CBSE Class 12 Chemistry Amines MCQs Set 09

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

Explanation:
1. C₆H₅NH₂ + 3Br₂ Br₂/H₂O
⟶ 2,4,6-Br₃C₆H₂NH₂ (2,4,6-tribromoaniline, white ppt.) + 3HBr

Teacher's Note:
1. –NH₂ activates the ring so strongly that all the o- and p-positions get brominated; protect it by acetylation to get a monobromo product.

Answer: (c) intermolecular hydrogen bonding is maximum in primary amines and absent in tertiary amines

Explanation:
1. The trend in boiling point can be explained on the basis of intermolecular hydrogen bonding which is maximum in primary amines.

Teacher's Note:
1. All three are isomers of C₄H₁₁N, so the difference must come from hydrogen bonding, not molecular mass.

Answer: (a) C₆H₅SO₂Cl

Explanation:
1. Benzenesulphonyl chloride (C₆H₅SO₂Cl) acts as Hinsberg’s reagent.

Teacher's Note:
1. Hinsberg’s reagent is used to distinguish 1°, 2° and 3° amines by the solubility of the sulphonamide formed.

Answer: (a) Anilinium ion has two stable canonical structures I and III.

Explanation:
1. II is not an acceptable canonical structure because nitrogen has 10 valence electrons in the structure. Anilinium ion exists in two canonical structures only which are I and III.

Teacher's Note:
1. In anilinium ion the nitrogen has no lone pair left, so it cannot form a C=N double bond; only the two Kekulé structures remain.

Answer: (c) N₂

Explanation:
1. CH₃–NH₂ + HNO₂ NaNO₂ + HCl
⟶ [CH₃N₂⁺Cl⁻] H₂O
⟶ CH₃OH + N₂ + HCl

Teacher's Note:
1. Brisk effervescence of N₂ with nitrous acid is a test for aliphatic primary amines.

Answer: (b) Benzanilide

Explanation:
1. C₆H₅NH₂ (aniline) + ClCOC₆H₅ (benzoyl chloride) + NaOH → C₆H₅NH–COC₆H₅ (benzanilide) + NaCl + H₂O

Teacher's Note:
1. Benzoylation in the presence of aqueous NaOH is called the Schotten–Baumann reaction.

Answer: (a) NaNO₂ + HCl, CuCN, H₂/Ni

Explanation:
1. C₆H₅NH₂ NaNO₂ + HCl
⟶
(0-4°C) C₆H₅N₂⁺Cl⁻ CuCN
⟶
or KCN C₆H₅CN H₂/Ni
⟶ C₆H₅CH₂NH₂ (benzylamine)

Teacher's Note:
1. Aniline (C₆) must gain one carbon to become benzylamine (C₇); the CN group supplies that carbon.

Answer: (d) Acetylation followed by nitration and hydrolysis.

Explanation:
1. –NH₂ group is oxidised on direct nitration hence –NH₂ group is blocked by acetylation and then nitration is carried out.
2. C₆H₅NH₂ CH₃COCl
⟶ C₆H₅NHCOCH₃ HNO₃
⟶
H₂SO₄ p-O₂NC₆H₄NHCOCH₃ H₂O/H⁺
⟶ p-O₂NC₆H₄NH₂

Teacher's Note:
1. Acetylation both protects –NH₂ from oxidation and reduces its activating effect, so the p-nitro product is obtained cleanly.

Answer: (a) Aniline

Explanation:
1. C₆H₅COOH PCl₅
⟶ C₆H₅COCl (benzoyl chloride, P) 2NH₃
⟶
−NH₄Cl C₆H₅CONH₂ (benzamide, Q) Br₂/KOH
⟶ C₆H₅NH₂ (aniline, R)

Teacher's Note:
1. Two moles of NH₃ are needed because the second one neutralises the HCl formed.

Answer: (b) NH₃

Explanation:
1. Higher the basicity, lower is the pK_b value. Since NH₃ is the weakest base, hence it has highest pK_b value.

Teacher's Note:
1. Highest pK_b means weakest base; all the alkylamines are stronger bases than ammonia.

Answer: (d) Preparation of primary amines using phthalimide – Victor Meyer’s synthesis

Explanation:
1. The synthesis of primary amines from phthalimide is known as Gabriel phthalimide synthesis.

Teacher's Note:
1. Gabriel synthesis gives only pure primary aliphatic amines; it cannot make aromatic amines.

Answer: (d) C₆H₅NHC₆H₅

Explanation:
1. Secondary amines do not give carbylamine test. C₆H₅NHC₆H₅ (diphenylamine) is a secondary amine.

Teacher's Note:
1. Both aliphatic and aromatic primary amines give the test; check whether N carries two H atoms.

Answer: (d) tetraethylammonium iodide.

Explanation:
1. NH₃ + C₂H₅I (excess) → [(C₂H₅)₄N]⁺I⁻ (quaternary salt)

Teacher's Note:
1. With excess alkyl halide alkylation continues until all four positions on N are alkylated; with excess NH₃ the primary amine is the main product.

Answer: (b) (C₆H₅)₃N

Explanation:
1. In (C₆H₅)₃N the lone pair on nitrogen is delocalised over three benzene rings and the nitrogen is also sterically crowded, so it is practically non-basic. It does not form a salt and is insoluble in dil. HCl.

Teacher's Note:
1. Triphenylamine is an exception among amines – three phenyl groups pull the lone pair into resonance so it hardly accepts a proton.

Answer: (c) p-toluidine (p-methylaniline)

Explanation:
1. Only –CH₃ group is electron donating hence it increases the electron density on nitrogen making it most basic.

Teacher's Note:
1. –NO₂ and –F withdraw electrons and lower basicity; –CH₃ releases electrons and raises it.

Answer: (a) propanenitrile

Explanation:
1. C₂H₅NH₂ HNO₂
⟶ C₂H₅OH (X) PCl₅
⟶ C₂H₅Cl (Y) KCN
⟶ C₂H₅CN (propanenitrile, Z)

Teacher's Note:
1. Nitrous acid converts an aliphatic primary amine into an alcohol with loss of N₂ – a common first step in such sequences.

Answer: (c) (CH₃CH₂)₃N

Explanation:
1. 3° amines do not react with Hinsberg reagent because they have no hydrogen atom on nitrogen.

Teacher's Note:
1. Hinsberg's reagent replaces an H on nitrogen, so an amine with no N–H cannot react.

Answer: (b) KOH/Br₂, CH₃COCl

Explanation:
1. C₆H₅CONH₂ KOH/Br₂
⟶ C₆H₅NH₂ CH₃COCl
⟶ C₆H₅NHCOCH₃

Teacher's Note:
1. Work backwards: acetanilide is acetylated aniline, and aniline comes from benzamide by Hoffmann degradation.

Answer: (b) substituted amides

Explanation:
1. RNH₂ + RCOCl → RNHCOR + HCl
2. RNH₂ + (RCO)₂O → RNHCOR + RCOOH (substituted amides)

Teacher's Note:
1. This acylation needs an H on nitrogen, so tertiary amines do not give it.

Answer: (d) In (4), –NO₂ is showing −R effect.

Explanation:
1. At meta position –NO₂ shows −I effect while at o- and p-positions it shows −R effect. So in (4) –NO₂ cannot show −R effect.

Teacher's Note:
1. Resonance effects act only at ortho and para positions; a meta substituent acts only through its inductive effect.

Answer: (c) NH₃ < (CH₃)₃N < CH₃NH₂ < (CH₃)₂NH

Explanation:
1. The value of pK_b describes the relative strength of the bases.
2.

Teacher's Note:
1. In water, trimethylamine is weaker than methylamine because its protonated form is poorly solvated and the N is crowded.

Answer: (c) CH₃NH₂

Explanation:
1. It is Hoffmann bromamide degradation reaction.
2. CH₃CONH₂ + Br₂ + 4NaOH \( \xrightarrow{\Delta} \) CH₃NH₂ + 2NaBr + Na₂CO₃ + 2H₂O
3. Acetamide → Methylamine

Teacher's Note:
1. In Hoffmann bromamide degradation the amine always has one carbon atom less than the amide, because the C=O carbon leaves as Na₂CO₃.

Answer: (b) Ethanamine

Explanation:
1. CH₃CH₂CONH₂ + Br₂ + 4NaOH \( \xrightarrow{\Delta} \) CH₃CH₂NH₂ + 2NaBr + Na₂CO₃ + 2H₂O
2. Propanamide → Ethanamine. This is Hoffman bromamide reaction.

Teacher's Note:
1. Count carbons: a 3-carbon amide gives a 2-carbon amine. Propanamine would need reduction (LiAlH₄), not Br₂/NaOH.

Answer: (d) N-Ethyl-N-methylethanamine

Explanation:
1. By alkylation reaction of methyl amine with two moles of ethyl chloride, N-Ethyl-N-methylethanamine is formed.
2. CH₃NH₂ + C₂H₅Cl → CH₃–NH–CH₂CH₃ \( \xrightarrow{\text{C}_{2}\text{H}_{5}\text{Cl}} \) CH₃–N(C₂H₅)–C₂H₅ (N-ethyl-N-methylethanamine)

Teacher's Note:
1. In IUPAC naming of amines the longest carbon chain on nitrogen is the parent (here ethyl, so “ethanamine”) and the other groups get the prefix N-.

Answer: (b) LiAlH₄ in ether

Explanation:
1. C₆H₅NO₂ is reduced to C₆H₅NH₂ by H₂(excess)/Pt, Fe/HCl and Sn/HCl.
2. With LiAlH₄/ether, nitrobenzene gives azobenzene, C₆H₅–N=N–C₆H₅.

Teacher's Note:
1. LiAlH₄ reduces aliphatic nitro compounds to amines, but aromatic nitro compounds give azo compounds with it.

Answer: (d) II < I < III

Explanation:
1. p-Nitroaniline (II) < Aniline (I) < p-Toluidine (III)
2. Electron withdrawing group decreases the basic strength while electron releasing groups increases the basic strength of aniline.

Teacher's Note:
1. Compare every substituted aniline with aniline itself: –NO₂ (electron withdrawing) lowers basicity, –CH₃ (electron releasing) raises it.

Answer: (d) LiAlH₄ in ether

Explanation:
1. C₆H₅CH(CH₃)CONH₂ + 4[H] \( \xrightarrow[\text{ether}]{\text{LiAlH}_{4}} \) C₆H₅CH(CH₃)CH₂NH₂ + H₂O
2. 2-Phenylpropanamide → 2-Phenylpropanamine
3. Whereas in Br₂/NaOH, a carbon is less from the amide to give amine, I₂/Red P is a strong reducing agent convert into alkanamine.

Teacher's Note:
1. When the amine must keep the same number of carbons as the amide, use LiAlH₄; when one carbon must be removed, use Br₂/NaOH.

Answer: (c) Benzene sulphonyl chloride

Explanation:
1. Hinsberg’s reagent which is used to test amines is benzene sulphonyl chloride, C₆H₅SO₂Cl.

Teacher's Note:
1. Hinsberg’s reagent must have a reactive S–Cl bond to react with N–H of amines; a sulphonamide has already reacted.

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

Explanation:
1. Aniline is a Lewis base and reacts with the Lewis acid AlCl₃ (the catalyst) to form a salt, C₆H₅NH₂⁺–AlCl₃⁻.
2. The positively charged nitrogen strongly deactivates the ring, so aniline does not undergo Friedel-Crafts reaction.

Teacher's Note:
1. The amine 'uses up' the catalyst and also deactivates the ring – both effects stop the Friedel-Crafts reaction.

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

Explanation:
1. The assertion is true. It is because hydrogen bonding is less pronounced in primary and secondary amines than that in alcohols or carboxylic acids and nitrogen is less electronegative than oxygen.
2. The reason is false: amines are readily soluble in organic solvents like alcohol, ether and benzene, and the solubility of higher amines in water is very low.

Teacher's Note:
1. Lower amines dissolve in water through hydrogen bonding, but solubility in water falls quickly as the alkyl part grows.

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

Explanation:
1. Acetylation decreases the electron density in the benzene ring and deactivates the ring hence control the reaction.
2. So the assertion is true but the reason is false.

Teacher's Note:
1. In acetanilide the nitrogen lone pair is shared with the C=O group as well, so it activates the ring less than –NH₂ does.

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

Explanation:
1. –NH₂ being o, p-directing group directs –SO₃H group to less hindered p-position.
2. The reason is true, but the electron-withdrawing nature of –SO₃H does not explain the formation of the p-product.

Teacher's Note:
1. The product, sulphanilic acid, exists as a zwitter ion – which is why it has a high melting point.

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

Explanation:
1. In alkylation, an amine can react with alkyl halide to form next higher class of amine caused by the presence of electron pair on nitrogen which makes amine to behave as nucleophile and alkyl halide thus undergo nucleophilic substitution reaction.
2. Whereas primary and secondary amines react with acid chlorides, anhydrides and esters to give monosubstituted amides as products. Due to involvement of lone pairs of nitrogen in delocalization within the carbonyl the charge density on N decreases, so it acts poorly as a nucleophile. Hence, acylation gives monosubstituted products.

Teacher's Note:
1. Acylation stops after one step for an electronic reason (lone pair delocalised into C=O), not a steric one; this is also why the –NHCOCH₃ group is a weaker activator than –NH₂.

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

Explanation:
1. Acetanilide is less basic than aniline as in amides the carbonyl group (C=O) is a stronger dipole than N–C dipole. Therefore, the ability of N–C group to act as H-bond acceptor (as a base) is restricted in the presence of a C=O dipole.
2. The lone pair on nitrogen is delocalised towards the carbonyl group, so electron density on nitrogen decreases; hence R explains A.

Teacher's Note:
1. Any group that pulls the nitrogen lone pair into resonance (C=O, SO₂, benzene ring) lowers basicity.

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

Explanation:
1. N,N-Diethylbenzene sulphonamide is insoluble in alkali because it has no acidic hydrogen on nitrogen.
2. Sulphonyl group attached to nitrogen atom is electron withdrawing group; this makes the N–H hydrogen acidic in N-alkylsulphonamides, but it does not explain the insolubility here, which is due to the absence of an N–H hydrogen.

Teacher's Note:
1. In Hinsberg’s test, solubility in alkali needs an N–H hydrogen made acidic by the SO₂ group; with two alkyl groups on N there is no such hydrogen.

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

Explanation:
1. Aromatic 1° amines cannot be prepared by Gabriel Phthalimide synthesis because in Gabriel Phthalimide synthesis, it involves S_N2 mechanism and aryl halides do not undergo nucleophilic substitution with anion formed by phthalimide.

Teacher's Note:
1. The C–X bond of aryl halides has partial double bond character, so the phthalimide anion cannot displace the halide.