Which statements about the Hofmann bromamide degradation of RCONH2 with Br2 + NaOH are correct?

An isocyanate R-N=C=O is an intermediate.

The group R migrates from carbon to nitrogen with retention of configuration.

The N-bromoamide is formed first and is deprotonated to give an anion.

The amine formed has the same number of carbon atoms as the amide.

An isocyanate R-N=C=O is an intermediate.

The group R migrates from carbon to nitrogen with retention of configuration.

The N-bromoamide is formed first and is deprotonated to give an anion.

JEE Advanced levelOne or more correct optionsDifficulty 3 of 3nc015

Which statements about the Hofmann bromamide degradation of $RCONH X_{2}$ with $Br X_{2} + NaOH$ are correct?

AThe amine formed has the same number of carbon atoms as the amide.
BAn isocyanate $R - N = C = O$ is an intermediate.
CThe group $R$ migrates from carbon to nitrogen with retention of configuration.
DThe N-bromoamide is formed first and is deprotonated to give an anion.

The sequence is N-bromination, deprotonation to the bromoamide anion, migration of

R

to nitrogen with loss of bromide (retention at

R

) to give

R - N = C = O

, and hydrolysis of the isocyanate to

RNH X_{2}

with loss of the carbonyl carbon as carbonate. Hence the amine has one carbon less than the amide, so the last statement is wrong.