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Chemistry · Amines

Reactions of Amines

The chemistry of amines is governed by one feature: the unshared electron pair on nitrogen, which makes amines both basic and nucleophilic. NCERT Class XII Unit 9 (§9.6) and NIOS Chapter 28 (§28.2.5) organise the reactions of amines around how many N–H hydrogens are present and, for aniline, around the powerful effect of the –NH2 group on the ring. This page works through acylation, the diagnostic reaction with nitrous acid, and electrophilic substitution in aniline — a cluster that NEET returns to almost every year through statement-matching and product-identification questions.

Why the Lone Pair Sets the Agenda

Every amine carries a trivalent, $sp^3$-hybridised nitrogen with one unshared electron pair occupying the fourth orbital. NCERT (§9.6) makes the point that two things follow directly from this: the electronegativity difference between nitrogen and hydrogen, and the availability of that lone pair, make amines reactive. The lone pair lets the amine behave as a Lewis base towards protons and as a nucleophile towards electron-poor carbon, such as the carbonyl carbon of an acid chloride.

The second organising idea is the number of N–H hydrogens. A primary amine has two, a secondary amine one, and a tertiary amine none. Because several reactions of amines proceed by replacing an N–H hydrogen, this count decides whether a given amine reacts at all, and what it gives. This is exactly why primary, secondary and tertiary amines diverge so sharply in acylation and in their behaviour with nitrous acid — and why those divergences make such reliable diagnostic tests.

The basic character of amines — the reaction with mineral acids to form ammonium salts — is treated in detail on the dedicated basicity of amines page. Here the focus is on the substitution and electrophilic reactions that NEET tests as products and sequences.

Acylation: Amines to Amides

In acylation, an N–H hydrogen of the amine is replaced by an acyl group ($\ce{R-C(=O)-}$), giving an amide. NCERT (§9.6) states that aliphatic and aromatic primary and secondary amines react with acid chlorides, anhydrides and esters by nucleophilic substitution. Tertiary amines, having no N–H to replace, do not give amides this way.

A primary amine reacting with an acid chloride gives an N-substituted amide and hydrogen chloride. NIOS (§28.2.5) writes the general reaction as:

$$\ce{R-NH2 + R'COCl -> R-NH-CO-R' + HCl}$$

With an acid anhydride the second acyl-bearing fragment leaves as a carboxylic acid. The classic example, given in both sources, is aniline acetylated by acetic anhydride (or acetyl chloride) to give acetanilide:

$$\ce{C6H5NH2 + (CH3CO)2O -> C6H5NHCOCH3 + CH3COOH}$$

Methanamine reacting with benzoyl chloride is an example of benzoylation, giving N-methylbenzamide:

$$\ce{CH3NH2 + C6H5COCl -> C6H5CONHCH3 + HCl}$$

NEET Trap

Acylation needs a base — and an N–H

NCERT specifies that acylation is carried out in the presence of a base stronger than the amine, such as pyridine. Pyridine removes the HCl liberated, shifting the equilibrium to the product side. Two common slips: assuming tertiary amines acylate (they cannot — no N–H), and forgetting that amines with carboxylic acids simply form salts at room temperature rather than amides.

Acylation: 1° and 2° amines only → amide. 3° amine → no amide. Acid + amine → salt, not amide.

Acylation is not only a transformation in its own right; it is also a protecting strategy. Converting the highly activating –NH2 of aniline into the milder –NHCOCH3 (acetanilide) lets chemists control electrophilic substitution on the ring — a point developed in the aniline section below.

Reaction with Nitrous Acid

Nitrous acid ($\ce{HNO2}$) is unstable and is generated in situ from a mineral acid and sodium nitrite ($\ce{NaNO2 + HCl -> HNO2 + NaCl}$). NCERT (§9.6) stresses that the three classes of amine react differently with it — which is precisely what makes this reaction a workhorse for distinguishing amine types and a recurring NEET theme.

Primary aliphatic amines → alcohol + N2

A primary aliphatic amine first forms an aliphatic (alkyl) diazonium salt. This ion is highly unstable: it loses nitrogen gas quantitatively and ends up, after reaction with water, as an alcohol. The quantitative evolution of nitrogen is used in the estimation of amino acids and proteins (NCERT §9.6).

$$\ce{R-NH2 + HNO2 -> [R-N2+ Cl-] ->[H2O] R-OH + N2 ^ + HCl}$$

For ethanamine, the product is ethanol (NIOS §28.2.5):

$$\ce{CH3CH2NH2 ->[NaNO2/HCl] [CH3CH2N2+ Cl-] ->[H2O] CH3CH2OH + N2 ^ + HCl}$$

Primary aromatic amines → stable diazonium salt

A primary aromatic amine such as aniline reacts with nitrous acid at 273–278 K to form a diazonium salt that is stable for a short time in cold solution. This conversion is called diazotisation, and the arenediazonium ion is stabilised by resonance with the benzene ring — the reason it survives where the alkyl analogue does not.

$$\ce{C6H5NH2 + NaNO2 + 2HCl ->[273-278\ K] C6H5N2+ Cl- + NaCl + 2H2O}$$

Secondary and tertiary amines

Secondary amines react with nitrous acid in a different manner — they give an N-nitrosamine, a yellow oily liquid. Tertiary amines do not give a diazonium product; with the mineral acid present they simply form a soluble ammonium salt. NCERT notes only that secondary and tertiary amines "react in a different manner", so the safe NEET takeaway is the contrast in the table below.

Figure 1 Amine + HNO₂ 1° aliphatic unstable salt R–OH + N₂↑ 1° aromatic 273–278 K Ar–N₂⁺ (stable) 2° amine N-nitrosamine 3° amine no diazonium (salt only) Only the aryldiazonium ion is stable enough to use Resonance with the ring is what saves it; the alkyl analogue collapses to a carbocation

Figure 1 — How the four amine types diverge with nitrous acid. The contrast between the unstable alkyldiazonium ion (1° aliphatic) and the resonance-stabilised arenediazonium ion (1° aromatic) is the single most-tested point of this reaction.

Goes Deeper

The arenediazonium salt formed here is the gateway to aryl halides, phenols, cyanides and azo dyes. Follow the full set of substitution and coupling reactions on diazonium salts.

Reaction with Grignard Reagents

Primary and secondary amines have an acidic N–H bond. A Grignard reagent ($\ce{R-MgX}$) is a powerful base, so when it meets an amine it simply abstracts that N–H proton instead of adding to anything: the alkyl group of the Grignard reagent leaves as an alkane, and the amine is converted to a magnesium amide salt.

$$\ce{R'-NH2 + R-MgX -> R-H + R'-NH-MgX}$$

This acid–base quenching is the practical reason Grignard reagents must be prepared and used under scrupulously dry, amine-free and water-free conditions: any species with an active hydrogen — water, alcohols, or amines — destroys the reagent. Tertiary amines, lacking an N–H, do not protonate a Grignard reagent and are often tolerated as solvents or bases in such reactions.

Electrophilic Substitution in Aniline

In aniline the –NH2 group is bonded directly to the ring, so the nitrogen lone pair is in conjugation with the benzene π-system. NCERT (§9.6) shows aniline as a resonance hybrid of five structures in which negative charge builds up at the ortho and para positions. The consequence is decisive: –NH2 is a powerful activating group and is ortho/para directing. The practical difficulty is that this activation is so strong that substitution is hard to stop at one position.

Bromination

Aniline reacts with bromine water at room temperature to give a white precipitate of 2,4,6-tribromoaniline — all three activated positions are substituted at once:

$$\ce{C6H5NH2 + 3Br2 (aq) -> C6H2(Br)3NH2 (v) + 3HBr}$$

To obtain a monobromo product the activation of –NH2 must be moderated. NCERT prescribes acetylation with acetic anhydride first: in acetanilide the nitrogen lone pair is partly delocalised onto the acetyl oxygen, so –NHCOCH3 is a weaker activator than –NH2. Bromination then stops at the para position, and acid hydrolysis regenerates the amine as p-bromoaniline.

Figure 2 Aniline –NH₂ (strong) (CH₃CO)₂O Acetanilide –NHCOCH₃ (mild) Br₂ p-Bromoacetanilide mono, para H₃O⁺ p-Bromo- aniline CONTROLLED (acetyl protection) ↑ Aniline free –NH₂ Br₂ water, RT 2,4,6-Tribromoaniline white ppt, no control UNCONTROLLED — all three o/p sites substitute at once ↓

Figure 2 — Acetyl protection. Acetanilide's milder –NHCOCH₃ group allows clean mono-bromination at para, then hydrolysis recovers the amine. Without protection, free aniline gives the tri-substituted product.

Nitration

Direct nitration of aniline is messy. The strongly acidic nitrating mixture protonates aniline to the anilinium ion; the positively charged –NH3+ group is electron withdrawing and meta directing. As a result, alongside the ortho and para nitroanilines a substantial amount of the meta product appears, and the free amine is also lost to tarry oxidation. The clean route again uses acetylation: nitration of acetanilide gives mainly the para isomer, which is hydrolysed to p-nitroaniline.

NEET Trap

"o/p-directing" vs the meta product of nitration

The –NH2 group is genuinely ortho/para directing. The meta product appears only because, in strong acid, the species actually present is the anilinium ion, whose –NH3+ group is meta directing. The directing power belongs to the protonated form, not to neutral aniline. This exact distinction is the basis of NEET 2018 Q.47.

Neutral aniline → o/p directing. Anilinium ion (strong acid) → m directing.

Sulphonation

Aniline reacts with concentrated sulphuric acid to form anilinium hydrogensulphate, which on heating at 453–473 K rearranges to p-aminobenzenesulphonic acid, commonly called sulphanilic acid, as the major product:

$$\ce{C6H5NH2 ->[H2SO4] C6H5NH3+ HSO4- ->[453-473\ K] p\text{-}H2N\text{-}C6H4\text{-}SO3H}$$

Sulphanilic acid is unusual in that it exists largely as an internal salt — a dipolar zwitterion in which the basic –NH2 is protonated by the acidic –SO3H, giving $\ce{^{+}H3N-C6H4-SO3^-}$.

Why Aniline Fails Friedel-Crafts

Despite being strongly activated towards bromination, nitration and sulphonation, aniline does not undergo the Friedel-Crafts reaction (neither alkylation nor acylation). The reason, given verbatim in NCERT (§9.6), is the catalyst. Friedel-Crafts uses the Lewis acid $\ce{AlCl3}$; aniline's basic nitrogen donates its lone pair to AlCl3 and forms a salt. Nitrogen now carries a positive charge, so it behaves as a strong deactivating group, and the ring is too electron-poor to react with the electrophile.

The same lone pair that makes aniline a vigorous substrate for bromination is what disqualifies it from Friedel-Crafts: tied up as a salt with AlCl3, it inverts from activator to deactivator.

Reaction Behaviour by Amine Class

The reactions above sort cleanly by the number of N–H hydrogens. The table consolidates the class-dependent outcomes that NEET tests as statement-matching and product-identification items. Note that two diagnostic tests — the carbylamine (isocyanide) test for primary amines and the Hinsberg test with benzenesulphonyl chloride — are covered on their own page and are listed here only for orientation.

ReactionPrimary (1°)Secondary (2°)Tertiary (3°)
Acylation (acid chloride / anhydride) N-substituted amide N,N-substituted amide No reaction (no N–H)
With HNO2 (aliphatic) Alcohol + N2 (via unstable salt) N-nitrosamine Soluble salt; no diazonium
With HNO2 (aromatic) Stable diazonium salt at 273–278 K N-nitrosamine No diazonium
With Grignard RMgX Quenches reagent (N–H abstracted) Quenches reagent (N–H abstracted) No N–H; tolerated
Carbylamine test Positive (isocyanide, foul smell) Negative Negative
Worked Example

Identify the major product C: $\ce{C6H5NH2 ->[NaNO2/HCl][273\ K] A ->[H2O][warm] B}$, and state what happens if instead the diazonium solution from aniline is treated with hot water above 283 K.

Aniline first diazotises to benzenediazonium chloride, $\ce{C6H5N2+ Cl-}$ (A), which is stable in the cold (273–278 K). On warming the aqueous solution above about 283 K the diazonium group is replaced by –OH and the salt is hydrolysed to phenol (B): $\ce{C6H5N2+ Cl- + H2O -> C6H5OH + N2 + HCl}$. The contrast with a primary aliphatic amine is the lesson: there, the diazonium ion never survives the cold step at all and the alcohol forms directly.

Quick Recap

Reactions of Amines — at a glance

  • Acylation: 1° and 2° amines + acid chloride/anhydride → amide; needs a base (pyridine) to mop up HCl. Tertiary amines have no N–H, so no amide.
  • Nitrous acid: 1° aliphatic → unstable diazonium → alcohol + N2; 1° aromatic → stable diazonium salt at 273–278 K; 2° → N-nitrosamine; 3° → salt only.
  • Grignard: the acidic N–H of 1° and 2° amines protonates RMgX, destroying the reagent.
  • Aniline EAS: –NH2 is strongly activating and o/p-directing; bromine water gives 2,4,6-tribromoaniline; acetylation moderates the ring for mono-substitution.
  • Nitration in strong acid gives meta product too, because aniline becomes the m-directing anilinium ion; sulphonation gives sulphanilic acid.
  • No Friedel-Crafts: aniline + AlCl3 forms a salt; the positively charged N deactivates the ring.

NEET PYQ Snapshot — Reactions of Amines

Real NEET previous-year questions on acylation, nitrous acid and aniline substitution. Solutions condensed from the official keys.

NEET 2022 · Q.69

Statement I: Primary aliphatic amines react with HNO2 to give unstable diazonium salts. Statement II: Primary aromatic amines react with HNO2 to form diazonium salts which are stable even above 300 K.

  1. Both Statement I and Statement II are incorrect.
  2. Statement I is correct but Statement II is incorrect.
  3. Statement I is incorrect but Statement II is correct.
  4. Both Statement I and Statement II are correct.
Answer: (2)

Statement I is correct: the alkyldiazonium ion is unstable and collapses to an alcohol with loss of N2. Statement II is wrong on the temperature — arenediazonium salts are stable only in the cold, at 273–278 K, not above 300 K.

NEET 2018 · Q.47

Nitration of aniline in strong acidic medium also gives m-nitroaniline because:

  1. In spite of substituents, the nitro group always goes only to the m-position
  2. In electrophilic substitution reactions the amino group is meta directive
  3. In absence of substituents the nitro group always goes to m-position
  4. In strongly acidic medium aniline is present as the anilinium ion
Answer: (4)

In strong acid, aniline is protonated to the anilinium ion. The –NH3+ group is electron-withdrawing and meta-directing, so a significant m-nitroaniline fraction forms alongside the o/p products.

NEET 2024 · Q.62

Statement I: Aniline does not undergo Friedel-Crafts alkylation reaction. Statement II: Aniline cannot be prepared through Gabriel synthesis.

  1. Both Statement I and Statement II are true
  2. Both Statement I and Statement II are false
  3. Statement I is correct but Statement II is false
  4. Statement I is incorrect but Statement II is true
Answer: (1)

Statement I is true: aniline forms a salt with the Lewis acid AlCl3, so the positively charged nitrogen deactivates the ring and Friedel-Crafts fails. Statement II is also true (aromatic primary amines cannot be made by Gabriel synthesis), so both statements are correct.

NEET 2025 · Q.72

Statement I: Benzenediazonium salt is prepared by the reaction of aniline with nitrous acid at 273–278 K. It decomposes easily in the dry state. Statement II: Insertion of iodine into the benzene ring is difficult and hence iodobenzene is prepared through the reaction of benzenediazonium salt with KI.

  1. Statement I is incorrect but Statement II is correct
  2. Both Statement I and Statement II are correct
  3. Both Statement I and Statement II are incorrect
  4. Statement I is correct but Statement II is incorrect
Answer: (2)

Aniline diazotises at 273–278 K to benzenediazonium chloride, which does decompose in the dry state. Direct iodination of the ring is difficult, so iodobenzene is made from the diazonium salt with KI. Both statements are correct.

FAQs — Reactions of Amines

The conceptual edges NEET probes most often in this topic.

Why does a primary aliphatic amine give an alcohol with nitrous acid, while a primary aromatic amine gives a stable salt?

Both first form a diazonium ion. The alkyldiazonium ion has no way to stabilise the positive charge, so it loses N2 almost instantly to give a carbocation that ends up as an alcohol, with quantitative evolution of nitrogen. The arenediazonium ion is stabilised by resonance with the benzene ring, so at 273–278 K it survives long enough to be a usable benzenediazonium salt.

Why must the NH2 group of aniline be acetylated before mono-substitution?

The free NH2 group is so strongly activating that bromine water substitutes all three of the ortho and para positions in one step, giving 2,4,6-tribromoaniline. Acetylation converts NH2 into NHCOCH3; the nitrogen lone pair is then partly drawn towards the acetyl oxygen by resonance, so the ring is less activated. Substitution can then be stopped at one position, after which the amide is hydrolysed back to the amine.

Why does aniline not undergo the Friedel-Crafts reaction?

Friedel-Crafts alkylation and acylation use the Lewis acid AlCl3 as catalyst. Aniline's basic nitrogen forms a salt with AlCl3, putting a positive charge on nitrogen. The positively charged nitrogen now acts as a strong electron-withdrawing, deactivating group, so the ring is too unreactive for Friedel-Crafts substitution.

Why does nitration of aniline in strong acid give a substantial amount of the meta product?

In the strongly acidic nitrating mixture, aniline is protonated to the anilinium ion. The positively charged NH3+ group is electron-withdrawing and meta-directing, so along with the ortho and para nitroanilines a significant amount of m-nitroaniline is formed.

Which amines undergo acylation, and what is the role of pyridine?

Only primary and secondary amines undergo acylation, because the reaction replaces an N–H hydrogen by an acyl group to give an amide; tertiary amines have no N–H. Pyridine, a base stronger than the amine, neutralises the HCl liberated from the acid chloride, shifting the equilibrium towards the amide product.

What product does sulphonation of aniline ultimately give?

Aniline first forms anilinium hydrogensulphate with concentrated sulphuric acid. On heating at 453–473 K this rearranges to p-aminobenzenesulphonic acid, commonly called sulphanilic acid, as the major product. Sulphanilic acid exists largely as a dipolar zwitterion.

Related Topics
Amines — Chapter Home All subtopics, methods and reactions in one place. Hinsberg & Carbylamine Tests The two diagnostic tests for 1°, 2° and 3° amines. Diazonium Salts Sandmeyer, Gattermann and coupling reactions of Ar–N₂⁺. Basicity of Amines Why the lone pair makes amines basic, and the aqueous orders. Preparation of Amines Reduction, ammonolysis, Gabriel and Hoffmann routes. Structure & Classification Pyramidal nitrogen and the 1°/2°/3° framework.