JEE Mains Chemistry · Amines

Preparation: Reduction, Ammonolysis and Gabriel Synthesis

Amines are made by reducing nitro compounds, nitriles and amides, by letting ammonia displace a halide, or by Gabriel synthesis when a pure primary amine is needed.

All Amines notes JEE Mains Chemistry strategy

Why this matters

Nineteen PYQs, four numerical, three from 2026. Eight ask which reagent reduces a nitro compound, nitrile or amide to an amine, and what the carbon count becomes; four test ammonolysis of alkyl halides; seven test what Gabriel synthesis can and cannot make, often by counting isomers.

Concept 1 of 3: Reduction routes to amines: nitro compounds, nitriles and amides

Three nitrogen groups can be reduced to an amine. A nitro group becomes

\mathrm{NH_2}

on the same carbon, which is how aryl amines are made. A nitrile and an amide both become

\mathrm{CH_2NH_2}

. The carbon count tells you which route was used: going through a nitrile adds one carbon to the halide you started from.

Definition

Nitro compounds: a metal in acid ( $\mathrm{Sn/HCl}$ , $\mathrm{Fe/HCl}$ , $\mathrm{Zn/HCl}$ ) or hydrogen over a catalyst (Pd, Pt, Raney Ni). Iron scrap with HCl is preferred in industry: the $\mathrm{FeCl_2}$ formed hydrolyses and releases HCl, so only a little acid is needed.
Nitriles: $\mathrm{LiAlH_4}$ , $\mathrm{H_2/Ni}$ or $\mathrm{Na(Hg)/C_2H_5OH}$ give $\mathrm{RCH_2NH_2}$ . Since $\mathrm{RCN}$ comes from $\mathrm{RX + KCN}$ , this route lengthens the chain by one carbon.
Amides: $\mathrm{LiAlH_4}$ then water gives $\mathrm{RCH_2NH_2}$ , with no loss of carbon.
$\mathrm{SnCl_2/HCl}$ on a nitrile stops at the imine, which hydrolyses to an aldehyde (Stephen reduction). It does not give an amine.

Starting compound	Reagent	Product	Carbon count
$\mathrm{C_6H_5NO_2}$	$\mathrm{Sn/HCl}$ , $\mathrm{Fe/HCl}$ or $\mathrm{H_2/Pd}$	$\mathrm{C_6H_5NH_2}$	Unchanged
$\mathrm{RC{\equiv}N}$	$\mathrm{LiAlH_4}$ , $\mathrm{H_2/Ni}$ or $\mathrm{Na(Hg)/C_2H_5OH}$	$\mathrm{RCH_2NH_2}$	One more than the halide RX
$\mathrm{RCONH_2}$	$\mathrm{LiAlH_4}$ , then $\mathrm{H_2O}$	$\mathrm{RCH_2NH_2}$	Unchanged
$\mathrm{RC{\equiv}N}$	$\mathrm{SnCl_2/HCl}$ , then $\mathrm{H_3O^+}$	$\mathrm{RCHO}$ , not an amine	Unchanged
$\mathrm{RCONH_2}$	$\mathrm{Br_2}$ and NaOH (Hofmann)	$\mathrm{RNH_2}$	One fewer

Reduction keeps every carbon; only the Hofmann route drops one.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2021 · Paper 22 · Q54Moderate

Example 1 · Amines · Preparation: Reduction, Ammonolysis and Gabriel Synthesis

The total number of reagents from those given below, that can convert nitrobenzene into aniline is . (Integer answer) I.

Sn - HCl

II.

Sn - {NH}_{4}OH

III.

Fe - HCl

IV.

Zn - HCl

H_{2} - Pd

VI.

H_{2}

- Raney Nickel

Nitrobenzene needs acid or a catalyst to reach aniline

A metal reduces nitrobenzene to aniline only in acid. In neutral solution the reduction stops at intermediate stages; for example

\mathrm{Zn/NH_4Cl}

gives N-phenylhydroxylamine.

Two amide reactions, two carbon counts

\mathrm{LiAlH_4}

reduces

\mathrm{RCONH_2}

\mathrm{RCH_2NH_2}

and keeps every carbon.

\mathrm{Br_2}

with NaOH turns the same amide into

\mathrm{RNH_2}

, one carbon shorter.

Drill 7 more on reduction routes to amines: nitro compounds, nitriles and amides

Concept 2 of 3: Ammonolysis of alkyl halides

Ammonia has a lone pair and can displace a halide from an alkyl halide by SN2. The trouble is that the amine it makes is a better nucleophile than ammonia, so it attacks more halide. The result is a mixture of primary, secondary and tertiary amines and the quaternary salt.

Definition

The halide is heated with ethanolic ammonia in a sealed tube. Each amine is formed as its salt, since the HX released protonates it.
Strong base (NaOH) liberates the free amine from its salt: $\mathrm{RNH_3^+X^- + NaOH \to RNH_2 + NaX + H_2O}$ .
A large excess of ammonia favours the primary amine; excess halide carries the reaction to the quaternary salt $\mathrm{R_4N^+X^-}$ .
Reactivity of the halide: RI > RBr > RCl. Aryl halides do not react under these conditions.
An α-halo acid with ammonia gives the amino acid in good yield: the product exists as a zwitterion, whose $\mathrm{NH_3^+}$ has no lone pair to attack further.

Successive alkylation in ammonolysis

\mathrm{NH_3 \xrightarrow{RX} RNH_2 \xrightarrow{RX} R_2NH \xrightarrow{RX} R_3N \xrightarrow{RX} R_4N^+X^-}

Worked example

1-Bromopropane is heated with ethanolic ammonia in a sealed tube and the mixture is then shaken with NaOH solution. What products can form, and how is the primary amine made the main product?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2021 · Paper 7 · Q43Moderate

Example 2 · Amines · Preparation: Reduction, Ammonolysis and Gabriel Synthesis

Which of the following reaction is an example of ammonolysis?

Ammonolysis rarely gives one amine

Each amine formed is a better nucleophile than ammonia and reacts again. Expect a mixture of 1°, 2° and 3° amines and the quaternary salt; only a large excess of ammonia makes the primary amine dominant.

Ammonolysis breaks a C–X bond

Ammonolysis means ammonia replacing a halogen on an alkyl or benzyl carbon. Acylation of an amine, reduction of a nitrile or protonation of aniline by HCl is not ammonolysis.

Drill 3 more on ammonolysis of alkyl halides

Concept 3 of 3: Gabriel phthalimide synthesis of primary amines

Phthalimide has one N–H between two carbonyl groups, which makes it acidic. Its anion attacks an alkyl halide once, and then the nitrogen has no hydrogen left to react again. Hydrolysis releases a single primary amine, with no over-alkylation. Because the key step is SN2, it only works on halides that allow SN2.

Definition

Phthalimide + ethanolic KOH gives potassium phthalimide; its anion is stabilised by both carbonyl groups.
The anion displaces halide from a primary, benzylic or allylic halide (SN2) to give an N-alkylphthalimide.
Hydrolysis with aqueous NaOH (or reaction with hydrazine) releases $\mathrm{RNH_2}$ .
It makes primary amines only, and never aryl amines: aryl halides do not undergo SN2 with the phthalimide anion.
A tertiary halide undergoes elimination instead, so a tertiary alkyl amine cannot be made this way.
Phthalimide itself comes from phthalic acid (or its anhydride) heated with ammonia.

Gabriel phthalimide synthesis

\text{phthalimide} \xrightarrow{\mathrm{KOH}} \text{K-phthalimide} \xrightarrow{\mathrm{RX}} \text{N-alkylphthalimide} \xrightarrow{\mathrm{NaOH(aq)}} \mathrm{RNH_2} + \text{phthalate}

Worked example

How many of the amines with molecular formula

\mathrm{C_7H_9N}

can be made by Gabriel phthalimide synthesis?

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2023 · 6 April 2023 · Q149Moderate

Example 3 · Amines · Preparation: Reduction, Ammonolysis and Gabriel Synthesis

Number of isomeric aromatic amines with molecular formula

C_{8}H_{11}\text{ }N

, which can be synthesized by Gabriel Phthalimide synthesis is__

Gabriel gives no aryl amines

4-Methoxyaniline, aniline or any amine with

\mathrm{NH_2}

on a ring carbon cannot be made by Gabriel synthesis, because an aryl halide does not undergo SN2.

Gabriel gives no secondary amines

After one alkylation the nitrogen of the phthalimide carries no hydrogen, so only one alkyl group can be attached. The product after hydrolysis is always a primary amine.

Drill 6 more on gabriel phthalimide synthesis of primary amines

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)

Ammonolysis of alkyl halides
Successive alkylation in ammonolysis
$\mathrm{NH_3 \xrightarrow{RX} RNH_2 \xrightarrow{RX} R_2NH \xrightarrow{RX} R_3N \xrightarrow{RX} R_4N^+X^-}$
Gabriel phthalimide synthesis of primary amines
Gabriel phthalimide synthesis
$\text{phthalimide} \xrightarrow{\mathrm{KOH}} \text{K-phthalimide} \xrightarrow{\mathrm{RX}} \text{N-alkylphthalimide} \xrightarrow{\mathrm{NaOH(aq)}} \mathrm{RNH_2} + \text{phthalate}$

Reference tables (1)

Reduction routes to amines: nitro compounds, nitriles and amides5 rows

Starting compound	Reagent	Product	Carbon count
$\mathrm{C_6H_5NO_2}$	$\mathrm{Sn/HCl}$ , $\mathrm{Fe/HCl}$ or $\mathrm{H_2/Pd}$	$\mathrm{C_6H_5NH_2}$	Unchanged
$\mathrm{RC{\equiv}N}$	$\mathrm{LiAlH_4}$ , $\mathrm{H_2/Ni}$ or $\mathrm{Na(Hg)/C_2H_5OH}$	$\mathrm{RCH_2NH_2}$	One more than the halide RX
$\mathrm{RCONH_2}$	$\mathrm{LiAlH_4}$ , then $\mathrm{H_2O}$	$\mathrm{RCH_2NH_2}$	Unchanged
$\mathrm{RC{\equiv}N}$	$\mathrm{SnCl_2/HCl}$ , then $\mathrm{H_3O^+}$	$\mathrm{RCHO}$ , not an amine	Unchanged
$\mathrm{RCONH_2}$	$\mathrm{Br_2}$ and NaOH (Hofmann)	$\mathrm{RNH_2}$	One fewer

Reduction keeps every carbon; only the Hofmann route drops one.

Watch out for (6)

Nitrobenzene needs acid or a catalyst to reach aniline
→ Reduction routes to amines: nitro compounds, nitriles and amides
Two amide reactions, two carbon counts
→ Reduction routes to amines: nitro compounds, nitriles and amides
Ammonolysis rarely gives one amine
→ Ammonolysis of alkyl halides
Ammonolysis breaks a C–X bond
→ Ammonolysis of alkyl halides
Gabriel gives no aryl amines
→ Gabriel phthalimide synthesis of primary amines
Gabriel gives no secondary amines
→ Gabriel phthalimide synthesis of primary amines

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 19 questions