JEE Chemistry Amines MCQs Set 31

Answer: (b) It is soluble in aqueous HCl

Explanation:
1. Sulphanilic acid exists as a zwitterion (inner salt) \( \ce{H3N+-C6H4-SO3^-} \), with a high melting/decomposition point and poor solubility in organic solvents.
2. It dissolves in NaOH by forming the sodium salt.
3. Because the strongly acidic sulphonic group has already transferred its proton, the amino group is already protonated and the compound is only sparingly soluble in dilute acid.

Teacher's Note:
1. Zwitterionic compounds behave like salts: high decomposition point, insoluble in organic solvents.
2. Sulphanilic acid is used to prepare azo dyes such as methyl orange.

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

Explanation:
1. Coupling is an electrophilic substitution by the weak electrophile \( \ce{ArN2+} \), so only strongly activated rings react.
2. Activating power: \( \ce{O^-} \) > \( \ce{NH2} \) > OH.
3. The anilinium ion has a positive \( \ce{NH3+} \) group, which is deactivating, so it is least reactive.
4. Order: phenoxide > aniline > phenol > anilinium.

Teacher's Note:
1. The diazonium ion is a weak electrophile and needs an activated ring.
2. Phenols couple in mildly alkaline medium (phenoxide), amines in mildly acidic medium.

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

Explanation:
1. The diazonium ion is the electrophile; electron-withdrawing groups make it more electrophilic.
2. Two nitro groups (IV) are better than one (III), which is better than H (I).
3. A methyl group donates electrons and makes the cation less electrophilic (II least).
4. Order: IV > III > I > II.

Teacher's Note:
1. Electron-withdrawing groups on the diazonium ring increase coupling reactivity.
2. Electron-donating groups decrease it.

Answer: (d) \( \ce{C6H5NH2} \)

Explanation:
1. Aniline with chloroform and KOH gives phenyl isocyanide [A], \( \ce{C6H5NC} \) (carbylamine reaction).
2. Isocyanides are hydrolysed by aqueous acid to a primary amine and formic acid.
3. So [B] is aniline.

Teacher's Note:
1. Carbylamine reaction product, hydrolysed, regenerates the primary amine.
2. Isocyanide hydrolysis: R-NC + 2H2O gives RNH2 + HCOOH.

Answer: (b) 2,6-diiodo-4-nitroaniline

Explanation:
1. In ICl, iodine is the electrophilic (positive) end and chlorine the negative end.
2. The amino group activates the ortho positions (para is blocked by the nitro group).
3. So both ortho positions are iodinated, giving 2,6-diiodo-4-nitroaniline.

Teacher's Note:
1. I-Cl is polarised as I(delta+)-Cl(delta-), so it iodinates.
2. The amino group directs to ortho/para positions; the para position is already occupied.

Answer: (a) Benzylamine

Explanation:
1. Benzylamine is an aliphatic-type amine (pKb about 4.7).
2. In aniline the lone pair is delocalised into the ring (pKb about 9.4).
3. In p-nitroaniline and acetanilide delocalisation is even greater (pKb about 13 and 14).
4. The lowest pKb corresponds to the strongest base, benzylamine.

Teacher's Note:
1. Lower pKb means stronger base.
2. Delocalisation of the N lone pair into a ring or carbonyl lowers basicity.

Answer: (a) I < II < III

Explanation:
1. A nitro group lowers basicity (higher pKb).
2. At the para position the nitro group withdraws electrons by both -I and -M effects; at meta only by -I.
3. Basicity: aniline > m-nitroaniline > p-nitroaniline, so pKb increases in the order I < II < III.

Teacher's Note:
1. Para-nitro lowers basicity most because of resonance (-M) as well as inductive effect.
2. Higher pKb means weaker base.

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

Explanation:
1. Electron-donating groups (OCH3, CH3) increase basicity; OCH3 donates more than CH3 in the para position.
2. Electron-withdrawing groups decrease basicity; \( \ce{NO2} \) (-I and -M) lowers it more than \( \ce{CF3} \) (-I only).
3. Order: II > I > III > IV.

Teacher's Note:
1. Donor groups raise Kb; acceptors lower it.
2. Compare para-substituent effects: OCH3 > CH3 > H > CF3 > NO2.

Answer: (b) I > IV > III > II

Explanation:
1. Methylamine is an aliphatic amine (conjugate acid pKa about 10.6).
2. Tetrahydroisoquinoline is a secondary aliphatic amine, pKa about 9.4, slightly lower than methylamine.
3. Tetrahydroquinoline has N on the aromatic ring, similar to N-alkyl aniline (about 5).
4. Aniline (about 4.6) is the weakest.
5. Order: I > IV > III > II.

Teacher's Note:
1. Amines with N attached directly to benzene ring are much weaker bases.
2. N-alkylation of aniline slightly increases basicity.

Answer: (c) Hofmann-Martius rearrangement

Explanation:
1. Heating N-alkylaniline hydrochloride (e.g. at 200-300 °C) moves the alkyl group from nitrogen to the ortho/para position of the ring.
2. This is the Hofmann-Martius rearrangement.
3. Fries migration concerns phenolic esters; Hofmann degradation concerns amides.

Teacher's Note:
1. Hofmann-Martius: N-alkylaniline hydrochloride gives o- and p-alkylaniline.
2. Do not confuse it with Hofmann bromamide degradation.

Answer: (c) Balz-Schiemann reaction

Explanation:
1. Aniline is diazotised with \( \ce{NaNO2/HCl} \) to give benzenediazonium chloride.
2. With fluoroboric acid \( \ce{HBF4} \) the diazonium tetrafluoroborate precipitates.
3. On heating it decomposes to fluorobenzene, \( \ce{BF3} \) and \( \ce{N2} \); this is the Balz-Schiemann reaction.

Teacher's Note:
1. Fluorine and iodine cannot be introduced by the Sandmeyer method.
2. Use HBF4 for F, KI for I.

Answer: (b) p-hydroxyazobenzene

Explanation:
1. In weakly alkaline solution phenol is present as the phenoxide ion, a strongly activated nucleophile.
2. The diazonium ion attacks the para position (azo coupling).
3. The product is the orange dye p-hydroxyazobenzene, \( \ce{C6H5-N=N-C6H4-OH} \).

Teacher's Note:
1. Azo coupling with phenols: pH 9-10, para position.
2. With amines: pH 4-5.

Answer: (d) p-methylbenzylamine

Explanation:
1. Carbylamine test is given only by primary amines.
2. N,N-dimethylaniline is tertiary; N,2,4-trimethylaniline and N-methyl-o-methylaniline are secondary.
3. p-Methylbenzylamine is a primary amine (-CH2NH2), so it gives the test.

Teacher's Note:
1. Check that N bears two hydrogens (primary amine).
2. Both aromatic and aliphatic primary amines respond.

Answer: (c) 4-Aminophenol

Explanation:
1. Paracetamol is N-(4-hydroxyphenyl)acetamide.
2. It is made by acetylating the amino group of 4-aminophenol (p-aminophenol) with acetic anhydride.
3. The amino group is more nucleophilic than the phenolic OH, so N-acetylation occurs.

Teacher's Note:
1. Paracetamol = 4-hydroxyacetanilide.
2. Amine acylation is faster than phenol acylation.

Answer: (b) Aromatic amines are stronger bases than ammonia

Explanation:
1. Alkyl groups donate electrons, so alkylamines are stronger bases than \( \ce{NH3} \).
2. In aromatic amines the lone pair is delocalised into the ring, so they are weaker bases than ammonia (aniline pKb 9.4 versus 4.75 for ammonia).
3. So statement (b) is wrong; the others are correct.

Teacher's Note:
1. Aniline is a weaker base than ammonia.
2. Resonance stabilises the free aromatic amine more than its ion.

Answer: (d) all of the above

Explanation:
1. Benzenediazonium chloride couples with phenols in alkaline medium.
2. Phenol, cresol (with a free para or ortho position) and resorcinol are all activated phenols.
3. Each forms a coloured azo dye (orange/red/yellow).

Teacher's Note:
1. Azo dyes are coloured because of the extended conjugation through -N=N-.
2. Coupling needs an activated aromatic ring.

Answer: (c) nitrobenzene

Explanation:
1. The electronegative nitrogen in the ring withdraws electrons, deactivating the ring strongly.
2. In acidic nitrating media pyridine is also protonated, making it even more deactivated.
3. Substitution occurs at position 3, like meta substitution in nitrobenzene, and needs vigorous conditions.

Teacher's Note:
1. Pyridine is a pi-deficient heterocycle.
2. Pyridine is far less reactive than benzene in electrophilic substitution.

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

Explanation:
1. A diazonium ion is a better electrophile when its ring carries electron-withdrawing groups.
2. Donor strength: \( \ce{N(CH3)2} \) > \( \ce{OCH3} \) > \( \ce{CH3} \), so the cation is less electrophilic with stronger donors.
3. Reactivity: IV < II < I < III.

Teacher's Note:
1. Resonance donation into the N2+ group reduces its electrophilicity.
2. NMe2 is the strongest donor here.

Answer: (c) around 10

Explanation:
1. Coupling with phenols needs the more nucleophilic phenoxide ion, so mild alkalinity (pH 9-10) is used.
2. At very high pH (12) the diazonium ion is converted into unreactive diazotate.
3. At pH 4-5, coupling with amines is best, not phenols.

Teacher's Note:
1. Phenols: pH 9-10. Aromatic amines: pH 4-5.
2. Strongly alkaline medium destroys the diazonium ion.

Answer: (c) p-toluidine

Explanation:
1. A brilliant orange-red azo dye needs a stable aromatic diazonium salt.
2. Only a primary aromatic amine (p-toluidine) gives such a diazonium salt at 0-5 °C.
3. Benzylamine (aliphatic) gives an unstable diazonium that decomposes; secondary amines give nitrosamines; tertiary amines give a ring C-nitroso compound.

Teacher's Note:
1. Diazotisation then coupling with beta-naphthol is the standard test for primary aromatic amines.
2. Aliphatic amines do not give this test.

Answer: (c) \( \ce{C6H5NC ->[LiAlH4]} \)

Explanation:
1. Schmidt reaction of benzoic acid gives aniline.
2. Acid hydrolysis of phenyl isocyanide gives aniline and formic acid.
3. LiAlH4 reduces the isocyanide to the secondary amine N-methylaniline, \( \ce{C6H5NHCH3} \), not aniline.
4. Hofmann degradation of benzamide gives aniline.

Teacher's Note:
1. Reduction of isocyanides gives N-methyl secondary amines.
2. Hydrolysis of isocyanides gives primary amines.

Answer: (c) N-Methylaniline

Explanation:
1. Diazotisation needs a primary aromatic amine (\( \ce{ArNH2} \)).
2. Aniline, p-toluidine and p-chloroaniline are all primary and give diazonium salts.
3. N-Methylaniline is secondary and gives an N-nitrosamine, not a diazonium salt.

Teacher's Note:
1. Primary: diazonium; secondary: N-nitrosamine; tertiary: C-nitroso (p-nitroso) compound.
2. This is the basis of the nitrous acid test for amines.