The amide shown is treated with $Br X_{2}$ /aqueous $NaOH$ and the product is then treated with cold $NaNO X_{2}$ / $HCl$ . Select all the correct statements.

AThe migration step gives an isocyanate intermediate
BThe Hofmann product has one carbon fewer than the amide
CThe final step gives an arenediazonium salt
DThe Hofmann product is a benzylamine

Answer: A, B, C

The amide shown is 4-methylbenzamide. Hofmann rearrangement goes through an aryl isocyanate and loses one carbon, giving 4-methylaniline (p-toluidine), a primary aromatic amine, not a benzylamine. This amine gives the 4-methylbenzenediazonium salt with cold nitrous acid.

Practice Amines and Aromatic Nitrogen Compounds More on Hofmann bromamide and Gabriel synthesis on drawn substrates

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The amide shown is treated with BrX2​/aqueous NaOH and the product is then treated with cold NaNOX2​/HCl. Select all the correct statements.

Answer: A, B, C

More questions on Hofmann bromamide and Gabriel synthesis on drawn substrates

The amide shown is treated with $Br X_{2}$ /aqueous $NaOH$ and the product is then treated with cold $NaNO X_{2}$ / $HCl$ . Select all the correct statements.