Amines NEET PYQReal NEET & AIPMT 2005–2021 questions, year-tagged, with solutions
By the RankUp team, founded by Dr. Deepak Sahu (MBBS Gold Medalist, NEET PG AIR 1731) · Updated
Every Amines question from the NEET & AIPMT papers on this site (2005–2021), newest first. Each is tagged with its paper. Tap an option to check your answer and read the explanation.
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Amines previous year questions
28 questions. Tap an option to check your answer.
Q1NEET 2021
Identify the compound that will react with Hinsberg's reagent to give a solid which dissolves in alkali.
Not quite — the answer is C.
A primary amine (CH₃CH₂NH₂) reacts with benzenesulphonyl chloride to give N-ethylbenzenesulphonamide, whose N–H is acidic, so it dissolves in alkali. The secondary amine gives a sulphonamide without N–H (insoluble in alkali), and tertiary amines and nitroalkanes do not react.
Q2NEET 2021
The reagent 'R' in the given sequence of chemical reaction is : 2,4,6-Tribromoaniline —(NaNO₂, HCl, 0–5°C)→ 2,4,6-Tribromobenzenediazonium chloride —R→ 1,3,5-Tribromobenzene
Not quite — the answer is B.
Ethanol (like H₃PO₂) reduces an arenediazonium salt, replacing –N₂⁺Cl⁻ by –H and itself being oxidised to ethanal. This converts 2,4,6-tribromobenzenediazonium chloride into 1,3,5-tribromobenzene. Water would give a phenol and CuCN/KCN a nitrile.
Q3NEET 2020 · Q102
Which of the following amine will give the carbylamine test ?
Not quite — the answer is A.
The carbylamine (isocyanide) test, heating with CHCl₃ and alcoholic KOH, is given only by primary amines, so aniline gives foul-smelling phenyl isocyanide. N-methylaniline and N-ethylaniline are secondary and N,N-dimethylaniline is tertiary, so none of them respond.
Q4NEET 2019 · Q90
The correct order of the basic strength of methyl substituted amines in aqueous solution is:
Not quite — the answer is A.
In water, basicity depends on inductive effect, solvation of the ammonium ion and steric hindrance together; for methyl amines this gives (CH₃)₂NH > CH₃NH₂ > (CH₃)₃N. Trimethylamine as strongest is the gas-phase order, where solvation does not act.
Q5NEET 2018 · Q55
Nitration of aniline in strong acidic medium also gives m-nitroaniline because
Not quite — the answer is D.
In strongly acidic nitrating mixture, aniline is protonated to the anilinium ion C₆H₅NH₃⁺, whose –NH₃⁺ group is electron-withdrawing and meta-directing. The free –NH₂ group itself is ortho/para-directing.
Q6NEET 2017 · Q14
The correct increasing order of basic strength for the following compounds is: (I) aniline, C₆H₅NH₂ (II) p-nitroaniline, p-O₂N–C₆H₄–NH₂ (III) p-toluidine, p-CH₃–C₆H₄–NH₂
Not quite — the answer is D.
The electron-donating –CH₃ at the para position increases electron density on nitrogen, so p-toluidine is more basic than aniline, while the –NO₂ group withdraws the lone pair by resonance and makes p-nitroaniline the weakest base. Hence II < I < III.
Q7NEET 2017 · Q45
Which of the following reactions is appropriate for converting acetamide to methanamine?
Not quite — the answer is B.
CH₃CONH₂ + Br₂ + 4NaOH → CH₃NH₂ + Na₂CO₃ + 2NaBr + 2H₂O: the Hofmann bromamide (hypobromite) degradation gives an amine with one carbon fewer, exactly acetamide to methanamine. Gabriel synthesis starts from an alkyl halide, Stephens reduction gives aldehydes from nitriles, and the carbylamine reaction is a test for primary amines.
Q8NEET 2016 Phase 2 · Q81
Which one of the following nitro-compounds does not react with nitrous acid?
Not quite — the answer is C.
Nitroalkanes react with nitrous acid through the hydrogen on the carbon carrying the nitro group (primary give nitrolic acids, secondary give pseudonitroles). 2-Methyl-2-nitropropane is a tertiary nitro compound with no α-hydrogen, so it does not react. The other compounds all have at least one α-H.
Q9NEET 2016 Phase 2 · Q85
A given nitrogen-containing aromatic compound A reacts with Sn/HCl, followed by HNO₂ to give an unstable compound B. B, on treatment with phenol, forms a beautiful coloured compound C with the molecular formula C₁₂H₁₀N₂O. The structure of compound A is
Not quite — the answer is B.
Sn/HCl reduces nitrobenzene to aniline; HNO₂ converts aniline to the unstable benzenediazonium salt, which couples with phenol to give p-hydroxyazobenzene, C₆H₅–N=N–C₆H₄OH (C₁₂H₁₀N₂O). Aniline needs no reduction, so Sn/HCl would be pointless as the first step.
Q10NEET 2016 Phase 1 · Q88
The correct statement regarding the basicity of arylamines is:
Not quite — the answer is A.
In aniline the nitrogen lone pair is delocalised into the benzene ring by resonance, so it is less available to accept a proton; arylamines are therefore weaker bases than alkylamines, whose alkyl groups push electron density onto nitrogen. The nitrogen in arylamines is not sp-hybridised.
Q11AIPMT 2015
The electrolytic reduction of nitrobenzene in strongly acidic medium produces:
Not quite — the answer is B.
In strongly acidic medium nitrobenzene is reduced electrolytically to phenylhydroxylamine, which rearranges in acid to p-aminophenol. Aniline is the product of complete reduction in mildly acidic conditions (e.g. Sn/HCl), and azo/azoxy compounds form in alkaline medium.
Q12AIPMT 2014
Which of the following will be most stable diazonium salt RN₂⁺X⁻?
Not quite — the answer is A.
Benzenediazonium salts are stabilised by resonance of the –N₂⁺ group with the ring and are stable below about 5°C. Aliphatic and benzyl diazonium salts lack this delocalisation and decompose immediately, losing N₂.
Q13AIPMT 2014
In the following reaction, the product (A) is: Benzenediazonium chloride (C₆H₅N₂⁺Cl⁻) + Aniline (C₆H₅NH₂) —H⁺→ (A) Yellow dye
Not quite — the answer is C.
In mildly acidic medium the diazonium ion couples with aniline at the para position (electrophilic substitution) to give p-aminoazobenzene, a yellow dye. The N-coupled diazoamino compound is not the dye described.
Q14NEET 2013
In the reaction 2-bromo-4-nitrobenzenediazonium chloride —(A)→ m-bromonitrobenzene A is
Not quite — the answer is C.
Removing the diazonium group at C-1 leaves Br at C-2 and NO₂ at C-4, which are meta to each other, so the reagent must replace –N₂⁺ by H. Hypophosphorous acid (H₃PO₂) with water does this. Cu₂Cl₂ would introduce Cl and H⁺/H₂O would give a phenol.
Q15AIPMT 2012 Mains
An organic compound C₃H₉N (A), when treated with nitrous acid, gave an alcohol and N₂ gas was evolved. (A) on warming with CHCl₃ and caustic potash gave (C) which on reduction gave isopropylmethylamine. Predict the structure of (A).
Not quite — the answer is A.
Evolving N₂ with nitrous acid and giving a carbylamine (isocyanide) with CHCl₃/KOH both show A is a primary amine. Reducing (CH₃)₂CH–NC gives (CH₃)₂CH–NH–CH₃ (isopropylmethylamine), so A is isopropylamine, (CH₃)₂CH–NH₂. n-Propylamine would give n-propylmethylamine instead.
Q16AIPMT 2011 Mains
An organic compound 'A' on treatment with NH₃ gives 'B' which on heating gives 'C'. 'C' when treated with Br₂ in the presence of KOH produces ethylamine. Compound 'A' is
Not quite — the answer is B.
Hoffmann bromamide degradation removes one carbon, so ethylamine must come from propanamide (compound 'C'), which forms on heating ammonium propanoate ('B'), made from propanoic acid ('A') and NH₃. Acetic acid would give methylamine and butanoic acid would give propylamine.
Q17AIPMT 2011 Mains
Which of the following compounds is most basic?
Not quite — the answer is D.
In benzylamine the –NH₂ is attached to a CH₂ group, so its lone pair is not delocalised into the ring and it behaves like an alkylamine. In aniline the lone pair is delocalised into the ring, a para –NO₂ withdraws it further, and in acetanilide it is also delocalised into the C=O group, so these are much weaker bases.
Q18AIPMT 2011 Prelims
What is the product obtained in the following reaction? C₆H₅NO₂ —(Zn, NH₄Cl)→ ?
Not quite — the answer is B.
Nitrobenzene in neutral medium (Zn dust with NH₄Cl) is reduced only as far as N-phenylhydroxylamine, C₆H₅NHOH. Aniline forms in acidic medium (Sn/HCl or Fe/HCl), and azo/azoxy compounds in alkaline reductions.
Q19AIPMT 2011 Prelims
In a set of reactions m-bromobenzoic acid (A) gave a product D: A —(SOCl₂)→ B —(NH₃)→ C —(NaOH, Br₂)→ D Identify the product D.
Not quite — the answer is D.
SOCl₂ converts the acid to m-bromobenzoyl chloride, NH₃ converts it to m-bromobenzamide, and Br₂/NaOH (Hoffmann bromamide degradation) removes the carbonyl carbon to give m-bromoaniline. The amide is only the intermediate C, not the final product.
Q20AIPMT 2010 Mains · Q108
Match the compounds given in List-I with their characteristic reactions given in List-II. Select the correct option. List-I: a. CH₃CH₂CH₂CH₂NH₂; b. CH₃C≡CH; c. CH₃CH₂COOCH₃; d. CH₃CH(OH)CH₃ List-II: (i) Alkaline hydrolysis; (ii) With KOH (alcohol) and CHCl₃ produces bad smell; (iii) Gives white ppt. with ammoniacal AgNO₃; (iv) With Lucas reagent cloudiness appears after 5 minutes
Not quite — the answer is C.
n-Butylamine, a primary amine, gives the foul-smelling isocyanide with CHCl₃/alcoholic KOH; propyne, a terminal alkyne, gives a white precipitate with ammoniacal AgNO₃; the ester undergoes alkaline hydrolysis; 2-propanol, a secondary alcohol, turns cloudy with Lucas reagent in about 5 minutes. So a(ii), b(iii), c(i), d(iv). Assigning the silver test to the amine misses that only terminal alkynes have an acidic H.
Q21AIPMT 2010 Prelims · Q83
Aniline in a set of the following reactions yielded a coloured product Y: C₆H₅NH₂ —(NaNO₂/HCl, 273–278 K)→ X —(N,N-dimethylaniline)→ Y The structure of Y would be
Not quite — the answer is B.
Aniline with NaNO₂/HCl at 273–278 K gives benzenediazonium chloride (X). This couples with N,N-dimethylaniline at the para position to give the yellow azo dye p-(dimethylamino)azobenzene, C₆H₅–N=N–C₆H₄–N(CH₃)₂. The other structures do not contain the unchanged phenyl ring from aniline joined to the dimethylamino ring.
Q22AIPMT 2010 Prelims · Q95
Acetamide is treated with the following reagents separately. Which one of these would yield methyl amine?
Not quite — the answer is B.
Hofmann bromamide degradation: CH₃CONH₂ + Br₂ + 4NaOH → CH₃NH₂, losing one carbon. PCl₅ or hot conc. H₂SO₄ dehydrate acetamide to acetonitrile, and soda lime does not give methylamine.
Q23AIPMT 2009 Prelims · Q178
Predict the product : C₆H₅–NHCH₃ + NaNO₂ + HCl → Product
Not quite — the answer is D.
N-methylaniline is a secondary amine; with HNO₂ (from NaNO₂ + HCl) the N–H hydrogen is replaced by –N=O, giving N-nitroso-N-methylaniline, C₆H₅N(CH₃)–N=O. Ring nitrosation happens with tertiary aromatic amines such as N,N-dimethylaniline.
Q24AIPMT 2008 Prelims · Q98
In a reaction of aniline a coloured product C was obtained. C₆H₅NH₂ (A) —(NaNO₂, HCl)→ B —(C₆H₅N(CH₃)₂, cold)→ C The structure of C would be
Not quite — the answer is A.
Diazotisation of aniline gives benzenediazonium chloride, the intermediate B. In the cold it couples with N,N-dimethylaniline at the activated para position, giving the coloured azo dye p-(dimethylamino)azobenzene. The –NH–NH– (hydrazo) compound has no N=N chromophore, so it would not be the coloured coupling product.
Q25AIPMT 2007 Prelims · Q51
Which one of the following on reduction with lithium aluminium hydride yields a secondary amine?
Not quite — the answer is A.
LiAlH₄ reduces methyl isocyanide CH₃–N≡C to CH₃–NH–CH₃, a secondary amine. Acetamide and methyl cyanide both give ethylamine (primary), and nitroethane also gives a primary amine.
Q26AIPMT 2006 Prelims · Q55
In a set of reactions propionic acid yielded a compound D. CH₃CH₂COOH —(SOCl₂)→ B —(NH₃)→ C —(KOH, Br₂)→ D The structure of D would be :
Not quite — the answer is D.
SOCl₂ converts the acid to propanoyl chloride, CH₃CH₂COCl, and NH₃ then gives propanamide, CH₃CH₂CONH₂. Br₂/KOH carries out the Hofmann bromamide degradation, removing one carbon to give ethylamine, CH₃CH₂NH₂, as the final compound. Propylamine would keep all three carbons, and propanamide is only the intermediate before the Hofmann step.
Q27AIPMT 2006 Prelims · Q88
Which of the following is more basic than aniline ?
Not quite — the answer is D.
In benzylamine, C₆H₅CH₂NH₂, the CH₂ group separates nitrogen from the ring, so its lone pair is not delocalised and it behaves like an alkylamine, more basic than aniline. In diphenylamine and triphenylamine the lone pair is spread over two or three rings, and the nitro group in p-nitroaniline withdraws electrons, so all three are weaker than aniline.
Q28AIPMT 2005 Prelims · Q55
Aniline in a set of reactions yielded a product D. C₆H₅NH₂ —(NaNO₂/HCl)→ A —(CuCN)→ B —(H₂/Ni)→ C —(HNO₂)→ D The structure of the product D would be :
Not quite — the answer is D.
Aniline → benzenediazonium chloride → (CuCN) benzonitrile → (H₂/Ni) benzylamine C₆H₅CH₂NH₂ → (HNO₂) benzyl alcohol C₆H₅CH₂OH. Benzylamine is only intermediate C; nitrous acid converts this primary aliphatic amine to the alcohol.
How often Amines came up
Questions from this chapter in each paper on this site.
- NEET 2021: 2
- NEET 2020: 1
- NEET 2019: 1
- NEET 2018: 1
- NEET 2017: 2
- NEET 2016 Phase 2: 2
- NEET 2016 Phase 1: 1
- AIPMT 2015: 1
- AIPMT 2014: 2
- NEET 2013: 1
- AIPMT 2012 Mains: 1
- AIPMT 2011 Mains: 2
- AIPMT 2011 Prelims: 2
- AIPMT 2010 Mains: 1
- AIPMT 2010 Prelims: 2
- AIPMT 2009 Prelims: 1
- AIPMT 2008 Prelims: 1
- AIPMT 2007 Prelims: 1
- AIPMT 2006 Prelims: 2
- AIPMT 2005 Prelims: 1
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