An optically active compound [X] (Molecular formula \( C_8H_{11}N \)) reacts with \( CHCl_3/ \) Ethanolic KOH on heating to form an isocyanide. On reaction with \( NaNO_2/HCl \), it yields an alcohol with liberation of \( N_2 \). Identify the compound [X].
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Carbylamine test confirms primary amine, while reaction with nitrous acid giving alcohol and \( N_2 \) indicates primary aliphatic amineCheck optical activity by identifying a chiral carbon.
Step 1: Use the carbylamine test.
The compound reacts with \( CHCl_3 \) and ethanolic KOH on heating to form an isocyanide.
This is the carbylamine reaction, which is given only by primary amines.
Therefore, compound [X] must be a primary amine. Step 2: Use reaction with nitrous acid.
The reagent \( NaNO_2/HCl \) generates nitrous acid \( HNO_2 \) in situ.
Primary aliphatic amines react with nitrous acid to give alcohols with liberation of nitrogen gas.
The reaction is:
\[
RNH_2 + HNO_2 \rightarrow ROH + N_2 + H_2O
\] Step 3: Check optical activity.
The compound [X] is optically active, so it must contain a chiral carbon atom.
\( 1\text{-Phenylethanamine} \) has the structure:
\[
C_6H_5-CH(NH_2)-CH_3
\]
The carbon attached to \( C_6H_5 \), \( NH_2 \), \( CH_3 \), and \( H \) has four different groups, so it is chiral and optically active. Step 4: Verify molecular formula.
For \( 1\text{-Phenylethanamine} \):
\[
C_6H_5-CH(NH_2)-CH_3
\]
The molecular formula is:
\[
C_8H_{11}N
\]
This matches the given molecular formula. Step 5: Check other options.
2-Ethylaniline, 3,4-dimethylaniline, and 2,4-dimethylaniline are aromatic amines but they do not have a chiral carbon atom.
Also, aromatic primary amines with \( NaNO_2/HCl \) form diazonium salts rather than directly giving alcohol with liberation of \( N_2 \) under the same condition. Step 6: Conclusion.
Thus, the compound [X] is \( 1\text{-Phenylethanamine} \).
Therefore:
\[
\boxed{1\text{-Phenylethanamine}}
\]
