JEE Chemistry Amines MCQs Set 32

Answer: (c) (C₂H₅)₂NH > C₂H₅NH₂ > NH₃

Explanation:
1. Basic strength in water depends on the stability of the ammonium ion formed: the +I effect of alkyl groups, solvation by hydrogen bonding, and steric hindrance.
2. Ethyl groups are electron releasing, so ethylamine is stronger than ammonia, and diethylamine (two ethyl groups) is stronger still.
3. Order: (C₂H₅)₂NH > C₂H₅NH₂ > NH₃.

Teacher's Note:
1. For ethyl groups the secondary amine is the strongest in water.
2. Alkyl groups push electron density towards nitrogen.

Answer: (a) n-butylamine

Explanation:
1. The Gabriel synthesis makes only primary amines: potassium phthalimide is alkylated (S_N2) and then hydrolysed.
2. The alkyl halide must be a primary, unhindered halide. n-Butyl bromide works; tertiary halides eliminate, and neopentyl halides are too hindered for S_N2.
3. Triethylamine is a tertiary amine and cannot be made this way.

Teacher's Note:
1. Gabriel: only primary aliphatic amines, and not aromatic amines.
2. The S_N2 step needs an unhindered primary halide.

Answer: (c) 4-(phenylazo)-1-naphthol

Explanation:
1. In alkaline medium, 1-naphthol exists as the naphthoxide ion, which is a very strong activator.
2. The electrophile ArN₂⁺ couples at the para position (C-4) to the oxygen.
3. Product: 4-(phenylazo)-1-naphthol, an orange-red azo dye.

Teacher's Note:
1. Phenols couple at the para position (ortho if para is blocked).
2. Alkaline pH is needed to form the phenoxide ion.

Answer: (a) 4-aminoazobenzene

Explanation:
1. Diazotisation gives benzenediazonium chloride.
2. Coupling with phenols is fastest in mildly alkaline medium, and coupling with aromatic amines is fastest in mildly acidic medium.
3. The solution here is acidic (dilute HCl), so the diazonium ion couples with aniline at the para position: 4-aminoazobenzene (yellow dye).

Teacher's Note:
1. Amines couple in weakly acidic solution; phenols in weakly basic solution.
2. In strong acid the amine is protonated and does not couple.

Answer: (c) C₆H₅SO₂Cl

Explanation:
1. Hinsberg's reagent is benzenesulphonyl chloride, C₆H₅SO₂Cl.
2. It is used to distinguish primary, secondary and tertiary amines.

Teacher's Note:
1. Primary amines give a sulphonamide soluble in alkali; secondary give an insoluble one; tertiary do not react.
2. SOCl₂ is thionyl chloride.

Answer: (a) HCONH–CH₂CH₂CH(OH)CH₃

Explanation:
1. The amino group is a much better nucleophile than the hydroxyl group, so it attacks the carbonyl carbon of the ester (acyl nucleophilic substitution).
2. Loss of ethoxide (as ethanol) gives the N-formyl derivative, N-(3-hydroxybutyl)formamide.
3. With only one equivalent of the ester, the product is the mono-formylated amide.

Teacher's Note:
1. N is more nucleophilic than O in neutral or basic conditions.
2. One equivalent means one acyl group is introduced.

Answer: (b) NH₂NH₂

Explanation:
1. Hydrazinolysis of N-ethylphthalimide with NH₂NH₂ gives phthalhydrazide and ethylamine, the second step of the Gabriel synthesis.
2. Water alone does not cleave the imide, and NaBH₄ and CaH₂ are not suitable.

Teacher's Note:
1. Alkaline hydrolysis or hydrazine liberates the amine.
2. Hydrazinolysis is milder than alkaline hydrolysis.

Answer: (b) (i) HCl/H₂O (ii) NaBH₄

Explanation:
1. Catalytic hydrogenation and LiAlH₄ both reduce the nitrile (–C≡N) to –CH₂NH₂. SnCl₂/HCl reduces it to the iminium salt, which NaBH₄ reduces to the amine.
2. In (b), HCl/H₂O first hydrolyses the nitrile to benzoic acid, and NaBH₄ does not reduce a carboxylic acid to an amine (it does not reduce acids at all).
3. So (b) is not a valid method.

Teacher's Note:
1. Hydrolysis of a nitrile gives an acid, not an amine.
2. NaBH₄ is a mild reagent and does not reduce acids.

Answer: (c) B < D < A < C

Explanation:
1. A smaller pK_b means a stronger base. Basicity increases with the electron density on nitrogen.
2. OCH₃ is a strong +R donor and CH₃ a weaker +I donor; F withdraws electrons by –I more than it donates by +R, and NO₂ withdraws strongly (–I, –R).
3. Basicity: OCH₃ > CH₃ > F > NO₂, so pK_b order: B < D < A < C.

Teacher's Note:
1. Donors raise basicity (lower pK_b); acceptors lower it.
2. Compare the substituent effects through the ring.

Answer: (c) 4-aminoazobenzene

Explanation:
1. This is an azo coupling: the diazonium ion is an electrophile and aniline's para position is highly activated by the –NH₂ group.
2. Product: p-aminoazobenzene (a yellow dye), C₆H₅N=N–C₆H₄NH₂.

Teacher's Note:
1. The para position is preferred for steric reasons.
2. Azo compounds have the –N=N– link between two rings.

Answer: (a) RCHO

Explanation:
1. DIBAL-H is a selective, bulky reducing agent that adds one hydride to the nitrile to give an imine-aluminium complex.
2. Aqueous work-up hydrolyses the imine to the aldehyde RCHO.
3. (Stronger reagents such as LiAlH₄ would go on to give the amine.)

Teacher's Note:
1. DIBAL-H stops at the aldehyde stage for nitriles and esters.
2. The imine is hydrolysed during work-up.

Answer: (d) III > I > II

Explanation:
1. Basicity depends on the availability of the nitrogen lone pair, which falls as the s-character of the orbital rises.
2. Piperidine: sp³ N (25% s), the lone pair is freely available (strongest base). Pyridine: sp² N (33% s), lone pair in an sp² orbital (moderate). Pyrrole: the lone pair is part of the aromatic sextet, so it is not available (weakest base).
3. Order: III > I > II.

Teacher's Note:
1. A lone pair that is part of an aromatic system is not basic.
2. More s-character holds the lone pair closer to nitrogen.

Answer: (d) D < C < A < B

Explanation:
1. Aniline is the weakest base because the lone pair is delocalised into the ring.
2. In water, secondary amines are the strongest aliphatic amines (+I effect with good solvation). Primary amines come next and tertiary amines are weaker because of steric hindrance and poor solvation.
3. Order: aniline < trimethylamine < ethylamine < N-methylethanamine, that is D < C < A < B.

Teacher's Note:
1. In water, 2° > 3° ≈ 1° for methyl/ethyl amines depending on solvation.
2. Arylamines are always much weaker than alkylamines.

Answer: (b) only the –NH₂ group

Explanation:
1. The nitrogen of the amino group is less electronegative than oxygen and holds its lone pair less tightly, so it is a stronger nucleophile.
2. With one equivalent of the acylating agent at room temperature, the –NH₂ group is acetylated selectively, giving the amide (N-acetyl derivative).

Teacher's Note:
1. Amines react faster than alcohols in acylation.
2. Only an excess of the reagent and more forcing conditions would acetylate the –OH as well.

Answer: (a) the N-terminal –NH₂ and the serine –OH only

Explanation:
1. Acetic anhydride acetylates amines and alcohols: the free N-terminal –NH₂ becomes an acetamide and the serine –OH becomes an acetate ester.
2. The side-chain amide –CONH₂ is not acetylated, because the lone pair on its nitrogen is delocalised over the C=O group.
3. So two groups react.

Teacher's Note:
1. Amide nitrogens are not nucleophilic.
2. 'Excess' reagent means both nucleophilic groups react.

Answer: (c) benzoic acid

Explanation:
1. C₆H₇N is aniline. It is formed from an amide by Hofmann bromamide degradation (Br₂/KOH), which loses one carbon.
2. So B is benzamide (C₇H₇NO), made by heating ammonium benzoate, which comes from benzoic acid and NH₃.
3. So A is benzoic acid (C₇H₆O₂).

Teacher's Note:
1. Work backwards: amine, then amide (one carbon more), then acid.
2. Hydroxybenzaldehydes do not form amides with ammonia.

Answer: (b) N-(4-nitrophenyl)benzamide

Explanation:
1. The electrophile is NO₂⁺. The ring attached to nitrogen is activated by the +R effect of the –NH– lone pair (ortho/para director).
2. The other ring is deactivated by the –R effect of the C=O group.
3. Attack occurs on the activated ring, mainly at the para position (less hindered): N-(4-nitrophenyl)benzamide.

Teacher's Note:
1. The ring joined to N is the activated one.
2. Para is preferred over ortho because of steric hindrance.

Answer: (b) CH₃CH₂CH₂CONH₂

Explanation:
1. A positive carbylamine test means the product is a primary amine. Hofmann bromamide degradation converts a primary amide RCONH₂ into RNH₂ with one carbon less.
2. C₃H₉N as a primary amine is propylamine, CH₃CH₂CH₂NH₂, so X has four carbons: butanamide, CH₃CH₂CH₂CONH₂.

Teacher's Note:
1. Hofmann degradation removes the carbonyl carbon.
2. Only primary amines give the carbylamine test.

Answer: (c) B < A < C < D

Explanation:
1. Reactivity towards alkyl halides (S_N2) depends on the availability of the nitrogen lone pair.
2. In phthalimide the nitrogen is attached to two C=O groups and its lone pair is least available. In benzamide the lone pair is delocalised over one C=O.
3. In 2-aminobenzonitrile the lone pair is partly withdrawn by the electron-withdrawing –CN group, while aniline's lone pair is the most available.
4. Order: B < A < C < D.

Teacher's Note:
1. More electron-withdrawing groups on nitrogen means a weaker nucleophile.
2. Amides are much poorer nucleophiles than amines.

Answer: (c) aniline

Explanation:
1. In the strongly acidic nitrating mixture, aniline is protonated to the anilinium ion, –NH₃⁺, which is a strongly deactivating, meta-directing group.
2. So a significant amount (about 47%) of the meta-nitro product forms. Phenol, phenyl acetate and acetanilide keep an ortho/para-directing group.

Teacher's Note:
1. Protonation changes –NH₂ (o/p) into –NH₃⁺ (m).
2. Acetylation avoids this problem by reducing the basicity of nitrogen.

Answer: (c) four moles of NaOH and one mole of Br₂

Explanation:
1. RCONH₂ + Br₂ + 4NaOH gives RNH₂ + Na₂CO₃ + 2NaBr + 2H₂O.
2. So four moles of NaOH and one mole of Br₂ are used per mole of amine.

Teacher's Note:
1. Balance the equation to count the reagents: Na on the right is 2 + 2 = 4.
2. The amine has one carbon less than the amide.

Answer: (c) 4-methylbenzonitrile

Explanation:
1. Diazotisation gives 4-methylbenzenediazonium chloride (D).
2. Sandmeyer reaction with CuCN replaces the diazonium group by –CN: E = 4-methylbenzonitrile (p-tolunitrile).

Teacher's Note:
1. The Sandmeyer reaction uses CuCl, CuBr or CuCN.
2. The ring methyl group stays in place.

Answer: (d) an alkyl isocyanide

Explanation:
1. This is the carbylamine (isocyanide) reaction: RNH₂ + CHCl₃ + 3KOH gives RNC + 3KCl + 3H₂O.
2. The product is an alkyl isocyanide (carbylamine) with a foul smell.

Teacher's Note:
1. The test is specific to primary amines.
2. In an isocyanide the alkyl group is bonded to nitrogen: R–N≡C.

Answer: (a) (CH₃)₂NH

Explanation:
1. The smaller the pK_b, the stronger the base.
2. In water the order is (CH₃)₂NH > CH₃NH₂ > (CH₃)₃N > C₆H₅NH₂, because of +I effect, steric hindrance and solvation of the ammonium ion.
3. So dimethylamine has the smallest pK_b.

Teacher's Note:
1. Remember: in water, methylamines follow 2° > 1° > 3° > NH₃.
2. pK_b = –log K_b.

Answer: (c) CH₃CH₂COOH

Explanation:
1. The last step is a Hofmann bromamide reaction, so C is an amide with one more carbon than the amine: CH₃CH₂CONH₂ (propanamide).
2. B is the ammonium salt CH₃CH₂COO^–NH₄⁺, which on heating loses water to give the amide.
3. So A is propanoic acid.

Teacher's Note:
1. Work backwards one step at a time.
2. Ammonium salts of acids give amides when heated.

Answer: (c) 5

Explanation:
1. Each acetylation of –NH₂ replaces one H (1 u) by COCH₃ (43 u), a net increase of 42 u.
2. Total increase = 390 − 180 = 210 u, so the number of –NH₂ groups = 210 ÷ 42 = 5.

Teacher's Note:
1. Find the mass change per acetyl group first.
2. Divide the total change by the change per group.

Answer: (b) benzenediazonium chloride and benzonitrile

Explanation:
1. Diazotisation of aniline at 278 K (0–5 °C) gives benzenediazonium chloride (A).
2. Sandmeyer reaction with CuCN replaces N₂⁺ by CN to give benzonitrile (B).

Teacher's Note:
1. Low temperature keeps the diazonium salt from decomposing.
2. CuCN introduces the nitrile group.

Answer: (c) benzenediazonium chloride and fluorobenzene

Explanation:
1. A is benzenediazonium chloride. Adding HBF₄ precipitates the diazonium fluoroborate, which on heating decomposes to fluorobenzene, N₂ and BF₃ (Balz–Schiemann reaction).

Teacher's Note:
1. Balz–Schiemann is the standard route to aryl fluorides.
2. Direct fluorination of benzene is too violent.