JEE Chemistry Amines MCQs Set 23

Answer: (c) labelled aniline and unlabelled 4-nitroaniline

Explanation:
1. The Hofmann bromamide reaction is an intramolecular rearrangement: the aryl group migrates from carbon to nitrogen within the same molecule.
2. The nitrogen atom of each amide stays with its own aryl group, so no cross products form.
3. Labelled benzamide gives labelled aniline and 4-nitrobenzamide gives unlabelled 4-nitroaniline.

Teacher's Note:
1. No free isocyanate fragments exchange between molecules.
2. The nitrogen that ends up on the ring is the original amide nitrogen.

Answer: (d)

Explanation:
1. Acetylation protects the amino group as NHCOCH₃, a weaker activator.
2. Bromine goes ortho to the amide group because the para position holds the methyl group.
3. Acid hydrolysis removes the acetyl group to leave 2-bromo-4-methylaniline.

Teacher's Note:
1. Acyl groups are removed by acid or base hydrolysis.
2. Substitution is directed by the strongest activating group present.

Answer: (a) aniline

Explanation:
1. Aniline is brominated in all three activated positions to give 2,4,6-tribromoaniline (Y).
2. Diazotisation gives the diazonium salt (Z), and boiling ethanol replaces the diazonium group with hydrogen.
3. The result is 1,3,5-tribromobenzene, so X is aniline.

Teacher's Note:
1. Diazonium salt plus ethanol or hypophosphorous acid removes the nitrogen group.
2. The amino group is used to direct the bromines and then removed.

Answer: (c) p-tolyl isocyanide

Explanation:
1. p-Toluidine is a primary aromatic amine and gives the carbylamine reaction.
2. Dichlorocarbene from chloroform and KOH is attacked by the amine nitrogen and two molecules of HCl are lost.
3. The product is p-tolyl isocyanide, an offensive-smelling liquid.

Teacher's Note:
1. Isocyanide formation is the test for primary amines.
2. The nitrogen, not carbon, is attached to the ring in an isocyanide.

Answer: (d) nitrobenzene

Explanation:
1. Catalytic reduction of an isocyanide R-NC gives R-NH-CH₃, so R is phenyl isocyanide.
2. Phenyl isocyanide comes from the carbylamine reaction of aniline (Q).
3. Aniline is obtained by reducing nitrobenzene (P).

Teacher's Note:
1. Work backwards from the final product.
2. An isocyanide reduces to an N-methyl secondary amine.

Answer: (d) benzene

Explanation:
1. Aniline gives benzenediazonium chloride.
2. H₃PO₂ reduces the diazonium salt, replacing N₂+ with hydrogen.
3. The product is benzene, with nitrogen gas evolved.

Teacher's Note:
1. H₃PO₂ and ethanol both replace a diazonium group by hydrogen.
2. The aromatic ring is untouched.

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

Explanation:
1. The amino group is a powerful activator for ortho and para substitution.
2. In water, a polar medium, bromination cannot be stopped at one stage.
3. All three activated positions are brominated, giving 2,4,6-tribromoaniline as a white precipitate.

Teacher's Note:
1. Protect the amino group by acetylation if a mono-bromo product is wanted.
2. Bromine water also decolourises quickly with phenol.

Answer: (d) Primary amines

Explanation:
1. The reaction needs an N-H₂ group to form the dithiocarbamate and then lose H₂S to give R-N=C=S.
2. Secondary amines stop at the dithiocarbamate stage and tertiary amines do not react.
3. Only primary amines give mustard oils.

Teacher's Note:
1. Isothiocyanate R-N=C=S needs a nitrogen with only one carbon substituent.
2. Same selectivity as the carbylamine reaction.

Answer: (d) CH₃CH₂NH₂

Explanation:
1. Silver cyanide is covalent and attacks through nitrogen, so P is ethyl isocyanide C₂H₅NC.
2. Acid hydrolysis of an isocyanide gives formic acid and the primary amine.
3. Q is ethylamine (C₂H₅NH₂).

Teacher's Note:
1. AgCN gives isocyanides and KCN gives nitriles.
2. Isocyanides hydrolyse to a primary amine and formic acid.

Answer: (d) Schotten-Baumann reaction

Explanation:
1. Benzoylation of an amine or an alcohol with benzoyl chloride in the presence of aqueous alkali is the Schotten-Baumann reaction.
2. The product here is benzanilide.

Teacher's Note:
1. The base removes HCl and drives the reaction.
2. Perkin and Claisen reactions form carbon-carbon bonds.

Answer: (b) N,N-dimethylaniline

Explanation:
1. The carbylamine test needs a primary amino group.
2. The last three are primary amines, aromatic or aliphatic.
3. N,N-Dimethylaniline is tertiary and does not respond.

Teacher's Note:
1. Count the hydrogens on nitrogen.
2. Benzylamine is a primary aliphatic amine.

Answer: (c) fluorobenzene

Explanation:
1. C₆H₅N₂+ BF₄- (B) is a stable solid that precipitates from the mixture.
2. On heating it loses N₂ and BF₃.
3. This is the Balz-Schiemann reaction and the product is fluorobenzene.

Teacher's Note:
1. Direct fluorination of arenes is violent, so this is the usual route to aryl fluorides.
2. Heating the tetrafluoroborate salt gives the fluoroarene.

Answer: (c)

Explanation:
1. NBS brominates the benzylic position to give 4-bromobenzyl bromide.
2. Phthalimide anion displaces bromide from the benzylic carbon (Gabriel synthesis, first step) to give the N-substituted phthalimide.
3. The ring bromine is not displaced; the product is N-(4-bromobenzyl)phthalimide.

Teacher's Note:
1. Aryl halides do not take part in nucleophilic substitution under these conditions.
2. Hydrazine or alkali hydrolysis would be needed to free the amine.

Answer: (d) CH₃CH₂N≡C

Explanation:
1. A is ethanenitrile; reduction gives ethylamine (B), a primary amine.
2. Ethylamine gives the carbylamine reaction with chloroform and KOH.
3. The foul-smelling product C is ethyl isocyanide.

Teacher's Note:
1. Offensive smell indicates an isocyanide.
2. Nitriles are reduced to primary amines.

Answer: (d) dimethylamine

Explanation:
1. Secondary amines give N-nitrosoamines, which are yellow oils.
2. Dimethylamine is the only secondary amine listed.
3. It gives N-nitrosodimethylamine.

Teacher's Note:
1. Primary amines evolve nitrogen; tertiary aliphatic amines only form salts.
2. Ammonia gives nitrogen.

Answer: (c) Gabriel phthalimide synthesis

Explanation:
1. Ammonolysis of an alkyl halide gives a mixture of primary, secondary and tertiary amines and the quaternary salt.
2. The phthalimide nitrogen can be alkylated only once, so the Gabriel synthesis gives only the primary amine.

Teacher's Note:
1. Gabriel synthesis is selective for primary amines.
2. It fails with aryl halides.

Answer: (b) C₆H₅CH₂NH₂

Explanation:
1. Only a primary amine gives a sulphonamide RSO₂NHR' having an acidic N-H.
2. That product dissolves in alkali as its sodium salt.
3. Benzylamine is the primary amine in the list; diethylamine gives an insoluble product and the amides do not react.

Teacher's Note:
1. Hinsberg test: primary amine product dissolves in alkali.
2. Secondary amine product has no acidic hydrogen.

Answer: (a) bromination of acetanilide followed by hydrolysis

Explanation:
1. Direct bromination gives the 2,4,6-tribromo compound.
2. Acetylation lowers the activating power of nitrogen and its bulk favours the para position.
3. Hydrolysis of p-bromoacetanilide gives p-bromoaniline.

Teacher's Note:
1. Protection followed by deprotection is the usual route to mono-substitution.
2. Para is favoured over ortho because of the size of the acetamido group.

Answer: (b) Yes, in a strongly acidic medium

Explanation:
1. In strong acid, aniline is protonated to the anilinium ion C₆H₅NH₃+.
2. The NH₃+ group is strongly electron-withdrawing and deactivating, so it directs the electrophile to the meta position.

Teacher's Note:
1. This explains the m-nitroaniline formed in nitration with the acid mixture.
2. A positively charged substituent is meta-directing.

Answer: (c) +1

Explanation:
1. Oxidation of a tertiary amine by a peracid gives an amine oxide, (C₂H₅)₃N+-O-.
2. Nitrogen has four bonds and no lone pair, so its formal charge is +1 (the oxygen carries -1).

Teacher's Note:
1. Formal charge = valence electrons - non-bonding electrons - number of bonds.
2. Amine oxides are dipolar.

Answer: (c)

Explanation:
1. A clear solution with nitrous acid at 0 C shows a primary aromatic amine giving a diazonium salt.
2. Sandmeyer cyanation and hydrolysis give a toluic acid and permanganate oxidises the methyl group to a second carboxyl group.
3. Only phthalic acid (from the 2-substituted compound) dehydrates on heating, to phthalic anhydride; so X is 2-methylaniline (o-toluidine).

Teacher's Note:
1. Dehydration to an anhydride needs two carboxyl groups on adjacent carbons.
2. Secondary amines and benzylamine do not give a clear solution with a stable diazonium salt.

Answer: (c)

Explanation:
1. Both amino groups are diazotised; heating with dilute acid replaces the diazonium groups by hydroxyl, giving 2-hydroxybenzyl alcohol.
2. Hot HI converts the benzylic alcohol into the benzylic iodide, whereas the phenolic OH is not replaced.
3. The product is 2-(iodomethyl)phenol.

Teacher's Note:
1. Benzylic alcohols react with HI by substitution; phenols do not.
2. Aliphatic diazonium ions give alcohols.

Answer: (b)

Explanation:
1. In the strongly acidic nitrating mixture the amine is protonated to C₆H₅NH+(CH₃)₂.
2. The ammonium group is electron-withdrawing and meta-directing.
3. The major product is 3-nitro-N,N-dimethylaniline.

Teacher's Note:
1. Protonation changes the directing effect of nitrogen.
2. A single nitration is expected under these conditions.