Question

An organic compound [X] (molecular formula \(C_5H_{11}NO\)) when reacted with \(Br_2/\text{aq. NaOH}\) yielded [Y] which reacts with \(CHCl_3\) and ethanolic KOH to produce a foul smelling compound. Compound [Y] also reacts with HONO to produce Butan-1-ol with liberation of \(N_2(g)\). Identify [X].

• A. \(CH_3-(CH_2)_3-CONH_2\)
• B. \(CH_3-(CH_2)_2-CO-CH_2-NH_2\)
• C. \(CH_3-CH_2-CO-(CH_2)_2-NH_2\)
• D. \(CH_3-(CH_2)_2-CH(NH_2)-CHO\)

Hint:

Hofmann bromamide degradation converts a primary amide into a primary amine with one carbon atom less. Primary amines give carbylamine test and aliphatic primary amines form alcohols with HONO.

Answer 1

The Correct Option is A

Step 1: Identify the reaction with \(Br_2/\text{aq. NaOH}\).
The reagent \(Br_2/\text{aq. NaOH}\) gives Hofmann bromamide degradation reaction.
In this reaction, a primary amide converts into a primary amine with one carbon atom less.
\[ RCONH_2 \xrightarrow{Br_2/NaOH} RNH_2 \]

Step 2: Use molecular formula of [X].
The compound [X] has molecular formula:
\[ C_5H_{11}NO. \]
A primary amide with this formula can be:
\[ CH_3-(CH_2)_3-CONH_2. \]
This is pentanamide.

Step 3: Find product [Y] from Hofmann bromamide reaction.
If [X] is pentanamide:
\[ CH_3-(CH_2)_3-CONH_2 \xrightarrow{Br_2/NaOH} CH_3-(CH_2)_3-NH_2. \]
So, [Y] is butan-1-amine.
\[ Y = CH_3CH_2CH_2CH_2NH_2. \]

Step 4: Confirm with carbylamine test.
[Y] reacts with \(CHCl_3\) and ethanolic KOH to produce a foul smelling compound.
This is the carbylamine reaction, given only by primary amines.
\[ RNH_2 + CHCl_3 + 3KOH \rightarrow RNC + 3KCl + 3H_2O. \]
The foul smelling compound is an isocyanide.
Thus, [Y] must be a primary amine.

Step 5: Confirm with HONO reaction.
Primary aliphatic amines react with nitrous acid to form alcohol with liberation of nitrogen gas.
\[ RNH_2 + HONO \rightarrow ROH + N_2 + H_2O. \]
Since [Y] gives Butan-1-ol, [Y] must be butan-1-amine.
\[ CH_3CH_2CH_2CH_2NH_2 \xrightarrow{HONO} CH_3CH_2CH_2CH_2OH + N_2 + H_2O. \]

Step 6: Trace back [X].
Since Hofmann bromamide degradation decreases carbon chain by one, [X] must be the amide containing one more carbon than butan-1-amine.
Therefore, [X] is pentanamide:
\[ CH_3-(CH_2)_3-CONH_2. \]

Step 7: Final conclusion.
The compound [X] is:
\[ \boxed{CH_3-(CH_2)_3-CONH_2} \]