.