JEE Chemistry Amines MCQs Set 18

Answer: (b) Triethylamine

Explanation:
1. A tertiary amine has three carbon groups attached to nitrogen.
2. Triethylamine has three ethyl groups.
3. 1-Methylcyclohexylamine and tert-butylamine are primary amines (the carbon bearing N is tertiary, not the nitrogen), and N-methylaniline is secondary.

Teacher's Note:
1. The class of an amine depends on the number of carbon groups on nitrogen, not on the carbon type.

Answer: (d) N-methylprop-2-en-1-amine

Explanation:
1. The longest chain containing the amine carbon is the three-carbon allyl chain, prop-2-en-1-amine.
2. The methyl group on nitrogen is shown by the locant N, giving N-methylprop-2-en-1-amine.

Teacher's Note:
1. In IUPAC names of secondary amines the smaller group on N becomes an N-substituent.

Answer: (c) \(\ce{(CH3)2NH}\)

Explanation:
1. Alkyl groups raise the electron density on N (+I) and the ammonium ion is stabilised by hydration.
2. Dimethylamine is more basic than methylamine in water.
3. The cyano group and the phenyl group withdraw electrons and reduce basicity.

Teacher's Note:
1. In water, for methylamines the order is secondary > primary > tertiary > NH3.

Answer: (a) aniline

Explanation:
1. In aniline the lone pair on nitrogen is delocalised into the benzene ring, so it is much less available for protonation.
2. The other three are aliphatic amines with localised lone pairs.

Teacher's Note:
1. Resonance delocalisation of the lone pair makes aryl amines weaker bases than alkyl amines.

Answer: (b) \(\ce{LiAlH4}\) in ether

Explanation:
1. Catalytic hydrogenation, Fe/HCl and Sn/HCl convert nitroarenes cleanly to anilines.
2. \(\ce{LiAlH4}\) reduces aromatic nitro compounds to azo compounds, not amines.

Teacher's Note:
1. Remember: LiAlH4 gives azobenzene from nitrobenzene; metals in acid give aniline.

Answer: (d) \(\ce{LiAlH4}\) in ether

Explanation:
1. The product has the same number of carbon atoms as the amide, so the C=O must be reduced to CH2.
2. \(\ce{LiAlH4}\) does this.
3. \(\ce{Br2/NaOH}\) (Hofmann) would give an amine with one carbon fewer.

Teacher's Note:
1. Reduction retains carbon count; Hofmann degradation loses one carbon.

Answer: (b) \(\ce{(CH3)2NH}\)

Explanation:
1. Reactivity towards dilute HCl is a measure of basic strength in aqueous solution.
2. Dimethylamine is the strongest of these bases in water (+I effect together with good solvation of the cation).

Teacher's Note:
1. Trimethylamine is less basic in water because its ammonium ion has only one N-H bond for hydration.

Answer: (c) phenol

Explanation:
1. The conjugate acids have pKa values of about 4.6 (anilinium), 10.6 (cyclohexylammonium), about -2 (protonated cyclohexanol) and about -6.7 (protonated phenol).
2. The lower the pKa of the conjugate acid, the weaker the base.
3. Phenol is the weakest because its oxygen lone pair is delocalised into the ring.

Teacher's Note:
1. Alcohols and phenols are very weak bases; the lone pair of phenol is also involved in resonance.

Answer: (d) pyrrolidine

Explanation:
1. Pyrrolidine is a secondary aliphatic amine with a localised lone pair and alkyl groups that donate electrons.
2. In pyrrole the lone pair is part of the aromatic sextet, and in aniline it is delocalised into the ring, so both are very weak bases.

Teacher's Note:
1. A lone pair that is part of an aromatic system is not available for protonation.

Answer: (d) \(\ce{CH3CH2CH3}\)

Explanation:
1. Primary and secondary amines form intermolecular hydrogen bonds and boil higher than hydrocarbons of similar molar mass.
2. Tertiary amines have no N-H bond and do not associate, but are still polar.
3. Propane is non-polar with no hydrogen bonding, so it is the most volatile.

Teacher's Note:
1. Volatility is highest for molecules with the weakest intermolecular forces.

Answer: (a) (i), (ii) and (iii) only

Explanation:
1. Nitrile reduction, amide reduction and the Gabriel synthesis all keep the carbon chain intact.
2. The Hofmann bromamide reaction removes the carbonyl carbon as carbonate, so the amine has one carbon fewer.

Teacher's Note:
1. Hofmann degradation is the only common amine synthesis that shortens the chain by one carbon.

Answer: (a) (i), (ii) and (iii) only

Explanation:
1. Tin or iron with hydrochloric acid and catalytic hydrogenation reduce nitrobenzene completely to aniline.
2. Zinc with aqueous ammonium chloride stops at N-phenylhydroxylamine.

Teacher's Note:
1. Neutral reducing conditions (Zn/NH4Cl) give hydroxylamine; acidic metals give the amine.

Answer: (a) (ii) and (iii) only

Explanation:
1. Hypophosphorous acid and ethanol replace the diazonium group by hydrogen.
2. \(\ce{SnCl2/HCl}\) reduces the diazonium salt to phenylhydrazine, and \(\ce{LiAlH4}\) is not used for this purpose.

Teacher's Note:
1. Deamination of diazonium salts uses H3PO2 or ethanol (the latter may give some phenyl ether).

Answer: (a) (i) and (ii) only

Explanation:
1. The Gabriel synthesis gives only primary aliphatic amines from primary alkyl halides.
2. Isobutylamine and 2-phenylethylamine qualify.
3. N-Methylbenzylamine is secondary, and aryl halides do not react with phthalimide anion, so aniline cannot be made.

Teacher's Note:
1. Gabriel phthalimide synthesis is limited to primary alkyl amines.

Answer: (a) (i) and (ii) only

Explanation:
1. Acetylation protects the amino group; the acetamido group is an ortho/para director and, being bulky, gives mainly the para nitro compound.
2. Direct nitration oxidises aniline and, in strongly acidic medium, forms the anilinium ion, giving a mixture that includes the meta isomer.

Teacher's Note:
1. Protect the amino group by acetylation before nitration or bromination to control the product.

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

Explanation:
1. The Hofmann bromamide (degradation) reaction needs an amide with two hydrogens on nitrogen, so primary amides only; the Assertion is true.
2. In aqueous solution, secondary aliphatic amines are generally more basic than primary amines, so the Reason is false.

Teacher's Note:
1. The Hofmann reaction requires the N-H2 group, since the isocyanate forms after loss of both hydrogens.

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

Explanation:
1. The strongly electron-withdrawing sulphonyl group makes the N-H hydrogen acidic, so the sulphonamide of a primary amine dissolves in alkali by forming a salt.
2. The Reason explains the Assertion.

Teacher's Note:
1. This is the basis of the Hinsberg test for primary amines.

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

Explanation:
1. The sulphonamide of a secondary amine has no hydrogen on nitrogen and cannot form a soluble salt, so it is insoluble in alkali; the Assertion is true.
2. The Reason is also true, but the insolubility is due to the absence of an N-H hydrogen, not to the withdrawing nature of the sulphonyl group.

Teacher's Note:
1. In the Hinsberg test, secondary amines give an alkali-insoluble sulphonamide.

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

Explanation:
1. In acetanilide the lone pair on nitrogen is delocalised towards the carbonyl group (amide resonance), so the electron density on N decreases and basicity is lower.
2. The Reason correctly explains the Assertion.

Teacher's Note:
1. Amides are far weaker bases than amines.

Answer: (c) Schotten-Baumann reaction

Explanation:
1. Acylation of amines (or alcohols) with an acid chloride in the presence of aqueous alkali is the Schotten-Baumann reaction.
2. The base neutralises the HCl formed and drives the reaction forward.

Teacher's Note:
1. Do not call it 'acetylation' since the acyl group here is benzoyl.

Answer: (a) (i), (ii) and (iv) only

Explanation:
1. Hofmann bromamide degradation of benzamide gives aniline, catalytic hydrogenation of nitrobenzene gives aniline, and acid hydrolysis of an isocyanide gives the primary amine (and formic acid).
2. The Gabriel synthesis fails with aryl halides.

Teacher's Note:
1. Acid hydrolysis of an isocyanide RNC gives RNH2 and HCOOH.

Answer: (a) \(\ce{H3PO2}\) and \(\ce{H2O}\)

Explanation:
1. Hypophosphorous acid reduces the diazonium salt and replaces the \(\ce{-N2+}\) group by hydrogen.
2. Warming with \(\ce{H+/H2O}\) gives a phenol, and \(\ce{Cu2Cl2}\) gives the aryl chloride (Sandmeyer).

Teacher's Note:
1. Replacing the amino group by hydrogen through diazotisation is useful for making substitution patterns that cannot be made directly.

Answer: (a) (ii), (iii) and (iv) only

Explanation:
1. (i) is wrong: a tertiary aromatic amine undergoes C-nitrosation at the para position and does not form a diazonium salt.
2. (ii) A primary aliphatic amine gives an alcohol and nitrogen.
3. (iii) A primary amine gives the sulphonamide (Hinsberg).
4. (iv) A secondary amine gives an N-nitrosamine.

Teacher's Note:
1. Know the three outcomes of nitrous acid: primary gives N2 (and alcohol or diazonium), secondary gives nitrosamine, tertiary aromatic gives C-nitroso compound.

Answer: (a) \(\ce{(CH3)2CH-NH2}\)

Explanation:
1. Nitrous acid gives N2 and an alcohol, so A is a primary amine.
2. The carbylamine reaction gives an isocyanide C = R-NC.
3. Reduction of R-NC gives R-NH-CH3; since this is isopropylmethylamine, R = isopropyl.
4. So A is propan-2-amine.

Teacher's Note:
1. Reduction of an isocyanide RNC gives the N-methyl secondary amine RNHCH3.

Answer: (a) N-phenylhydroxylamine

Explanation:
1. Zinc in neutral ammonium chloride solution reduces the nitro group only to the hydroxylamine stage: \(\ce{C6H5NO2 -> C6H5NHOH}\).

Teacher's Note:
1. Acidic reduction gives aniline, neutral reduction gives phenylhydroxylamine, alkaline reduction gives azo/azoxy products.