Hoffmann Bromamide Degradation Reaction: Amide to Amine (One Carbon Lost)
Chemistry · Amines · NEET
Hoffmann bromamide degradation changes a primary amide (R-CONH2) into a primary amine (R-NH2) using bromine and a strong base (Br2 + NaOH or KOH). The new amine has ONE carbon less than the amide you started with, because the C=O carbon is lost. Memory hook: "Amide loses its carbonyl carbon, amine walks out shorter by one C."
Hoffmann bromamide degradation: Br2 + NaOH turns a primary amide into a primary amine through an isocyanate intermediate. The carbonyl carbon is lost, so the amine has one carbon less than the amide.
Your doubts, answered
What is the product of the Hoffmann bromamide degradation?
You always get a PRIMARY amine (R-NH2). Start from a primary amide R-CONH2, add Br2 with NaOH or KOH, and the product is R-NH2. Important: the amine has one fewer carbon than the amide. Example: CH3CONH2 (2 carbons) gives CH3NH2 (1 carbon).
Why does one carbon get lost in this reaction?
The carbon that is lost is the carbonyl carbon (the C of C=O in the amide). During the reaction this carbon leaves as carbonate/CO after passing through an isocyanate (R-N=C=O) intermediate. So the nitrogen stays but its old carbonyl neighbour is gone, making the amine shorter by one carbon. This is the single most tested point in NEET.
What reagent is used in the Hoffmann bromamide reaction?
Bromine (Br2) plus a strong aqueous base: NaOH or KOH. In NEET you will see it written as Br2/NaOH, Br2/KOH, or 'Br2 + aqueous NaOH'. All three mean the same reaction. It is also called the Hoffmann hypobromite or hypobromamide reaction because Br2 + base first makes hypobromite (BrO-).
Is Hoffmann bromamide the same as Hoffmann elimination?
No. This is a common trap. Hoffmann BROMAMIDE degradation = amide to amine (uses Br2 + base, loses one carbon). Hoffmann ELIMINATION = a quaternary ammonium hydroxide is heated to give the least-substituted alkene (uses heat, no Br2). Same name 'Hoffmann', totally different reactions. For amides, think bromamide.
Does it work on secondary or tertiary amides?
No. It only works on PRIMARY amides, which have the -CONH2 group with two N-H bonds. Secondary (R-CONHR') and tertiary amides do not have the free N-H needed to make the N-Br intermediate, so the reaction fails. If NEET gives you R-CONHR', do not choose Hoffmann bromamide.
How do I convert acetamide to methanamine?
Use Br2/KOH (Hoffmann bromamide). Acetamide CH3CONH2 has 2 carbons; methanamine CH3NH2 has 1 carbon. Because Hoffmann bromamide removes exactly one carbon (the carbonyl C), it is the correct method. This exact conversion was asked in NEET 2017.
⚠️ The NEET trap ✗ CH3CONH2 gives CH3CH2NH2 (ethanamine) — assuming the carbon count stays the same. ✓ CH3CONH2 gives CH3NH2 (methanamine) — one carbon LESS, because the carbonyl carbon is lost. 🧠 Hoffmann bromamide ALWAYS shrinks the chain by one carbon. If the option keeps the same number of carbons (like LiAlH4 reduction does), it is wrong for this reaction.
Real NEET questions
NEET 2017
Which of the following reactions is appropriate for converting acetamide (CH3CONH2) to methanamine (CH3NH2)?
A · Carbylamine reaction
B · Hoffmann hypobromamide (bromamide) reaction ✓
C · Stephen's reaction
D · Gabriel's phthalimide synthesis
Solution: Acetamide has 2 carbons; methanamine has only 1 carbon. So the correct method must remove one carbon. Hoffmann bromamide degradation (Br2/KOH) turns a primary amide into a primary amine with the loss of one carbon: CH3CONH2 to CH3NH2. Carbylamine is only a TEST for primary amines, Stephen's reaction reduces nitriles to aldehydes, and Gabriel synthesis makes amines from alkyl halides (not from amides). Answer: (B).
NEET 2023
Which of the following reactions will NOT give a primary amine as the product?
A · CH3CONH2 (Br2/KOH) to Product
B · CH3CN (i) LiAlH4 (ii) H3O+ to Product
C · CH3NC (i) LiAlH4 (ii) H3O+ to Product ✓
D · CH3CONH2 (i) LiAlH4 (ii) H3O+ to Product
Solution: Reduction of an isocyanide CH3NC gives a SECONDARY amine (CH3NHCH3), because the nitrogen ends up bonded to two carbons. So (C) does NOT give a primary amine. The others all give primary amines: (A) Hoffmann bromamide CH3CONH2 to CH3NH2, (B) nitrile reduction CH3CN to CH3CH2NH2, (D) amide reduction CH3CONH2 to CH3CH2NH2. Note how Hoffmann bromamide (A) loses a carbon while LiAlH4 on the same amide (D) keeps all carbons. Answer: (C).
NEET 2024
Identify the major product C: CH3CH2CH2I (NaCN) to A (OH-/NaOH, partial hydrolysis) to B (Br2/NaOH) to C
A · butylamine
B · butanamide
C · alpha-bromobutanoic acid
D · propylamine ✓
Solution: Step 1: n-propyl iodide + NaCN gives butanenitrile CH3CH2CH2CN (A, 4 carbons). Step 2: partial alkaline hydrolysis stops at the amide, butanamide CH3CH2CH2CONH2 (B). Step 3: Br2/NaOH is the Hoffmann bromamide degradation, which removes one carbon to give a primary amine: CH3CH2CH2NH2, which is propylamine (3 carbons). Butylamine (A) would wrongly keep 4 carbons. Because Hoffmann bromamide always drops the carbonyl carbon, the answer is propylamine (D).
Solved Amines NEET PYQs
Try the real previous-year questions from this chapter — each with the answer and a full solution.
What type of amine does Hoffmann bromamide degradation always give?
It always gives a primary amine (R-NH2). It can never make a secondary or tertiary amine because the nitrogen keeps its two hydrogens as N-H bonds in the final amine.
What is the intermediate in the Hoffmann bromamide reaction?
The key intermediate is an isocyanate (R-N=C=O). It forms after an N-bromoamide and a nitrene-like rearrangement, then hydrolyses in base to give the amine and carbonate. NEET usually only tests that a carbon is lost, so remembering 'isocyanate intermediate, carbon leaves as carbonate' is enough.
Does the amine keep the same number of carbons as the amide?
No. The amine has exactly one fewer carbon than the amide, because the carbonyl carbon of the amide leaves during the reaction. This carbon loss is the favourite NEET trap.
Can I use Br2 with NaOH or must it be KOH?
Both NaOH and KOH work. NEET writes it as Br2/NaOH or Br2/KOH interchangeably. Any strong aqueous base plus bromine drives the same degradation.
Why is this reaction useful for making primary amines?
Unlike ammonolysis of alkyl halides (which gives a messy mixture of primary, secondary and tertiary amines), Hoffmann bromamide gives a clean, pure primary amine. That is why it is grouped with the reliable primary-amine methods for NEET.
