Organic Compounds Containing Nitrogen

The amino group is among the most powerfully activating, ortho-para directing substituents in aromatic chemistry. Yet nitrating aniline gives roughly a third meta product. How?

Because the substrate being nitrated is not aniline.

A nitrating mixture is concentrated sulphuric acid, and aniline is a base. It is protonated almost completely to the anilinium ion, whose substituent is $-{\text{NH}}_3^{+}$ — a positively charged group with no lone pair to donate, strongly deactivating and meta directing.

The product is a mixture, because some unprotonated aniline is always present and reacts far faster than the anilinium ion. The synthetic fix is to acetylate first: acetanilide's nitrogen is much less basic, so it survives the acid, and the amide is still ortho-para directing but moderated. Nitration then gives clean para product, and hydrolysis restores the amine.

Reagents can change the substrate before they react with it, and Advanced tests exactly that kind of reasoning throughout this chapter.

1. Making amines: which route gives which

Direct alkylation of ammonia is useless preparatively, because each product is a better nucleophile than the starting material, so the reaction runs on to give a mixture of primary, secondary, tertiary amines and the quaternary salt. Three routes avoid that, and each has its own carbon bookkeeping:

$\text{Hofmann}:{\text{RCONH}}_2\to {\text{RNH}}_2\left(\text{one carbon less}\right)$ $\text{nitrile reduction}:\text{RCN}\to {\text{RCH}}_2{\text{NH}}_2\left(\text{one carbon more}\right)$ $\text{Gabriel}:\text{RX}\to {\text{RNH}}_2\left(\text{same count}\right)$

Gabriel's synthesis fails for aryl halides, since it depends on a backside displacement that cannot occur at an $s{p}^2$ aromatic carbon.

Illustration 1

Suggest how to convert ethanamide into (a) ethylamine and (b) methylamine.

(b) Hofmann bromamide degradation with bromine and alkali gives methylamine, losing one carbon as carbonate.

(a) Reduce the amide instead, with lithium aluminium hydride, to give ethylamine with the carbon count intact.

The same starting material gives two different amines, and which one depends entirely on whether the carbonyl carbon is removed or reduced. Reading the carbon count of the answer identifies the required route immediately.

2. Telling primary, secondary and tertiary apart

The Hinsberg test is the most informative because it separates all three. A primary amine gives a sulphonamide that still has an $\text{N}-\text{H}$ bond, and that hydrogen is acidic enough for alkali to remove, so the product dissolves. A secondary amine's sulphonamide has no such hydrogen and stays as a solid. A tertiary amine has no hydrogen on nitrogen at all and cannot form a sulphonamide.

The carbylamine test is specific to primary amines, aliphatic or aromatic, and is remembered by its smell.

Illustration 2

Three unlabelled bottles contain propan-$1$-amine, N-methylethanamine and trimethylamine. Distinguish them with a single reagent.

Use benzenesulphonyl chloride followed by alkali.

Propan-$1$-amine gives a sulphonamide with an $\text{N}-\text{H}$ bond, which alkali deprotonates, so it dissolves.

N-methylethanamine gives a sulphonamide with no $\text{N}-\text{H}$ bond, which remains as an insoluble solid.

Trimethylamine has no hydrogen on nitrogen and does not react at all, so it remains as a separate liquid layer.

One reagent, three distinct observations. No other single test separates all three classes.

Illustration 3

Suggest a route from benzene to phenylmethanamine, and a second route to aniline, noting the carbon count in each.

Phenylmethanamine: chlorinate the side chain of toluene to give benzyl chloride, displace with cyanide to give phenylacetonitrile, and reduce with lithium aluminium hydride. The nitrile route adds a carbon, so toluene's seven become eight.

Aniline: nitrate benzene and reduce the nitro group with tin and hydrochloric acid. The carbon count is unchanged at six.

Nitrogen can be attached to a ring or to a side chain, and the routes are entirely different. Nitration attaches it to the ring directly; the nitrile route attaches it one carbon out.

3. Diazonium salts: the synthetic hub

Aromatic primary amines with nitrous acid at $0$ to $5^{\circ }$C give diazonium salts, which are the most versatile intermediates in aromatic chemistry. Aliphatic diazonium salts decompose at once, which is why the reaction is restricted to aryl amines.

The Sandmeyer reactions use copper(I) halides or cyanide; the Gattermann variant uses copper powder with the hydrogen halide. Iodide needs no catalyst at all, and fluoride is introduced through the tetrafluoroborate salt. Warming with water gives a phenol, and hypophosphorous acid removes the group entirely, replacing it with hydrogen.

That last reaction is what makes the diazonium route so powerful: a nitrogen substituent can be introduced purely to direct another group into position, and then removed.

Illustration 4

Suggest a synthesis of $1,3,5$-tribromobenzene from benzene.

Direct bromination cannot give this pattern, since bromine is ortho-para directing and would never place three groups all meta to one another.

Nitrate benzene, then reduce the nitro group to give aniline. The amino group is powerfully ortho-para directing, so brominating in water gives $2,4,6$-tribromoaniline immediately.

Diazotise with nitrous acid at $0$ to $5^{\circ }$C, then treat with hypophosphorous acid to replace the diazonium group by hydrogen.

The amino group was never wanted in the product. It was installed purely to direct the three bromines, then removed — which is the classic use of the diazonium route.

4. Coupling reactions and dyes

A diazonium ion is a weak electrophile, so it attacks only strongly activated rings — phenols and aromatic amines. The product is an azo compound, intensely coloured because the extended conjugation across the two rings absorbs in the visible.

The pH matters. Coupling with a phenol is done in mildly alkaline solution, which converts it to the more activated phenoxide without destroying the diazonium ion. Coupling with an amine is done in mildly acidic solution, acidic enough to keep the diazonium ion stable but not so acidic that the amine is fully protonated.

Attack occurs at the para position, or ortho if para is blocked.

Illustration 5

Explain why coupling of benzenediazonium chloride with phenol is carried out at pH $9$ to $10$ rather than in strong alkali or in acid.

In acid the phenol is not deprotonated, so the ring is not activated enough for the weakly electrophilic diazonium ion to attack.

In strong alkali the diazonium ion itself is converted to a diazotate, which is not electrophilic at all.

At pH $9$ to $10$ a useful concentration of phenoxide coexists with intact diazonium ion, and coupling proceeds.

Both reagents must survive the same conditions. Optimising a coupling is a matter of finding the pH window in which neither is destroyed.

Illustration 6

Suggest a synthesis of $m$-bromoaniline from benzene, and explain why the obvious order of steps fails.

Brominating first and then nitrating fails, because bromine is ortho-para directing and the nitro group would arrive at the wrong positions.

Nitrate benzene first to give nitrobenzene. The nitro group is meta directing, so brominating now places the bromine correctly.

Finally reduce the nitro group with tin and hydrochloric acid to give $m$-bromoaniline.

The nitrogen must be introduced as a nitro group and reduced last. Introducing it as an amine first would give the ortho and para products, since the amino group is a powerful ortho-para director.

5. Reduction of nitro compounds: the medium decides

The same nitro group gives entirely different products depending on the conditions:

The pattern is that acidic conditions carry the reduction all the way to the amine, neutral conditions stop it half way, and alkaline conditions allow two molecules to couple before reduction completes. The electrolytic case is the odd one: the hydroxylamine formed first rearranges under strong acid to $p$-aminophenol.

Illustration 7

Nitrobenzene is reduced under three different conditions to give aniline, N-phenylhydroxylamine and azobenzene. State the conditions for each and explain the pattern.

Aniline: tin and hydrochloric acid, or iron with hydrochloric acid. Acidic conditions supply protons freely and drive the reduction to completion.

N-Phenylhydroxylamine: zinc dust with ammonium chloride in neutral solution. Without a plentiful proton supply the reduction stalls at the intermediate stage.

Azobenzene: sodium arsenite or zinc with sodium hydroxide. In alkali two partially reduced molecules condense before reduction can finish.

The nitro group is the same in all three. Only the medium differs, which is why these products are so often set as a single comparison question.

6. Aniline's other restrictions

Beyond nitration, aniline is unusable in two further reactions, and both failures have the same origin.

Friedel-Crafts reactions fail because the nitrogen lone pair coordinates to the aluminium chloride catalyst. The resulting complex bears a positive charge on nitrogen, which deactivates the ring exactly as the anilinium ion does.

Bromination succeeds but cannot be controlled: aniline is so activated that bromine water immediately gives the $2,4,6$-tribromide. Monobromination requires acetylation first, which moderates the donation enough for a single substitution.

Acetylation is therefore the standard protecting strategy throughout aniline chemistry, and hydrolysis afterwards restores the amine.

Illustration 8

Suggest how to prepare $p$-bromoaniline from aniline.

Direct bromination gives $2,4,6$-tribromoaniline, because the ring is too activated to stop at one substitution.

Acetylate the amine first with ethanoic anhydride to give acetanilide. The amide nitrogen donates far less, since its lone pair is partly delocalised onto the carbonyl.

Brominate the acetanilide; the moderated ring now gives clean monosubstitution, mainly para.

Hydrolyse the amide with aqueous acid or alkali to recover the amine as $p$-bromoaniline.

Protection, react, deprotect. This three-step pattern recurs whenever a group is too reactive to be used directly.

Illustration 9

Explain why an aliphatic primary amine cannot be diazotised usefully, while an aromatic one can.

Both form a diazonium ion initially.

An aliphatic diazonium ion has an excellent leaving group, molecular nitrogen, attached to an $s{p}^3$ carbon, so it loses nitrogen immediately to give a carbocation. That cation then does whatever carbocations do, giving alcohols, alkenes and rearranged products.

An aromatic diazonium ion would have to leave behind an aryl cation, which is far too unstable to form at low temperature, so the ion survives.

The stability of the leaving carbon is what makes the difference, and it is why the whole synthetic hub exists only for aromatic amines.

Illustration 10

Give the products when propan-$1$-amine and N-methylpropan-$1$-amine each react with nitrous acid.

Propan-$1$-amine is primary and aliphatic. It gives an unstable diazonium ion that loses nitrogen at once, producing propan-$1$-ol together with rearranged and eliminated products, and brisk effervescence of nitrogen.

N-Methylpropan-$1$-amine is secondary, so it cannot form a diazonium ion. Instead the nitrogen is nitrosated to give a yellow oily N-nitrosoamine.

The visible difference is the gas. Brisk effervescence identifies a primary aliphatic amine, while a yellow oil separating out identifies a secondary one.

Illustration 11

Arrange in order of increasing basicity: aniline, $p$-toluidine, $p$-nitroaniline, cyclohexylamine.

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

Aniline follows, its lone pair delocalised into the ring.

$p$-Toluidine has an electron-donating methyl group opposing that delocalisation slightly.

Cyclohexylamine is strongest, having no ring to delocalise into at all.

Order: $p$-nitroaniline $<$ aniline $<$ $p$-toluidine $<$ cyclohexylamine.

Any conjugation with the ring costs basicity, because protonation forces the lone pair out of that conjugation.

Illustration 12

Suggest a synthesis of benzoic acid from aniline.

Diazotise aniline with nitrous acid at $0$ to $5^{\circ }$C.

Treat the diazonium salt with copper(I) cyanide, a Sandmeyer reaction, to give benzonitrile.

Hydrolyse the nitrile with aqueous acid to give benzoic acid.

The carbon count has increased by one. The nitrile route is the standard way to lengthen a chain by a single carbon, in aromatic and aliphatic chemistry alike.

Illustration 13

Explain why methylamine is a stronger base than ammonia but aniline is much weaker, using the same principle for both.

Basicity depends on how available the nitrogen lone pair is.

In methylamine an electron-donating methyl group pushes density towards nitrogen, making the pair more available and the amine a stronger base than ammonia.

In aniline the lone pair is delocalised into the aromatic ring, so it is substantially less available. Protonating also destroys that delocalisation, adding a further energetic cost.

Availability of the lone pair is the single controlling quantity, and it accounts for a difference of about five orders of magnitude between the two directions.

Summary

• Nitration of aniline gives meta product because the acid protonates it first, and $-{\text{NH}}_3^{+}$ is deactivating and meta directing.
• The fix is to acetylate, nitrate, then hydrolyse — the standard protect, react, deprotect pattern.
• Direct alkylation of ammonia is useless: each product is a better nucleophile than the last.
• Carbon bookkeeping: Hofmann loses one carbon, nitrile reduction gains one, Gabriel keeps the count.
• Gabriel's synthesis fails for aryl halides, since backside attack at an $s{p}^2$ carbon is impossible.
• Introduce ring nitrogen as a nitro group when meta substitution is wanted, and reduce it last.
• Hinsberg separates all three amine classes; carbylamine is specific to primary amines.
• Nitrous acid: primary aliphatic gives brisk nitrogen, secondary gives a yellow oily nitrosamine, tertiary gives a salt.
• Only aromatic amines give useful diazonium salts, because an aryl cation is too unstable to form.
• The diazonium hub: Sandmeyer for $\text{Cl}$, $\text{Br}$ and $\text{CN}$; $\text{KI}$ needs no catalyst; ${\text{HBF}}_4$ gives fluoride; water gives phenol; ${\text{H}}_3{\text{PO}}_2$ gives hydrogen.
• Replacing the group by hydrogen lets an amine be installed purely to direct, then removed — the route to $1,3,5$-tribromobenzene.
• Coupling needs mild alkali for phenols and mild acid for amines, so that both partners survive.
• Nitro reduction: acidic gives the amine, neutral gives the hydroxylamine, alkaline gives coupled products.
• Friedel-Crafts fails on aniline because the lone pair coordinates to the aluminium chloride, deactivating the ring.