JEE Mains Chemistry · Amines

Acylation, Nitrous Acid and Hofmann Elimination

Reactions at the amine nitrogen: acylation puts an acyl group on it, nitrous acid drives it off as nitrogen gas, and Hofmann elimination removes it as trimethylamine to leave an alkene.

All Amines notes JEE Mains Chemistry strategy

Why this matters

Fourteen PYQs, six numerical, three from 2026. Eight are acylation, most of them mass and yield problems; four test nitrous acid on primary aliphatic amines, usually through the volume of nitrogen released; one tests an amine salt and one a Hofmann elimination.

Concept 1 of 2: Acylation of amines: products, stoichiometry and yield

A primary or secondary amine attacks an acid chloride or anhydride and swaps one N–H for an acyl group. The product is an amide, one to one with the amine. Each acetyl group replaces one hydrogen, so the molar mass rises by exactly 42 per acetylation. A tertiary amine has no N–H and cannot be acylated.

Definition

Reagents: acetic anhydride or acetyl chloride, with pyridine to take up the acid and push the equilibrium forward.
Benzoylation with $\mathrm{C_6H_5COCl}$ in aqueous NaOH is the Schotten–Baumann reaction; aniline gives benzanilide, $\mathrm{C_6H_5NHCOC_6H_5}$ (M = 197).
Acetylation of aniline gives acetanilide, $\mathrm{C_6H_5NHCOCH_3}$ (M = 135). One mole of amine gives one mole of amide.
Each acetylation adds $\mathrm{C_2H_2O}$ = 42 g/mol; the number of acetylated groups = (product M − starting M)/42.
With two groups competing, the more nucleophilic one reacts first: an alkyl $\mathrm{NH_2}$ before an aryl or amide nitrogen, and $\mathrm{NH_2}$ before a phenolic OH.
The amide nitrogen is much less basic and less activating than the amine, which is why acetylation is used to protect aniline.

Acetylation of an amine and the mass it adds

\mathrm{RNH_2 + (CH_3CO)_2O \to RNHCOCH_3 + CH_3COOH} \qquad \Delta M = +42\ \mathrm{g\,mol^{-1}}\ \text{per acetyl group}

Worked example

9.0 g of ethanamine is acetylated completely with acetic anhydride. What mass of N-ethylacetamide forms?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 27 Jan 2024 · Q149Moderate

Example 1 · Amines · Acylation, Nitrous Acid and Hofmann Elimination

9.3\text{ }g

of aniline is subjected to reaction with excess of acetic anhydride to prepare acetanilide. The mass of acetanilide produced if the reaction is

100\%

completed is

\_\_\_\_

\times10^{- 1}\text{ }g

. (Given molar mass in

gmol^{- 1}\text{ }N:14,O:16,C

12,H:1)

Acylation is one to one per nitrogen

One mole of aniline gives one mole of acetanilide, whatever the excess of anhydride. The excess reagent does not add a second acetyl group to the amide nitrogen under normal conditions.

The better nucleophile is acylated first

In a molecule with an alkyl

\mathrm{NH_2}

and an amide or aryl nitrogen, one equivalent of anhydride acylates the alkyl

\mathrm{NH_2}

. In 4-aminophenol it acylates the

\mathrm{NH_2}

, not the OH.

Drill 7 more on acylation of amines: products, stoichiometry and yield

Concept 2 of 2: Amine salts, nitrous acid on primary aliphatic amines, and Hofmann elimination

Three reactions change what is on the nitrogen. An acid simply protonates it to a salt, and alkali gives the amine back. Nitrous acid turns a primary aliphatic amine into a diazonium ion so unstable that it loses nitrogen gas at once, leaving an alcohol; the gas is released mole for mole, which makes these reactions easy to count. Exhaustive methylation turns the nitrogen into a good leaving group,

\mathrm{N(CH_3)_3}

, and heating with base eliminates it to give an alkene.

Definition

Salts: $\mathrm{RNH_2 + HCl \to RNH_3^+Cl^-}$ . The salt dissolves in water and its solution is acidic; NaOH liberates the amine as a separate layer.
Nitrous acid ( $\mathrm{NaNO_2 + HCl}$ , cold) on a primary aliphatic amine gives an alkyldiazonium ion that decomposes: one mole of $\mathrm{N_2}$ per mole of amine, plus the alcohol. At STP, 1 mol of $\mathrm{N_2}$ occupies 22.4 L.
The alcohol can be identified further: propan-2-amine gives propan-2-ol, which oxidises to acetone and gives the iodoform test.
A primary aromatic amine gives a stable diazonium salt at 273–278 K instead (see the diazonium pages).
Hofmann elimination: excess $\mathrm{CH_3I}$ converts the amine to $\mathrm{R{-}N^+(CH_3)_3}$ ; moist $\mathrm{Ag_2O}$ gives the hydroxide, and heating (or a strong base such as $\mathrm{C_2H_5O^-}$ ) eliminates $\mathrm{(CH_3)_3N}$ .
The less substituted alkene is major (Hofmann rule): the bulky, positively charged leaving group makes the base take the most accessible, most acidic β-hydrogen.

Nitrous acid on a primary aliphatic amine

\mathrm{RNH_2 \xrightarrow{NaNO_2,\ HCl,\ cold} [RN_2^+Cl^-] \xrightarrow{H_2O} ROH + N_2 + HCl} \qquad n(\mathrm{N_2}) = n(\mathrm{RNH_2})

Worked example

1.18 g of propan-1-amine is treated with

\mathrm{NaNO_2}

and dilute HCl in the cold, then warmed. What volume of nitrogen is released at STP?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 4 April 2024 · Q53Moderate

Example 2 · Amines · Acylation, Nitrous Acid and Hofmann Elimination

Xg

of ethylamine is subjected to reaction with

NaNO_{2}/HCl

followed by water; evolved dinitrogen gas which occupied

2.24\text{ }L

volume at STP.

X

is _______

\times10^{- 1}\text{ }g

Aliphatic and aromatic primary amines differ with nitrous acid

A primary aliphatic amine loses

\mathrm{N_2}

at once and gives an alcohol. A primary aromatic amine at 273–278 K gives a diazonium salt that stays in solution; it loses

\mathrm{N_2}

only on warming.

Hofmann elimination is not Saytzeff elimination

Dehydrohalogenation of an alkyl halide usually gives the more substituted alkene. Elimination from a quaternary ammonium salt gives the less substituted alkene as the major product.

Drill 5 more on amine salts, nitrous acid on primary aliphatic amines, and hofmann elimination

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)

Acylation of amines: products, stoichiometry and yield
Acetylation of an amine and the mass it adds
$\mathrm{RNH_2 + (CH_3CO)_2O \to RNHCOCH_3 + CH_3COOH} \qquad \Delta M = +42\ \mathrm{g\,mol^{-1}}\ \text{per acetyl group}$
Amine salts, nitrous acid on primary aliphatic amines, and Hofmann elimination
Nitrous acid on a primary aliphatic amine
$\mathrm{RNH_2 \xrightarrow{NaNO_2,\ HCl,\ cold} [RN_2^+Cl^-] \xrightarrow{H_2O} ROH + N_2 + HCl} \qquad n(\mathrm{N_2}) = n(\mathrm{RNH_2})$

Watch out for (4)

Acylation is one to one per nitrogen
→ Acylation of amines: products, stoichiometry and yield
The better nucleophile is acylated first
→ Acylation of amines: products, stoichiometry and yield
Aliphatic and aromatic primary amines differ with nitrous acid
→ Amine salts, nitrous acid on primary aliphatic amines, and Hofmann elimination
Hofmann elimination is not Saytzeff elimination
→ Amine salts, nitrous acid on primary aliphatic amines, and Hofmann elimination

Test yourself on Amines

20 past JEE Mains questions from this chapter, timed at 48 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Take the Amines chapter test Drill all 14 questions