"Organic Compounds Containing Nitrogen" — Practice

Question 1

Suggest a synthesis of $p$-nitroaniline from aniline, explaining why direct nitration is unsatisfactory.

Accepted Answer

Direct nitration protonates the amine, and $-	ext{NH}_3^+$ is deactivating and meta directing, so a mixture containing substantial meta product results.

Acetylate the aniline with ethanoic anhydride to give acetanilide. The amide nitrogen is far less basic and survives the acid.

Nitrate the acetanilide; the amide is still ortho-para directing but moderated, and the bulky group favours para.

Hydrolyse the amide with aqueous acid to recover $p$-nitroaniline.

Question 2

Starting from propanamide, give routes to methylamine, ethylamine and propylamine.

Accepted Answer

**Ethylamine** has one carbon fewer than propanamide, so use the Hofmann bromamide degradation with bromine and alkali.

**Propylamine** has the same count, so reduce the amide directly with lithium aluminium hydride.

**Methylamine** has two fewer, so no single step suffices. Degrade to ethylamine, convert to ethanamide by a longer sequence, and degrade again.

The carbon count of the target against the starting material identifies the route in each case before any mechanism is considered.

Question 3

Explain why the Gabriel phthalimide synthesis gives a pure primary amine while direct alkylation of ammonia does not.

Accepted Answer

In direct alkylation the primary amine formed is a **better** nucleophile than ammonia, so it competes for the remaining halide and gives the secondary amine, and so on to the quaternary salt.

In the Gabriel synthesis the nitrogen is held in a phthalimide ring, so once one alkyl group is attached the nitrogen carries no further hydrogen and cannot be alkylated again.

Hydrolysis then releases the primary amine cleanly.

The method protects the nitrogen from over-alkylation by construction rather than by controlling conditions.

Question 4

Give a route from bromoethane to propan-$1$-amine, and one from bromoethane to ethanamine.

Accepted Answer

**Propan-$1$-amine** has three carbons against bromoethane's two, so a carbon must be added. Displace bromide with cyanide to give propanenitrile, then reduce with lithium aluminium hydride or with hydrogen over nickel.

**Ethanamine** has the same count, so use the Gabriel synthesis: alkylate potassium phthalimide with bromoethane and hydrolyse.

Direct reaction with ammonia would also give ethanamine but contaminated with the secondary, tertiary and quaternary products.

The cyanide route is the standard one-carbon extension in aliphatic chemistry.

Question 5

Distinguish between aniline, N-methylaniline and N,N-dimethylaniline using two tests.

Accepted Answer

**Carbylamine test:** aniline alone gives the foul-smelling phenyl isocyanide. Neither of the others reacts.

**Hinsberg test:** aniline gives a sulphonamide with an acidic $	ext{N}-	ext{H}$, which dissolves in alkali. N-methylaniline gives a sulphonamide with no such hydrogen, which remains as a solid. N,N-dimethylaniline has no hydrogen on nitrogen and does not react at all.

The Hinsberg test alone would suffice, distinguishing all three.

The carbylamine test provides a rapid confirmation for the primary amine.

Question 6

Arrange in order of increasing basicity in water: ammonia, methylamine, dimethylamine, aniline, $p$-nitroaniline.

Accepted Answer

$p$-Nitroaniline is weakest: the nitro group withdraws the already delocalised lone pair still further.

Aniline follows, its lone pair delocalised into the ring so that protonation costs the resonance stabilisation.

Ammonia has no alkyl assistance at all.

Methylamine gains inductive donation, and dimethylamine gains more while still retaining a hydrogen for solvation of its cation.

Order: $p$-nitroaniline $<$ aniline $<$ ammonia $<$ methylamine $<$ dimethylamine.

Question 7

Predict the observations when propan-$2$-amine, N-ethylethanamine and triethylamine are each treated with cold nitrous acid.

Accepted Answer

**Propan-$2$-amine** is primary and aliphatic. It forms an unstable diazonium ion that loses nitrogen immediately, giving brisk **effervescence** and a mixture of propan-$2$-ol, propene and rearranged products.

**N-ethylethanamine** is secondary. It cannot form a diazonium ion, and the nitrogen is nitrosated instead, giving a **yellow oil** that separates from the solution.

**Triethylamine** is tertiary with no hydrogen on nitrogen, so it simply forms a soluble salt with the acid and nothing else happens.

The gas and the oil are the two diagnostic observations.

Question 8

Give reagents to convert benzenediazonium chloride into chlorobenzene, iodobenzene, fluorobenzene, phenol and benzene.

Accepted Answer

**Chlorobenzene:** copper(I) chloride in hydrochloric acid, a Sandmeyer reaction. Copper powder with the acid, the Gattermann variant, also works.

**Iodobenzene:** simply warm with potassium iodide. No catalyst is needed, since iodide is a sufficiently good nucleophile.

**Fluorobenzene:** treat with fluoroboric acid to precipitate the tetrafluoroborate salt, then heat it. This is the Balz-Schiemann reaction.

**Phenol:** warm the aqueous solution. **Benzene:** treat with hypophosphorous acid, which replaces the group by hydrogen.

Question 9

Suggest a synthesis of $1,3,5$-tribromobenzene from benzene, explaining why direct bromination cannot work.

Accepted Answer

Bromine is ortho-para directing, so successive brominations of benzene would never place three bromines all meta to one another.

Nitrate benzene and reduce with tin and hydrochloric acid to give aniline.

The amino group is powerfully ortho-para directing, so treating with bromine water gives $2,4,6$-tribromoaniline at once and without a catalyst.

Diazotise at $0$ to $5^{\circ}$C and treat with hypophosphorous acid, which replaces the diazonium group by hydrogen and leaves $1,3,5$-tribromobenzene.

Question 10

Explain why coupling of a diazonium salt with aniline is carried out in mildly acidic solution rather than in neutral or strongly acidic solution.

Accepted Answer

The diazonium ion is stable only in acidic solution; in neutral or alkaline conditions it decomposes or is converted to an unreactive diazotate.

Aniline must remain unprotonated to be nucleophilic, since the anilinium ion is strongly deactivated.

Strong acid would protonate essentially all of the aniline, and neutral conditions would destroy the diazonium ion.

Mildly acidic conditions, around pH $4$ to $5$, keep the diazonium ion intact while leaving a useful fraction of free aniline. Coupling then occurs at the para position.

Question 11

Nitrobenzene is reduced under four different conditions. Give the product in each and explain the pattern.

Accepted Answer

**Tin with hydrochloric acid:** aniline. Acidic conditions supply protons freely and complete the reduction.

**Zinc with ammonium chloride, neutral:** N-phenylhydroxylamine, the reduction stalling at the intermediate stage.

**Zinc with sodium hydroxide, alkaline:** hydrazobenzene, two partially reduced molecules having condensed first.

**Electrolytic in concentrated sulphuric acid:** $p$-aminophenol, the hydroxylamine formed first having rearranged under the strong acid.

One functional group, four products, distinguished only by the medium.

Question 12

Suggest a synthesis of $m$-nitrophenol from benzene.

Accepted Answer

A hydroxyl group is ortho-para directing, so introducing it first would place the nitro group wrongly.

Nitrate benzene twice to give $m$-dinitrobenzene, the nitro group being meta directing.

Reduce **one** nitro group selectively with ammonium sulphide to give $m$-nitroaniline.

Diazotise at $0$ to $5^{\circ}$C and warm the solution with water, which replaces the diazonium group by hydroxyl and gives $m$-nitrophenol.

Practice — "Organic Compounds Containing Nitrogen"

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