JEE Mains Chemistry · Amines

Hofmann Bromamide Degradation

A primary amide with bromine and alkali loses its carbonyl carbon as carbonate and gives a primary amine one carbon shorter, through an isocyanate.

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Why this matters

Seventeen PYQs, one numerical, six from 2026. Six test the balanced equation, the isocyanate intermediate, the migrating group and which amides react; eleven hide the reaction inside a multistep sequence, often after a Grignard carboxylation, and ask for a structure or a mass.

Concept 1 of 2: Hofmann bromamide degradation: equation, intermediates and scope

Bromine in alkali puts a bromine on the amide nitrogen. Base then removes the second N–H, bromide leaves, and the nitrogen is left with only six electrons. The group on the carbonyl carbon moves across to it, giving an isocyanate, R–N=C=O. Water adds to the isocyanate and carbon dioxide leaves, so the amine has one carbon fewer than the amide.

Definition

Overall: one $\mathrm{Br_2}$ and four NaOH per mole of amide; by-products $\mathrm{Na_2CO_3}$ , NaBr and water.
Steps: $\mathrm{RCONH_2 \to RCONHBr}$ (N-bromoamide) $\to$ its anion $\to$ R migrates as $\mathrm{Br^-}$ leaves $\to$ $\mathrm{R{-}N{=}C{=}O}$ (isocyanate) $\to$ $\mathrm{RNH_2 + CO_2}$ , and $\mathrm{CO_2}$ ends as carbonate in the alkali.
The migrating group can be alkyl or aryl. It moves to an electron-deficient nitrogen and keeps its configuration.
Only an unsubstituted (primary) amide $\mathrm{RCONH_2}$ reacts: it needs two N–H hydrogens. An N-substituted amide $\mathrm{RCONHR'}$ does not give an amine.
The product is always a primary amine.

Hofmann bromamide degradation

\mathrm{RCONH_2 + Br_2 + 4NaOH \to RNH_2 + Na_2CO_3 + 2NaBr + 2H_2O}

Worked example

3-Phenylpropanamide,

\mathrm{C_6H_5CH_2CH_2CONH_2}

, is warmed with bromine and aqueous NaOH. Name the product and say where the carbonyl carbon goes.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 8 Apr 2026 Shift 2 · Q43Moderate

Example 1 · Amines · Hofmann Bromamide Degradation

Which statements are True? (A) In Hoffmann bromamide degradation, 4 moles of NaOH and 2 moles of

{Br}_{2}

are consumed per mole of an amide (B) Hoffmann bromamide reaction is not given by alkyl amides. (C) Primary amines can be synthesized by Hoffmann bromamide degradation. (D) Secondary amide on reaction with

{Br}_{2}

and NaOH will give secondary amine. (E) The by-products of Hoffmann degradation are

{Na}_{2}{CO}_{3},NaBr

and

H_{2}O

. Choose the correct answer from the options given below :

Benzamide gives aniline, not benzylamine

Hofmann degradation removes the carbonyl carbon.

\mathrm{C_6H_5CONH_2}

gives

\mathrm{C_6H_5NH_2}

. Benzylamine would come from reducing benzamide with

\mathrm{LiAlH_4}

Aryl groups migrate too

The group that moves from carbon to nitrogen can be alkyl or aryl. A statement that only an alkyl group migrates is false; benzamide reacts perfectly well.

One bromine, four hydroxides

The balanced equation uses one

\mathrm{Br_2}

and four NaOH per amide. Two hydroxides neutralise the two HBr equivalents and two more trap

\mathrm{CO_2}

as carbonate.

Drill 5 more on hofmann bromamide degradation: equation, intermediates and scope

Concept 2 of 2: Hofmann degradation in multistep sequences

Most questions bury the Hofmann step inside a chain. The usual chain goes halide, Grignard, carboxylic acid, amide, amine. The Grignard step adds one carbon and the Hofmann step removes one, so the amine ends up where the halogen started, with the same number of carbons. On a benzene ring the

\mathrm{NH_2}

appears exactly where the

\mathrm{CONH_2}

was.

Definition

Acid to amide: heat with $\mathrm{NH_3}$ (via the ammonium salt), or $\mathrm{SOCl_2}$ then $\mathrm{NH_3}$ .
Halide to acid: Mg in dry ether, then $\mathrm{CO_2}$ , then $\mathrm{H_3O^+}$ . This adds one carbon.
Hofmann removes one carbon. So $\mathrm{RX \to RCOOH \to RCONH_2 \to RNH_2}$ replaces X by $\mathrm{NH_2}$ with no net change in carbons.
Ring position is kept: 3-chlorobenzamide gives 3-chloroaniline.
The amine can then be taken on: $\mathrm{CHCl_3/KOH}$ gives the isocyanide; $\mathrm{HNO_2}$ gives an alcohol (aliphatic) or a diazonium salt (aryl).
Acid hydrolysis of an isocyanide gives the amine back, with formic acid: $\mathrm{RNC + 2H_2O \xrightarrow{H^+} RNH_2 + HCOOH}$ . A nitrile instead gives $\mathrm{RCOOH}$ .

Halide to amine with the same carbon count

\mathrm{RX \xrightarrow{Mg,\ ether} RMgX \xrightarrow{CO_2,\ H_3O^+} RCOOH \xrightarrow{NH_3,\ \Delta} RCONH_2 \xrightarrow{Br_2,\ NaOH} RNH_2}

Worked example

1-Bromobutane is treated with (i) Mg in dry ether, (ii)

\mathrm{CO_2}

then

\mathrm{H_3O^+}

, (iii)

\mathrm{NH_3}

with heat, (iv)

\mathrm{Br_2}

and NaOH. Identify each product.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 30 January 2024 · Q42Moderate

Example 2 · Amines · Hofmann Bromamide Degradation

The final product A, formed in the following multistep reaction sequence is:

The Grignard route does not lengthen the amine

\mathrm{CO_2}

adds one carbon, but the Hofmann step removes it again. Starting from bromobenzene you end at aniline, not benzylamine.

Isocyanides and nitriles hydrolyse differently

Acid hydrolysis of

\mathrm{R{-}NC}

gives

\mathrm{RNH_2}

and formic acid; acid hydrolysis of

\mathrm{R{-}CN}

gives

\mathrm{RCOOH}

. The atom bonded to R decides.

Drill 10 more on hofmann degradation in multistep sequences

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Hofmann bromamide degradation: equation, intermediates and scope
Hofmann bromamide degradation
$\mathrm{RCONH_2 + Br_2 + 4NaOH \to RNH_2 + Na_2CO_3 + 2NaBr + 2H_2O}$
Hofmann degradation in multistep sequences
Halide to amine with the same carbon count
$\mathrm{RX \xrightarrow{Mg,\ ether} RMgX \xrightarrow{CO_2,\ H_3O^+} RCOOH \xrightarrow{NH_3,\ \Delta} RCONH_2 \xrightarrow{Br_2,\ NaOH} RNH_2}$

Watch out for (5)

Benzamide gives aniline, not benzylamine
→ Hofmann bromamide degradation: equation, intermediates and scope
Aryl groups migrate too
→ Hofmann bromamide degradation: equation, intermediates and scope
One bromine, four hydroxides
→ Hofmann bromamide degradation: equation, intermediates and scope
The Grignard route does not lengthen the amine
→ Hofmann degradation in multistep sequences
Isocyanides and nitriles hydrolyse differently
→ Hofmann degradation in multistep sequences

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