JEE Chemistry Amines MCQs Set 33

Answer: (a) \( \ce{R-C(=O+H)-NH2} \) (O-protonated)

Explanation:
1. In an amide the nitrogen lone pair is delocalised into the carbonyl group, so nitrogen is not basic.
2. Protonation at oxygen gives a resonance-stabilised cation in which the positive charge is shared by O and N.
3. Protonation at nitrogen would destroy this delocalisation.

Teacher's Note:
1. Amides protonate on oxygen, not nitrogen.
2. The O-protonated cation is stabilised by resonance.

Answer: (c) The aci form is more acidic than the nitro form

Explanation:
1. The nitro form is a C-H acid with pKa about 10.2.
2. The aci form has a real O-H group on a N+ centre and is a much stronger acid (pKa about 3.2).
3. So the aci form is the stronger acid.

Teacher's Note:
1. Aci-nitro forms are stronger acids because the O-H is attached to the positively charged nitrogen.
2. Only nitroalkanes with alpha-hydrogen show this tautomerism.

Answer: (c) III < I < II

Explanation:
1. Ethane is non-polar (b.p. about -89 °C).
2. Ethyl nitrite is a polar ester with b.p. about 17 °C.
3. Nitroethane is highly polar (large dipole moment, due to the charge-separated nitro group) and boils at about 114 °C.
4. Order: ethane < ethyl nitrite < nitroethane.

Teacher's Note:
1. Nitro compounds have high dipole moments and relatively high boiling points.
2. Alkyl nitrites are much more volatile than their nitro isomers.

Answer: (d) p-dinitrobenzene

Explanation:
1. Acetanilide and benzamide have polar amide groups and nonzero dipole moments.
2. In m-dinitrobenzene the two bond dipoles are at 120 °C and add up to a non-zero value.
3. In p-dinitrobenzene the two equal C-NO2 dipoles point in opposite directions and cancel, giving zero dipole moment.

Teacher's Note:
1. Symmetrical para-disubstituted benzenes with identical groups are non-polar.
2. Dipole moments are vector quantities.

Answer: (a) ortho

Explanation:
1. Br2/Fe brominates 4-nitrotoluene ortho to the methyl group (activating, directing) which is meta to the nitro group, giving 2-bromo-4-nitrotoluene.
2. Sn/HCl reduces the nitro group to amino; diazotisation and H3PO2 remove the amino group (as N2), giving 2-bromotoluene.
3. KMnO4 oxidises the methyl group to COOH, giving 2-bromobenzoic acid (ortho).

Teacher's Note:
1. The NH2 group is used as a temporary directing group and then removed by diazotisation and H3PO2.
2. The methyl group controls the position of bromine.

Answer: (c) The rate of nitration of benzene is greater than that of hexadeuterobenzene

Explanation:
1. Nitration is electrophilic substitution by \( \ce{NO2+} \).
2. The rate-determining step is attack of the electrophile on the ring, not loss of H+, so there is no kinetic isotope effect: \( \ce{C6H6} \) and \( \ce{C6D6} \) react at nearly the same rate.
3. Statement (c) claims a different rate and is false.

Teacher's Note:
1. No primary kinetic isotope effect is seen in nitration, sulphonation or halogenation.
2. Methyl is an activating group, so toluene reacts faster than benzene.

Answer: (b) I and II only

Explanation:
1. H2/Ni and Sn/HCl (acidic medium, catalytic hydrogenation) reduce nitrobenzene completely to aniline.
2. In basic medium (Zn/NaOH) bimolecular products (azoxy-, azo-, hydrazo-benzene) are obtained.
3. LiAlH4 reduces aromatic nitro compounds to azo compounds.

Teacher's Note:
1. Acidic/catalytic reduction gives the amine.
2. Neutral medium: hydroxylamine. Alkaline medium: azoxy, azo and hydrazo compounds.

Answer: (c) carbene

Explanation:
1. The diazo ketone \( \ce{R-CO-CH-N2+} \) loses \( \ce{N2} \) from the carbon atom.
2. This leaves a carbon with only six valence electrons and no charge, an acyl carbene (ketocarbene).
3. The carbene then rearranges (Wolff rearrangement) to a ketene.

Teacher's Note:
1. Loss of N2 from a diazo compound gives a carbene.
2. Wolff rearrangement is used in the Arndt-Eistert homologation.

Answer: (a) \( \ce{1-cyclopropylpropanenitrile, (cyclopropyl)CH(CH3)CN} \)

Explanation:
1. The alpha-hydrogens of a nitrile are acidic because of the electron-withdrawing \( \ce{C#N} \) group.
2. The strong base sodamide removes an alpha-proton to give a carbanion.
3. The carbanion attacks methyl bromide by SN2, giving the alpha-methylated nitrile.

Teacher's Note:
1. Nitrile alpha-carbons can be alkylated through the carbanion.
2. Sodamide in liquid ammonia is a strong base.

Answer: (c) acetaldehyde

Explanation:
1. Partial reduction of the nitrile (e.g. with SnCl2/HCl, Stephen reduction) gives the imine \( \ce{CH3CH=NH} \) (X).
2. Boiling water hydrolyses the imine to the aldehyde and ammonia.
3. So Y is acetaldehyde.

Teacher's Note:
1. Stephen reaction: RCN + SnCl2 + HCl, then H3O+, gives RCHO.
2. Complete reduction would give the primary amine.

Answer: (b) 3-methoxyacetophenone

Explanation:
1. The Grignard reagent adds to the C#N carbon to form an imine magnesium salt.
2. Acid hydrolysis converts the imine to a ketone with loss of ammonia.
3. The product is 3-methoxyacetophenone, \( \ce{3-CH3O-C6H4-COCH3} \).

Teacher's Note:
1. Nitriles with RMgX then H3O+ give ketones.
2. Methoxy group is unaffected.

Answer: (a) \( \ce{C6H5-N=O} \)

Explanation:
1. Reduction in acid proceeds stepwise: \( \ce{C6H5NO2 -> C6H5NO -> C6H5NHOH -> C6H5NH2} \).
2. Nitrosobenzene and phenylhydroxylamine are intermediates.
3. Azoxy-, azo- and hydrazo-benzene are formed only under basic conditions through condensation.

Teacher's Note:
1. Acidic medium: nitroso, then hydroxylamine, then amine.
2. Basic medium: bimolecular products.

Answer: (c) 3-nitroaniline

Explanation:
1. Ammonium hydrosulphide (Zinin reduction) is a mild selective reducing agent.
2. It reduces only one nitro group of a polynitro compound.
3. So m-dinitrobenzene gives m-nitroaniline.

Teacher's Note:
1. Selective reduction of one nitro group: NH4HS, (NH4)2S, Na2S.
2. Sn/HCl would reduce both.

Answer: (a) nucleophilic substitution

Explanation:
1. The nitro group strongly deactivates the ring to electrophiles but activates ortho and para positions to nucleophiles.
2. Hydroxide attacks the ring and replaces H (via a Meisenheimer-type complex).
3. So the reaction is nucleophilic aromatic substitution.

Teacher's Note:
1. Strong electron-withdrawing groups make a ring susceptible to nucleophilic attack at o/p positions.
2. Here H is replaced by OH (oxidative substitution).

Answer: (c) \( \ce{(CH3)2CHNO2} \)

Explanation:
1. A primary nitroalkane gives a nitrolic acid, soluble in alkali with a red colour.
2. A secondary nitroalkane gives a pseudonitrole, which is insoluble in alkali and blue in solution/melt.
3. A tertiary nitroalkane has no alpha-H and does not react.
4. 2-Nitropropane is the secondary nitroalkane; (b) is primary.

Teacher's Note:
1. Nitrous acid test: 1° red, 2° blue, 3° no reaction.
2. Count alpha-hydrogens on the carbon bearing the nitro group.

Answer: (a) \( \ce{(CH3)2CHNO2} \)

Explanation:
1. Boiling a secondary nitroalkane with hydrochloric acid hydrolyses its aci form (Nef reaction) to the ketone and nitrous oxide.
2. Primary nitroalkanes give carboxylic acids (with hydroxylamine) under hot acid.
3. The tertiary nitro compound does not react as it has no alpha-H.

Teacher's Note:
1. Nef reaction: secondary nitroalkane gives ketone; primary gives aldehyde (or acid under strong acid).
2. 2-Nitropropane gives acetone.

Answer: (d) \( \ce{(CH3)3CNO2} \)

Explanation:
1. Nitro-aci tautomerism requires a hydrogen on the carbon attached to the nitro group (alpha-H).
2. Nitromethane, phenylnitromethane and nitroethane all have alpha-H.
3. Tertiary-butyl nitro has none.

Teacher's Note:
1. No alpha-H, no tautomerism.
2. Same reason a tertiary nitro compound does not react with HONO.

Answer: (b) Nitrobenzene

Explanation:
1. Aldehydes, alpha-hydroxy ketones and hydroxylamines are reducing agents toward \( \ce{Ag(NH3)2+} \).
2. Nitrobenzene is an oxidising (not reducing) substance.

Teacher's Note:
1. Tollens' reagent oxidises easily oxidisable groups: CHO, alpha-hydroxy ketone, NHOH.
2. Nitro compounds are not oxidised.

Answer: (c) nitroethane

Explanation:
1. The nitrite ion is ambident (can attack through N or O).
2. Covalent silver nitrite attacks through nitrogen, giving the nitro compound as the major product.
3. Ethyl nitrite is formed as the minor product (KNO2 gives mainly the nitrite).

Teacher's Note:
1. AgNO2 gives nitroalkane, KNO2/NaNO2 gives alkyl nitrite.
2. Ambident nucleophile: N versus O attack.

Answer: (b) \( \ce{Zn/NH4Cl} \)

Explanation:
1. In neutral medium (zinc with aqueous ammonium chloride) nitro compounds are reduced only as far as hydroxylamines.
2. Zn/HCl and Sn/HCl give the amine (methylamine).
3. Zn/NaOH gives bimolecular products in the aromatic series.

Teacher's Note:
1. Neutral medium: Zn/NH4Cl stops at N-alkyl hydroxylamine.
2. Acidic medium gives the amine.

Answer: (c) \( \ce{Zn/NaOH-CH3OH} \)

Explanation:
1. In alkaline medium, nitrobenzene is reduced by zinc via nitroso and hydroxylamine intermediates which condense to azoxy and azo compounds.
2. With excess zinc in alkaline alcohol the azo compound is further reduced to hydrazobenzene.

Teacher's Note:
1. Zn/NaOH-alcohol, prolonged, gives hydrazobenzene.
2. Zn/NH4Cl gives phenylhydroxylamine.

Answer: (b) Both A and R are true, but R is not the correct explanation of A.

Explanation:
1. Isocyanides are oxidised by ozone to isocyanates, so the assertion is true.
2. Methyl isocyanate (MIC) was the gas released in the Bhopal tragedy, so the reason is also true.
3. But the reason does not explain why ozone converts the isocyanide into the isocyanate.

Teacher's Note:
1. Both statements are facts but are unrelated in cause.
2. MIC is \( \ce{CH3-N=C=O} \).

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. The strongly electron-withdrawing nitro group deactivates the ring, so it does not undergo Friedel-Crafts reaction.
2. Because it is inert to Friedel-Crafts conditions yet dissolves reagents such as AlCl3, it is used as a solvent for Friedel-Crafts reactions.
3. So the reason explains the assertion.

Teacher's Note:
1. Deactivated rings do not undergo Friedel-Crafts reactions.
2. Nitrobenzene is a useful polar solvent for such reactions.

Answer: (d) A is false, but R is true.

Explanation:
1. Aryl halides have a C-Cl bond with partial double-bond character and are very unreactive to nucleophilic substitution, so the assertion is false.
2. Cyanide is indeed a strong nucleophile, so the reason is true.
3. Benzonitrile is made from benzenediazonium chloride and CuCN (Sandmeyer reaction).

Teacher's Note:
1. Aryl halides do not react with KCN under normal conditions.
2. Use the Sandmeyer reaction to get aryl nitriles.