JEE (Advanced) · Chemistry, Class 12
Amines: JEE Advanced PYQ, 2007–2026
Amines (Chemistry, Class 12) has been asked 26 times in the 40 JEE (Advanced) papers from 2007–2026: about 0.7 questions a paper, 3% of all Chemistry questions. It is the 15th most-asked of 31 Chemistry chapters. The two 2026 papers had 1 from it. Last checked 28 October 2026.
Questions per year, both papers
Year by year.
| Year | Questions |
|---|---|
| 2007 | 0in 2 papers |
| 2008 | 1in 2 papers |
| 2009 | 0in 2 papers |
| 2010 | 1in 2 papers |
| 2011 | 2in 2 papers |
| 2012 | 0in 2 papers |
| 2013 | 0in 2 papers |
| 2014 | 2in 2 papers |
| 2015 | 2in 2 papers |
| 2016 | 2in 2 papers |
| 2017 | 2in 2 papers |
| 2018 | 3in 2 papers |
| 2019 | 1in 2 papers |
| 2020 | 2in 2 papers |
| 2021 | 1in 2 papers |
| 2022 | 2in 2 papers |
| 2023 | 3in 2 papers |
| 2024 | 0in 2 papers |
| 2025 | 1in 2 papers |
| 2026 | 1in 2 papers |
Each year has two papers, and the number of questions has fallen over time, so compare per paper and expect some drift.
Inside the chapter
Topics asked most.
- Diazonium salts14
- Classification and identification of amines9
- Basicity of amines3
Try it
10 Amines questions to practise.
Our own questions, written in the JEE Advanced pattern and checked, each with an explanation. Pick an option to see the answer.
Q1 · 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.
Q2 · 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.
Q3 · 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.
Q4 · 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.
Q5 · 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.
Q6 · 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.
Q7 · Chemistry · Amines
Which of the following reaction sequences, starting from nitrobenzene, gives 1-bromo-3-chlorobenzene as the main product?
Answer: (A)
Plan the route by the directing effects. Bromine must be introduced while the meta-directing –NO2 is present, so brominate first to get 1-bromo-3-nitrobenzene. Then reduce –NO2 to –NH2 (Sn/HCl). Diazotise the amine at 273–278 K, and replace the diazonium group with Cl using CuCl/HCl (Sandmeyer). The Cl takes the position of the original –NO2, which is meta to Br. The other routes either use the wrong order of substitution, skip the reduction, or over-brominate.
Q8 · Chemistry · Amines
50.0 mL of 0.200 M methylamine solution (pKb = 3.36 at 298 K, pKw = 14.00) is mixed with 25.0 mL of 0.200 M hydrochloric acid. Assuming volumes are additive and that the amine and its salt do not undergo any other reaction, what is the pH of the resulting solution?
Answer: (A)
Moles of CH₃NH₂ = 50.0 × 0.200 = 10.0 mmol. Moles of HCl = 25.0 × 0.200 = 5.0 mmol. HCl reacts completely: CH₃NH₂ + H⁺ → CH₃NH₃⁺. This leaves 5.0 mmol CH₃NH₂ and 5.0 mmol CH₃NH₃⁺ in the same volume, so their concentrations are equal and it is a basic buffer. For a base buffer, pOH = pKb + log([salt]/[base]) = 3.36 + log 1 = 3.36. So pH = 14.00 − 3.36 = 10.64. Equivalently, pH = pKa of CH₃NH₃⁺ = 10.64.
Q9 · Chemistry · Amines
Ethanamine, N-ethylethanamine and N,N-diethylethanamine are each treated separately with NaNO2 and dilute HCl at 273–278 K. Which option correctly describes what is observed for each, in that order?
Answer: (A)
Nitrous acid (NaNO2 + HCl) distinguishes the three classes of aliphatic amine. Primary: RNH2 gives an unstable diazonium salt that decomposes at once to an alcohol and N2, seen as effervescence. Secondary: R2NH gives an N-nitrosoamine, a yellow oily liquid. Tertiary: R3N has no N–H and only dissolves as the trialkylammonium nitrite salt, with no gas and no oil. Ethanamine, N-ethylethanamine and N,N-diethylethanamine are primary, secondary and tertiary respectively, so the first option is correct.
Q10 · Chemistry · Amines
Consider all amines of molecular formula C4H11N (primary, secondary and tertiary). Counting each enantiomer as a separate stereoisomer, how many of these stereoisomers liberate nitrogen gas when treated with NaNO2 and dilute HCl at 273–278 K? (Ignore nitrogen inversion.)
Answer: (B)
Only primary aliphatic amines react with nitrous acid to give an unstable diazonium salt that decomposes to an alcohol and N2 gas. Secondary amines give N-nitrosoamines (yellow oils) and tertiary amines give water-soluble nitrite salts, with no gas. The primary C4H11N isomers are: (1) CH3CH2CH2CH2NH2, (2) CH3CH2CH(NH2)CH3, (3) (CH3)2CHCH2NH2 and (4) (CH3)3CNH2. Of these, only (2) has a stereocentre, so it exists as two enantiomers. Each of the other three has no stereocentre, so the total is 3 + 2 = 5. Tert-butylamine still gives N2 because it is a primary amine (the tertiary label refers only to its alkyl group).
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All 26 real JEE Advanced questions from Amines.
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About Amines in JEE Advanced.
How many questions come from Amines in JEE Advanced?
Amines (Chemistry, Class 12) has been asked 26 times in the 40 JEE (Advanced) papers from 2007–2026: about 0.7 questions a paper, 3% of all Chemistry questions. It is the 15th most-asked of 31 Chemistry chapters. The two 2026 papers had 1 from it.
Which topics of Amines are asked most in JEE Advanced?
Diazonium salts (14), Classification and identification of amines (9), Basicity of amines (3), counted over the papers from 2007–2026.
Is Amines important for JEE Advanced 2027?
It is in the syllabus in force, so it can be asked. By past papers it ranks 15 of 31 in Chemistry; use that to decide how much time it gets, not whether to study it.
Where can I solve Amines JEE Advanced previous year questions?
On Nikaalle, free: sign up and practise every past JEE Advanced 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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