NEET (UG) · Chemistry, Class 12
Amines: NEET PYQ, 2018–2026
Amines (Chemistry, Class 12) has been asked 12 times in the 7 NEET papers from 2018, 2020, 2021, 2023, 2024, 2025 and 2026: about 1.7 questions a paper, 4% of all Chemistry questions. It is the 13th most-asked of 21 Chemistry chapters. NEET 2026 had 2 from it. Last checked 28 October 2026.
Questions per paper
Year by year.
| Year | Questions |
|---|---|
| 2018 | 1 |
| 2020 | 1 |
| 2021 | 2 |
| 2023 | 2 |
| 2024 | 2 |
| 2025 | 2 |
| 2026 | 2 |
From 2021 to 2024 each paper had 200 questions (a Section B with a choice); the others had 180.
Inside the chapter
Topics asked most.
- Classification and identification of amines6
- Basicity of amines3
- Diazonium salts3
Try it
10 Amines questions to practise.
Our own questions, written in the NEET pattern and checked, each with an explanation. Pick an option to see the answer.
Q1 · Chemistry · Amines
An aqueous solution of benzenediazonium chloride is warmed. The main organic product is:
Answer: (B)
When the temperature of a solution of a diazonium salt rises to about 283 K, it is hydrolysed: C₆H₅N₂⁺Cl⁻ + H₂O → C₆H₅OH + N₂ + HCl. This is why diazonium salts are prepared cold.
Q2 · Chemistry · Amines
The pKa values of the conjugate acids of methylamine and aniline are about 10.6 and 4.6 respectively. In a buffer of pH 7.4 at 298 K, which statement about the dominant form of each amine is correct?
Answer: (A)
Use the relation [BH+]/[B] = 10^(pKa − pH). For methylamine: 10.6 − 7.4 = 3.2, so [CH3NH3+]/[CH3NH2] ≈ 1.6 × 10^3 and the amine is mainly protonated. For aniline: 4.6 − 7.4 = −2.8, so [C6H5NH3+]/[C6H5NH2] ≈ 1.6 × 10^−3 and the amine is mainly unprotonated. The stronger base (higher pKa of conjugate acid) is far more protonated at a given pH.
Q3 · Chemistry · Amines
Benzene is to be converted into m-bromophenol (3-bromophenol) in good regiochemical purity. Which of the following reagent sequences, applied in the order written, achieves this?
Answer: (A)
A phenol cannot be obtained with a meta bromine by direct bromination of phenol, because OH directs ortho/para. The way round this is to put a meta director (NO2) on the ring first, then convert it to NH2, and finally to OH through the diazonium salt. Step 1: nitration gives nitrobenzene. Step 2: NO2 directs the incoming Br to the meta position, giving m-bromonitrobenzene. Step 3: Sn/HCl reduces NO2 to NH2, giving m-bromoaniline. Step 4: NaNO2/HCl at 273–278 K gives the diazonium chloride. Step 5: warming with water replaces N2+ by OH, giving m-bromophenol. Reversing steps 1 and 2 gives the para isomer. Reducing before brominating makes NH2 an ortho/para director. Using H3PO2 instead of warm water gives bromobenzene.
Q4 · Chemistry · Amines
In aqueous solution at 298 K, which is the stronger base: aniline (C6H5NH2) or acetanilide (C6H5NHCOCH3)?
Answer: (A)
Principle: basicity depends on how available the nitrogen lone pair is for protonation. Aniline's lone pair is partly delocalised into the benzene ring, so aniline is weaker than alkyl amines. In acetanilide, the nitrogen lone pair is also delocalised into the carbonyl group, which makes it much less available. Therefore acetanilide is a weaker base than aniline.
Q5 · Chemistry · Amines
Iodobenzene is usually prepared from aniline through its diazonium salt rather than by treating benzene directly with I2. The main reason direct iodination with I2 alone is unsatisfactory is that:
Answer: (A)
Direct iodination, C6H6 + I2 ⇌ C6H5I + HI, is reversible and slow. HI is a good reducing agent and converts C6H5I back to C6H6, so the reaction needs an oxidising agent to proceed. Aryl iodides are therefore made by diazotising aniline and warming the diazonium salt with aqueous KI. This reaction needs no copper catalyst and goes to completion with loss of N2.
Q6 · Chemistry · Amines
Benzenediazonium chloride is to be coupled with phenol and, separately, with aniline to give azo dyes. Which of the following correctly pairs each coupling partner with a suitable pH of the reaction medium?
Answer: (A)
The diazonium ion is a weak electrophile, so the coupling partner must be highly activated, but the diazonium ion itself must survive. For phenol, a mildly alkaline medium (pH 9–10) gives phenoxide, which is far more reactive than phenol, while the diazonium ion still persists. Much higher pH converts ArN₂⁺ into non-electrophilic diazotate. For aniline, a mildly acidic medium (pH 4–5) keeps most of the amine unprotonated and the diazonium ion stable. In strong acid, aniline becomes anilinium, which is deactivated. Only the first option satisfies both requirements.
Q7 · Chemistry · Amines
The carbylamine (isocyanide) test is given by:
Answer: (D)
Heated with chloroform and ethanolic KOH, primary amines give foul-smelling isocyanides (carbylamines). Secondary and tertiary amines do not, so the test identifies primary amines.
Q8 · Chemistry · Amines
Primary aromatic amines give arenediazonium salts that survive in solution at 273–278 K, whereas primary aliphatic amines under the same conditions give alkanediazonium salts that decompose at once with loss of N2. What is the main reason the arenediazonium ion is the more stable one?
Answer: (A)
Stability of an ion depends on how well its charge is dispersed. In ArN2⁺ the π electrons of the ring interact with the N≡N⁺ group, giving resonance structures in which the positive charge sits on ring carbons. In RN2⁺ there is no such delocalisation, so N2 is lost immediately, giving a carbocation and then mixtures of products. Resonance is therefore the reason. The +I effect and steric shielding do not apply, and the aryl cation is itself unstable.
Q9 · Chemistry · Amines
Four benzenediazonium chlorides, each carrying a single para substituent (–OCH3, –CH3, –H, –NO2), are separately added under identical conditions (273–278 K, pH about 9–10) to equal amounts of phenoxide ion. The coupling to form the p-hydroxyazobenzene dye is an electrophilic substitution in which the terminal nitrogen of the diazonium ion attacks the ring. Which diazonium ion couples fastest?
Answer: (A)
Coupling is an electrophilic substitution in which the arenediazonium ion is the electrophile and the phenoxide ion is the electron-rich nucleophile. The rate therefore depends on how electron-deficient the diazonium group is. Electron-withdrawing groups (–NO2) increase its electrophilicity, so coupling is faster. Electron-donating groups (–OCH3 by +M, –CH3 by +I and hyperconjugation) push electron density towards the N2+ group and reduce its electrophilicity. The order of reactivity is p-NO2 > H > p-CH3 > p-OCH3, so the p-nitrobenzenediazonium ion couples fastest. The mistake to avoid is applying the 'donor activates' rule to the electrophile instead of the nucleophile.
Q10 · Chemistry · Amines
Which of the following is a secondary amine?
Answer: (A)
An amine is primary, secondary or tertiary according to the number of alkyl or aryl groups on nitrogen. In N-methylethanamine, CH₃NHC₂H₅, nitrogen carries two alkyl groups and one hydrogen, so it is secondary.
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All 12 real NEET questions from Amines.
The questions from the actual papers, with the official answer and an explanation for each — chapter by chapter, or as full timed papers.
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About Amines in NEET.
How many questions come from Amines in NEET?
Amines (Chemistry, Class 12) has been asked 12 times in the 7 NEET papers from 2018, 2020, 2021, 2023, 2024, 2025 and 2026: about 1.7 questions a paper, 4% of all Chemistry questions. It is the 13th most-asked of 21 Chemistry chapters. NEET 2026 had 2 from it.
Which topics of Amines are asked most in NEET?
Classification and identification of amines (6), Basicity of amines (3), Diazonium salts (3), counted over the papers from 2018–2026.
Is Amines important for NEET 2027?
It is in the syllabus in force, so it can be asked. By past papers it ranks 13 of 21 in Chemistry; use that to decide how much time it gets, not whether to study it.
Where can I solve Amines NEET previous year questions?
On Nikaalle, free: sign up and practise every past NEET question from Amines with an explanation for each, or sit the full papers timed. The ten questions on this page are our own, written in the same pattern.
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