JEE Chemistry Amines MCQs Set 26

Answer: (b) boiling point of II > I

Explanation:
1. 4-Methylpiperidine is a secondary amine with an N-H bond and forms intermolecular hydrogen bonds.
2. N-Methylpiperidine is tertiary, has no N-H and cannot hydrogen bond to itself.
3. So II boils higher.

Teacher's Note:
1. Hydrogen bonding raises boiling point.
2. Both compounds have the same molar mass.

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

Explanation:
1. Piperidine is a saturated secondary amine with a localised lone pair and is the strongest base.
2. Morpholine is weaker because of the electron-withdrawing oxygen.
3. Pyridine's lone pair is in an sp2 orbital.
4. Pyrrole's lone pair is part of the aromatic sextet, so it is a very weak base.

Teacher's Note:
1. Localised lone pair means strongest base.
2. Pyrrole's lone pair is used for aromaticity.

Answer: (c) Sandmeyer

Explanation:
1. Treatment of the diazonium salt with \( \ce{CuCl/HCl} \) replaces the diazonium group by chlorine.
2. This is the Sandmeyer reaction.

Teacher's Note:
1. Sandmeyer uses cuprous halides.
2. Gattermann uses copper powder.

Answer: (a) methyl cyanide

Explanation:
1. \( \ce{P2O5} \) dehydrates an aldoxime to a nitrile.
2. \( \ce{CH3CH=NOH} \) loses water to give \( \ce{CH3C#N} \), methyl cyanide.

Teacher's Note:
1. Aldoxime dehydration gives nitrile.
2. Carbon count is unchanged.

Answer: (d) \( \ce{CH3CHO} \)

Explanation:
1. Sodium and ethanol reduce the nitrile to ethylamine (A).
2. Nitrous acid gives ethanol (B).
3. Dehydrogenation over copper at 573 K gives acetaldehyde (C).

Teacher's Note:
1. Mendius reduction, then diazotisation, then dehydrogenation.
2. Copper at 573 K converts 1° alcohols to aldehydes.

Answer: (b) ethyl isocyanide, \( \ce{CH3CH2N#C} \)

Explanation:
1. A is \( \ce{CH3CN} \) and its reduction gives ethylamine (B).
2. Nitrous acid gives ethanol, confirming B.
3. The carbylamine reaction of B gives ethyl isocyanide (C), which has a foul smell.

Teacher's Note:
1. Carbylamine test is positive for primary amines.
2. The product is an isocyanide.

Answer: (b) 4-ethylaniline and phenol

Explanation:
1. The ethyl-substituted ring comes from the diazonium salt, so 4-ethylaniline is diazotised.
2. It couples at the para position of phenol, giving the hydroxy-substituted ring.

Teacher's Note:
1. Azo dye = diazonium salt plus activated aromatic compound.
2. Each ring comes from one reactant.

Answer: (c) cyclopentanecarbonyl chloride + \( \ce{CH3CH2NH2} \) (pyridine), then (i) \( \ce{LiAlH4} \) (ii) \( \ce{H2O} \)

Explanation:
1. Options a and b are reductive aminations that give N-ethylcyclopentylamine.
2. Option d acetylates the amine and reduces the amide C=O to CH2, also giving the same product.
3. Option c gives N-ethylcyclopentanecarboxamide, whose reduction gives N-ethyl(cyclopentylmethyl)amine, with an extra CH2.

Teacher's Note:
1. Check the carbon count between ring and nitrogen.
2. The acid chloride route adds a carbon.

Answer: (a) I-C; II-C; III-B; IV-A

Explanation:
1. Amides and nitriles are essentially neutral in water.
2. Triethylamine is basic and phenol is weakly acidic.

Teacher's Note:
1. Amide nitrogen lone pair is delocalised.
2. Nitrile nitrogen lone pair is in an sp orbital.

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

Explanation:
1. Amidines are strong bases because the protonated form is a resonance-stabilised amidinium ion.
2. Dimethylamine (pKb about 3.3) is slightly stronger than ethylamine (pKb about 3.4).
3. Acetamide is very weakly basic.

Teacher's Note:
1. Resonance stabilisation of the conjugate acid raises basicity.
2. 2° > 1° in water for simple alkylamines.

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

Explanation:
1. Bromine water gives 2,4,6-tribromoaniline.
2. Diazotisation, then \( \ce{HBF4} \), gives the diazonium tetrafluoroborate.
3. Heating it (Balz-Schiemann reaction) replaces N2+ by fluorine.

Teacher's Note:
1. Balz-Schiemann introduces fluorine.
2. The three bromines remain.

Answer: (a) 2-cyclopropylpropanenitrile

Explanation:
1. \( \ce{NaNH2} \) removes an alpha hydrogen next to the CN group to form a carbanion.
2. The carbanion is alkylated by \( \ce{CH3Br} \), giving \( \ce{C3H5-CH(CH3)-CN} \).

Teacher's Note:
1. Alpha hydrogens of nitriles are acidic.
2. Alkylation occurs at the alpha carbon.

Answer: (b) benzene

Explanation:
1. Nitrobenzene (W) is reduced to aniline (X).
2. Diazotisation gives the diazonium salt (Y), and \( \ce{H3PO2} \) replaces it by hydrogen, giving benzene.

Teacher's Note:
1. The sequence returns to benzene.
2. \( \ce{H3PO2} \) is a reducing deamination reagent.

Answer: (c) piperidine

Explanation:
1. On heating, ammonium chloride is lost and the free amine group displaces the other ammonium group intramolecularly.
2. A six-membered saturated ring containing one nitrogen forms, which is piperidine.

Teacher's Note:
1. Intramolecular cyclisation forms six-membered ring.
2. Piperidine has five carbons and one nitrogen.

Answer: (b) I, III and IV are aromatic

Explanation:
1. Pyrrole uses its nitrogen lone pair in the aromatic sextet, and protonation at nitrogen would remove that lone pair, so II is not aromatic.
2. Pyridine keeps its lone pair outside the ring, so protonation does not affect the sextet and the pyridinium ion remains aromatic.

Teacher's Note:
1. Protonation of pyrrole destroys aromaticity.
2. Pyridine protonates at the in-plane lone pair.

Answer: (b) heating \( \ce{CH3CH2N+(CH3)3 OH-} \)

Explanation:
1. Hofmann elimination needs a hydroxide counter ion, which acts as the base.
2. The quaternary hydroxide on heating gives ethylene, trimethylamine and water.
3. The iodide salt just decomposes by substitution, and ethylamine on heating does not lose \( \ce{NH3} \).

Teacher's Note:
1. Quaternary ammonium hydroxide on heating gives the Hofmann alkene.
2. Halide salts do not give Hofmann elimination.

Answer: (d) an alkyl group with its electron pair to an electron-deficient N atom

Explanation:
1. The N-bromoamide anion loses bromide, and the alkyl group migrates with its bonding pair from carbon to the electron-deficient nitrogen.
2. This gives an isocyanate that is hydrolysed to the amine.

Teacher's Note:
1. Rearrangement similar to carbocation shift.
2. Migration is with the electron pair and retention of configuration.

Answer: (c) \( \ce{(CH3)3CND2} \)

Explanation:
1. Hofmann bromamide reaction converts the amide to the amine with one carbon less.
2. In \( \ce{OD-/D2O} \) the N-H hydrogens are exchanged, giving \( \ce{(CH3)3CND2} \).

Teacher's Note:
1. Amide to amine, one carbon lost.
2. Deuterated medium exchanges the N-H hydrogens.

Answer: (b) N,N,3,3-tetramethylbutan-1-amine

Explanation:
1. Strong oxidation gives 3,3-dimethylbutanoic acid (A) and thionyl chloride gives the acid chloride (B).
2. Dimethylamine gives the N,N-dimethylamide (C).
3. \( \ce{LiAlH4} \) reduces the amide C=O to CH2, giving \( \ce{(CH3)3CCH2CH2N(CH3)2} \) (D).

Teacher's Note:
1. Amide reduction gives an amine with the same number of carbons.
2. Count the carbons from the starting alcohol.

Answer: (c) Schiff's base

Explanation:
1. A primary amine condenses with an aldehyde to give an imine, \( \ce{C6H5CH=NC6H5} \).
2. Such imines are called Schiff bases.

Teacher's Note:
1. Imine = Schiff base.
2. Schiff's reagent is a different substance used for aldehydes.

Answer: (d) III, IV

Explanation:
1. Butan-2-amine has a chiral carbon and gives stable enantiomers.
2. In the aziridine, nitrogen inversion is very slow because of ring strain, so the N-stereocentre is configurationally stable and the C-2 carbon is also chiral.
3. Acyclic amines I and II undergo rapid nitrogen inversion, and II is also achiral.

Teacher's Note:
1. Ring strain slows nitrogen inversion.
2. Rapid inversion prevents resolution.