Multiple Choice Questions (MCQs) for Class 12 Chemistry: Unit 9 Amines
Access targeted multiple-choice questions for Unit 9 Amines designed to align with the latest CBSE academic syllabus for Class 12 Chemistry. These objective practice sets help students evaluate their conceptual understanding and improve exam readiness.
Practice Unit 9 Amines MCQs for Class 12 Chemistry
Access the complete set of multiple-choice questions for Unit 9 Amines below. This focused format allows students to isolate specific topics for thorough review and uninterrupted practice alongside official CBSE textbooks.
Multiple Choice Questions
CH2=CH2 Br2
⟶
CCl4 X NaCN
⟶ Y LiAlH4
⟶ Z
(a) X = CH2Br–CH2Br, Y = CH3CH2CH2CN, Z = CH3CH2CH2CH2NH2
(b) X = CH2Br–CH2Br, Y = CH3CH2CN, Z = CH3CH2CH2NH2
(c) X = CH3CH2Br, Y = CH3CH2CN, Z = CH3CH2CH2NH2
(d) X = CH2Br–CH2Br, Y = NCCH2CH2CN, Z = H2NCH2CH2CH2CH2NH2
Show Answer & Explanation
Answer: (d) X = CH2Br–CH2Br, Y = NCCH2CH2CN, Z = H2NCH2CH2CH2CH2NH2
Explanation:
1. CH2=CH2 Br2
⟶
CCl4 CH2Br–CH2Br (X) NaCN
⟶ NCCH2CH2CN (Y) LiAlH4
⟶ H2NCH2CH2CH2CH2NH2 (Z, 1,4-diaminobutane)
Teacher's Note:
1. Both bromine atoms of 1,2-dibromoethane are replaced by CN, so every later step happens at both ends of the chain.
(a) aniline
(b) chlorobenzene
(c) benzamide
(d) benzoyl chloride.
Show Answer & Explanation
Answer: (a) aniline
Explanation:
1. C6H5COOH (benzoic acid) SOCl2
⟶ C6H5COCl (benzoyl chloride, X) NH3
⟶ C6H5CONH2 (benzamide, Y) Br2
⟶
KOH C6H5NH2 (aniline, Z)
Teacher's Note:
1. The last step is Hoffmann bromamide degradation, which removes the C=O carbon and leaves NH2 directly on the ring.
(a) 2° R–Br + NH3
(b) 2° R–Br + NaCN followed by H2/Pt
(c) 1° R–NH2 + RCHO followed by H2/Pt
(d) 1° R–Br (2 moles) + potassium phthalimide followed by H3O+/heat.
Show Answer & Explanation
Answer: (c) 1° R–NH2 + RCHO followed by H2/Pt
Explanation:
1. R–NH2 + RCHO Reductive
⟶
amination [R–N=CHR] H2/Pt
⟶ R–NH–CH2–R (2° amine)
Teacher's Note:
1. Reductive amination puts exactly one new alkyl group on the nitrogen, so it avoids the mixture of amines that ammonolysis gives.
(a) p-Aminoacetophenone
(b) Sulphanilic acid
(c) p-Nitroaminobenzene
(d) p-Methoxyphenol
Show Answer & Explanation
Answer: (b) Sulphanilic acid
Explanation:
1. Sulphanilic acid, p-H2N–C6H4–SO3H, has an acidic –SO3H group and a basic –NH2 group in the same molecule. Proton transfer gives the zwitter ion p-H3N+–C6H4–SO3−.
2. p-H2NC6H4SO3H ⇌ p-H3N+C6H4SO3− (zwitter ion)
Teacher's Note:
1. A zwitter ion needs both an acidic group and a basic group in one molecule; check for both before choosing.
(a) (C2H5)3N
(b) (C2H5)2NH
(c) C2H5NH2
(d) C3H7NHC2H5
Show Answer & Explanation
Answer: (c) C2H5NH2
Explanation:
1. Only primary amines will give carbylamine reaction.
2. C2H5NH2 + CHCl3 + 3KOH (alc.) Heat
⟶ C2H5NC + 3KCl + 3H2O
Teacher's Note:
1. Carbylamine test needs two H atoms on nitrogen, so it works for 1° amines (aliphatic or aromatic) only.
(a) CH3NH2
(b) C2H5NH2
(c) C3H7NH2
(d) C6H5NH2
Show Answer & Explanation
Answer: (d) C6H5NH2
Explanation:
1. Aryl amines are less basic than alkyl amines because the phenyl group exerts −I effect and the lone pair of electrons on nitrogen is in conjugation, resulting in decrease in electron density on nitrogen atom for protonation.
Teacher's Note:
1. Any amine whose nitrogen lone pair is in resonance with a benzene ring is weaker than the simple alkylamines.
C4H9NH2, (C2H5)2NH, C2H5N(CH3)2
(a) C2H5N(CH3)2 > (C2H5)2NH > C4H9NH2
(b) (C2H5)2NH > C2H5N(CH3)2 > C4H9NH2
(c) C4H9NH2 > (C2H5)2NH > C2H5N(CH3)2
(d) (C2H5)2NH > C4H9NH2 > C2H5N(CH3)2
Show Answer & Explanation
Answer: (c) C4H9NH2 > (C2H5)2NH > C2H5N(CH3)2
Explanation:
1. The boiling points of amines vary in the order of primary > secondary > tertiary.
2. This is because the extent of intermolecular hydrogen bonding decreases as the number of H atoms on nitrogen decreases.
Teacher's Note:
1. For isomeric amines, count the H atoms on N: two (1°) > one (2°) > none (3°) gives the boiling point order.
(a) they have highest molecular mass
(b) they do not form hydrogen bonds
(c) they are more polar in nature
(d) they are most basic in nature.
Show Answer & Explanation
Answer: (b) they do not form hydrogen bonds
Explanation:
1. Primary and secondary amines can form hydrogen bonds whereas tertiary amines fail to do so. Hence, their boiling points are lowest.
Teacher's Note:
1. Isomers have the same molecular mass, so mass cannot explain differences in their boiling points.
(a) CH2=CH–CH2–CH=CH2
(b) (CH3)2N–CH=CH–CH2–CH2–CH3
(c) (CH3)2N–CH2–CH2–CH2–CH=CH2
(d) (CH3)2N–CH2–CH=CH–CH2–CH3
Show Answer & Explanation
Answer: (c) (CH3)2N–CH2–CH2–CH2–CH=CH2
Explanation:
1. OH− removes a hydrogen from the β-carbon (next to the ring carbon attached to N+) and the C–N bond breaks, with loss of H2O (Hofmann elimination).
2. The ring opens and the double bond forms at the end of the chain: (CH3)2N–CH2CH2CH2CH=CH2.
Teacher's Note:
1. In Hofmann elimination the ring opens, so nitrogen stays at one end of the chain and the new C=C appears at the other end.
(a) Hoffmann
(b) Claisen
(c) Perkin
(d) Kekule.
Show Answer & Explanation
Answer: (a) Hoffmann
Explanation:
1. Amides are converted into primary amines with one carbon atom less by treatment with Br2 and NaOH/KOH; this is Hoffmann bromamide degradation.
2. RCONH2 + Br2 + 4NaOH → RNH2 + Na2CO3 + 2NaBr + 2H2O
Teacher's Note:
1. Remember that Hoffmann bromamide degradation shortens the carbon chain by one carbon.
(a) N-ethylbenzenesulphonamide, insoluble in alkali
(b) N, N-diethylbenzenesulphonamide, soluble in alkali
(c) N, N-diethylbenzenesulphonamide, insoluble in alkali
(d) N-ethylbenzenesulphonamide, soluble in alkali.
Show Answer & Explanation
Answer: (d) N-ethylbenzenesulphonamide, soluble in alkali.
Explanation:
1. C6H5SO2Cl + H–NH–C2H5 → C6H5SO2–NH–C2H5 + HCl
2. N-Ethylbenzenesulphonamide still has an H on nitrogen, which is acidic due to the strongly electron-withdrawing sulphonyl group, so it is soluble in alkali.
Teacher's Note:
1. The H left on nitrogen in the sulphonamide of a 1° amine is what makes it dissolve in KOH; a 2° amine product has no such H.
(a) +I effect
(b) steric effect
(c) solvation effect
(d) all of these.
Show Answer & Explanation
Answer: (d) all of these.
Explanation:
1. In aqueous solution, the basic strength of alkyl amines is decided by the combined effect of the +I effect of alkyl groups, steric hindrance and solvation (H-bonding) of the protonated amine.
Teacher's Note:
1. This combination is why the order in water (2° > 1° > 3° for methylamines) differs from the order in the gas phase.
(a) Butanamide
(b) Propanamide
(c) Acetamide
(d) Benzamide
Show Answer & Explanation
Answer: (b) Propanamide
Explanation:
1. CH3CH2CONH2 (propanamide) Br2/NaOH
⟶
(NaOBr) CH3CH2NH2 + NaBr + Na2CO3 + H2O
Teacher's Note:
1. To find the amide, add one carbon to the amine you want: ethylamine (C2) comes from a C3 amide.
(a) chlorobenzene
(b) bromobenzene
(c) o-bromochlorobenzene (1-bromo-2-chlorobenzene)
(d) o-dibromobenzene (1,2-dibromobenzene)
Show Answer & Explanation
Answer: (c) o-bromochlorobenzene (1-bromo-2-chlorobenzene)
Explanation:
1. o-ClC6H4NH2 NaNO2
⟶
HCl o-ClC6H4N2+Cl− CuBr
⟶
HBr o-ClC6H4Br
Teacher's Note:
1. In Sandmeyer reaction only the –N2+ group is replaced; the other substituents on the ring stay where they are.
(a) p-CH3C6H4–CN
(b) p-CH3C6H4–NC
(c) p-CH3C6H4–NHCH3
(d) p-CH3C6H4–N2+Cl−
Show Answer & Explanation
Answer: (b) p-CH3C6H4–NC
Explanation:
1. p-CH3C6H4NH2 (p-toluidine, 4-methylaniline) + CHCl3 + 3KOH (alc.) Δ
⟶ p-CH3C6H4NC + 3KCl + 3H2O
Teacher's Note:
1. The carbylamine product is an isocyanide (–NC, nitrogen attached to the ring), not a cyanide (–CN).
(a) N, N-dialkyl acetamide
(b) N, N-dialkyl amine
(c) N, N-dialkyl amide
(d) acetyl dialkyl amine.
Show Answer & Explanation
Answer: (a) N, N-dialkyl acetamide
Explanation:
1. R2NH (secondary amine) + CH3COCl (acetyl chloride) → R2NCOCH3 (N, N-dialkyl acetamide or dialkylethanamide)
Teacher's Note:
1. The base is added to remove the HCl formed, which shifts the equilibrium towards the amide.
(a) aniline gets protonated with strong acids to give anilinium ion which is m-directing
(b) nitration requires nitric acid which oxidises –NH2 to –NO2 group
(c) electrophile NO2+ is a m-directing group
(d) benzene ring exerts +I effect and deactivates the ring.
Show Answer & Explanation
Answer: (a) aniline gets protonated with strong acids to give anilinium ion which is m-directing
Explanation:
1. Anilinium ion formed by protonation of aniline deactivates o- and p-positions, hence substitution takes place at m-position.
2. C6H5NH2 H+
⟶ C6H5NH3+ (anilinium ion) HNO3
⟶
H2SO4 m-O2NC6H4NH3+ −H+
⟶ m-O2NC6H4NH2 (m-nitroaniline)
Teacher's Note:
1. –NH3+ has no lone pair and pulls electrons by −I effect, so it behaves like a typical meta-directing group.
(a) CH3NH2
(b) CH3NHCH3
(c) CH3CONH2
(d) (CH3)3N
Show Answer & Explanation
Answer: (c) CH3CONH2
Explanation:
1. Amines are basic in nature while amides are amphoteric in nature.
2. In CH3CONH2 the lone pair on N is delocalised over the C=O group, so it is a very weak base and its N–H is weakly acidic.
Teacher's Note:
1. Amides react both with strong acids and with strong bases (e.g. Na/Hg forms salts), which is why they are called amphoteric.
(a) nitrogen forms three sp3 hybridised sigma bonds with carbon atoms of methyl groups and there is one non-bonding electron pair
(b) nitrogen forms three sp2 hybridised sigma bonds with carbon atoms of methyl groups and fourth orbital forms pi bond
(c) nitrogen has five valencies which are arranged in pyramidal shape.
(d) the unpaired electron present on nitrogen is delocalised.
Show Answer & Explanation
Answer: (a) nitrogen forms three sp3 hybridised sigma bonds with carbon atoms of methyl groups and there is one non-bonding electron pair
Explanation:
1. Nitrogen forms three sp3 hybridised sigma bonds with carbon atoms of methyl groups and has a non-bonding electron pair in the fourth sp3 orbital. Thus (CH3)3N has pyramidal shape.
Teacher's Note:
1. The lone pair compresses the C–N–C angle to about 108°, slightly less than the tetrahedral 109.5°.
(a) C6H5NO2
(b) C6H5CH2NH2
(c) C6H5CONH2
(d) C6H5NHCH3
Show Answer & Explanation
Answer: (c) C6H5CONH2
Explanation:
1. Amides on reaction with Br2 and KOH will give amine containing one carbon atom less. This reaction is called Hoffmann bromamide degradation.
2. C6H5CONH2 + Br2 + 4KOH Δ
⟶ C6H5NH2 + 2KBr + K2CO3 + 2H2O
Teacher's Note:
1. Only a primary amide (–CONH2) can undergo this reaction; look for the C=O attached to NH2.
(a) In aqueous solution (CH3)2NH is a stronger base than (CH3)3N.
(b) Secondary amines show carbylamine reaction.
(c) Nitrogen gas is evolved when ethylamine is treated with nitrous acid.
(d) Secondary amines can show metamerism.
Show Answer & Explanation
Answer: (b) Secondary amines show carbylamine reaction.
Explanation:
1. Only primary amines show carbylamine reaction. Hence statement (b) is not true.
Teacher's Note:
1. Read 'not true' questions slowly; here three statements are correct facts and only one is false.
(a) aromatic primary amines
(b) aromatic secondary amines
(c) aromatic tertiary amines
(d) aromatic amides.
Show Answer & Explanation
Answer: (a) aromatic primary amines
Explanation:
1. C6H5NO2 Sn/HCl
⟶ C6H5NH2
Teacher's Note:
1. The nitrogen of –NO2 is attached to one carbon only, so its reduction can give only a primary amine.
(a) C6H5NH2 < NH3 < (CH3)2NH < CH3NH2
(b) CH3NH2 < (CH3)2NH < NH3 < C6H5NH2
(c) C6H5NH2 < NH3 < CH3NH2 < (CH3)2NH
(d) (CH3)2NH < CH3NH2 < NH3 < C6H5NH2
Show Answer & Explanation
Answer: (c) C6H5NH2 < NH3 < CH3NH2 < (CH3)2NH
Explanation:
1. C6H5NH2 < NH3 < CH3NH2 < (CH3)2NH
2. Aniline is weakest due to resonance; alkyl groups (+I effect) make methylamines stronger than NH3, and in water (CH3)2NH is the strongest of these.
Teacher's Note:
1. Check the pKb values: aniline 9.38, NH3 4.75, CH3NH2 3.38, (CH3)2NH 3.27 – lower pKb means stronger base.
(a) benzamides
(b) ethanamides
(c) imides
(d) imines.
Show Answer & Explanation
Answer: (a) benzamides
Explanation:
1. Primary amines react with benzoyl chloride to give benzamides and the reaction is known as benzoylation.
2. CH3NH2 (methanamine) + C6H5COCl (benzoyl chloride) → CH3NHCOC6H5 (N-methylbenzamide) + HCl
Teacher's Note:
1. Acylation with acetyl chloride gives ethanamides; with benzoyl chloride it gives benzamides.
⟶ X Ni/H2
⟶ Y Acetic
⟶
anhydride Z
Z in the above reaction is
(a) CH3CH2CH2NHCOCH3
(b) CH3CH2CH2NH2
(c) CH3CH2CH2CONHCH3
(d) CH3CH2CH2CONHCOCH3
Show Answer & Explanation
Answer: (a) CH3CH2CH2NHCOCH3
Explanation:
1. CH3CH2Cl NaCN
⟶ CH3CH2CN (X) Ni/H2
⟶ CH3CH2CH2NH2 (Y) (CH3CO)2O
⟶ CH3CH2CH2NHCOCH3 (Z)
Teacher's Note:
1. The cyanide step adds one carbon, so the amine formed has one carbon more than the alkyl halide.
(a) sulphanilic acid
(b) benzenesulphonic acid
(c) aniline
(d) anthranilic acid.
Show Answer & Explanation
Answer: (a) sulphanilic acid
Explanation:
1. C6H5NH3+HSO4− 453-473 K
⟶ p-H2NC6H4SO3H (sulphanilic acid) Rearrangement
⟶ p-H3N+C6H4SO3− (zwitter ion)
Teacher's Note:
1. Sulphanilic acid is the para product, because –NH2 directs to the less hindered p-position.
(a) CH3CH2CH2NH2
(b) (CH3)3N
(c) CH3–NH–CH2CH3
(d) CH3–CH(NH2)–CH3
Show Answer & Explanation
Answer: (c) CH3–NH–CH2CH3
Explanation:
1. Since the compound reacts with benzenesulphonyl chloride to give a product which is insoluble in alkali, it shows there is no H attached to N in the product. Hence, the compound X is a secondary amine.
2. CH3–NH–C2H5 + C6H5SO2Cl → CH3–N(C2H5)–SO2C6H5 (N-ethyl-N-methylbenzenesulphonamide) + HCl
Teacher's Note:
1. A 3° amine does not react at all with Hinsberg's reagent, so 'reacts but insoluble in alkali' points only to a 2° amine.
Assertion–Reason Questions
Assertion (A): Ortho substituted anilines are usually weaker bases than anilines.
Reason (R): This is due to ortho effect.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (a) Both A and R are true, and R is the correct explanation of A.
Explanation:
1. Ortho effect is a consequence of steric and electronic factors.
2. Hence, ortho substituted anilines are usually weaker bases than aniline, whatever the nature of the substituent.
Teacher's Note:
1. o-Toluidine is weaker than aniline even though –CH3 releases electrons – a classic example of the ortho effect.
Assertion (A): In Hoffmann bromamide reaction, the amine formed has one carbon atom less than the parent 1° amide.
Reason (R): N-methyl acetamide undergoes Hoffmann bromamide reaction.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (c) A is true, but R is false.
Explanation:
1. Only primary amines can be prepared from amides (RCONH2) by treating with Br2 and KOH. Thus, N-methyl acetamide i.e., CH3CONHCH3 does not undergo Hoffmann bromamide reaction.
Teacher's Note:
1. Hoffmann bromamide reaction needs two H atoms on the amide nitrogen; a secondary amide such as CH3CONHCH3 does not react.
Assertion (A): In strongly acidic solutions, aniline becomes more reactive towards electrophilic reagents.
Reason (R): The amino group being completely protonated in strongly acidic solution, the lone pair of electrons on the nitrogen is no longer available for resonance.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (d) A is false, but R is true.
Explanation:
1. In strongly acidic medium, aniline gets protonated and so the lone pair of electrons is not available to produce +E or +M effects. On the other hand, the –NH3+ group exerts strong −I effect and thus it causes the deactivation of the ring.
2. Hence the assertion is false and the reason is true.
Teacher's Note:
1. Protonation turns the activating –NH2 into the deactivating –NH3+; that is why aniline becomes less reactive in strong acid.
Assertion (A): Ammonia is more basic than water.
Reason (R): Nitrogen is less electronegative than oxygen.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (a) Both A and R are true, and R is the correct explanation of A.
Explanation:
1. Ammonia is more basic than water. It is because nitrogen being less electronegative than oxygen, has a greater tendency to donate electrons.
Teacher's Note:
1. The less electronegative atom holds its lone pair less tightly, so it donates the pair to a proton more readily.
Assertion (A): Me3N reacts with BF3 whereas Ph3N does not.
Reason (R): The electron pair on nitrogen atom in Ph3N is delocalised in the benzene ring and is not available for boron in BF3.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (a) Both A and R are true, and R is the correct explanation of A.
Explanation:
1. Me3N is a Lewis base and donates its lone pair to the electron deficient boron of BF3 to form the adduct Me3N→BF3.
2. In Ph3N the lone pair on nitrogen is delocalised over the three benzene rings, so it is not available for donation to BF3. Hence the reason correctly explains the assertion.
Teacher's Note:
1. BF3 is a Lewis acid, so its reaction with an amine is a test of how available the nitrogen lone pair is.
Assertion (A): Aniline is a weaker base than cyclohexylamine.
Reason (R): Aniline undergoes halogenation even in the absence of a catalyst.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (b) Both A and R are true, but R is not the correct explanation of A.
Explanation:
1. Aniline exists as resonance hybrid. As a result of resonance, the lone pair of electrons on nitrogen gets delocalized over the benzene ring and thus, is less easily available for protonation than in case of cyclohexylamine where no such resonance takes place.
2. The reason is also true (the ring is highly activated by –NH2), but it does not explain the assertion.
Teacher's Note:
1. Both statements come from the same resonance of the lone pair, but the reason describes ring reactivity, not basicity.
Assertion (A): Controlled nitration of aniline at low temperature mainly gives m-nitroaniline.
Reason (R): In acidic medium, –NH2 group gets converted into m-directing group.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (a) Both A and R are true, and R is the correct explanation of A.
Explanation:
1. Under acidic condition, aniline gets protonated to anilinium ion (–NH3+ group). This is deactivating and m-directing group. Thus, controlled nitration of aniline mainly gives m-nitroaniline.
Teacher's Note:
1. Direct nitration of aniline gives a mixture with about 47% m-nitroaniline, together with p-nitroaniline and oxidation products.
Assertion (A): Ammonolysis of alkyl halides involves the reaction between alkyl halides and alcoholic ammonia.
Reason (R): Reaction can be used to prepare 1°, 2°, 3° amines and finally quaternary ammonium salts.
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation
Answer: (b) Both A and R are true, but R is not the correct explanation of A.
Explanation:
1. Alkyl halides react with alcoholic ammonia (in a sealed tube at about 373 K) to give amines; this is ammonolysis.
2. RX + NH3 → RNH2 RX
⟶ R2NH RX
⟶ R3N RX
⟶ R4N+X−
3. Both statements are true, but the reason only states what products can form; it does not explain the assertion.
Teacher's Note:
1. The main drawback of ammonolysis is that it gives a mixture of amines; a large excess of NH3 favours the primary amine.
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