"Organic Compounds Containing Nitrogen (Amines)" — Practice

Question 1

Arrange aniline, 4-nitroaniline, 4-methylaniline, ammonia and methylamine in increasing order of basicity, with reasons.

Accepted Answer

4-Nitroaniline is weakest, since its lone pair is delocalised into the ring and the nitro group withdraws density further by both induction and resonance. Aniline follows with the ring delocalisation alone, then 4-methylaniline, where the donating methyl group partly compensates. Ammonia is more basic than all three aromatic amines because its lone pair is entirely localised on nitrogen. Methylamine is strongest, having a localised lone pair plus a donating alkyl group. The order is 4-nitroaniline, aniline, 4-methylaniline, ammonia, methylamine.

Question 2

Explain why aniline is roughly ten thousand times less basic than methylamine.

Accepted Answer

Basicity depends on the availability of the nitrogen lone pair to accept a proton. In aniline the nitrogen is attached directly to a benzene ring and its lone pair is delocalised into the ring's pi system rather than sitting on nitrogen, so protonating it would destroy that delocalisation, which costs energy. A supporting factor is that aniline's nitrogen is attached to an $sp^2$ carbon, more electronegative than methylamine's $sp^3$ carbon, which pulls density away. The $\mathrm{p}K_b$ values are 9.38 against 3.38, a factor of ten thousand. The same argument taken further makes an amide, whose lone pair is delocalised into a carbonyl, essentially non-basic at all.

Question 3

In water, dimethylamine is a stronger base than trimethylamine even though the inductive effect predicts the reverse. Explain.

Accepted Answer

Three effects compete and only one favours trimethylamine. The inductive effect does favour it, since three methyl groups donate more density to nitrogen than two. But basicity in water depends on the stability of the protonated cation, and that cation is stabilised by hydrogen bonding to surrounding water. The dimethylammonium ion has two N-H bonds available for such bonding while the trimethylammonium ion has only one, and steric crowding by three methyl groups hinders both protonation and solvation. In the gas phase, with no solvent to vote, the order reverts to exactly what induction predicts: tertiary above secondary above primary above ammonia.

Question 4

How would you distinguish primary, secondary and tertiary amines in a single test?

Accepted Answer

Use the Hinsberg test with benzenesulphonyl chloride in alkali. A primary amine gives a sulphonamide that retains one N to H bond, made acidic by the two electron-withdrawing sulphonyl oxygens, so the product dissolves in the alkali. A secondary amine gives a sulphonamide with no N to H bond, which cannot lose a proton and stays as an insoluble solid. A tertiary amine has no N to H bond at all and does not react, so no product forms. Three distinct outcomes from one reagent.

Question 5

Suggest a route from propanamide to ethanamine, and from bromoethane to propan-1-amine.

Accepted Answer

Count carbons first. Propanamide has three and ethanamine two, so a carbon must be lost: use Hofmann bromamide degradation with $\mathrm{Br_2}$ and alkali, in which the carbonyl carbon departs as carbonate and the primary amine has one carbon fewer. Bromoethane has two carbons and propan-1-amine three, so a carbon must be gained: treat bromoethane with alcoholic KCN to give propanenitrile, then reduce with $\mathrm{LiAlH_4}$. Counting carbons turns this into a two-second decision, and a conversion that changes the carbon count has already named its own route.

Question 6

Why does the Gabriel phthalimide synthesis give a pure primary amine when ammonolysis of a halide does not, and why does it fail for aniline?

Accepted Answer

Ammonolysis fails to stop because the primary amine formed is itself nucleophilic and attacks more halide, giving secondary and tertiary amines and finally the quaternary salt. In the Gabriel synthesis the phthalimide nitrogen carries only one hydrogen, so it can be alkylated once and no further, and hydrolysis then releases a single pure primary amine. It fails for aromatic amines because the alkylation step is an SN2 displacement, and an aryl halide cannot undergo backside attack at an sp2 ring carbon. Aniline is made by reducing nitrobenzene instead.

Question 7

Give the product of reducing nitrobenzene under acidic, neutral and alkaline conditions.

Accepted Answer

With tin and hydrochloric acid, or by catalytic hydrogenation, the reduction goes all the way and gives aniline. With zinc dust and ammonium chloride in neutral solution it stops part way and gives phenylhydroxylamine. With zinc dust and sodium hydroxide in alkaline solution two molecules couple, giving hydrazobenzene. Electrolytic reduction in strongly acidic solution gives 4-aminophenol by rearrangement. The pattern is that acidic and vigorous conditions complete the reduction while neutral and alkaline ones stop short or allow coupling.

Question 8

Why does bromoethane give propanenitrile with potassium cyanide but ethyl isocyanide with silver cyanide?

Accepted Answer

Because the two reagents present the cyanide differently. Potassium cyanide is ionic, so the free cyanide ion is available and attacks through its carbon atom, which carries the greater share of the negative charge, giving a carbon to carbon bond and hence a nitrile. Silver cyanide is largely covalent, so the carbon is committed to the silver and only the nitrogen lone pair is free to attack, giving a carbon to nitrogen bond and hence an isocyanide. The products differ chemically too: a nitrile hydrolyses to a carboxylic acid, an isocyanide to a primary amine and methanoic acid.

Question 9

How would you prepare 4-bromoaniline from aniline?

Accepted Answer

Treating aniline directly with bromine water is useless, since the lone pair activates the ring so strongly that all three ortho and para positions brominate at once, giving 2,4,6-tribromoaniline. Acetylate the amine first with ethanoic anhydride to give acetanilide, which delocalises the lone pair into the amide carbonyl and moderates the activation so only one substitution occurs, directed para by the bulky acetamido group. Brominate to give 4-bromoacetanilide, then hydrolyse with acid or alkali to remove the acetyl group and release 4-bromoaniline. The same protection is needed before nitration, and for a further reason: the nitrating mixture would otherwise protonate aniline to the meta-directing anilinium ion.

Question 10

Explain why Friedel-Crafts alkylation cannot be carried out on aniline.

Accepted Answer

Because the nitrogen lone pair is a better Lewis base than the ring is a nucleophile, so it binds to the aluminium chloride catalyst as soon as the two are mixed. Two things then go wrong at once. The catalyst is consumed and can no longer generate the carbocation electrophile the reaction needs. And the nitrogen now carries a positive charge, which converts the strongly activating amino group into a strongly deactivating one, so even a free electrophile would find the ring unreactive. The problem cannot be solved by using more catalyst.

Question 11

Why are aromatic diazonium salts isolable at low temperature while aliphatic ones are not?

Accepted Answer

Because the positive charge on the diazonium group can be delocalised into an aromatic ring, spreading it over several atoms and lowering the energy of the ion substantially. No aliphatic system offers such delocalisation, so an aliphatic diazonium ion has its full positive charge on one nitrogen and loses nitrogen gas immediately on formation. Even the aromatic salt is only marginally stable and decomposes above about 278 K, which is why diazotisation is carried out in an ice bath at 273 to 278 K and the salt is used at once without isolation.

Question 12

Iodobenzene cannot be made by direct iodination of benzene. Give a route from benzene and explain why the diazonium method succeeds.

Accepted Answer

Nitrate benzene to nitrobenzene, reduce with tin and hydrochloric acid to aniline, diazotise with sodium nitrite and hydrochloric acid at 273 K, then treat with potassium iodide. Direct iodination fails because iodine is the least reactive halogen and the reaction is reversible, with the hydrogen iodide produced reducing the product back. The diazonium route succeeds because the leaving group departs as nitrogen gas, which is exceptionally stable and escapes from the solution, making the step irreversible. No catalyst is even needed for the iodide substitution.

Practice — "Organic Compounds Containing Nitrogen (Amines)"

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