JEE Chemistry Amines MCQs Set 03

Answer: (b) benzoylation with benzoyl chloride in the presence of aqueous NaOH

Explanation:
1. In the Schotten-Baumann reaction the amine is treated with benzoyl chloride in aqueous NaOH.
2. The product is the N-substituted benzamide, \( \ce{RNHCOC6H5} \).
3. NaOH removes the HCl formed and drives the reaction forward.

Teacher's Note:
1. Do not confuse it with simple acetylation.
2. Remember the key words: benzoyl chloride and aqueous base.

Answer: (a) \( \ce{CH3NCO} \), \( \ce{CH3NC} \)

Explanation:
1. \( \ce{CH3NCO} \) is hydrolysed by KOH to \( \ce{CH3NH2} \) and carbonate.
2. \( \ce{CH3NC} \) is hydrolysed by dilute acid to \( \ce{CH3NH2} \) and formic acid.
3. \( \ce{CH3CN} \) gives acetic acid on hydrolysis, and an oxime does not give an amine on simple hydrolysis. So only the first pair works.

Teacher's Note:
1. Isocyanides and isocyanates keep the C-N bond on nitrogen, so hydrolysis frees the amine.
2. Nitriles give carboxylic acids instead.

Answer: (b) Nitrobenzene

Explanation:
1. The nitro group is strongly electron withdrawing and deactivates the ring towards electrophilic attack.
2. Toluene and phenol have activating groups, so they nitrate faster than benzene.
3. Hence nitrobenzene is the hardest to nitrate.

Teacher's Note:
1. Activating groups speed up electrophilic substitution and deactivating groups slow it down.
2. Nitrobenzene needs fuming nitric acid and heat to react further.

Answer: (a) solid product

Explanation:
1. The nitro group is meta directing.
2. The product is m-dinitrobenzene, which is a pale yellow crystalline solid.

Teacher's Note:
1. Meta directors send the incoming group to the 3-position.
2. m-Dinitrobenzene has a melting point near 90 C.

Answer: (b) \( \ce{H3PO2} \)

Explanation:
1. Sn/HCl reduces nitrobenzene to aniline (A).
2. Aniline with \( \ce{NaNO2}/HCl \) at 0-5 C gives benzenediazonium chloride (B).
3. Hypophosphorous acid reduces the diazonium group to hydrogen, giving benzene.

Teacher's Note:
1. Hypophosphorous acid (or ethanol) replaces \( \ce{N2+} \) by H.
2. Methanol is not used for this step in these question types.

Answer: (b) D.N.B.

Explanation:
1. Nitrobenzene is nitrated at the meta position.
2. The product is m-dinitrobenzene (D.N.B.).
3. TNT comes from toluene, DDT from chlorobenzene and chloral, and TEL is tetraethyl lead.

Teacher's Note:
1. D.N.B. stands for dinitrobenzene.
2. Further nitration to trinitrobenzene needs even more drastic conditions.

Answer: (b) 1,2-dinitrobenzene

Explanation:
1. Para isomers are symmetrical, so the bond moments cancel and the dipole moment is zero.
2. Between the two ortho isomers, the C-NO2 bond moment is much larger than the C-Cl bond moment.
3. So 1,2-dinitrobenzene has the largest dipole moment.

Teacher's Note:
1. Symmetrical para disubstituted benzenes with identical groups have zero dipole moment.
2. For ortho isomers, a more polar substituent gives a larger resultant.

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

Explanation:
1. The sequence is amide to N-bromoamide (\( \ce{RCONHBr} \)), then to the bromamide anion, then rearrangement to isocyanate \( \ce{RNCO} \).
2. Hydrolysis of the isocyanate gives \( \ce{RNH2} \) and carbonate.
3. An isocyanide \( \ce{RNC} \) is never formed in this reaction.

Teacher's Note:
1. Isocyanate (R-N=C=O) is the key intermediate, not isocyanide.
2. Do not mix up the two N-C compounds.

Answer: (c) Mendius reaction

Explanation:
1. \( \ce{P2O5} \) dehydrates the amide to methyl cyanide, \( \ce{CH3CN} \) (A).
2. Reduction of the nitrile with Na/ethanol gives ethylamine (B).
3. The reduction of alkyl cyanides to primary amines by nascent hydrogen is the Mendius reaction.

Teacher's Note:
1. Mendius reaction means nitrile reduced to primary amine with Na/alcohol.
2. Stephen reaction stops at the aldehyde stage with SnCl2/HCl.

Answer: (a) (a) and (b)

Explanation:
1. The amide is first brominated to N-bromoacetamide, \( \ce{CH3CONHBr} \).
2. Base removes a proton and the anion rearranges to methyl isocyanate, \( \ce{CH3NCO} \).
3. Hydrolysis of the isocyanate then gives \( \ce{CH3NH2} \). So (a) and (b) are the intermediates.

Teacher's Note:
1. The two isolable-type intermediates are the N-bromoamide and the isocyanate.
2. The R group migrates from carbon to nitrogen in the rearrangement.

Answer: (c) \( \ce{R-CN + H2O} \) with \( \ce{H+} \)

Explanation:
1. Ammonolysis of an alkyl halide gives amines.
2. Reduction of an oxime or of an amide with \( \ce{LiAlH4} \) gives a primary amine.
3. Acid hydrolysis of a nitrile gives a carboxylic acid (via the amide), not an amine.

Teacher's Note:
1. Reduction of nitrogen compounds gives amines, but hydrolysis of a nitrile gives an acid.
2. Always check whether the reagent is a reducing agent.

Answer: (a) aniline treated with \( \ce{CH3I} \), and then again with excess \( \ce{CH3I} \)

Explanation:
1. Aniline reacts with \( \ce{CH3I} \) to give N-methylaniline (secondary amine).
2. A second treatment with \( \ce{CH3I} \) gives N,N-dimethylaniline, a tertiary amine.
3. Sn/HCl reduction of nitrobenzene gives only aniline, a primary amine.

Teacher's Note:
1. Each alkylation adds one alkyl group on nitrogen.
2. Excess alkyl halide pushes the reaction on to the tertiary amine and the quaternary salt.

Answer: (c) \( \ce{C2H5NC} \)

Explanation:
1. Nitroethane, acetaldoxime and methyl cyanide all give \( \ce{C2H5NH2} \) on reduction.
2. Reduction of an isocyanide gives a secondary amine, \( \ce{C2H5NHCH3} \) (N-methylethylamine).
3. So ethylamine cannot be obtained from \( \ce{C2H5NC} \).

Teacher's Note:
1. Isocyanide reduction always gives an N-methyl secondary amine.
2. This is a favourite exam point.

Answer: (c) imides

Explanation:
1. Gabriel synthesis starts with phthalimide, which is an imide.
2. Its potassium salt reacts with an alkyl halide, and hydrolysis releases the primary amine.

Teacher's Note:
1. Phthalimide is a cyclic imide with two C=O groups on one nitrogen.
2. Remember: imide in, primary amine out.

Answer: (c) primary aliphatic amines

Explanation:
1. The N-alkylphthalimide is formed by SN2 attack on an alkyl halide.
2. Aryl halides do not undergo this substitution, so aromatic amines cannot be made.
3. Hydrolysis gives only pure primary aliphatic amines.

Teacher's Note:
1. Aryl halides are inert to SN2, so aniline cannot be made by this route.
2. The method gives no 2 degree or 3 degree amine contamination.

Answer: (c) ammonolysis as well as nucleophilic substitution

Explanation:
1. Ammonia is the nucleophile that replaces bromide, so the reaction is nucleophilic substitution.
2. Because the reagent cleaving the C-Br bond is ammonia, it is also called ammonolysis.

Teacher's Note:
1. Ammonolysis is the cleavage of a bond by ammonia.
2. It is a nucleophilic substitution in mechanism.

Answer: (c) formation of hydrogen bonds with water

Explanation:
1. The N-H and the lone pair on nitrogen form hydrogen bonds with water molecules.
2. Low molar mass helps, but the cause of solubility is the hydrogen bonding.

Teacher's Note:
1. Solubility of amines falls as the alkyl group gets larger.
2. Hydrogen bonding with water is the main reason for solubility.

Answer: (d) Schiff's base

Explanation:
1. A primary amine and a ketone lose water to form an imine.
2. The imine \( \ce{R2C=N-R'} \) is the Schiff's base.

Teacher's Note:
1. Aldehydes and ketones both give Schiff's bases.
2. Secondary amines give enamines instead.

Answer: (b) nitrene and carbene

Explanation:
1. In the Hofmann bromamide reaction the loss of bromide from the bromamide anion is described as giving a nitrene-type species that rearranges to the isocyanate.
2. In the carbylamine reaction, alkaline chloroform gives dichlorocarbene, \( \ce{:CCl2} \), which reacts with the amine.

Teacher's Note:
1. Hofmann bromamide reaction: nitrene (acyl nitrene) route.
2. Carbylamine and Reimer-Tiemann reactions both involve dichlorocarbene.

Answer: (c) acetyl chloride and dimethylamine

Explanation:
1. Acylation of a secondary amine gives an N,N-disubstituted amide.
2. Acetyl chloride with dimethylamine gives \( \ce{CH3CON(CH3)2} \), N,N-dimethylacetamide.

Teacher's Note:
1. The alkyl groups on nitrogen of the amide come from the amine.
2. Primary amines give N-alkylamides.

Answer: (b) diethylamine

Explanation:
1. Only primary amines (those with an -NH2 group) give the carbylamine test.
2. Isopropylamine, tert-butylamine and sec-butylamine are all primary amines.
3. Diethylamine is a secondary amine and does not respond.

Teacher's Note:
1. Carbylamine test = primary amines only.
2. Look for the \( \ce{-NH2} \) group in each option.

Answer: (b) benzylamine

Explanation:
1. Benzylamine has the \( \ce{-NH2} \) on a side-chain carbon, so it behaves like an aliphatic primary amine.
2. With nitrous acid it gives an unstable diazonium salt that is replaced by OH, forming benzyl alcohol.
3. Aniline gives a stable diazonium salt at low temperature.

Teacher's Note:
1. Aliphatic-type primary amines give alcohols with nitrous acid.
2. Aromatic primary amines give diazonium salts at 0-5 C.

Answer: (a) aniline

Explanation:
1. Electrolytic reduction in dilute acid medium gives aniline.
2. In strongly acidic medium (conc. \( \ce{H2SO4} \)) it gives N-phenylhydroxylamine, which rearranges to p-aminophenol.

Teacher's Note:
1. Weak acid leads to full reduction to aniline.
2. Strong acid leads to p-aminophenol through rearrangement.

Answer: (b) acetylation

Explanation:
1. In acid, aniline is protonated to the anilinium ion, which gives meta products and is also oxidised.
2. Acetylation gives acetanilide, which is less strongly activating, prevents oxidation and gives mostly the para product.

Teacher's Note:
1. Acetylation reduces the activating power of the amino group.
2. After nitration the acetyl group is removed by hydrolysis.

Answer: (b) steam distillation

Explanation:
1. Aniline is steam volatile and immiscible with water.
2. Steam distillation separates it from non-volatile impurities and tarry matter.

Teacher's Note:
1. Steam distillation is used for compounds that decompose at their normal boiling point.
2. Aniline darkens on standing due to oxidation.