JEE Chemistry Amines MCQs Set 21

Answer: (b)

Explanation:
1. The diazonium ion attacks the position para to the strongly activating dimethylamino group.
2. The azo (-N=N-) group joins the two rings.
3. The product is 4-(dimethylamino)azobenzene, a yellow dye.

Teacher's Note:
1. Azo coupling is an electrophilic substitution by the diazonium ion.
2. The coupling takes place at the para position.

Answer: (d)

Explanation:
1. X is 4-nitrobenzenediazonium chloride, a good electrophile because of the nitro group.
2. The methoxy group directs the substitution to its para position (ortho is hindered).
3. Y is 4-methoxy-4'-nitroazobenzene.

Teacher's Note:
1. A nitro group on the diazonium ring increases its electrophilicity.
2. Coupling with anisole gives the para azo compound.

Answer: (d) benzaldehyde

Explanation:
1. The Sandmeyer cyanation gives benzonitrile (X).
2. DIBAL-H reduces a nitrile to an imine-aluminium complex, which is hydrolysed by water to the aldehyde.
3. Y is benzaldehyde.

Teacher's Note:
1. DIBAL-H stops at the aldehyde stage when used at low temperature.
2. Complete reduction to the amine needs LiAlH4.

Answer: (c) 3

Explanation:
1. The three isomers (o-, m- and p-toluidine) each give a diazonium salt.
2. Hypophosphorous acid replaces the diazonium group by H in every case, giving toluene.
3. The answer is 3.

Teacher's Note:
1. Follow the transformations one at a time and count atoms or add atomic masses.
2. Integer-type answers are best checked by writing the formula of the final product.

Answer: (c) 94

Explanation:
1. Sn/HCl gives aniline, diazotisation gives the diazonium salt, and warming with water gives phenol, Z.
2. C₆H₅OH: 72 + 6 + 16 = 94.

Teacher's Note:
1. Follow the transformations one at a time and count atoms or add atomic masses.
2. Integer-type answers are best checked by writing the formula of the final product.

Answer: (d) 0

Explanation:
1. Excess Sn/HCl reduces both nitro groups to amino groups: benzene-1,3-diamine.
2. Both amino groups are diazotised and then removed by H₃PO₂.
3. The final product is benzene, which has no nitrogen.

Teacher's Note:
1. Follow the transformations one at a time and count atoms or add atomic masses.
2. Integer-type answers are best checked by writing the formula of the final product.

Answer: (c)

Explanation:
1. Bromination of nitrobenzene gives 3-bromonitrobenzene (F), reduction gives 3-bromoaniline (G).
2. Diazotisation gives the diazonium salt (H) and warming with water replaces it by OH.
3. I is 3-bromophenol.

Teacher's Note:
1. Meta orientation is set by the nitro group.
2. Diazonium hydrolysis gives phenol.

Answer: (c) B, C and D only

Explanation:
1. Coupling requires an activated aromatic ring.
2. alpha-Naphthol, N,N-dimethylaniline and phenol are all strongly activated; benzaldehyde is deactivated by the CHO group.

Teacher's Note:
1. The diazonium ion is a weak electrophile.
2. The ring must carry OH or an amino group.

Answer: (c) N-bromobenzamide

Explanation:
1. Benzamide reacts with Br₂/NaOH to give N-bromobenzamide.
2. Base removes a proton to give the bromamide anion, which loses bromide as the phenyl group migrates to nitrogen, giving phenyl isocyanate.
3. A carbene or carbocation is not involved, and the intermediate is an isocyanate, not an isocyanide.

Teacher's Note:
1. The Hofmann rearrangement proceeds through N-bromoamide and isocyanate.
2. Isocyanate is hydrolysed and loses carbon dioxide.

Answer: (a) p-hydroxyazobenzene

Explanation:
1. The phenoxide ion is a strong nucleophile at the para position.
2. The diazonium ion attacks it in an azo coupling, forming 4-hydroxyazobenzene.

Teacher's Note:
1. Phenoxide couples at the para position.
2. Azo compounds contain -N=N-.

Answer: (b)

Explanation:
1. Reduction gives 4-chloroaniline; acetylation gives the acetanilide (NHAc), which does not poison the Friedel-Crafts catalyst.
2. The isopropyl cation enters ortho to the NHAc group because the para position is occupied by Cl.
3. Hydrolysis then gives 4-chloro-2-isopropylaniline.

Teacher's Note:
1. A free amino group cannot be used in Friedel-Crafts alkylation, but an acetylated one can.
2. The acyl group is removed by hydrolysis.

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

Explanation:
1. The aryl group migrates to nitrogen with its electron pair, so electron-releasing substituents on the migrating ring speed up the reaction.
2. Methyl is electron-releasing (fastest), H is neutral, chloro is weakly electron-withdrawing, nitro is strongly electron-withdrawing (slowest).
3. The order is II > I > IV > III.

Teacher's Note:
1. Migrating group acts as a nucleophile in the rearrangement step.
2. Electron-releasing groups increase migratory aptitude.

Answer: (a) (IV)(iii)(R)

Explanation:
1. Ethylbenzene with bromine and a metal halide (Lewis acid, formed from Al) undergoes electrophilic aromatic substitution, so (IV)(iii)(R) is correct.
2. p-Fluoronitrobenzene with NaOH is nucleophilic substitution (P), not R; bromine and UV give free radical substitution (Q), not R; and the nitrating mixture does not give nucleophilic substitution of a bromoalkane.

Teacher's Note:
1. Match the reagent type to the mechanism: Lewis acid for electrophilic, UV for radical, base for nucleophilic.
2. Check all three columns together.

Answer: (d) Assertion is false but Reason is true

Explanation:
1. In strong acid aniline is converted to the anilinium ion, C₆H₅NH₃+.
2. The -NH₃+ group is electron-withdrawing and meta-directing; the lone pair can no longer activate the ring.
3. So the ring is less reactive: the Assertion is false and the Reason is true.

Teacher's Note:
1. Protonation removes the +M effect of the amino group.
2. Direct nitration of aniline therefore gives meta product too.

Answer: (c) and

Explanation:
1. The Hofmann rearrangement is intramolecular: the aryl group migrates from carbon to the nitrogen of the same molecule.
2. The 3-deuterio amide therefore gives 3-deuterioaniline with ordinary nitrogen, and the 15N amide gives aniline with 15N.
3. No crossover of labels occurs between the two molecules.

Teacher's Note:
1. An intramolecular rearrangement keeps each label with its own molecule.
2. The carbonyl carbon is lost as carbonate.

Answer: (d) formation of the isocyanate by migration of the aryl group

Explanation:
1. Deprotonation and bromination are fast equilibria.
2. The slow step is the concerted loss of bromide with migration of the aryl group from carbon to nitrogen, which gives the isocyanate.
3. Hydrolysis of the isocyanate to the amine is fast.

Teacher's Note:
1. The slowest step involves bond breaking and migration at the same time.
2. Electron-releasing groups on the ring speed this step.

Answer: (c) CH₃-N+≡C-

Explanation:
1. This is the carbylamine reaction of a primary amine.
2. Dichlorocarbene reacts with the amine and two HCl are lost, leaving the isocyanide group.
3. The product is methyl isocyanide, CH₃-N+≡C-.

Teacher's Note:
1. Isocyanide: carbon is bonded to nitrogen; nitrile: carbon is bonded to carbon.
2. The isocyanide has a very unpleasant smell.

Answer: (b)

Explanation:
1. Only a primary aromatic amine gives a stable diazonium salt at 0-5 degrees C.
2. p-Toluidine gives the diazonium ion, which couples with beta-naphthol to give an orange-red azo dye.
3. N,N-Dimethylaniline gives a ring-nitroso compound, N-methylaniline gives an N-nitrosoamine, and benzylamine gives an unstable aliphatic diazonium ion.

Teacher's Note:
1. Dye formation needs a stable aryl diazonium salt.
2. Beta-naphthol test is a standard test for primary aromatic amines.

Answer: (c) optically inactive meso-tartaric acid and an optically inactive (racemic) pair of tartaric acids

Explanation:
1. P forms an anhydride, so it is maleic acid (cis); Q is fumaric acid (trans).
2. Syn hydroxylation of the cis isomer gives meso-tartaric acid, which is optically inactive by internal compensation.
3. Syn hydroxylation of the trans isomer gives equal amounts of the two enantiomers, a racemic mixture, which is optically inactive.

Teacher's Note:
1. Dilute cold KMnO4 adds two OH groups on the same side (syn).
2. A racemic mixture shows no net optical rotation.

Answer: (d) 4

Explanation:
1. I is the Gattermann-Koch reaction, which formylates benzene.
2. II is hydrolysis of a gem-dihalide, giving the aldehyde.
3. III is the Rosenmund reduction.
4. IV is the partial reduction of an ester to an aldehyde by DIBAL-H at low temperature.
5. All four give benzaldehyde.

Teacher's Note:
1. Know the named routes to aromatic aldehydes.
2. DIBAL-H at -78 degrees C stops at the aldehyde.