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Predict the product:
A
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B
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C
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D
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Detailed Solution
$NaNO_2$ and HCl generate nitrous acid, $HNO_2$, in situ.
N-methylaniline is a secondary amine (the nitrogen carries one H).
Secondary amines, aliphatic or aromatic, react with nitrous acid to give N-nitrosoamines, which separate as yellow oily liquids: $R_2NH + HONO \rightarrow R_2N-N=O + H_2O$
$C_6H_5NHCH_3 + HONO \rightarrow C_6H_5-N(CH_3)-N=O + H_2O$
Ring nitrosation (at the para position) is characteristic of tertiary aromatic amines, not secondary ones.
Hence the product is N-nitroso-N-methylaniline, $C_6H_5-N(CH_3)-N=O$.
N-methylaniline is a secondary amine (the nitrogen carries one H).
Secondary amines, aliphatic or aromatic, react with nitrous acid to give N-nitrosoamines, which separate as yellow oily liquids: $R_2NH + HONO \rightarrow R_2N-N=O + H_2O$
$C_6H_5NHCH_3 + HONO \rightarrow C_6H_5-N(CH_3)-N=O + H_2O$
Ring nitrosation (at the para position) is characteristic of tertiary aromatic amines, not secondary ones.
Hence the product is N-nitroso-N-methylaniline, $C_6H_5-N(CH_3)-N=O$.