Amines Class 12 Notes - CBSE Chemistry Chapter 13 (Free PDF)

Chapter summary

Amines are organic derivatives of ammonia in which one, two or three hydrogen atoms are replaced by alkyl or aryl groups, giving primary, secondary and tertiary amines whose chemistry is driven by the lone pair on nitrogen. The chapter covers their classification and nomenclature, preparation routes (reduction, ammonolysis, Gabriel and Hoffmann bromamide), physical properties, basicity trends, characteristic reactions, and the versatile aryl diazonium salts. It is a high-yield NEET topic because basicity comparisons, identification tests and diazonium conversions appear almost every year.

Chapter notes
🃏 Flash Cards: Amines

Class 12 Chemistry · Amines – swipe through all 10 cards to understand the whole chapter.

🔬Start here1/10

What Amines Are

Amines are ammonia (NH3) with one or more H atoms replaced by alkyl/aryl groups.

1°: R−NH2 · 2°: R2NH · 3°: R3N

N is sp3, pyramidal, with a lone pair that drives basicity & nucleophilicity.

  • Class = how many CARBONS attach to N (not the carbon type)
  • tert-butylamine & benzylamine are 1° — only one C on N
  • The lone pair is the engine of every reaction in this chapter
🏷️Naming2/10

Nomenclature

IUPAC drops the -e of the alkane and adds -amine; groups on N take an N- prefix.

CH3NH2 → methanamine · C6H5NH2 → aniline

Common names list groups + ‘amine’ as one word (e.g. ethylmethylamine).

  • (CH3)2CH−NH2 → propan-2-amine (1°)
  • C2H5−NH−CH3 → N-methylethanamine
  • Aniline = aminobenzene = benzenamine (fixed name)
⚗️Key process3/10

Preparing Amines

Five NEET routes — and three of them quietly change the carbon count.

R−NO2 →(Sn/HCl or H2/Ni) R−NH2

Ammonolysis (R−X + NH3) is non-selective → gives a 1°/2°/3°/quaternary mixture.

  • Nitrile R−C≡N →(H2/Ni) R−CH2−NH2 — GAINS one carbon
  • Amide →(LiAlH4) amine — SAME carbon count
  • NaNO2/HCl is diazotisation, NOT reduction — don’t confuse
🧪Two traps4/10

Gabriel vs Hoffmann

Two famous routes to pure 1° amines, each with a classic exam trap.

R−CONH2 →(Br2/KOH) R−NH2 (one C fewer)

Hoffmann expels the carbonyl carbon as carbonate → amine has ONE C less than the amide.

  • Gabriel → pure 1° amine; FAILS for aryl amines (no SN2 on aryl halides)
  • Hoffmann bromamide → loses one carbon
  • Propanamide (3C) →Hoffmann→ ethanamine (2C)
🌡️Physical5/10

Boiling Point & Solubility

Boiling point tracks the number of N–H bonds available for hydrogen bonding.

1° amine > 2° amine > 3° amine (similar molar mass)

3° amines boil lowest — no N–H to donate, not because they’re less polar.

  • Across families: alcohol > amine > ether ≈ alkane (O more electronegative than N)
  • 3° amines still dissolve in water — lone pair accepts H-bond (N···H−O)
  • Solubility falls as the carbon chain grows; lower amines smell fishy
💪Core trend6/10

Basicity in Water

A base donates its N lone pair to a proton; in water, solvation breaks the simple trend.

(CH3)2NH > CH3NH2 > (CH3)3N > NH3

Lower pK_b (higher K_b) = stronger base. Gas-phase order is the regular 3°>2°>1°>NH3.

  • +I effect of alkyl groups makes aliphatic amines stronger than NH3
  • 3° drops below 1° in water: poor solvation + steric crowding
  • Aryl > don’t apply +I here — see the next card
🟢Aromatic7/10

Why Aniline Is Weak

In aniline the N lone pair is delocalised into the benzene ring, so it’s less available.

aliphatic amine > NH3 > aniline

Weakness is from resonance delocalisation, not any inductive effect.

  • Electron-withdrawing −NO2 on the ring → even weaker base
  • Electron-donating −CH3 / −OCH3 → stronger base
  • p-nitroaniline ≪ aniline < p-toluidine
🔥ID tests8/10

Identifying the Class

Two tests pin down whether an amine is 1°, 2° or 3°.

R−NH2 + CHCl3 + 3KOH(alc.) → R−NC (foul isocyanide)

Carbylamine test is POSITIVE for 1° amines only; needs ALCOHOLIC KOH.

  • Hinsberg (C6H5SO2Cl): 1° → soluble in KOH, 2° → insoluble, 3° → no reaction
  • 3° amines can’t be acylated (no N–H) and give no carbylamine test
  • 1° & 2° amines acylate with acid chlorides/anhydrides → amides
🟤Aniline ring9/10

Reactions of Aniline

The −NH2 group is strongly activating and o/p-directing, so the ring is very reactive.

aniline + Br2 water → 2,4,6-tribromoaniline (instant ppt)

To stop at mono-substitution, acetylate first (−NH2 → −NHCOCH3), then hydrolyse.

  • 1° aromatic amine + HNO2 at 0–5 °C → stable diazonium salt
  • 1° aliphatic amine + HNO2 → alcohol + brisk N2 gas
  • Acetylation also protects N from oxidation
🔌Synthetic hub10/10

Diazonium Salts

Aryl diazonium salt acts like a universal adapter — −N2⁺ swaps for almost any group.

C6H5NH2 →(NaNO2/HCl, 273–278K) C6H5N2⁺Cl⁻

Cold (0–5 °C) is non-negotiable; warming gives phenol + N2. Only ARYL salts are usable.

  • Sandmeyer: −Cl/−Br (CuCl/CuBr), −CN (CuCN); −I needs only KI (no Cu)
  • −F via Balz–Schiemann (HBF4, heat); −OH via warm water; −H via H3PO2
  • Coupling with phenol/aniline → azo dye (−N=N−) at para position
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📝 Practice Amines — 10 NEET PYQs
Real previous-year questions · with answers & solutions
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Tap an option to check your answer and see the worked solution. Every question is a real NEET previous-year question.
Q1NEET 2020
Reaction of propanamide (CH₃CH₂CONH₂) with ethanolic sodium hydroxide and bromine gives:
Correct answer: A. This is the Hoffmann bromamide degradation: R-CONH₂ →[Br₂, NaOH] R-NH₂, with loss of the carbonyl carbon as carbonate. Propanamide (3 C) therefore gives a 2-carbon amine, ethanamine (ethylamine).
🔎 See the full step-by-step solution in the app →
Q2NEET 2019
The correct order of the basic strength of methyl-substituted amines in aqueous solution is:
Correct answer: A. In water, basicity is governed by the interplay of +I effect, steric crowding and solvation of the cation. For methylamines the experimental order is 2° > 1° > 3° > NH₃, i.e. (CH₃)₂NH > CH₃NH₂ > (CH₃)₃N > NH₃.
🔎 See the full step-by-step solution in the app →
Q3NEET 2019
The secondary amine that reacts with Hinsberg’s reagent (benzenesulphonyl chloride) to give a product that is INSOLUBLE in alkali is:
Correct answer: B. In the Hinsberg test a 2° amine forms an N,N-dialkyl sulphonamide that has no N–H, so it cannot lose a proton to alkali and is insoluble in KOH. (CH₃CH₂)₂NH (diethylamine) is the 2° amine; 1° amines give KOH-soluble products and 3° amines do not react.
🔎 See the full step-by-step solution in the app →
Q4NEET 2018
Nitration of aniline in strongly acidic medium also gives a significant amount of m-nitroaniline because:
Correct answer: D. In strong acid, aniline is protonated to the anilinium ion. The -NH₃⁺ group is electron-withdrawing and m-directing, so a substantial fraction of nitration occurs at the meta position (alongside the o/p product from any free aniline). The -NH₂ group itself is o/p-directing.
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Q5NEET 2017
Which of these reactions is appropriate for converting acetamide (CH₃CONH₂) to methanamine (CH₃NH₂)?
Correct answer: B. Going from a 2-carbon amide to a 1-carbon amine means losing one carbon. Only the Hoffmann bromamide degradation (Br₂/NaOH) does this: CH₃CONH₂ → CH₃NH₂. Gabriel adds an intact alkyl group; Stephen’s reduces nitriles to aldehydes.
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Q6NEET 2017
The correct increasing order of basic strength for aniline (I), p-nitroaniline (II) and p-toluidine (III, p-CH₃) is:
Correct answer: D. -NO₂ (electron-withdrawing) removes electron density from N, making p-nitroaniline (II) the weakest base; aniline (I) is intermediate; -CH₃ (electron-donating, +I/hyperconjugation) makes p-toluidine (III) the strongest. So II < I < III.
🔎 See the full step-by-step solution in the app →
Q7NEET 2016
In the reaction CH₃CH₂CH₂Br + NaCN → CH₃CH₂CH₂CN + NaBr, the reaction will be fastest in:
Correct answer: C. This S_N2 substitution is fastest in a polar aprotic solvent. DMF does not hydrogen-bond to (and thus does not cage) the CN⁻ nucleophile, so CN⁻ stays ‘naked’ and highly reactive. Protic solvents (water, alcohols) solvate CN⁻ and slow the reaction.
🔎 See the full step-by-step solution in the app →
Q8NEET 2015
The number of structural isomers (amines) possible from the molecular formula C₃H₉N is:
Correct answer: A. C₃H₉N amines are: propan-1-amine (CH₃CH₂CH₂NH₂), propan-2-amine ((CH₃)₂CHNH₂), N-methylethanamine (CH₃CH₂NHCH₃, 2°) and N,N-dimethylmethanamine (trimethylamine, (CH₃)₃N, 3°) — four structural isomers.
🔎 See the full step-by-step solution in the app →
Q9NEET 2015
Aniline cannot be prepared by which of the following methods?
Correct answer: D. Gabriel synthesis needs S_N2 substitution of an alkyl halide by the phthalimide anion. Chlorobenzene (an aryl halide) will not undergo this because the C–Cl bond has partial double-bond character (resonance), so aniline cannot be made this way. The other three routes do give aniline.
🔎 See the full step-by-step solution in the app →
Q10NEET 2014
Which of the following will be the most stable diazonium salt, R-N₂⁺X⁻?
Correct answer: B. An aryl diazonium ion is stabilised by resonance delocalisation of the positive charge into the benzene ring directly attached to the diazonium nitrogen. C₆H₅-N₂⁺ is therefore far more stable than alkyl or benzyl diazonium ions, which lack this direct ring conjugation and decompose at once.
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Frequently Asked Questions

What are amines and how are they classified?

Amines are derivatives of ammonia (NH3) in which one or more hydrogen atoms are replaced by alkyl or aryl groups. They are classified as primary (R-NH2), secondary (R2NH) or tertiary (R3N) based on how many carbon atoms are attached to the nitrogen, not on the type of carbon.

How are aryl diazonium salts prepared and why must it be done cold?

A primary aromatic amine like aniline is treated with sodium nitrite and HCl (nitrous acid) at 273 to 278 K (0 to 5 degrees C) in a reaction called diazotisation, giving C6H5N2+Cl-. The low temperature is essential because warm solutions decompose the diazonium salt into phenol and nitrogen gas.

Why is aniline a weaker base than aliphatic amines and even ammonia?

In aniline the lone pair on nitrogen is delocalised into the benzene ring through resonance, so it is less available to accept a proton. This makes aniline a weaker base than NH3 and much weaker than aliphatic amines, which are strengthened by the electron-donating inductive effect of alkyl groups.

What is the difference between the Gabriel and Hoffmann bromamide methods?

The Gabriel phthalimide synthesis gives pure primary amines but fails for aromatic amines because aryl halides do not undergo the required nucleophilic substitution. The Hoffmann bromamide reaction converts an amide to a primary amine using Br2 and KOH and produces an amine with one carbon fewer than the starting amide.

Is Amines important for NEET and which parts are most tested?

Yes, Amines is a regularly tested NEET chapter from Class 12 Organic Chemistry. The most frequently asked areas are basicity comparisons in water versus gas phase, identification tests like carbylamine and Hinsberg, and the substitution and coupling reactions of diazonium salts.

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