JEE Chemistry Amines MCQs Set 09

Answer: (b) Hofmann bromamide reaction

Explanation:
1. Carbylamine and Reimer-Tiemann reactions involve dichlorocarbene.
2. Friedel-Crafts uses a carbocation or acylium ion.
3. Hofmann bromamide rearrangement proceeds via an electron-deficient nitrogen (nitrene) species.

Teacher's Note:
1. Carbene intermediates have a divalent carbon, nitrenes a monovalent nitrogen.

Answer: (d) all of these

Explanation:
1. In the strongly acidic nitrating mixture aniline is partly protonated to the anilinium ion, which is a meta director.
2. The unprotonated aniline gives ortho and para products.
3. So a mixture of o-, m- and p-nitroaniline is obtained along with oxidation products.

Teacher's Note:
1. Direct nitration of aniline gives a mixture, which is why the amine group is acetylated first.

Answer: (a) \(\ce{C6H5-N=N-Cl}\) (covalent)

Explanation:
1. Benzenediazonium chloride is an ionic salt, written \(\ce{[C6H5N2]+Cl-}\) or \(\ce{C6H5-N+#N Cl-}\).
2. The covalent azo-chloride form \(\ce{C6H5-N=N-Cl}\) does not represent the actual structure.

Teacher's Note:
1. The diazonium group is a positively charged N2+ attached to the ring.
2. The salt is ionic, which is why it is soluble in water.

Answer: (c) p-bromoacetanilide

Explanation:
1. The NHCOCH3 group is an ortho/para director and is bulky.
2. Bromination therefore takes place mainly at the para position, giving p-bromoacetanilide.

Teacher's Note:
1. Bulky acetylamino group blocks the ortho positions sterically.
2. The para isomer is the major product.

Answer: (a) \(\ce{ArX + KCN}\)

Explanation:
1. Aryl halides do not undergo nucleophilic substitution with KCN because the C-X bond has partial double-bond character.
2. Sandmeyer cyanation of diazonium salts and dehydration of amides (with \(\ce{P2O5}\) or \(\ce{SOCl2}\)) give aromatic nitriles.

Teacher's Note:
1. Aryl halides are very unreactive towards nucleophiles.
2. Diazonium salts are the usual route to aryl cyanides.

Answer: (b) o-, p- and o-, p-directing

Explanation:
1. Diazotisation needs X = NH2, and warming the diazonium salt with water gives Y = OH.
2. Both NH2 and OH are ortho/para directing groups.

Teacher's Note:
1. NH2 and OH are strongly activating, o/p-directing groups.

Answer: (a) 2,4,6-trinitrobenzoic acid

Explanation:
1. Decarboxylation of 2,4,6-trinitrobenzoic acid gives 1,3,5-trinitrobenzene (TNB), a high explosive.
2. The other acids give less hazardous products (dinitrobenzene, aniline, phenol).

Teacher's Note:
1. Trinitro-aromatic compounds such as TNT and TNB are explosives.

Answer: (c) \(\ce{RCH2NH2}\)

Explanation:
1. Na/C2H5OH supplies nascent hydrogen.
2. The nitrile is reduced to the primary amine: \(\ce{RCN + 4[H] -> RCH2NH2}\) (Mendius reaction).

Teacher's Note:
1. The amine formed has one more carbon than the alkyl group R.

Answer: (b) nitrobenzene

Explanation:
1. Nitrobenzene is strongly deactivated and does not undergo Friedel-Crafts reaction itself.
2. It dissolves the reagents and the AlCl3 complex, so it is a useful solvent.
3. Toluene and benzene would react, and aniline would form a complex with AlCl3.

Teacher's Note:
1. A Friedel-Crafts solvent must itself be inert to electrophilic attack.

Answer: (a) a halide group

Explanation:
1. In the Sandmeyer reaction a diazonium salt reacts with CuCl, CuBr or CuCN to give aryl chloride, bromide or cyanide.
2. The diazonium group is replaced by a halide (or cyanide) group.

Teacher's Note:
1. Sandmeyer uses cuprous salts, while the Gattermann reaction uses Cu powder with HX.

Answer: (b) nitrobenzene

Explanation:
1. The nitro group strongly withdraws electrons from the ring and deactivates it.
2. Toluene and phenol have activating groups, and benzene is in between.
3. Nitrobenzene is hence the most difficult to nitrate.

Teacher's Note:
1. Electron-withdrawing groups slow electrophilic substitution.
2. Activating groups speed it up.

Answer: (b) o-dinitrobenzene

Explanation:
1. Para-disubstituted benzenes with identical groups have zero dipole moment because the bond moments cancel.
2. In the ortho isomers the moments add partially.
3. The nitro group has a much larger group moment than Cl, so o-dinitrobenzene has the largest dipole moment (about 6 D).

Teacher's Note:
1. Symmetrical para compounds with identical groups are nonpolar.
2. Larger bond moments in the ortho arrangement give larger net dipoles.

Answer: (d) \(\ce{CH3CH2NH2}\)

Explanation:
1. Propionic acid gives propionyl chloride (B), then propanamide (C).
2. Hofmann bromamide degradation removes one carbon, giving ethanamine \(\ce{CH3CH2NH2}\) (D).

Teacher's Note:
1. Hofmann degradation lowers the carbon count by one.
2. Do not confuse it with the Gabriel synthesis.

Answer: (b) \(\ce{C6H5-N=N-C6H4-N(CH3)2}\) (para-azo compound)

Explanation:
1. Aniline gives benzenediazonium chloride (B).
2. The diazonium ion is a weak electrophile and attacks the activated para position of N,N-dimethylaniline.
3. The product is the azo dye 4-(dimethylamino)azobenzene, \(\ce{C6H5-N=N-C6H4-N(CH3)2}\).

Teacher's Note:
1. Coupling with aromatic amines or phenols gives azo dyes with the N=N link at the para position.
2. Azo compounds are coloured because of extended conjugation.

Answer: (b) N-nitroso-N-methylaniline, \(\ce{C6H5-N(CH3)-N=O}\)

Explanation:
1. N-Methylaniline is a secondary amine.
2. Secondary amines react with nitrous acid to give N-nitrosoamines, \(\ce{R2N-N=O}\).
3. The product is N-nitroso-N-methylaniline, a yellow oil.

Teacher's Note:
1. Nitrosoamines are carcinogenic.
2. Compare with the 3° aromatic amine, which gives a ring C-nitroso compound.

Answer: (a) N-phenylhydroxylamine

Explanation:
1. In neutral medium (Zn/NH4Cl) the reduction stops at the hydroxylamine stage.
2. The product is \(\ce{C6H5NHOH}\), which reduces Tollens' reagent (Mullikan-Barker test).

Teacher's Note:
1. Neutral medium gives hydroxylamine, acidic medium gives aniline, alkaline medium gives bimolecular products.

Answer: (c) \(\ce{(CH3)2CH-NH2}\)

Explanation:
1. Evolution of nitrogen with HNO2 shows A is a primary amine.
2. Carbylamine reaction gives the isocyanide C, and reduction of an isocyanide R-NC gives R-NHCH3.
3. Isopropylmethylamine \(\ce{(CH3)2CH-NHCH3}\) means R is isopropyl, so A is isopropylamine \(\ce{(CH3)2CH-NH2}\).

Teacher's Note:
1. Work backwards from the reduction product to the isocyanide and then to the primary amine.

Answer: (c) 1,3-dinitrobenzene

Explanation:
1. The nitro group is a meta director and a deactivator.
2. So the second nitro group goes to the meta position, giving 1,3-dinitrobenzene.

Teacher's Note:
1. Meta directors: NO2, CN, COOH, CHO, SO3H.

Answer: (b) \(\ce{C6H5OH}\)

Explanation:
1. Aniline gives benzenediazonium salt.
2. On warming with water or dilute acid it is hydrolysed to phenol with loss of nitrogen.

Teacher's Note:
1. Keep the diazonium salt below 5°C to avoid phenol formation.

Answer: (b) \(\ce{C6H5N2+X-}\)

Explanation:
1. The aryl diazonium ion is stabilised by delocalisation of the positive charge into the benzene ring.
2. Alkyl diazonium salts decompose immediately even at low temperature.
3. Benzyl diazonium is also unstable because it has no delocalisation with the ring.

Teacher's Note:
1. Only aryl diazonium salts are stable at 0-5°C.
2. Delocalisation is the reason for their stability.

Answer: (d) p-aminophenol

Explanation:
1. In strongly acidic medium N-phenylhydroxylamine is formed first.
2. It undergoes Bamberger rearrangement in the acid to give p-aminophenol.

Teacher's Note:
1. Strongly acidic: p-aminophenol. Weakly acidic: aniline.