Amines and Nitrogen Compounds Questions and Answers

Answer: C

The highest pK_b belongs to benzenamine (aniline).

A higher pK_b means a weaker base. In aniline the nitrogen lone pair is delocalised into the benzene ring, so it is much less available for protonation; aniline's pK_b is about 9.38.

N-Methylaniline is slightly stronger (pK_b about 9.30) because the methyl group donates electrons. The aliphatic amines are far stronger bases, with pK_b near 3.3-3.4 for methanamine, ethanamine and N-methylmethanamine.

The order of base strength in water among the aliphatic amines (secondary > primary > tertiary for methyl groups) does not affect this question, because every aliphatic amine is far stronger than an aromatic one.

Answer: C

The correct combination is (a)-(iii), (b)-(iv), (c)-(i), (d)-(ii).

• (a) Benzenesulphonyl chloride is Hinsberg's reagent, used to tell primary, secondary and tertiary amines apart.
• (b) An amide is converted to an amine with one carbon fewer by Hofmann's bromamide reaction (Br₂ and NaOH).
• (c) A primary amine heated with chloroform and alcoholic KOH gives a foul-smelling isocyanide: the carbylamine reaction.
• (d) Diethylamine, (C₂H₅)₂NH, has two alkyl groups on nitrogen, so it is a secondary amine.

The option with (c)-(ii) and (d)-(i) swaps the last two pairs, and the others misplace Hinsberg's reagent.

Answer: E

The reagent X is \(KI\).

Iodide ion is a good enough nucleophile to replace the diazonium group directly, simply by warming the diazonium salt with potassium iodide:

\[ \begin{aligned} &C_{6}H_{5}N_{2}^{+}Cl^{-} + KI \\ &\rightarrow C_{6}H_{5}I + KCl + N_{2} \end{aligned} \]

Copper(I) salts are needed for chloride, bromide and cyanide (Sandmeyer), not for iodide. \(I_{2}\), AgI and HI are not the reagents used.

Answer: C

It takes 3 moles of alkyl halide.

Each alkylation replaces one hydrogen on nitrogen and the last one adds a fourth group. A primary amine has two N-H hydrogens:

\[ \begin{aligned} RNH_{2} &\rightarrow R_{2}NH \rightarrow R_{3}N \\ &\rightarrow R_{4}N^{+}X^{-} \end{aligned} \]

with one mole of RX per step, three in all. Ammonia would need four; a secondary amine two; a tertiary amine one.

This is why ammonolysis of an alkyl halide gives a mixture of primary, secondary and tertiary amines and the quaternary salt unless the reagents are carefully proportioned.

Answer: B

The order is (i) > (ii) > (iii).

The three amines are isomers (\(C_{4}H_{11}N\)), so the difference comes from hydrogen bonding. Butan-1-amine (primary) has two N-H hydrogens for intermolecular hydrogen bonds, N-ethylethanamine (secondary) has one, and N,N-dimethylethanamine (tertiary) has none.

Boiling points follow: primary > secondary > tertiary (about 350, 329 and 310 K). Orders putting the tertiary amine above the others ignore that it cannot hydrogen-bond with itself.

Answer: C

X and Y are Benzylchloride and Benzylamine.

\(C_{7}H_{7}Cl\) with a reactive (side-chain) chlorine is benzyl chloride, \(C_{6}H_{5}CH_{2}Cl\). Ammonolysis gives benzylamine, \(C_{6}H_{5}CH_{2}NH_{2}\) (\(C_{7}H_{9}N\)), and two methylations give \(C_{6}H_{5}CH_{2}N(CH_{3})_{2}\), N,N-dimethylphenylmethanamine.

Chlorobenzene is \(C_{6}H_{5}Cl\) and aniline \(C_{6}H_{7}N\), so they do not fit the formulas (and aryl halides resist ammonolysis). Benzoyl chloride is \(C_{7}H_{5}OCl\).

Answer: D

The order is \(\mathrm{(CH_{3})_{2}NH \gt CH_{3}NH_{2} \gt (CH_{3})_{3}N \gt NH_{3}}\).

In water, basic strength depends on three factors together: the electron-donating (+I) effect of methyl groups, which increases basicity; solvation of the protonated amine by hydrogen bonding, which is greatest when it has more N-H bonds; and steric hindrance, which grows with more methyl groups.

Dimethylamine has the best balance and is strongest. Trimethylamine, despite three methyl groups, has a poorly solvated cation with only one N-H, so it falls below methylamine. Ammonia, with no +I effect, is weakest. The order by +I effect alone, as in the gas phase, would be (CH₃)₃N \(\gt\) (CH₃)₂NH \(\gt\) CH₃NH₂ \(\gt\) NH₃, which is not the aqueous order.

Answer: E

Hinsberg's reagent is benzenesulphonyl chloride, C₆H₅SO₂Cl.

It distinguishes primary, secondary and tertiary amines. Primary amines give a sulphonamide that dissolves in alkali (it still has an acidic N-H); secondary amines give one that does not; tertiary amines do not react.

HCl/ZnCl₂ is the Lucas reagent for alcohols, Sn/HCl reduces nitro compounds to amines, ammoniacal AgNO₃ is Tollens' reagent for aldehydes, and FeBr₃ is a Lewis acid catalyst for bromination.

Answer: B

The highest boiling point belongs to CH₃CH₂CH₂CH₂NH₂ (n-butylamine).

Two factors decide the order: molar mass, and hydrogen bonding through N-H bonds. Primary amines have two N-H hydrogens, secondary amines one and tertiary amines none, so for amines of the same molar mass the order is primary > secondary > tertiary.

n-Butylamine, diethylamine and ethyldimethylamine are isomers (C4H11N): n-butylamine boils at about 78 °C, diethylamine at about 56 °C, and the tertiary amine at about 37 °C, since it cannot hydrogen bond with itself. Ethylamine and n-propylamine are primary but lighter, so they boil lower.

Answer: D

This is the Gattermann reaction.

Replacing the diazonium group by a halogen using copper powder with the halogen acid (here Cu and HBr) is the Gattermann reaction: C₆H₅N₂Cl + HBr (with Cu) gives C₆H₅Br, N₂ and CuCl.

The Sandmeyer reaction does the same job with a copper(I) salt, CuBr in HBr, rather than copper metal. Hoffmann bromamide degradation turns an amide into an amine, the Gabriel synthesis makes primary amines from phthalimide, and Hinsberg's reagent (benzenesulphonyl chloride) distinguishes primary, secondary and tertiary amines.