Practice MCQs for Class 12 Chemistry Unit 9 Amines
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Multiple Choice Questions
(a) amide
(b) imide
(c) secondary amine
(d) imine
Show Answer & Explanation
Answer: (a) amide
Explanation:
1. C2H5NH2 + (CH3CO)2O → CH3CONHC2H5 (N-Ethylacetamide) + CH3COOH
Teacher's Note:
1. This is acylation: the acyl group replaces an H on nitrogen, so 1° and 2° amines react but 3° amines do not.
(a) Aniline, C6H5NH2
(b) Piperidine, C5H10NH
(c) Cyclohexylamine, C6H11NH2
(d) CH3NH2
Show Answer & Explanation
Answer: (a) Aniline, C6H5NH2
Explanation:
1. In aniline the lone pair on nitrogen is delocalised into the benzene ring by resonance, so it is less available for protonation.
2. The other three are aliphatic amines whose lone pair is localised, so aniline is the weakest base.
Teacher's Note:
1. Cyclohexylamine and aniline look similar on paper, but only the aromatic ring pulls the lone pair into resonance.
⟶ A Δ
⟶ B Br2/KOH
⟶ C
The structure of ‘C’ would be
(a) CH3CH2CONH2
(b) CH3CH2NHCH3
(c) CH3CH2NH2
(d) CH3CH2CH2NH2
Show Answer & Explanation
Answer: (c) CH3CH2NH2
Explanation:
1. A is the ammonium salt CH3CH2COONH4, which on heating gives the amide B, CH3CH2CONH2.
2. Hoffmann bromamide degradation of B gives C, CH3CH2NH2, an amine with one carbon atom less.
Teacher's Note:
1. Hoffmann bromamide degradation always removes one carbon – count carbons in the amide and the product to check.
⟶
HCl B C6H5N(CH3)2
⟶
cold C
The structure of ‘C’ would be
(a) o-CH3C6H4–N=N–C6H4CH3-o
(b) o-CH3C6H4–NH–NH–C6H4–N(CH3)2-p
(c) C6H5–N=N–C6H4–N(CH3)2-p
(d) C6H5–N=N–CH2–N(CH3)–C6H5
Show Answer & Explanation
Answer: (c) C6H5–N=N–C6H4–N(CH3)2-p
Explanation:
1. Aniline with NaNO2/HCl at 273–278 K gives benzenediazonium chloride (B).
2. B couples with N,N-dimethylaniline at the para position to give p-(dimethylamino)azobenzene (C), a yellow dye.
Teacher's Note:
1. Coupling takes place at the para position of the activated ring; the –N=N– azo link is retained.
(a) reduction of nitrobenzene with H2/Pd in ethanol
(b) potassium salt of phthalimide treated with chlorobenzene
(c) hydrolysis of phenyl isocyanide with acidic solution
(d) degradation of benzamide with bromine in alkaline medium
Show Answer & Explanation
Answer: (b) potassium salt of phthalimide treated with chlorobenzene
Explanation:
1. Gabriel phthalimide synthesis needs nucleophilic substitution of the halide by the phthalimide anion.
2. Aryl halides such as chlorobenzene do not undergo nucleophilic substitution easily, so aromatic primary amines cannot be made this way.
Teacher's Note:
1. Gabriel synthesis gives only primary aliphatic amines – a favourite exam statement.
CH3CH2NH2 + CHCl3 + 3KOH → A + B + 3H2O
The compounds A and B are respectively:
(a) CH3CH2CONH2 and 3KCl
(b) CH3CH2NC and K2CO3
(c) CH3CH2NC and 3KCl
(d) CH3CH2CN and 3KCl
Show Answer & Explanation
Answer: (c) CH3CH2NC and 3KCl
Explanation:
1. This is the carbylamine reaction of a primary amine.
2. CH3CH2NH2 + CHCl3 + 3KOH → CH3CH2NC + 3KCl + 3H2O; the foul-smelling ethyl isocyanide is formed.
Teacher's Note:
1. Only primary amines give the carbylamine test; balance it by noting that 3 Cl atoms end up as 3KCl.
(a) (CH3)2NH
(b) C6H5NH2
(c) CH3NH2
(d) (CH3)3N
Show Answer & Explanation
Answer: (a) (CH3)2NH
Explanation:
1. In aqueous solution the order of basicity of methylamines is (CH3)2NH > CH3NH2 > (CH3)3N, due to combined inductive, solvation and steric effects.
2. Aniline is the weakest base. The strongest base, (CH3)2NH, has the smallest pKb.
Teacher's Note:
1. Smaller pKb means stronger base; in water the secondary methylamine is the strongest.
(a) Phenol, C6H5OH
(b) Phenyl acetate, C6H5OCOCH3
(c) Aniline, C6H5NH2
(d) Acetanilide, C6H5NHCOCH3
Show Answer & Explanation
Answer: (c) Aniline, C6H5NH2
Explanation:
1. In the strongly acidic nitrating mixture, aniline is protonated to the anilinium ion, C6H5NH3+.
2. The –NH3+ group is meta-directing, so about 47% m-nitroaniline is formed.
Teacher's Note:
1. Protect the amino group by acetylation when you want mainly the para nitro product.
C6H5NH2 Br2/H2O
⟶ A NaNO2
⟶
HCl B (i) HBF4
⟶
(ii) Δ C
(a) 4-nitro-2-fluoro-N-bromoaniline
(b) 2-bromo-4-fluorotoluene
(c) 4-bromo-2,6-difluoroaniline
(d) 1,3,5-tribromo-2-fluorobenzene
Show Answer & Explanation
Answer: (d) 1,3,5-tribromo-2-fluorobenzene
Explanation:
1. Aniline with bromine water gives 2,4,6-tribromoaniline (A), which is diazotised to the diazonium salt (B).
2. B with HBF4 gives the diazonium fluoroborate, which on heating gives 2,4,6-tribromofluorobenzene, i.e. 1,3,5-tribromo-2-fluorobenzene (C).
Teacher's Note:
1. Balz–Schiemann reaction (HBF4, then heat) is the way to put F on a benzene ring.
p-CH3C6H4CONH2 (i) NaOH, Br2
⟶
(ii) C6H5COCl A
the structure of product A is:
(a) p-CH3C6H4–NH–CO–C6H5
(b) p-CH3C6H4–CO–O–CO–C6H5
(c) C6H5–NH–CO–C6H4–CH3-p
(d) p-CH3C6H4–CO–NH–CO–C6H5
Show Answer & Explanation
Answer: (a) p-CH3C6H4–NH–CO–C6H5
Explanation:
1. Hoffmann bromamide degradation of p-toluamide gives p-toluidine, p-CH3C6H4NH2.
2. Benzoylation of p-toluidine with C6H5COCl gives the amide p-CH3C6H4NHCOC6H5.
Teacher's Note:
1. In Hoffmann degradation the –CONH2 carbon is lost and N becomes attached directly to the ring.
(a) N,N-Dimethylaniline
(b) 2,4-Dimethylaniline
(c) N-Methyl-o-methylaniline
(d) N-Methylbenzylamine
Show Answer & Explanation
Answer: (b) 2,4-Dimethylaniline
Explanation:
1. The carbylamine test is given only by primary amines (aliphatic or aromatic).
2. 2,4-Dimethylaniline is a primary amine; the others are secondary or tertiary amines.
Teacher's Note:
1. Look at the nitrogen, not the ring: methyl groups on the ring do not change a primary amine into a secondary one.
(a) C6H5NO2 Sn/HCl
⟶
(b) C6H5CONH2 NaOBr
⟶
(c) C6H5CN H+/H2O
⟶
(d) Phthalimide (i) KOH
⟶
(ii) C6H5CH2Br, (iii) aq. OH−
Show Answer & Explanation
Answer: (d) Phthalimide (i) KOH
⟶
(ii) C6H5CH2Br, (iii) aq. OH−
Explanation:
1. Gabriel phthalimide synthesis: potassium phthalimide reacts with benzyl bromide to give N-benzylphthalimide, which on alkaline hydrolysis gives benzylamine, C6H5CH2NH2.
2. Sn/HCl gives aniline, NaOBr (Hoffmann) gives aniline, and acid hydrolysis of C6H5CN gives benzoic acid.
Teacher's Note:
1. Benzyl halides react readily in Gabriel synthesis, unlike aryl halides.
CH3–C≡N → CH3CH2NH2
(a) H3O+ (hydrolysis)
(b) Ammoniacal AgNO3
(c) LiAlH4
(d) NaBH4
Show Answer & Explanation
Answer: (c) LiAlH4
Explanation:
1. Nitriles are reduced to primary amines by LiAlH4 (or by H2/Ni or Pt).
2. NaBH4 is too mild to reduce the C≡N group, and acid hydrolysis gives a carboxylic acid.
Teacher's Note:
1. Reduction of a nitrile adds a CH2 to the amine: R–CN gives R–CH2NH2, so the carbon count stays the same.
(a) aniline, m-nitroaniline
(b) m-toluidine, p-toluidine
(c) p-chloroaniline, aniline
(d) aniline, p-aminophenol
Show Answer & Explanation
Answer: (a) aniline, m-nitroaniline
Explanation:
1. Lower pKb means stronger base. The electron-withdrawing –NO2 group makes m-nitroaniline a weaker base than aniline, so aniline has the lower pKb.
2. p-Toluidine and p-aminophenol are stronger bases than m-toluidine and aniline respectively, and p-chloroaniline is weaker than aniline.
Teacher's Note:
1. Electron-releasing groups (CH3, OH) at the para position raise basicity; electron-withdrawing groups (NO2, Cl) lower it.
(a) H3PO2/H2O
(b) Br2; HBr
(c) CuBr/HBr
(d) Br2, CCl4
Show Answer & Explanation
Answer: (c) CuBr/HBr
Explanation:
1. Sandmeyer reaction: C6H5N2+Cl− + CuBr/HBr → C6H5Br + N2.
2. H3PO2 replaces the diazonium group by H, giving benzene.
Teacher's Note:
1. Sandmeyer uses Cu(I) halides; Gattermann uses Cu powder with HX – both give aryl halides.
| Column I | Column II |
|---|---|
| (A) Ammonolysis | (p) Amine with lesser no. of carbon atoms |
| (B) Gabriel phthalimide synthesis | (q) Detection of primary amine |
| (C) Hoffmann bromamide degradation | (r) Reaction of phthalimide with KOH and R–X |
| (D) Carbylamine reaction | (s) Reaction of alkyl halides with NH3 |
(a) A–s, B–r, C–p, D–q
(b) A–p, B–q, C–r, D–s
(c) A–r, B–s, C–p, D–q
(d) A–s, B–r, C–q, D–p
Show Answer & Explanation
Answer: (a) A–s, B–r, C–p, D–q
Explanation:
1. Ammonolysis is the reaction of alkyl halides with NH3 (s); Gabriel synthesis uses phthalimide, KOH and R–X (r).
2. Hoffmann bromamide degradation gives an amine with one carbon atom less (p); the carbylamine reaction detects primary amines (q).
Teacher's Note:
1. Link each named reaction to its reagent or use – these four are the most asked preparation/test reactions of amines.
| Column I | Column II |
|---|---|
| (A) Benzene sulphonyl chloride | (p) Zwitter ion |
| (B) Sulphanilic acid | (q) Hinsberg reagent |
| (C) Alkyl diazonium salts | (r) Dyes |
| (D) Aryl diazonium salts | (s) Conversion to alcohols |
(a) A–p, B–q, C–s, D–r
(b) A–q, B–p, C–s, D–r
(c) A–q, B–p, C–r, D–s
(d) A–s, B–r, C–q, D–p
Show Answer & Explanation
Answer: (b) A–q, B–p, C–s, D–r
Explanation:
1. Benzenesulphonyl chloride is Hinsberg reagent (q); sulphanilic acid exists as a zwitter ion (p).
2. Alkyl diazonium salts are unstable and are converted to alcohols (s); aryl diazonium salts are used to make azo dyes (r).
Teacher's Note:
1. Sulphanilic acid is high-melting and insoluble in organic solvents because it is a dipolar ion, H3N+–C6H4–SO3−.
(a) Nitrobenzene reacts with Zn dust and NH4Cl to produce aniline
(b) Nitrobenzene reacts with LiAlH4 to produce phenylhydroxylamine
(c) Nitrobenzene reacts with Fe and HCl to produce nitrosobenzene
(d) Nitrobenzene reacts with Zn dust and NH4Cl to produce phenylhydroxylamine
Show Answer & Explanation
Answer: (d) Nitrobenzene reacts with Zn dust and NH4Cl to produce phenylhydroxylamine
Explanation:
1. Reduction of nitrobenzene with Zn dust and NH4Cl (neutral medium) gives N-phenylhydroxylamine.
2. C6H5NO2 Zn/NH4Cl
⟶
C6H5NHOH
Teacher's Note:
1. The product of nitrobenzene reduction depends on the medium: acidic (Fe/HCl, Sn/HCl) gives aniline, neutral (Zn/NH4Cl) gives phenylhydroxylamine, LiAlH4 gives azobenzene.
(a) Nitrobenzene
(b) Toluene
(c) Cumene
(d) Xylene
Show Answer & Explanation
Answer: (a) Nitrobenzene
Explanation:
1. –NO2 is a powerful deactivating group. It reduces the electron density in the benzene ring considerably. As a result, it does not undergo F.C. reactions.
Teacher's Note:
1. This is why nitrobenzene is often used as a solvent for Friedel-Crafts reactions.
(a) 5
(b) 2
(c) 3
(d) 4
Show Answer & Explanation
Answer: (d) 4
Explanation:
1. Four structural isomers are possible. These are :
2. (i) CH3CH2CH2NH2 (Propan-1-amine) (ii) CH3–CH(NH2)–CH3 (Propan-2-amine)
3. (iii) CH3–NH–CH2CH3 (N-Methylethanamine) (iv) (CH3)3N (N,N-Dimethylmethanamine)
Teacher's Note:
1. Count 1°, 2° and 3° amines separately so that none is missed: 2 + 1 + 1 = 4.
(a) 5
(b) 6
(c) 7
(d) 8
Show Answer & Explanation
Answer: (d) 8
Explanation:
1. Eight structural isomers are possible:
2. 1° amines (4): CH3CH2CH2CH2NH2, CH3CH2CH(NH2)CH3, (CH3)2CHCH2NH2, (CH3)3CNH2
3. 2° amines (3): CH3NHCH2CH2CH3, CH3NHCH(CH3)2, CH3CH2NHCH2CH3
4. 3° amine (1): (CH3)2NCH2CH3
Teacher's Note:
1. Students often forget (CH3)3CNH2 and CH3NHCH(CH3)2; check every carbon skeleton.
CH3–CN Na(Hg) + C2H5OH
⟶
X
(a) CH3–CONH2
(b) CH3CH2–NH2
(c) C2H6
(d) CH3–NH–CH3
Show Answer & Explanation
Answer: (b) CH3CH2–NH2
Explanation:
1. Nascent hydrogen from sodium amalgam and ethanol reduces the nitrile to a primary amine (Mendius reduction).
2. CH3–C≡N + 4[H] → CH3CH2NH2
Teacher's Note:
1. Reduction of a nitrile gives a 1° amine with the same number of carbon atoms, so it ascends the series compared with the alkyl halide used to make the nitrile.
(a) NaBH4
(b) H2–Pd/C
(c) LiAlH4
(d) Zn–Hg/HCl
Show Answer & Explanation
Answer: (c) LiAlH4
Explanation:
1. LiAlH4 reduces the >C=O group of an amide to –CH2–, converting the amide into the corresponding amine.
2. C6H5CONHCH3 LiAlH4
⟶
C6H5CH2NHCH3
Teacher's Note:
1. NaBH4 is too mild to reduce amides; amide → amine needs LiAlH4.
(a) Acetamide
(b) Propanamide
(c) Formamide
(d) Methyl cyanide
Show Answer & Explanation
Answer: (b) Propanamide
Explanation:
1. Hofmann bromamide reaction gives an amine with one carbon atom less than the amide.
2. CH3CH2CONH2 + Br2 + 4KOH → CH3CH2NH2 + K2CO3 + 2KBr + 2H2O
Teacher's Note:
1. For a 2-carbon amine by Hofmann bromamide reaction, start with a 3-carbon amide.
Benzene Conc. HNO3
⟶
Conc. H2SO4 X LiAlH4
⟶
Y
(a) C6H5NH2
(b) C6H5NHOH
(c) C6H5–N=N–C6H5
(d) C6H5–NH–NH–C6H5
Show Answer & Explanation
Answer: (c) C6H5–N=N–C6H5
Explanation:
1. X is nitrobenzene. Nitrobenzene on reduction with LiAlH4 gives azobenzene.
2. 2 C6H5NO2 LiAlH4
⟶
C6H5N=NC6H5
Teacher's Note:
1. Two molecules of nitrobenzene combine during LiAlH4 reduction, so the product has two benzene rings joined by –N=N–.
Cyclohexyl bromide (bromocyclohexane) + NH3 →
(a) Cyclohexylamine, C6H11–NH2
(b) Dicyclohexylamine, C6H11–NH–C6H11
(c) N-Bromocyclohexylamine, C6H11–NHBr
(d) Cyclohexene
Show Answer & Explanation
Answer: (d) Cyclohexene
Explanation:
1. 2° Alkyl halides preferably undergo dehydrohalogenation (elimination) to give alkenes.
2. Thus, option (d) is correct.
Teacher's Note:
1. With alkyl halides, NH3 can act as a nucleophile or as a base; the more substituted the carbon bearing the halogen, the more elimination competes.
(a) aniline
(b) benzylamine
(c) n-butylamine
(d) isobutylamine
Show Answer & Explanation
Answer: (a) aniline
Explanation:
1. Aniline being an arylamine cannot be prepared by Gabriel phthalimide reaction while all others being aliphatic primary amines can be easily prepared by this method.
Teacher's Note:
1. Aryl halides do not undergo nucleophilic substitution with the phthalimide anion, so aromatic 1° amines cannot be made this way.
(a) CH3CONH2 Br2/KOH
⟶
(b) CH3CN LiAlH4
⟶
(c) CH3NC LiAlH4
⟶
(d) CH3CONH2 LiAlH4
⟶
Show Answer & Explanation
Answer: (c) CH3NC LiAlH4
⟶
Explanation:
1. LiAlH4 reduces isocyanides to 2° amines, i.e.,
2. CH3NC LiAlH4
⟶
CH3NHCH3
Teacher's Note:
1. Nitriles (–C≡N) reduce to 1° amines, but isocyanides (–N≡C) reduce to 2° amines because the N is already bonded to carbon on both sides.
Assertion–Reason Questions
Assertion (A): Propanamide on treatment with bromine and potassium hydroxide gives propanamine.
Reason (R): During this reaction, one carbon of amide is lost as carbonate ion.
(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. Correct Assertion. Propanamide on treatment with Br2 and KOH gives ethanamine.
2. CH3CH2CONH2 + Br2 + 4KOH → CH3CH2NH2 + K2CO3 + 2KBr + 2H2O. So A is false but R is true.
Teacher's Note:
1. The carbonyl carbon of the amide leaves as CO32−, so the amine always has one carbon fewer.
Assertion (A): Gabriel phthalimide reaction can be used to prepare aniline.
Reason (R): Bromobenzene does not react with potassium phthalimide to form N-phenylphthalimide.
(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. Correct Assertion. Aniline cannot be prepared by Gabriel phthalimide reaction.
2. Bromobenzene, an aryl halide, does not undergo nucleophilic substitution with potassium phthalimide, so R is true and A is false.
Teacher's Note:
1. Gabriel synthesis works only for aliphatic 1° amines made from alkyl halides that undergo SN2.
Assertion (A): Aniline cannot be prepared by Gabriel phthalimide reaction.
Reason (R): Aryl halides do not undergo nucleophilic substitution reactions under ordinary laboratory conditions.
(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. Aniline would need the reaction of potassium phthalimide with C6H5Cl or C6H5Br.
2. Since aryl halides do not undergo nucleophilic substitution under ordinary conditions, this step fails; so R correctly explains A.
Teacher's Note:
1. The partial double-bond character of the C–X bond in aryl halides (resonance) is the root cause of their low reactivity towards nucleophiles.
Assertion (A): Primary aliphatic and aromatic primary amines are obtained by reduction of the corresponding nitro compounds with H2/Pd-C.
Reason (R): Aromatic 1° amines are less basic than aliphatic 1° amines.
(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. Correct explanation. Catalytic hydrogenation reduces –NO2 to –NH2 irrespective of the fact whether it is attached to an alkyl group or an aryl group.
2. R is also true but has nothing to do with the reduction.
Teacher's Note:
1. In A-R questions, check whether R actually answers “why” for A; here basicity has no bearing on how the amine is made.
Assertion (A): Bromocyclohexane on treatment with ammonia gives cyclohexene.
Reason (R): Secondary alkyl halides on treatment with a base predominantly undergo elimination 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: (a) Both A and R are true, and R is the correct explanation of A.
Explanation:
1. NH3 acts as a base and removes HBr from the 2° alkyl halide bromocyclohexane, giving cyclohexene.
2. Since 2° alkyl halides with a base mainly undergo elimination, R explains A.
Teacher's Note:
1. Ammonolysis works best with 1° alkyl halides; with 2° and especially 3° halides elimination competes strongly.
Assertion (A): Both aliphatic and aromatic 1° amines give carbylamine test.
Reason (R): In carbylamine reaction, a mixture of 1° amine and carbon tetrachloride is heated in presence of alcoholic KOH.
(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. Correct reason. A mixture of 1° amine and CHCl3 is heated in presence of alcoholic KOH.
2. R–NH2 + CHCl3 + 3KOH (alc.) Δ
⟶
R–NC + 3KCl + 3H2O. So A is true but R is false.
Teacher's Note:
1. The carbylamine test needs chloroform (CHCl3), which forms dichlorocarbene with alcoholic KOH; CCl4 does not.
Assertion (A): Methylamine is more basic than trimethylamine.
Reason (R): +I-effect of the three methyl groups predominates over the stability of the conjugate acid due to H-bonding.
(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. Correct R. Greater stability of the conjugate acid due to H-bonding over stability due to +I-effect of the three CH3 groups makes CH3NH2 more basic than (CH3)3N.
2. So A is true but R is false.
Teacher's Note:
1. In aqueous solution, solvation (H-bonding) of the ammonium ion often outweighs the +I-effect; the order for methylamines is 2° > 1° > 3°.
Assertion (A): o-Toluidine is a weaker base than p-toluidine.
Reason (R): At o-position CH3 group exerts less + I-effect than at p-position.
(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. Correct reason. Due to ortho effect, o-toluidine is a weaker base than p-toluidine.
2. So A is true but R is false.
Teacher's Note:
1. Almost any group at the ortho position (electron-donating or withdrawing) lowers the basicity of aniline – this is the ortho effect.
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You can get most exhaustive CBSE Class 12 Chemistry Amines MCQs Set 11 for free on StudiesToday.com. These MCQs for Class 12 Chemistry are updated for the 2026-27 academic session as per CBSE examination standards.
Yes, our CBSE Class 12 Chemistry Amines MCQs Set 11 include the latest type of questions, such as Assertion-Reasoning and Case-based MCQs. 50% of the CBSE paper is now competency-based.
By solving our CBSE Class 12 Chemistry Amines MCQs Set 11, Class 12 students can improve their accuracy and speed which is important as objective questions provide a chance to secure 100% marks in the Chemistry.
Yes, Chemistry MCQs for Class 12 have answer key and brief explanations to help students understand logic behind the correct option as its important for 2026 competency-focused CBSE exams.
Yes, you can also access online interactive tests for CBSE Class 12 Chemistry Amines MCQs Set 11 on StudiesToday.com as they provide instant answers and score to help you track your progress in Chemistry.