CBSE Class 12 Chemistry Amines MCQs Set 12

Answer: (a) Pentanamine

Explanation:
1. Boiling point of isomeric pentylamines decreases with branching. Thus, pentanamine has the higher b.p.

Teacher's Note:
1. Branching makes a molecule more compact, reducing surface area and van der Waals forces, so the straight-chain isomer boils highest.

Answer: (a) Benzylamine

Explanation:
1. In benzylamine (C₆H₅CH₂NH₂) the N is not attached directly to the ring, so its lone pair is not delocalised and it behaves like an aliphatic amine.
2. In aniline, acetanilide and p-nitroaniline the lone pair is delocalised (over the ring and/or C=O, NO₂), so they are much weaker bases.

Teacher's Note:
1. Check whether N is directly bonded to the benzene ring: if not (as in benzylamine), the amine is aliphatic in behaviour.

Answer: (c) C₆H₅NH₂

Explanation:
1. In aniline the lone pair on N is delocalised over the benzene ring and is less available for protonation, so aniline is even less basic than NH₃.
2. Alkylamines are more basic than NH₃ due to the +I-effect of alkyl groups.

Teacher's Note:
1. Remember the order: alkylamines > NH₃ > aniline.

Answer: (c) I < II < IV < III

Explanation:
1. In aqueous solution, the combined effect of +I-effect, solvation (H-bonding) of the ammonium ion and steric hindrance decides basicity.
2. For ethylamines the order is (C₂H₅)₂NH > (C₂H₅)₃N > C₂H₅NH₂ > NH₃, i.e. I < II < IV < III.

Teacher's Note:
1. For ethylamines in water the order is 2° > 3° > 1°; for methylamines it is 2° > 1° > 3°. Learn both.

Answer: (a) NH₂⁻ > OH⁻ > NH₃ > H₂O

Explanation:
1. Anions are stronger bases than neutral molecules, and N being less electronegative than O holds its lone pair less tightly.
2. Hence NH₂⁻ > OH⁻ > NH₃ > H₂O.

Teacher's Note:
1. A weaker acid has a stronger conjugate base: NH₃ is a weaker acid than H₂O, so NH₂⁻ is a stronger base than OH⁻.

Answer: (d) Acetanilide (C₆H₅NHCOCH₃)

Explanation:
1. Aniline undergoes acetylation (nucleophilic substitution at the C=O of the anhydride) on N, not on the ring.
2. C₆H₅NH₂ + (CH₃CO)₂O → C₆H₅NHCOCH₃ + CH₃COOH

Teacher's Note:
1. Acetylation of aniline happens at the –NH₂ group; this is used to protect –NH₂ and reduce its activating effect.

Answer: (a) o-Nitroaniline < p-nitroaniline < m-nitroaniline

Explanation:
1. Due to ortho-effect, o-nitroaniline is the weakest.
2. Amongst m-nitroaniline and p-nitroaniline, p-nitroaniline is weaker than m-nitroaniline because in p-nitroaniline both –R and –I-effect of the NO₂ group decrease the basicity while in m-nitroaniline, only the weaker –I-effect decreases the basicity.
3. Thus, the overall basicity increases in the order : o-nitroaniline < p-nitroaniline < m-nitroaniline, i.e., option (a) is correct.

Teacher's Note:
1. The –R effect of NO₂ operates only from o- and p-positions; from m-position only the –I effect acts.

Answer: (a) phenyl isocyanide

Explanation:
1. This is the carbylamine reaction:
2. C₆H₅NH₂ + CHCl₃ + 3KOH (alc.) Δ
⟶
  C₆H₅NC + 3KCl + 3H₂O
3. So A = phenyl isocyanide, B = KCl, C = H₂O.

Teacher's Note:
1. The carbylamine (isocyanide) test is given only by 1° amines and is recognised by the foul smell of the isocyanide.

Answer: (b) C₆H₅NO₂

Explanation:
1. B must be aniline, since only a 1° amine gives phenyl isocyanide with CHCl₃/KOH.
2. Aniline is obtained by reduction of nitrobenzene with Sn/HCl, so A = C₆H₅NO₂.

Teacher's Note:
1. Work backwards: the carbylamine step tells you B is a 1° amine, and Sn/HCl tells you A is a nitro compound.

Answer: (b) 2-Bromo-4-methylaniline (Br ortho to NH₂)

Explanation:
1. Acetylation converts –NH₂ into –NHCOCH₃. This group is a stronger o/p-director than –CH₃; the p-position is occupied, so Br enters ortho to –NHCOCH₃.
2. Acid hydrolysis then removes the acetyl group, giving 2-bromo-4-methylaniline.

Teacher's Note:
1. When two activating groups compete, the stronger activator (–NHCOCH₃ over –CH₃) decides the position of substitution.

Answer: (d) Sulphanilic acid.

Explanation:
1. Aniline first forms anilinium hydrogen sulphate, which on heating at 455-475 K rearranges to p-aminobenzenesulphonic acid (sulphanilic acid).

Teacher's Note:
1. At room temperature the product is only the salt (anilinium hydrogen sulphate); heating to 455-475 K is needed for sulphonation of the ring.

Answer: (b) Sulphanilic acid

Explanation:
1. Since –COOH being weakly acidic cannot transfer a H⁺ ion to the weakly basic –NH₂ group, therefore, p-aminobenzoic acid does not exist as a zwitterion.
2. In contrast, –SO₃H being strongly acidic transfers its proton to –NH₂ group and hence sulphanilic acid exists as a zwitterion.
3. HO₃S–C₆H₄–NH₂ → ⁻O₃S–C₆H₄–NH₃⁺ (Zwitterion)

Teacher's Note:
1. A zwitterion forms only when the acidic group is strong enough to protonate the basic group in the same molecule.

Answer: (b) diethylamine

Explanation:
1. 2° Amines react with nitrous acid to form N-nitrosoamines, which are yellow oily liquids.
2. (C₂H₅)₂NH + HNO₂ → (C₂H₅)₂N–N=O + H₂O

Teacher's Note:
1. “Secondary butylamine” is a 1° amine (the NH₂ is on a 2° carbon); only diethylamine here is a 2° amine.

Answer: (b) (CH₃)₂C(OH)–CH₂CH₃

Explanation:
1. (CH₃)₂CH–CH(NH₂)–CH₃ HNO₂
⟶
  (CH₃)₂CH–CH(N₂⁺)–CH₃ −N₂
⟶
  (CH₃)₂CH–C⁺H–CH₃ (2° carbocation)
2. 1, 2-Hydride shift
⟶
  (CH₃)₂C⁺–CH₂CH₃ (3° carbocation) H₂O
⟶
−H⁺ (CH₃)₂C(OH)–CH₂CH₃

Teacher's Note:
1. Aliphatic diazonium ions lose N₂ at once to give carbocations, which can rearrange to a more stable carbocation before water attacks.

Answer: (b) CH₄

Explanation:
1. The N–H hydrogen of ethylamine is acidic enough to react with the Grignard reagent, which gives the alkane of its own alkyl group.
2. C₂H₅NH₂ + CH₃MgBr → CH₄ + C₂H₅NHMgBr

Teacher's Note:
1. Grignard reagents are destroyed by any compound with an active H (on O, N or terminal C≡C), giving R–H from the Grignard part.

Answer: (d) Benzylamine

Explanation:
1. Only aromatic 1° amines undergo diazotisation. Benzylamine is a 1° amine and hence does not undergo diazotisation but forms benzyl alcohol with HNO₂
2. C₆H₅CH₂NH₂ + HNO₂ → C₆H₅CH₂OH + N₂ + H₂O

Teacher's Note:
1. Benzylamine has –NH₂ on a CH₂ group, not on the ring, so it behaves like an aliphatic 1° amine.

Answer: (a) halogen (Cl or Br)

Explanation:
1. In Sandmeyer reaction, the diazo group is replaced by Cl or Br using CuCl/HCl or CuBr/HBr.
2. C₆H₅N₂⁺Cl⁻ CuCl/HCl
⟶
  C₆H₅Cl + N₂

Teacher's Note:
1. Sandmeyer uses cuprous salts; Gattermann uses copper powder with HX for the same replacement.

Answer: (d) Benzyl alcohol

Explanation:
1. Coupling is an electrophilic substitution on a strongly activated aromatic ring (phenols, naphthols, aromatic amines).
2. In benzyl alcohol the –OH is on the CH₂ group, so the ring is not activated and no coupling occurs.

Teacher's Note:
1. For coupling the –OH or –NH₂ must be attached directly to the aromatic ring.

Answer: (d) Separation of a mixture of primary, secondary and tertiary amines

Explanation:
1. Hinsberg’s reagent (benzenesulphonyl chloride) gives an alkali-soluble sulphonamide with 1° amines, an alkali-insoluble sulphonamide with 2° amines and does not react with 3° amines.
2. This difference is used to distinguish and separate 1°, 2° and 3° amines.

Teacher's Note:
1. Remember the KOH solubility test: 1° sulphonamide dissolves (acidic N–H), 2° does not (no N–H).

Answer: (c) (C₂H₅)₃N

Explanation:
1. 3° Amines have no H atom on nitrogen, so they do not react with benzenesulphonyl chloride.

Teacher's Note:
1. Reaction with C₆H₅SO₂Cl needs an N–H bond; count the H atoms on N.

Answer: (d) Aniline.

Explanation:
1. X = aniline, which with bromine water gives 2,4,6-tribromoaniline (Y).
2. Y on diazotisation gives 2,4,6-tribromobenzenediazonium chloride (Z), which on boiling with ethanol is reduced to 1,3,5-tribromobenzene.

Teacher's Note:
1. Diazotisation of the amine followed by reduction (ethanol or H₃PO₂) removes the –NH₂ group after it has been used to direct substitution.

Answer: (d) anisole

Explanation:
1. When a diazonium salt is warmed with methanol, the predominant product is the aryl methyl ether (anisole); with ethanol, replacement by hydrogen (benzene) predominates.
2. C₆H₅N₂⁺HSO₄⁻ + CH₃OH Warm
⟶
  C₆H₅OCH₃ + N₂ + H₂SO₄

Teacher's Note:
1. Methanol mainly gives anisole, while ethanol mainly gives benzene; for removing the diazo group H₃PO₂ or ethanol is used.

Answer: (d) (CH₃)₃C–NH₂

Explanation:
1. 1° aliphatic amines containing tert-alkyl groups are easily oxidised to the corresponding nitro compound, i.e., option (d) is correct.

Teacher's Note:
1. With no H on the carbon carrying –NH₂, a tert-alkylamine cannot be oxidised to an imine or carbonyl compound, so KMnO₄ oxidises it to the nitro compound.

Answer: (c) 2, 4, 6-tribromofluorobenzene

Explanation:
1. Aniline Br₂–H₂O
⟶
  2,4,6-tribromoaniline NaNO₂/HCl
⟶
  2,4,6-tribromobenzenediazonium chloride HBF₄
⟶
−HCl 2,4,6-tribromobenzenediazonium tetrafluoroborate Δ
⟶
−N₂, −BF₃ 2, 4, 6-Tribromofluorobenzene

Teacher's Note:
1. Heating the dry diazonium fluoroborate (Balz-Schiemann reaction) is the usual way to put F on a benzene ring.

Answer: (a) benzenediazonium chloride and fluorobenzene

Explanation:
1. C₆H₅NH₂ NaNO₂
⟶
HCl, 278K C₆H₅N₂⁺Cl⁻ (A) HBF₄
⟶
−HCl C₆H₅N₂⁺BF₄⁻ Heat
⟶
−N₂, −BF₃ C₆H₅F (B)

Teacher's Note:
1. NaNO₂/HCl at 273-278 K always signals diazotisation of an aromatic 1° amine.

Answer: (c) C₆H₅–C₆H₅

Explanation:
1. C₆H₅NH₂ NaNO₂/HCl
⟶
273-278 K C₆H₅N₂Cl (A)
2. C₆H₅N₂Cl C₆H₆/NaOH
⟶
(Gomberg–Bachmann reaction) C₆H₅ – C₆H₅ (Diphenyl)

Teacher's Note:
1. Benzene with NaOH and a diazonium salt gives a biaryl (Gomberg-Bachmann); phenol with NaOH would instead give an azo dye.

Answer: (a) aniline

Explanation:
1. C₆H₅N₂Cl Zn/HCl
⟶
  C₆H₅NH₂ + NH₃

Teacher's Note:
1. Mild reduction (SnCl₂/HCl or Na₂SO₃) stops at phenylhydrazine; vigorous reduction with Zn/HCl breaks the N–N bond to give aniline and NH₃.

Answer: (a) Na₂S

Explanation:
1. Na₂S is a weak reducing agent, it reduces one of the NO₂ groups of m-dinitrobenzene to give m-nitroaniline.

Teacher's Note:
1. Selective reduction of one nitro group needs a mild reagent such as Na₂S, (NH₄)₂S or NH₄HS; Sn/HCl reduces both groups.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. In the gas phase there is no solvation, so basicity depends only on the +I-effect of the alkyl groups, which increases from 1° to 2° to 3°.
2. Hence R explains A.

Teacher's Note:
1. Gas phase order 3° > 2° > 1° > NH₃; in water this order is upset by solvation and steric effects.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Correct explanation. Nitrosonium ion obtained from nitrous acid reacts with aromatic amine to form diazonium salts.
2. R is true (aryl diazonium ions are stabilised by resonance), but it does not explain how they are formed.

Teacher's Note:
1. Aryl diazonium salts are stable only at 273-278 K, which is why diazotisation is done in an ice bath.

Answer: (d) A is false, but R is true.

Explanation:
1. Correct A. Benzenediazonium chloride reacts with CH₃OH to form anisole as the major product.
2. So A is false but R is true.

Teacher's Note:
1. With methanol the major product is anisole, with ethanol it is benzene; read the alcohol carefully.

Answer: (d) A is false, but R is true.

Explanation:
1. Correct A. Benzenediazonium bromide reacts with CuCl dissolved in HCl to form chlorobenzene.
2. So A is false but R is true.

Teacher's Note:
1. In the Sandmeyer reaction the halogen of the copper(I) halide, not the counter-ion of the diazonium salt, enters the ring.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. The –I-effect of F withdraws electrons from the NH₃⁺ group, weakening the N–H bonds, so p-fluoroanilinium ion loses a proton more easily.
2. Hence R correctly explains A.

Teacher's Note:
1. A more acidic anilinium ion means a weaker base: p-fluoroaniline is less basic than aniline.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Correct explanation. Coupling reaction is an example of an electrophilic substitution reaction in which diazonium cation acts as a weak electrophile while phenol or amine acts as the nucleophile.
2. Both A and R are true, but R only defines azo compounds and does not explain A.

Teacher's Note:
1. The diazonium ion is a weak electrophile, so it couples only with strongly activated rings (phenoxide ion, aromatic amines).

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. In strong acid, aniline is converted to C₆H₅NH₃⁺, which deactivates the ring and cannot react with the weakly electrophilic diazonium ion.
2. Hence R correctly explains A.

Teacher's Note:
1. Coupling with amines is done in mildly acidic medium (pH 4-5) and with phenols in mildly alkaline medium (pH 9-10).

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Correct explanation. Distinction is based upon the fact that 1° amines give benzenesulphonamides which are soluble in NaOH while benzenesulphonamides of 2° amines are not soluble in NaOH.
2. R is true but only names the reagent; it does not explain A.

Teacher's Note:
1. The N–H left in the sulphonamide of a 1° amine is acidic (strongly electron-withdrawing SO₂ group), which is why it dissolves in alkali.

Answer: (d) A is false, but R is true.

Explanation:
1. Correct A. Nitration of aniline with a mixture of HNO₃ + H₂SO₄ gives a mixture of o-, m- and p-nitroaniline.
2. In strongly acidic medium aniline is protonated to the anilinium ion, which is m-directing; so A is false and R is true.

Teacher's Note:
1. To get mainly p-nitroaniline, protect –NH₂ by acetylation first, nitrate, and then hydrolyse.