JEE Chemistry Amines MCQs Set 22

Answer: (d)

Explanation:
1. V is benzenediazonium chloride.
2. Coupling with 2-naphthoxide occurs at the 1-position (the most activated, alpha to the oxygen-bearing ring carbon).
3. W is 1-(phenylazo)-2-naphthol, an orange-red dye.

Teacher's Note:
1. The 1-position of 2-naphthol is the reactive site.
2. The azo group links the benzene ring to naphthalene.

Answer: (c)

Explanation:
1. Acetylation moderates the activating effect of the amino group.
2. KBrO₃/HBr generates bromine; bromination of acetanilide occurs para to the NHCOCH3 group.
3. Hydrolysis gives 4-bromoaniline; diazotisation and the Gattermann-type replacement (Cu/HBr) give 1,4-dibromobenzene.

Teacher's Note:
1. Bromination of acetanilide stops at mono-substitution at para.
2. Cu/HBr replaces the diazonium group by Br.

Answer: (a) A, B and D only

Explanation:
1. A: three molecules of acetone condense by aldol steps and dehydrate to mesitylene.
2. B: three propyne molecules trimerise on heating to give mesitylene.
3. C: the haloform reaction and decarboxylation give benzene, not mesitylene, so C is wrong.
4. D: Clemmensen reduction converts each CHO to CH3, giving mesitylene.

Teacher's Note:
1. Check what each sequence does to every substituent.
2. Haloform followed by decarboxylation removes the substituent completely.

Answer: (c) 495

Explanation:
1. A is benzoic acid (haloform reaction): 10 × 0.60 = 6 mol.
2. B is benzamide: 6 × 0.50 = 3 mol; C is aniline (Hofmann): 3 × 0.50 = 1.5 mol.
3. D is 2,4,6-tribromoaniline, C₆H₄Br₃N, molar mass 72 + 4 + 240 + 14 = 330 g/mol.
4. Mass = 1.5 × 330 = 495 g.

Teacher's Note:
1. Multiply the yields stepwise.
2. Three bromine atoms enter the ring of aniline.

Answer: (d) an isocyanide

Explanation:
1. This is the carbylamine reaction.
2. The product is an isocyanide (carbylamine), R-NC.

Teacher's Note:
1. Only primary amines react.
2. The test is positive by the foul smell.

Answer: (a)

Explanation:
1. This is the carbylamine reaction of a primary aromatic amine.
2. The product is the isocyanide, 4-methylphenyl isocyanide (CH₃C₆H₄-NC).

Teacher's Note:
1. Isocyanides have the nitrogen attached to the ring.
2. Diazonium salts need nitrous acid.

Answer: (b) from aniline by diazotisation followed by heating the diazonium salt with HBF₄

Explanation:
1. Direct fluorination is violent and not controllable.
2. Aryl halides do not undergo simple nucleophilic exchange with fluoride, and phenol is not converted to an aryl fluoride by HF.
3. The Balz-Schiemann reaction (diazonium fluoroborate heated) gives fluorobenzene.

Teacher's Note:
1. Balz-Schiemann: ArN2+ BF4- gives ArF + N2 + BF3.
2. Fluorine is too reactive for direct aromatic fluorination.

Answer: (d) C₂H₅NC and 3KCl

Explanation:
1. The carbylamine reaction forms an isocyanide and chloride ions.
2. C₂H₅NH₂ + CHCl₃ + 3KOH gives C₂H₅NC + 3KCl + 3H₂O.
3. Check: the three Cl of chloroform end up as 3KCl.

Teacher's Note:
1. Balance chlorine and potassium.
2. The product is an isocyanide, not a nitrile.

Answer: (b) a mixture of o- and p-bromotoluenes

Explanation:
1. Nitration of toluene gives o- and p-nitrotoluene.
2. Reduction gives the toluidines, diazotisation gives the diazonium salts and Cu(I) bromide replaces the diazonium group by Br.
3. The methyl group is unaffected, so o- and p-bromotoluenes are obtained.

Teacher's Note:
1. The positions of substituents set in the nitration step are retained.
2. Sandmeyer reaction replaces N2+ by Br.

Answer: (d) benzenediazonium chloride and fluorobenzene

Explanation:
1. Diazotisation gives benzenediazonium chloride (A).
2. With HBF₄ it gives the fluoroborate, which on heating loses N₂ and BF₃ to give fluorobenzene (B).

Teacher's Note:
1. Diazotisation is done at low temperature.
2. The Balz-Schiemann reaction gives aryl fluorides.

Answer: (d) benzenediazonium chloride and benzonitrile

Explanation:
1. Diazotisation gives benzenediazonium chloride.
2. CuCN replaces the diazonium group by CN to give benzonitrile (Sandmeyer).

Teacher's Note:
1. Cyanide enters through a Cu(I) salt.
2. Benzonitrile can be hydrolysed to benzoic acid.

Answer: (b) CH₃CH₂COOH

Explanation:
1. CH₃CH₂NH₂ comes from an amide with one more carbon by the Hofmann reaction: C is propanamide.
2. Propanamide is made by heating ammonium propanoate (B) formed from propanoic acid (A) and ammonia.

Teacher's Note:
1. Work backwards: the amine has one carbon less than the amide.
2. Acid plus ammonia gives the ammonium salt, which loses water to give the amide.

Answer: (d)

Explanation:
1. Diazotisation gives 4-methylbenzenediazonium chloride (D).
2. CuCN/KCN replaces N2+ by CN (Sandmeyer cyanation).
3. E is 4-methylbenzonitrile (p-tolunitrile).

Teacher's Note:
1. No hydrolysis step is present, so the nitrile is the final product.
2. Biphenyls form in the Gomberg reaction instead.

Answer: (b) four moles of NaOH and one mole of Br₂

Explanation:
1. RCONH₂ + Br₂ + 4NaOH gives RNH₂ + 2NaBr + Na₂CO₃ + 2H₂O.
2. Per mole of amine, four moles of NaOH and one mole of bromine are needed.

Teacher's Note:
1. Balance the equation: two NaOH are used in forming NaBr and Na2CO3 and two to form water.
2. The carbonyl carbon is lost as carbonate.

Answer: (d) aniline

Explanation:
1. Nitration uses a strongly acidic mixture in which aniline is protonated to anilinium ion.
2. The -NH₃+ group is meta-directing, so a significant amount of m-nitroaniline forms along with the ortho and para products.

Teacher's Note:
1. Protection by acetylation avoids this problem.
2. Phenols and esters of phenols give ortho/para products only.

Answer: (a) aniline

Explanation:
1. Mild reducing agents (SnCl₂/HCl, Na₂SO₃) stop at phenylhydrazine.
2. Zn/HCl is a stronger reducing system; it reduces the diazonium salt through phenylhydrazine all the way to aniline and ammonia.

Teacher's Note:
1. Mild reduction stops at phenylhydrazine; strong reduction gives aniline.
2. Nitrogen-nitrogen bond is broken in complete reduction.

Answer: (c) Sandmeyer

Explanation:
1. Replacement of the diazonium group by chlorine using CuCl/HCl is the Sandmeyer reaction.

Teacher's Note:
1. Sandmeyer: CuX/HX; Gattermann: Cu powder/HX.
2. Friedel-Crafts and Wurtz are carbon-carbon bond forming reactions.

Answer: (c)

Explanation:
1. The methyl group directs bromination ortho (the para position is held by NO2), and this position is meta to NO2: 2-bromo-4-nitrotoluene.
2. Reduction gives the amine, diazotisation gives the diazonium salt and CuBr replaces it by Br.
3. E is 2,4-dibromotoluene.

Teacher's Note:
1. Both substituents direct to the same position.
2. The side-chain methyl is not touched under these conditions.

Answer: (a) only heat

Explanation:
1. The diazonium salt reacts with HBF₄ to give the insoluble aryl diazonium tetrafluoroborate.
2. Heating the dry salt decomposes it to give ArF, N₂ and BF₃; no copper is needed (the Balz-Schiemann reaction).

Teacher's Note:
1. Copper salts are used for Cl, Br and CN, not for F.
2. Isolate the fluoroborate and heat it.

Answer: (b)

Explanation:
1. C is aniline (C₆H₇N), produced by Hofmann degradation of benzamide (B).
2. Benzamide comes from benzoic acid (A) with ammonia on heating, and benzoic acid is C₇H₆O₂.

Teacher's Note:
1. Hofmann degradation loses one carbon: C7 amide to C6 amine.
2. The other aromatic isomers do not give amides with ammonia.