Which of the following is true regarding Hoffmann bromamide reaction?

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

It is an intramolecular rearrangement involving cyclic T.S (transition state)

Aromatic amides containing deactivating groups are more reactive than aromatic amides containing activating groups

Reaction occurs through formation of alkyl isocyanate Intermediate

Reaction occurs through formation of alkyl isocyanide Intermediate

answer is A, B.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

In Hoffmann bromamide reaction, bromamide anion goes through rearrangement in a way that the ethyl group (or any R- group) which is bonded with the carbonyl carbon now bonds with the nitrogen instead. Simultaneously, the bromide anion formed leaves the compound. This leads to the formation of an isocyanate.

Watch 3-min video & get full concept clarity

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Online Course for IIT JEE

Best Online Course for NEET

Best Online Course for CBSE