Animated Solution for Chemistry - Organic Chemistry: A compound 'X' on treatment with ${\text{Br}}_2/\text{NaOH}$, provided ${\text{C}}_3{\text{H}}_9\text{N}$, which gives positive carbylamine test. Compound 'X' is

Imagine you are a chemical detective, and you are handed an unknown compound 'X'. You are told that it undergoes two very famous organic reactions. First, it reacts with bromine and sodium hydroxide (${\text{Br}}_2/\text{NaOH}$) to form a new compound with the molecular formula ${\text{C}}_3{\text{H}}_9\text{N}$. Then, this new compound is subjected to the Carbylamine test, and it gives a positive result. Our mission is to work backward and identify the original compound 'X'.

Let's start with the final clue: the positive Carbylamine test. This test is highly specific. When an amine is heated with chloroform (${\text{CHCl}}_3$) and ethanolic potassium hydroxide ($\text{KOH}$), only primary ($1^{\circ }$) amines will react to form an isocyanide ($\text{R}-\text{NC}$), which is notorious for its incredibly foul smell.

Since our intermediate compound ${\text{C}}_3{\text{H}}_9\text{N}$ gives a positive test, it must be a primary amine. Let's think about the possible structures for a primary amine with three carbon atoms. There are exactly two isomers: 1. n-propylamine: ${\text{CH}}_3{\text{CH}}_2{\text{CH}}_2{\text{NH}}_2$ 2. isopropylamine: ${\text{CH}}_3\text{CH}({\text{NH}}_2){\text{CH}}_3$

Secondary and tertiary amines will not give this test, so we can confidently rule out any options that would lead to them.

Now, let's step back to the first reaction. Compound 'X' reacts with ${\text{Br}}_2$ and $\text{NaOH}$ to produce our primary amine. This specific combination of reagents is the hallmark of the Hoffmann Bromamide Degradation reaction.

This reaction is a classic tool in organic synthesis used to step down a carbon chain. It converts a primary amide ($\text{R}-{\text{CONH}}_2$) into a primary amine ($\text{R}-{\text{NH}}_2$). The crucial detail here is the word degradation—the carbonyl carbon of the amide is lost as sodium carbonate (${\text{Na}}_2{\text{CO}}_3$).

If the resulting amine (${\text{C}}_3{\text{H}}_9\text{N}$) has exactly 3 carbon atoms, the original amide 'X' must have had $3+1=4$ carbon atoms.

Let's evaluate the given options: - (a) ${\text{CH}}_3{\text{COCH}}_2{\text{NHCH}}_3$: This is a ketone with a secondary amine group, not a primary amide. - (b) ${\text{CH}}_3{\text{CH}}_2{\text{CH}}_2{\text{CONH}}_2$: This is butanamide, a primary amide with exactly 4 carbon atoms. This perfectly fits our deduction! - (c) ${\text{CH}}_3{\text{CON(CH}}_3{)}_2$: This is a tertiary amide. - (d) ${\text{CH}}_3{\text{CH}}_2{\text{COCH}}_2{\text{NH}}_2$: This is a primary amine, not an amide.

Therefore, the unknown compound 'X' is butanamide. The elegance of organic chemistry lies in how these specific tests and reagents act as puzzle pieces, guiding us straight to the answer.