Why are aliphatic amines more basic than aniline?

The key factor is availability of the lone pair on nitrogen. In aliphatic amines (like methylamine, CH₃NH₂), the alkyl group donates electrons by the +I (inductive) effect, increasing electron density on nitrogen. This makes the lone pair more available to accept a proton, so aliphatic amines are stronger bases (pKb 3–4). In aniline (C₆H₅NH₂), the lone pair on nitrogen is delocalised into the benzene ring by conjugation (resonance). The lone pair is spread over the ring and is less available to accept a proton. This lowers the basicity significantly (pKb 9.4). Quantitatively: Kb of methylamine = 5 × 10⁻⁴, Kb of aniline = 4 × 10⁻¹⁰. The resonance withdrawal in aniline reduces basicity by six orders of magnitude.

Why is the basicity order 2° > 1° > 3° for aliphatic amines in water?

Purely by inductive effect, you would expect 3° > 2° > 1° (more alkyl groups donating more electrons). But the actual order in aqueous solution is 2° > 1° > 3°. Two opposing factors are at work: (1) Inductive effect: more alkyl groups push more electrons onto N, increasing lone pair availability. This predicts 3° > 2° > 1°. (2) Solvation (hydration): the ammonium ion (conjugate acid) is stabilised by hydrogen bonding with water. Primary ammonium R-NH₃⁺ has 3 N-H bonds and is heavily solvated. Secondary R₂NH₂⁺ has 2 N-H bonds — moderate solvation. Tertiary R₃NH⁺ has only 1 N-H bond — least solvated. Less solvation means the conjugate acid is less stable, which reduces apparent basicity of the tertiary amine. The balance of these two opposing effects gives 2° (best balance) > 1° > 3°.

What is the Hoffmann bromamide degradation? What product forms?

Hoffmann bromamide degradation converts a primary amide (RCONH₂) to a primary amine (RNH₂) with one fewer carbon. Reagents: Br₂ + NaOH (aqueous). Mechanism: (1) Br₂ and NaOH form NaOBr (or OBr⁻). (2) OBr⁻ bromaminates the amide nitrogen: RCONH₂ → RCONHBr. (3) A strongly basic N-bromo intermediate forms and the acyl-N bond rearranges in a Curtius-type 1,2-shift: R migrates from C to N with the bonding pair. An isocyanate (RNCO) is the intermediate. (4) RNCO is hydrolysed by NaOH to give RNH₂ + CO₂. Product: a primary amine (RNH₂) with one carbon fewer than the starting amide. This is important in NEET because it distinguishes primary amides and yields primary amines specifically. Example: CH₃CONH₂ + Br₂/NaOH → CH₃NH₂.

How does the Hinsberg test distinguish primary, secondary, and tertiary amines?

Reagent: benzenesulfonyl chloride (C₆H₅SO₂Cl) in aqueous KOH. Primary amine (RNH₂): reacts to give a sulfonamide (RNH-SO₂C₆H₅). This sulfonamide has one acidic N-H and dissolves in excess KOH (alkali soluble). On acidification, it precipitates. Secondary amine (R₂NH): reacts to give a disubstituted sulfonamide (R₂N-SO₂C₆H₅). This has NO acidic N-H and is insoluble in KOH — precipitate forms in alkaline solution and remains on acidification. Tertiary amine (R₃N): does NOT react with benzenesulfonyl chloride because the nitrogen has no N-H. No reaction. Summary: 1° gives alkali-soluble sulfonamide; 2° gives alkali-insoluble precipitate; 3° gives no reaction.

What is the carbylamine reaction and who does it identify?

Carbylamine reaction (also called isocyanide test): a primary amine reacts with chloroform (CHCl₃) and alcoholic KOH to give a foul-smelling isocyanide (RNC, also called carbylamine). Reagents: CHCl₃ + alc. KOH + primary amine. Observation: characteristic intensely unpleasant smell. Primary aliphatic and aromatic amines both give a positive carbylamine test. Secondary and tertiary amines do NOT react. This makes it a specific test for primary amines (both aliphatic and aromatic). In NEET questions, the carbylamine test is used to confirm whether a compound is a primary amine. The isocyanide is toxic, which is why the characteristic smell is diagnostic.

What are diazonium salts and what reactions do they undergo?

Diazonium salts (ArN₂⁺ X⁻) are formed when aromatic primary amines react with NaNO₂ + HCl at 0–5°C (diazotisation). The benzene diazonium chloride (C₆H₅N₂⁺Cl⁻) is relatively stable at low temperature and undergoes many useful reactions: (1) Sandmeyer reactions: replacement of N₂⁺ by Cl, Br (with CuCl/CuBr + HCl/HBr), CN (with CuCN), or I (with KI). (2) Balz-Schiemann reaction: N₂⁺ replaced by F via ArN₂⁺BF₄⁻ → ArF + N₂ + BF₃. (3) Reduction: N₂⁺ → ArH (deamination) using H₃PO₂. (4) Coupling reactions: the diazonium ion acts as an electrophile and couples with activated aromatic rings (phenol in alkaline medium, aniline) at the para position to give azo compounds (ArN=NAr'). These orange/red azo dyes are used in NEET synthesis questions. Diazonium salts are only stable at 0–5°C; above this temperature they decompose.

What is Gabriel phthalimide synthesis and what is it used for?

Gabriel phthalimide synthesis makes pure primary aliphatic amines without contamination by secondary or tertiary amines. Steps: (1) Phthalimide (a cyclic imide) is treated with KOH to form potassium phthalimide (the N-K salt). (2) Potassium phthalimide reacts with a primary alkyl halide (RX, usually 1°) by SN2 displacement to give N-alkylphthalimide. (3) N-alkylphthalimide is hydrolysed with NaOH (aq.) OR hydrazinolysis (NH₂NH₂) to give the primary amine (RNH₂) + phthalic acid (or phthalhydrazide with hydrazine). Why it is useful: since phthalimide has only one N-H, only one alkyl group can be introduced, guaranteeing a primary amine product. Phenyl halides (ArX) do not work because SN2 on Ar is not possible. NEET often asks which synthesis gives a pure primary amine — answer: Gabriel phthalimide.

How many NEET questions come from Amines and what are the most important topics?

Amines contributes 3–5 NEET questions per year from Class 12 Chemistry. The highest-frequency topics are: (1) Basicity order — aliphatic vs aromatic, 2° > 1° > 3° in water. Know the reason (inductive vs solvation). (2) Hoffmann bromamide degradation — identify that the product has one fewer carbon and is a primary amine. (3) Distinction tests — Hinsberg test (1°: alkali-soluble; 2°: alkali-insoluble; 3°: no reaction), carbylamine test (only 1°). (4) Diazonium chemistry — Sandmeyer reaction sequence (Cl, Br, CN, I replacement), Balz-Schiemann (F), coupling reactions (azo dyes). (5) Gabriel synthesis — makes pure 1° aliphatic amine. Learn the reagent–product pairs precisely, as NEET frequently tests "which reagent gives which product" in the form of match-the-following or single correct MCQs.

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AminesNEET Chemistry · Class 12 · NCERT Chapter 9

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13 NEET previous-year questions on Amines, each with the correct answer and a step-by-step solution. Sourced directly from official NEET papers across every booklet code.

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All (13)

Amines (13)

Potassium salt of phthalimide treated with chlorobenzene followed by hydrolysis with aqueous NaOH solution.

Hydrolysis of phenylisocyanide with acidic solution

Degradation of benzamide with bromine in alkaline solution

Reduction of nitrobenzene with H2 Pd⁄ in ethanol

Solution

Due to resonance C – Cl bond acquires double bond character.

Schiff base

Ketone

Carboxylic acid

Aromatic acid

Solution

> 𝐶 = 𝑂 + 𝑅 − 𝑁𝐻2 → > 𝐶 = 𝑁 − 𝑅 𝑆𝑐ℎ𝑖𝑓𝑓 𝑏𝑎𝑠𝑒

Aryl amines are generally less basic than alkyl amines because the nitrogen lone-pair electrons are delocalized by interaction with the aromatic ring π electron system.

Aryl amines are generally more basic than alkyl amines because the nitrogen lone-pair electrons are not delocalized by interaction with the aromatic ring π electron system.

Aryl amines are generally more basic than alkyl amines because of aryl group.

Aryl amines are generally more basic than alkyl amines, because the nitrogen atom in aryl amines is sp- hybridized.

Solution

Here lone pair is in conjugation with double bond so basic strength decreased R − N̈ H2 (No conjugation)

II < III < I

III < I < II

III < II < I

II < I < III

Solution

–NO2 has strong –R effect and –CH 3 shows +R effect. ∴ Order of basic strength is NO2 NH2 NH2 NH2 CH3 <<

Carbylamine reaction

Hoffmann hypobromamide reaction

Stephens reaction

Gabriels phthalimide synthesis

Solution

O CH – C – NH3 2 + Br + 4NaOH2 Δ CH3 – NH2 + 2NaBr + Na2CO3 + 3H2O This is Hoffmann Bromamide reaction.

H3O+ and H2F+, respectively

OH– and H2F+, respectively

H3O+ and F–, respectively

OH– and F–, respectively

Solution

The conjugate base of $H_{2} O$ is $OH^{-}$ , and the conjugate base of $HF$ is $F^{-}$ , as they each lose a proton. According to the Bronsted-Lowry theory, option (d) is correct.

Solution

The carbylamine test is given by primary amines. The compound in option (a) is a primary amine, so it will give the carbylamine test. NCERT XII chapter Amines states that primary amines react with chloroform and alcoholic potassium hydroxide to form isocyanides (carbylamines), which have a characteristic offensive smell.

Solution

The compound that reacts with Hinsberg’s reagent to form a solid that dissolves in alkali is a primary amine. Primary amines form a solid sulfonamide which is soluble in alkali, so option (c) is correct.

Both Statement I and Statement II are correct.

Both Statement I and Statement II are incorrect.

Statement I is correct but Statement II is incorrect.

Statement I is incorrect but Statement II is correct.

Solution

• Primary aliphatic amines react with HNO2 and give unstable diazonium salt which turns into alcohol 2HO– 2 2 2 2R – NH HNO R – N – Cl ROH N HCl+⎡⎤+ ⎯⎯ → ⎯⎯⎯ → + +⎣⎦ • Primary aromatic amines reacts with HNO2 and give stable diazonium salt which are stable at 273 to 278 K. 273–278 K – 6 5 2 2 6 5 2C H – NH HNO C H N Cl ++ ⎯⎯⎯⎯⎯ →

CH₃CONH₂ →(i) LiAlH₄, (ii) H₃O⁺ Product

CH₃CONH₂ → Br₂ / KOH Product

CH₃CN →(i) LiAlH₄, (ii) H₃O⁺ Product

CH₃NC →(i) LiAlH₄, (ii) H₃O⁺ Product

Solution

Option (d) involves the reduction of methyl isocyanide (CH₃NC) to a secondary amine, not a primary amine. The other options produce primary amines through reduction of nitriles or amides, so option (d) is the correct answer.

Both statement I and Statement II are true

Both Statement I and Statement II are false

Statement I is correct but Statement II is false

Statement I is incorrect but Statement II is true

Solution

• Aniline does not undergo Friedel -Crafts alkylation reaction due to salt formation with aluminium chloride, the Lewis acid, which is used as a catalyst. • Aniline (aromatic primary amine) cannot be prepared by Gabriel phthalimide synthesis because aryl halides do not undergo nucleophilic substitution with anion formed by phthalimide.

N-methylaniline > benzenamine > ethanamine > N-ethylethanamine

N-ethylethanamine > ethanamine > benzenamine > N-methylaniline

N-ethylethanamine > ethanamine > N-methylaniline > benzenamine

benzenamine > ethanamine > N-methylaniline > N-ethylethanamine

Solution

Sol. Lower is the value of $p K_{b}$ , higher is the basicity

Also aliphatic amines are stronger bases than aromatic amines.

$p K_{b}$ : Benzenamine > N-Methylaniline > Ethanamine > N-Ethylethanamine

Basic strength : N-Ethylethanamine > Ethanamine > N-Methylaniline > Benzenamine

(1) Both Statement I and Statement II are correct

(2) Both Statement I and Statement II are incorrect

(3) Statement I is correct but Statement II is incorrect

(4) Statement I is incorrect but Statement II is correct

Solution

Sol. Benzene diazonium chloride is prepared by the reaction of aniline with nitrous acid at 273-278 K.
Nitrous acid is produced in the reaction mixture by reaction of NaNO₂ with HCl.

$C_{6} H_{5} N H_{2} + N a N O_{2} + 2 H C l 273 - 278 K C_{6} H_{5} N_{2}^{+} C l^{-} + N a C l + 2 H_{2} O$

Benzene diazonium chloride decomposes easily in the dry state

Iodobenzene is prepared by shaking benzene diazonium salt with KI because direct insertion of iodine into benzene ring is difficult

AminesNEET Chemistry · Class 12 · NCERT Chapter 9

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AminesNEET Chemistry · Class 12 · NCERT Chapter 9

1NEET 2015Medium4 bookletsMethod by which Aniline cannot be prepared is: www.vedantu.com 60

2NEET 2016Medium4 bookletsThe product formed by the reaction of an aldehyde with a primary amine is:

3NEET 2016Medium4 bookletsThe correct statement regarding the basicity of aryl amines is:

4NEET 2017Medium5 bookletsThe correct increasing order of basic strength for the following compounds is NH2 NH2 NH2 NO2 CH3 (I) (II) (III)

5NEET 2017Medium5 bookletsWhich of the following reactions is appropriate for converting acetamide to methanamine?

6NEET 2019Medium4 bookletsConjugate base for Bronsted acids H2O and HF are

7NEET 2020Medium24 bookletsWhich of the following amine will give the carbylamine test ?

8NEET 2021Medium24 bookletsIdentify the compound that will react with Hinsberg’s reagent to give a solid which dissolves in alkali.

10NEET 2023Medium24 bookletsWhich of the following reactions will NOT give primary amine as the product?

11NEET 2024Medium24 bookletsGiven below are two statements: Statement I : Aniline does not undergo Friedel-Crafts alkylation reaction. Statement II : Aniline cannot be prepared through Gabriel synthesis . In the light of the above statements, choose the correct answer from the options given below:

12NEET 2025Medium4 bookletsThe correct order of decreasing basic strength of the given amines is:

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