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Intermediates formed during reaction of $RCONH_2$ with $Br_2$ and KOH are : -
A
RCONHBr and RNCO
B
$RNHCOBr$ and RNCO
C
$RNHBr$ and RCONHBr
D
$RCONBr_2$
Detailed Solution
This is the Hoffmann bromamide degradation, which converts an amide into a primary amine with one carbon less.
Step 1: Bromination of the amide nitrogen gives an N-bromoamide: $RCONH_2 + Br_2 + KOH \rightarrow RCONHBr + KBr + H_2O$
Step 2: The base removes the remaining N–H proton, and the R group migrates from carbon to nitrogen with loss of $Br^-$, forming an isocyanate: $RCONHBr \rightarrow R-N=C=O$
Step 3: Hydrolysis of the isocyanate gives the amine: $RNCO + 2KOH \rightarrow RNH_2 + K_2CO_3$
So the intermediates are RCONHBr (N-bromoamide) and RNCO (alkyl isocyanate).
Step 1: Bromination of the amide nitrogen gives an N-bromoamide: $RCONH_2 + Br_2 + KOH \rightarrow RCONHBr + KBr + H_2O$
Step 2: The base removes the remaining N–H proton, and the R group migrates from carbon to nitrogen with loss of $Br^-$, forming an isocyanate: $RCONHBr \rightarrow R-N=C=O$
Step 3: Hydrolysis of the isocyanate gives the amine: $RNCO + 2KOH \rightarrow RNH_2 + K_2CO_3$
So the intermediates are RCONHBr (N-bromoamide) and RNCO (alkyl isocyanate).