JEE Chemistry Amines MCQs Set 29

Answer: (b) Boiling point of II > I

Explanation:
1. Both are isomers with the same molar mass.
2. II has an N–H bond and forms intermolecular hydrogen bonds, whereas I is a tertiary amine without N–H; so II boils higher.

Teacher's Note:
1. Hydrogen bonding raises the boiling point.

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

Explanation:
1. Piperidine is a saturated secondary amine and the most basic.
2. Morpholine is less basic because of the electron withdrawing oxygen, pyridine has an sp² nitrogen, and pyrrole's lone pair is part of the aromatic sextet so it is least basic.

Teacher's Note:
1. Availability of the lone pair decides basicity.

Answer: (a) aniline

Explanation:
1. Strong reduction with Zn/HCl cleaves the N–N bond completely.
2. The product is aniline; mild reducing agents like SnCl₂ give phenylhydrazine.

Teacher's Note:
1. Strong reduction gives aniline, mild reduction gives phenylhydrazine.

Answer: (c) I, III

Explanation:
1. I: acetone oxime is reduced by LiAlH₄ to (CH₃)₂CHNH₂.
2. III: (CH₃)₂CHOH gives (CH₃)₂CHCl with PCl₅, and NH₃ then gives isopropylamine.
3. II gives n-propylamine and IV does not give an isomerised amine.

Teacher's Note:
1. Track the carbon skeleton through each route.

Answer: (d) CH₃CHO

Explanation:
1. Na/C₂H₅OH reduces the nitrile to CH₃CH₂NH₂ (A); HNO₂ converts it to CH₃CH₂OH (B).
2. Dehydrogenation over Cu at 573 K gives CH₃CHO (C).

Teacher's Note:
1. A primary alcohol is dehydrogenated to an aldehyde over copper.

Answer: (d) ethylamine salt and methanoic acid

Explanation:
1. R–NC + 2H₂O + H⁺ → R–NH₃⁺ + HCOOH.
2. For ethyl isocyanide this gives the ethylammonium salt and methanoic acid (formic acid).

Teacher's Note:
1. Isocyanides hydrolyse to a primary amine and formic acid.

Answer: (a) 2-chloroacetophenone (o-chlorophenyl methyl ketone)

Explanation:
1. Diazotisation then CuCN gives 2-chlorobenzonitrile.
2. A Grignard reagent adds to the nitrile and hydrolysis of the imine gives the ketone, 2-chloroacetophenone.

Teacher's Note:
1. Nitrile + RMgX then H₃O⁺ gives a ketone.

Answer: (b) 4-ethylaniline + phenol

Explanation:
1. The diazonium salt of 4-ethylaniline couples with phenol at the para position.
2. This gives CH₃CH₂–C₆H₄–N=N–C₆H₄–OH.

Teacher's Note:
1. The azo compound is built from a diazonium salt and an activated aromatic coupling partner.

Answer: (b) 1 > 3 > 2 > 4

Explanation:
1. Acetamidine is the strongest base because its protonated form is stabilised by delocalisation of the charge over two nitrogens.
2. Dimethylamine is more basic than ethylamine, and acetamide is almost neutral.

Teacher's Note:
1. Amidines are among the strongest neutral organic bases.

Answer: (d) aniline

Explanation:
1. Aniline is brominated to 2,4,6-tribromoaniline (Y).
2. Diazotisation and boiling with ethanol replace the diazonium group by H to give 1,3,5-tribromobenzene.

Teacher's Note:
1. Aniline plus bromine water gives the tribromo derivative.

Answer: (a) B > A > C

Explanation:
1. The weaker the conjugate base amine, the stronger the acid.
2. p-Nitroaniline is the weakest base and p-toluidine the strongest, so acidity is B > A > C.

Teacher's Note:
1. Stronger acid means weaker conjugate base.

Answer: (a) C₆H₅–N=O

Explanation:
1. In acidic medium nitrobenzene is reduced stepwise: nitroso compound, then phenylhydroxylamine, then aniline.
2. The intermediate is nitrosobenzene, C₆H₅NO; azo and hydrazo products form in neutral or alkaline media.

Teacher's Note:
1. Acidic reduction goes nitro → nitroso → hydroxylamine → amine.

Answer: (c) piperidine

Explanation:
1. Heating the diammonium salt eliminates NH₄Cl.
2. Intramolecular nucleophilic substitution closes a six-membered ring, giving piperidine.

Teacher's Note:
1. Intramolecular ring closure forms a stable six-membered ring.

Answer: (b) ethyltrimethylammonium hydroxide

Explanation:
1. Hofmann elimination needs the quaternary ammonium hydroxide: CH₃CH₂N⁺(CH₃)₃OH⁻ → CH₂=CH₂ + (CH₃)₃N + H₂O.
2. The iodide undergoes substitution instead, and ethylamine does not eliminate on heating.

Teacher's Note:
1. Hofmann elimination needs the hydroxide salt.

Answer: (d) both (b) and (c)

Explanation:
1. A secondary amine with an aldehyde or ketone that has α-hydrogen gives an enamine (C=C–N).
2. A primary amine gives an imine (C=N).

Teacher's Note:
1. Secondary amines give enamines.

Answer: (c) (CH₃)₃CND₂

Explanation:
1. Hofmann degradation gives (CH₃)₃C–NH₂ with loss of the carbonyl carbon.
2. In OD⁻/D₂O the exchangeable N–H hydrogens become N–D, giving (CH₃)₃CND₂.

Teacher's Note:
1. N-H hydrogens exchange with D₂O.

Answer: (a) I > II > III

Explanation:
1. The more s-character in the carbon next to the CH₂NH₂ group, the more electron withdrawing it is.
2. The butyl group (sp³) donates most, the butenyl (sp²) less, and the butynyl (sp) least, so basicity is I > II > III.

Teacher's Note:
1. Greater s-character of the attached carbon lowers basicity.

Answer: (c) Schiff's base

Explanation:
1. Condensation of a primary amine with an aldehyde gives an imine, C₆H₅CH=NC₆H₅.
2. Such an imine is called a Schiff's base.

Teacher's Note:
1. Primary amine + carbonyl compound gives a Schiff's base.