Answer key (D) & worked solution
Correct answer: Option D
Amines carries 4.1 of 100 Chemistry marks in NEET by past-paper share: 29 of the 707 tagged NEET Chemistry PYQs in the ViJEEta bank (NEET 2013–2026). It ranks 10 of 26 Chemistry chapters.
How the weight is computed: this chapter's tagged NEET PYQs divided by all tagged NEET Chemistry PYQs in the bank (29 of 707), times 100, with repeated questions counted once. The bank does not store the year of each NEET question, so none is printed. Computed from the bank at build time. Updated 19 Oct 2026.
16 facts the paper keeps asking, each with how it is asked. Copy it onto one page of your notebook: that page is this chapter's handbook.
N in amines is with one lone pair, so amines are pyramidal; the C–N–C angle in trimethylamine is , below .
Asked as: hybridisation and shape of , or why its bond angle is less than tetrahedral.
is methanamine, is N,N-diethylethanamine, and aniline is also accepted as benzenamine.
Asked as: pick the correct IUPAC name of a given or amine.
Nitro compounds are reduced best with Fe scrap + , because the formed hydrolyses and gives back , so only a little acid is needed.
Asked as: why is preferred over other metal/acid pairs.
Ammonolysis uses ethanolic in a sealed tube at . It gives a mixture of , and amines plus a quaternary salt. A large excess of makes the amine the major product. Halide reactivity: .
Asked as: the drawback of ammonolysis, or how to get mainly the amine.
Nitriles reduced by or /catalyst give amines with one carbon more, so this is an ascent of the amine series.
Asked as: a conversion such as (via , then reduction).
Gabriel phthalimide synthesis (ethanolic , then , then alkaline hydrolysis) gives only amines. It cannot give aromatic amines, because aryl halides do not undergo substitution with the phthalimide anion.
Asked as: why aniline cannot be made by Gabriel synthesis.
Hoffmann bromamide degradation (amide + + ) gives a amine with one carbon less than the amide. For example, benzamide gives aniline and butanamide gives propanamine.
Asked as: name the starting amide, or the product of .
Boiling points of isomeric amines follow , since amines have no N–H for H-bonding. Alcohols boil higher than amines: butan-1-ol is , butan-1-amine is .
Asked as: arrange in increasing b.p., e.g. , , .
values: is , aliphatic amines are to (stronger bases, effect), and aniline is (weaker base). A smaller means a stronger base.
Asked as: why the of aniline is more than that of methylamine.
In the gas phase basicity follows . In water: , and . The water order comes from the combined inductive effect, solvation and steric hindrance.
Asked as: the correct basic-strength order in aqueous solution.
Aniline has 5 resonance structures but the anilinium ion has only 2, so aniline is less basic than . Groups like and raise basicity, while , , and lower it.
Asked as: order aniline, p-nitroaniline and p-toluidine by basic strength.
In the carbylamine test, a amine (aliphatic or aromatic) heated with + ethanolic gives a foul-smelling isocyanide. and amines do not respond.
Asked as: a test for amines, or to tell methylamine from dimethylamine.
With ( + mineral acid), aliphatic amines release quantitatively and form alcohols. Aromatic amines at form diazonium salts (diazotisation).
Asked as: which isomers liberate with nitrous acid.
Hinsberg's reagent (): the sulphonamide has an acidic N–H and dissolves in alkali. The sulphonamide has no N–H and is insoluble in alkali. amines do not react. p-Toluenesulphonyl chloride is now used instead.
Asked as: distinguish or separate , and amines.
Bromine water gives a white precipitate of 2,4,6-tribromoaniline. Acetylation is used to tame the group. Direct nitration gives a substantial amount of meta product, because the anilinium ion is meta-directing. Aniline does not undergo Friedel–Crafts, because it forms a salt with .
Asked as: why nitration of aniline gives m-nitroaniline, or how to make p-bromoaniline.
with , or is the Sandmeyer reaction, which gives a better yield than Gattermann (Cu powder + ). gives iodobenzene. followed by heating gives aryl fluoride. or ethanol gives the arene. Warming to gives phenol. Coupling with phenol at the para position gives p-hydroxyazobenzene.
Asked as: identify the reagent or product for a given diazonium conversion.
In water, 3° amines are not the strongest bases. The order 3° > 2° > 1° holds only in the gas phase; in water the 2° amine is the strongest, and the methyl and ethyl series have different orders.
7 real previous-year questions from this chapter, as printed in the paper, each with its worked solution. Answer the first one here with no sign-in.
Correct answer: Option D
Correct answer: Option D
Correct answer: Option D
Correct answer: Option C
Correct answer: Option A
Correct answer: Option B
Correct answer: Option A
All 10 topics, by tagged PYQs, then practice questions. Counted from the bank. Updated 19 Oct 2026.
| Topic | PYQs | Practice questions |
|---|---|---|
| Chemical Properties of Amines | 10 | 150 |
| Properties of Aniline | 10 | 83 |
| Preparation of Amines | 5 | 63 |
| Introduction to Amines | 3 | 39 |
| Test of Amines | 1 | 30 |
| Physical Properties of Amines | 0 | 11 |
| Preparation of Aniline | 0 | 7 |
| Introduction to Aniline | 0 | 6 |
| Miscellaneous Examples | 0 | 2 |
| Uses of Amines | 0 | 2 |
Unknown is not zero. Until you answer at least 3 graded questions in Amines, ViJEEta shows – instead of a zero or a 50% default. Book exercises can raise chapter mastery up to 0.60, adaptive practice up to 0.75, and only a timed chapter test or mock can certify 0.80 and above.
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