JEE Chemistry Amines MCQs Set 05

Answer: (d) p-aminoazobenzene, \( \ce{C6H5-N=N-C6H4-NH2} \) with the amino group para to the azo group

Explanation:
1. In pH 4-5 coupling occurs at the para position of aniline.
2. The product, p-aminoazobenzene, is the yellow dye (aniline yellow).

Teacher's Note:
1. Coupling with amines is done at pH 4-5.
2. Coupling with phenols is done at pH 8-10.

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

Explanation:
1. Only aryl diazonium salts are stable at low temperatures, because the positive charge is delocalised into the ring.
2. Alkyl diazonium salts decompose at once to give nitrogen.

Teacher's Note:
1. Delocalisation into the benzene ring stabilises the diazonium ion.
2. Aliphatic diazonium ions lose \( \ce{N2} \) at once.

Answer: (b) potassium salt of phthalimide treated with chlorobenzene followed by hydrolysis with aqueous NaOH

Explanation:
1. Chlorobenzene is an aryl halide and is not attacked by the phthalimide anion, so Gabriel synthesis fails for aromatic amines.
2. The other three methods all give aniline.

Teacher's Note:
1. Gabriel synthesis works for aliphatic primary amines only.
2. Aryl halides are inert to SN2 attack.

Answer: (b) nitrobenzene

Explanation:
1. The diazonium group is replaced by the nitro group when the fluoroborate salt is heated with \( \ce{NaNO2} \) and copper.
2. The product is nitrobenzene.

Teacher's Note:
1. Diazonium fluoroborate is stable and isolable.
2. Heating it alone gives fluorobenzene (Balz-Schiemann reaction).

Answer: (a) \( \ce{(CH3)3C-NO2} \)

Explanation:
1. A tertiary nitroalkane has no hydrogen on the carbon carrying the nitro group.
2. Primary and secondary nitroalkanes react with nitrous acid (nitrolic acid and pseudonitrole).
3. So the tertiary compound does not react.

Teacher's Note:
1. Alpha hydrogen is needed for the nitrous acid reaction.
2. This is the classical test to separate nitroalkane classes.

Answer: (d) nitrobenzene

Explanation:
1. Compound C is p-hydroxyazobenzene, \( \ce{C6H5N=NC6H4OH} \), formed by coupling of benzenediazonium chloride (B) with phenol.
2. B comes from aniline by diazotisation, and aniline comes from Sn/HCl reduction of nitrobenzene.

Teacher's Note:
1. Check the formula of the azo dye to confirm.
2. \( \ce{C12H10N2O} \) is hydroxyazobenzene.

Answer: (a) Hofmann hypobromamide reaction

Explanation:
1. Hofmann degradation converts an amide into an amine with one carbon fewer.
2. Acetamide gives methanamine.
3. Gabriel starts from an alkyl halide, and carbylamine starts from a primary amine.

Teacher's Note:
1. Amide to amine with one carbon less is the Hofmann reaction.
2. Stephen reaction converts nitrile to aldehyde.

Answer: (a) \( \ce{CH3-CH=CH-CH2-CH2-NH-CH3} \)

Explanation:
1. \( \ce{LiAlH4} \) reduces the C=N bond of the imine to a secondary amine.
2. It does not normally reduce an isolated C=C bond.

Teacher's Note:
1. LiAlH4 attacks polar double bonds like C=O and C=N.
2. Non-polar C=C is left alone.

Answer: (d) \( \ce{CH3CH2COOH} \)

Explanation:
1. Ethylamine arises by Hofmann degradation of propanamide (C).
2. Propanamide arises by heating ammonium propanoate (B).
3. Ammonium propanoate comes from propanoic acid and ammonia, so A is propanoic acid.

Teacher's Note:
1. Work backwards: amine, then amide, then ammonium salt, then acid.
2. Carbon count: amide has one carbon more than the amine.

Answer: (b) A: 1 degree amine, B: methyl ketone, C: 1 degree amine, D: alcohol

Explanation:
1. Nitrile reduction gives a primary amine (A).
2. A Grignard reagent adds to the nitrile and hydrolysis gives a methyl ketone (B).
3. Hydrolysis of an isocyanide gives a primary amine (C).
4. A primary aliphatic amine with nitrous acid gives an alcohol (D).

Teacher's Note:
1. Nitriles: reduce for amine, Grignard for ketone, hydrolyse for acid.
2. Isocyanides: hydrolyse for amine, reduce for N-methyl amine.

Answer: (d) carbylaminoethane, ethanamine, ethyl methyl amine

Explanation:
1. AgCN is covalent and gives the isocyanide \( \ce{C2H5NC} \) (A).
2. Hydrolysis gives formic acid and ethylamine (B), and ethylamine in the carbylamine test gives A again.
3. Reduction of A gives the secondary amine \( \ce{C2H5NHCH3} \) (C).

Teacher's Note:
1. Covalent AgCN gives isocyanide, ionic KCN gives nitrile.
2. Isocyanide reduction gives an N-methyl amine.

Answer: (b) ethylamine

Explanation:
1. Nitrous acid converts ethylamine to ethanol (A).
2. \( \ce{PCl5} \) converts ethanol to ethyl chloride (B).
3. Ammonia converts ethyl chloride back to ethylamine (C).

Teacher's Note:
1. The sequence is a loop that returns to the starting amine.
2. Ammonolysis of alkyl halide gives amine.

Answer: (b) nitrobenzene, aniline, benzenediazonium chloride

Explanation:
1. Phenol arises from hydrolysis of benzenediazonium chloride (C).
2. Diazonium salt is made from aniline (B) by diazotisation.
3. Aniline comes from reduction of nitrobenzene (A).

Teacher's Note:
1. Phenol, diazonium salt, aniline, nitrobenzene is the standard chain.
2. Work backwards one step at a time.

Answer: (c) carbylamine reaction

Explanation:
1. Carbylamine reaction (isocyanide test) is positive only for primary amino groups.
2. The iodoform test detects methyl ketones and CH3CH(OH)-, Victor Meyer's test classes alcohols, and Liebermann's test detects phenols or nitroso compounds.

Teacher's Note:
1. Foul-smelling isocyanide is the sign of a primary amine.
2. Check the reagent: chloroform and alcoholic KOH.

Answer: (b) \( \ce{Ph-NO2 -> Ph-NH2} \)

Explanation:
1. In acid medium with a metal, nitrobenzene is fully reduced to aniline.
2. Hydrazobenzene, hydroxylamine and azobenzene need neutral or alkaline media.

Teacher's Note:
1. Acid medium gives aniline.
2. Alkaline medium gives bimolecular products such as azobenzene.

Answer: (b) \( \ce{-NO2} \)

Explanation:
1. Zn/\( \ce{NH4Cl} \) reduces a nitro compound to a hydroxylamine.
2. The hydroxylamine reduces Tollens' reagent to a black or grey silver precipitate.
3. This is the Mullikan-Barker test for nitro compounds.

Teacher's Note:
1. Zn/NH4Cl then Tollens' reagent is the Mullikan-Barker test.
2. It detects the nitro group.

Answer: (d) Both Assertion and Reason are false.

Explanation:
1. Trimethylamine has no N-H bond, so it cannot form intermolecular hydrogen bonds.
2. n-Propylamine does, so its boiling point is higher.
3. The assertion is false, and the reason is also false.

Teacher's Note:
1. Boiling point of isomeric amines: primary above secondary above tertiary.
2. Tertiary amines have no N-H.

Answer: (d) Both Assertion and Reason are false.

Explanation:
1. Pyrrole is an aromatic heterocyclic compound with a planar ring and six pi electrons.
2. It is not alicyclic.
3. Both statements are false.

Teacher's Note:
1. Alicyclic means cyclic but non-aromatic carbon rings.
2. Pyrrole is aromatic and heterocyclic.

Answer: (c) Assertion is true but the Reason is false.

Explanation:
1. Each of these has an alpha hydrogen on the carbon bearing the nitro group, so each is acidic enough to form a soluble sodium salt (aci form).
2. The assertion is true.
3. Their boiling points are higher, not lower, because their molar masses are larger, so the reason is false.

Teacher's Note:
1. Nitroalkanes with alpha-H dissolve in NaOH.
2. Higher homologues have higher boiling points.

Answer: (b) Both Assertion and Reason are true but the Reason is not a correct explanation of the Assertion.

Explanation:
1. \( \ce{CH3CN} \) is named acetonitrile or ethanenitrile, so the assertion is true.
2. The alpha hydrogens are acidic, so the reason is true as well.
3. The naming has nothing to do with the acidity, so the reason does not explain the assertion.

Teacher's Note:
1. Both are separate true facts.
2. Check whether the reason actually explains the assertion.

Answer: (c) Assertion is true but the Reason is false.

Explanation:
1. Ethanol reduces the diazonium group and is itself oxidised to acetaldehyde, so the assertion is true.
2. Ethanol is a reducing agent, not an oxidising agent, so the reason is false.

Teacher's Note:
1. Ethanol acts as a reducing agent here.
2. Hypophosphorous acid does the same job.

Answer: (a) Both Assertion and Reason are true and the Reason is a correct explanation of the Assertion.

Explanation:
1. Diazonium salts decompose above 5 C.
2. At higher temperature water attacks the salt and forms phenol, which is why low temperature is essential.
3. Both are true and the reason explains the assertion.

Teacher's Note:
1. Keep the diazotisation mixture ice cold.
2. Warming with water gives phenol.

Answer: (c) Assertion is true but the Reason is false.

Explanation:
1. A primary amine with \( \ce{CHCl3} \) and KOH gives an isocyanide, so the assertion is true.
2. The intermediate is dichlorocarbene, \( \ce{:CCl2} \), not methylene, so the reason is false.

Teacher's Note:
1. The carbene in this reaction is dichlorocarbene.
2. It is formed by alpha elimination from chloroform.

Answer: (b) Both Assertion and Reason are true but the Reason is not a correct explanation of the Assertion.

Explanation:
1. Excess ammonia favours the primary amine, so the assertion is true.
2. The reaction does in fact give all three amines and the quaternary salt, so the reason is true.
3. But the formation of a mixture does not explain why excess ammonia favours the primary amine.

Teacher's Note:
1. Excess ammonia means the haloalkane is more likely to meet ammonia than amine.
2. The mixture is the problem, not the explanation.