Organic Compounds Containing Nitrogen (Amines)

Ethanamide, $\mathrm{C}{\mathrm{H}}_3\mathrm{CON}{\mathrm{H}}_2$, has an $\mathrm{N}{\mathrm{H}}_2$ group. Add acid to it. Where does the proton go?

Nitrogen carries the lone pair and nitrogen is the less electronegative of the two, so it should hold its electrons more loosely and take the proton. The oxygen has lone pairs too, but oxygen grips them harder.

The proton goes to the oxygen.

The nitrogen lone pair in an amide is not sitting on nitrogen at all. It is delocalised into the carbonyl, and protonating the nitrogen would tear that delocalisation apart. Protonating the oxygen leaves it intact.

So the amide is barely basic in any sense. Compare the three:

Ethylamine is roughly ten billion times more basic than ethanamide, and the nitrogen atom is the same element in both.

That is the whole chapter. Nitrogen has a lone pair, and everything here is a question about whether it is available.

Two things take it away. Delocalisation into a ring or a carbonyl ties it up, which is why aniline is a weak base and an amide is not one at all. And bulk around the nitrogen physically blocks approach, which is where the surprising aqueous basicity order comes from.

1. Classification, Structure and Nomenclature

An amine is primary, secondary or tertiary by how many carbon groups are attached to nitrogen, not to carbon as for alcohols and halides, which is a routine source of confusion.

The nitrogen is $s{p}^3$ and pyramidal, with the lone pair in the fourth position exactly as in ammonia, and the C-N-C angle a little under tetrahedral.

In common nomenclature the alkyl groups are named and "amine" appended, as in ethylmethylamine. In IUPAC the amine is a suffix, so $\mathrm{C}{\mathrm{H}}_3\mathrm{C}{\mathrm{H}}_2\mathrm{N}{\mathrm{H}}_2$ is ethanamine, with N-prefixes for substituents on nitrogen.

Primary amines boil above comparable ethers and below comparable alcohols, since nitrogen is less electronegative than oxygen and its hydrogen bonds are weaker. Tertiary amines have no N-H bond, cannot hydrogen bond to each other at all, and boil lowest of the three.

Illustration 1

The C-N bond in an amine is 147 pm. In an amide it is 132 pm, and rotation about it is restricted enough to be observed by spectroscopy. Explain, and say what it means for proteins.

A single C-N bond is 147 pm and a double one would be about 127 pm. The amide's 132 pm sits much closer to the double.

The cause is the same delocalisation that killed the amide's basicity. The nitrogen lone pair flows into the carbonyl, so the real structure has substantial contribution from a form with a $\mathrm{C}={\mathrm{N}}^{+}$ double bond and a negative oxygen.

$\mathrm{R}-\mathrm{C}(=\mathrm{O})-\mathrm{N}{\mathrm{H}}_2⟷\mathrm{R}-\mathrm{C}(-{\mathrm{O}}^{-})=\mathrm{N}{\mathrm{H}}_2^{+}$

A partial double bond cannot be rotated without breaking the sideways overlap, so the whole $\mathrm{O}=\mathrm{C}-\mathrm{N}$ unit is locked planar.

The consequence is structural biology. Every peptide bond in every protein is an amide, so every one of them is flat and rigid. A protein chain cannot rotate freely along its backbone: it can only turn at the two bonds either side of each planar unit.

That restriction is what makes protein folding a tractable problem rather than an impossible one, and it is why the alpha helix and beta sheet exist as recurring shapes. One delocalisation explains the amide's basicity, its bond length and the architecture of every protein you contain.

Illustration 2

Classify tert-butylamine, $(\mathrm{C}{\mathrm{H}}_3{)}_3\mathrm{C}-\mathrm{N}{\mathrm{H}}_2$, and trimethylamine, $(\mathrm{C}{\mathrm{H}}_3{)}_3\mathrm{N}$. Both names suggest three methyl groups. Are both tertiary amines?

Count what is attached to the nitrogen, and nothing else.

In tert-butylamine the nitrogen carries one carbon group and two hydrogens, so it is a primary amine. The word tert is describing the carbon skeleton — the carbon that bears the nitrogen has three other carbons on it — and says nothing about the nitrogen at all.

In trimethylamine the nitrogen carries three carbon groups, so it is a genuine tertiary amine.

The trap is that the same vocabulary counts a different atom in different families. For alcohols and halides, primary through tertiary describes the carbon bearing the functional group. For amines it describes the nitrogen. So tert-butyl alcohol really is a tertiary alcohol while tert-butylamine is a primary amine, and the two names look identical.

This matters beyond nomenclature: the Hinsberg test and the carbylamine test both sort amines by how many hydrogens remain on nitrogen, so a wrong classification gives a wrong prediction immediately.

2. Preparation

Reduction of nitro compounds with tin and HCl, or by catalytic hydrogenation, is the standard route to aromatic amines, and is how aniline is made from nitrobenzene.

Ammonolysis of a halide replaces halogen with an amino group, but it does not stop: the primary amine produced is itself nucleophilic and reacts on, giving a mixture of primary, secondary and tertiary amines and the quaternary salt. A large excess of ammonia favours the primary product without giving it cleanly.

Reduction of nitriles or amides with $\mathrm{LiAl}{\mathrm{H}}_4$ gives a primary amine, and the nitrile route usefully adds a carbon.

Two routes that give a pure primary amine

Gabriel phthalimide synthesis uses potassium phthalimide, whose nitrogen carries only one hydrogen, so alkylation can happen only once. Hydrolysis then releases a pure primary amine. It fails for aromatic amines, because the alkylation step is an ${\mathrm{S}}_{\mathrm{N}}2$ displacement and an aryl halide will not undergo one.

Hofmann bromamide degradation converts an amide to a primary amine with one carbon fewer, using bromine and alkali, the carbonyl carbon leaving as carbonate.

Illustration 3

Convert butan-1-ol into pentan-1-amine, and butanamide into propan-1-amine.

Count carbons before choosing anything. That single step decides the route every time.

Butan-1-ol to pentan-1-amine: four carbons to five, so gain one. Only the nitrile route adds a carbon.

${\mathrm{C}}_4{\mathrm{H}}_9\mathrm{OH}\stackrel{\mathrm{HBr}}{\to }{\mathrm{C}}_4{\mathrm{H}}_9\mathrm{Br}\stackrel{\mathrm{KCN}}{\to }{\mathrm{C}}_4{\mathrm{H}}_9\mathrm{CN}\stackrel{\mathrm{LiAl}{\mathrm{H}}_4}{\to }{\mathrm{C}}_4{\mathrm{H}}_9\mathrm{C}{\mathrm{H}}_2\mathrm{N}{\mathrm{H}}_2$

Butanamide to propan-1-amine: four carbons to three, so lose one. Only Hofmann removes a carbon.

${\mathrm{C}}_3{\mathrm{H}}_7\mathrm{CON}{\mathrm{H}}_2\stackrel{\mathrm{B}{\mathrm{r}}_2,\mathrm{KOH}}{\to }{\mathrm{C}}_3{\mathrm{H}}_7\mathrm{N}{\mathrm{H}}_2+{\mathrm{K}}_2\mathrm{C}{\mathrm{O}}_3$

Trap. Count the carbons in the question before reading the options. A conversion that changes the carbon count has already named its own route, and no other reasoning is needed.

Nitro compounds and their selective reduction

The pattern is worth noticing rather than memorising as four facts: acidic and vigorous conditions take the reduction all the way, while neutral and alkaline conditions stop part way or let two molecules couple.

Cyanides and isocyanides

Potassium cyanide is ionic, so the free cyanide ion attacks through its carbon, which carries the greater electron density, giving a nitrile. Silver cyanide is largely covalent, so only the nitrogen lone pair is freely available, giving an isocyanide.

The products differ completely: a nitrile hydrolyses to a carboxylic acid, an isocyanide to a primary amine and methanoic acid. Isocyanides also carry the extremely offensive smell that makes the carbylamine test so recognisable.

Illustration 4

You need to make aniline and benzylamine. Which of the Gabriel phthalimide synthesis and the Hoffmann bromamide degradation can deliver each?

Gabriel cannot make aniline. Its key step is an ${\mathrm{S}}_{\mathrm{N}}2$ displacement on the halide by the phthalimide anion, and an aryl halide will not undergo ${\mathrm{S}}_{\mathrm{N}}2$ at all — the carbon is sp² and the ring blocks backside attack. Gabriel is restricted to amines whose nitrogen sits on an sp³ carbon.

Gabriel makes benzylamine easily, because benzyl bromide is an excellent ${\mathrm{S}}_{\mathrm{N}}2$ substrate, and the product is guaranteed primary since the phthalimide nitrogen carries only one replaceable hydrogen.

Hoffmann makes aniline, starting from benzamide. The degradation removes the carbonyl carbon, so ${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{CON}{\mathrm{H}}_2$ with seven carbons gives ${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{N}{\mathrm{H}}_2$ with six.

That carbon bookkeeping is the part worth carrying, and the figure above collects it. Hoffmann shortens the chain by one, nitrile reduction lengthens it by one, and nitro reduction and Gabriel leave it alone. A synthesis question that changes the carbon count has already told you which route it wants.

3. Basicity: The Central Topic

Aromatic against aliphatic

Aniline's $\mathrm{p}{K}_b$ is 9.38 against methylamine's 3.38, a factor of ten thousand.

The lone pair on aniline's nitrogen is delocalised into the ring, contributing to its pi system. A delocalised lone pair is not sitting on nitrogen waiting for a proton, and protonating it would destroy the delocalisation. Supporting this, aniline's nitrogen is attached to an $s{p}^2$ carbon, more electronegative than an $s{p}^3$ one.

Substituents follow directly. Donating groups such as methyl or methoxy make aniline more basic; withdrawing groups such as nitro make it far less so.

The aqueous order, and why it looks wrong

In the gas phase the order is exactly what induction predicts: tertiary > secondary > primary > ammonia.

In water, where $\mathrm{p}{K}_b$ values are measured, it changes. Three effects compete rather than one: induction favours more substitution, solvation of the cation favours less substitution because more N-H bonds mean better hydrogen bonding to water, and steric bulk hinders both protonation and solvation.

Illustration 5

The methylamine and ethylamine series give different aqueous orders. Explain why one swap occurs.

$\text{methyl: }(\mathrm{C}{\mathrm{H}}_3{)}_2\mathrm{NH}>\mathrm{C}{\mathrm{H}}_3\mathrm{N}{\mathrm{H}}_2>(\mathrm{C}{\mathrm{H}}_3{)}_3\mathrm{N}>\mathrm{N}{\mathrm{H}}_3$

$\text{ethyl: }({\mathrm{C}}_2{\mathrm{H}}_5{)}_2\mathrm{NH}>({\mathrm{C}}_2{\mathrm{H}}_5{)}_3\mathrm{N}>{\mathrm{C}}_2{\mathrm{H}}_5\mathrm{N}{\mathrm{H}}_2>\mathrm{N}{\mathrm{H}}_3$

The secondary amine leads in both, which is where induction and solvation happen to balance. What moves is the tertiary amine, from third place in the methyl series to second in the ethyl series.

An ethyl group is a better electron donor than a methyl group, having one more carbon contributing $+I$. So the inductive term is larger throughout the ethyl series, and the tertiary amine, which has three of them, gains most.

The solvation penalty is unchanged: a trialkylammonium ion has only one N-H bond whichever alkyl group it carries. So the same penalty is now being weighed against a bigger inductive gain, and the tertiary amine climbs one place.

The examinable point is never the exact order. It is that three effects compete, that only solvation explains why the gas-phase and aqueous orders differ, and that a question about trimethylamine in water is asking about solvation rather than induction.

Illustration 6

Cyclohexylamine has $\mathrm{p}{K}_b$ 3.3 and aniline 9.4 — a factor of roughly $10^6$ in basicity. Both are primary amines carrying a single six-membered ring. Account for the gap.

In cyclohexylamine the ring is a plain saturated alkyl group. The nitrogen is sp³ with its lone pair localised on the nitrogen alone and fully available, and the ring's weak electron-releasing effect makes it a slightly stronger base than ammonia itself at $\mathrm{p}{K}_b$ 4.75.

In aniline the nitrogen sits on a benzene ring, and its lone pair is delocalised into that ring. A lone pair spread over four atoms is not sitting on the nitrogen waiting to accept a proton.

Now look at what protonation costs. Forming the anilinium ion ties the lone pair up in a bond to hydrogen, which destroys the delocalisation entirely. The neutral amine enjoys a stabilisation the cation cannot have.

That is the general shape of every basicity argument in this chapter. Anything stabilising the base weakens it, and anything stabilising the conjugate acid strengthens it. Here both effects push the same way: aniline is stabilised and anilinium is not, so the equilibrium sits far to the left.

4. Reactions of Amines

Alkylation with a haloalkane gives the next higher amine and eventually the quaternary ammonium salt, the same lack of control as in ammonolysis. Acylation with an acid chloride or anhydride gives an amide, and works for primary and secondary amines but not tertiary ones, which have no N-H bond.

Three tests worth knowing cold

Carbylamine test: chloroform with alcoholic KOH. Primary amines only give an isocyanide, recognised by an extremely offensive smell.

Hinsberg test: benzenesulphonyl chloride separates all three classes in one operation.

Nitrous acid distinguishes aliphatic from aromatic primary amines. An aliphatic one gives an unstable diazonium salt that decomposes at once, releasing nitrogen visibly. An aromatic one at 273 to 278 K gives a salt that survives.

Illustration 7

Ethanamide has an $\mathrm{N}{\mathrm{H}}_2$ group. Predict its behaviour in the carbylamine test and the Hinsberg test.

Both give nothing, and for the reason this chapter opened on.

Every one of these tests requires the nitrogen to act as a nucleophile: carbylamine needs it to attack dichlorocarbene, Hinsberg needs it to attack a sulphonyl chloride. An amide's lone pair is delocalised into the carbonyl and is not available to attack anything.

So an amide is neither basic nor nucleophilic, despite carrying what looks like a primary amino group.

The practical use of that fact is the acetanilide strategy. Acetylating aniline converts it to an amide, deliberately switching the lone pair off, and hydrolysis afterwards switches it back on. The tests and the strategy are the same chemistry read in opposite directions.

Aniline and the ring

The lone pair activates the ring so strongly that aniline with bromine water gives 2,4,6-tribromoaniline immediately, with no catalyst.

Illustration 8

Nitrating aniline directly gives a substantial amount of the meta product, even though $\mathrm{N}{\mathrm{H}}_2$ is one of the strongest ortho-para directors known. Explain, and give a route to pure 4-nitroaniline.

The nitrating mixture is concentrated nitric acid in concentrated sulphuric acid. Aniline is a base, and in that medium it is protonated.

${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{N}{\mathrm{H}}_2+{\mathrm{H}}^{+}\to {\mathrm{C}}_6{\mathrm{H}}_5\mathrm{N}{\mathrm{H}}_3^{+}$

The anilinium ion is an entirely different substituent. Its nitrogen has no lone pair left to donate and carries a full positive charge, making it strongly deactivating and meta directing.

So the reaction mixture contains two species directing to opposite positions, and the product is a mixture with a large meta fraction.

The fix is to remove the amine's basicity before nitrating. Acetylate first.

${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{N}{\mathrm{H}}_2\stackrel{(\mathrm{C}{\mathrm{H}}_3\mathrm{CO}{)}_2\mathrm{O}}{\to }{\mathrm{C}}_6{\mathrm{H}}_5\mathrm{NHCOC}{\mathrm{H}}_3\stackrel{\mathrm{HN}{\mathrm{O}}_3/{\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4}{\to }\text{4-nitroacetanilide}\stackrel{{\mathrm{H}}_3{\mathrm{O}}^{+}}{\to }\text{4-nitroaniline}$

Acetanilide is not basic enough to be protonated, so it survives the nitrating mixture intact, and it is still activating enough to direct ortho and para, only more moderately. The bulky acetyl group also favours para over ortho.

The same protective strategy, for the same reason, gives monobromination instead of the tribromo product. Friedel-Crafts fails on aniline for a third variation of the theme: the lone pair binds the aluminium chloride catalyst, creating a positively charged nitrogen that both deactivates the ring and consumes the catalyst.

5. Diazonium Salts

Aromatic primary amine with $\mathrm{NaN}{\mathrm{O}}_2$ and HCl at 273 to 278 K gives a benzenediazonium salt.

Aromatic diazonium salts survive at low temperature and aliphatic ones do not, because the positive charge is delocalised into the ring, which no aliphatic system offers. Even so the salt decomposes above about 278 K, so it is made in an ice bath and used immediately.

Why they matter

The diazonium group is the most versatile leaving group in aromatic chemistry, because it departs as nitrogen gas: stable, gaseous and irreversible.

This solves what direct substitution cannot. Iodine and fluorine cannot be installed by electrophilic substitution at all, and the hypophosphorous entry deletes the amino group entirely, which means it can be installed purely to direct a substitution and then removed.

Illustration 9

Make 1,3,5-tribromobenzene from benzene. Direct bromination cannot do it. Why not, and what does?

Bromine is ortho-para directing, so once one bromine is on the ring the second goes ortho or para to it and the third follows suit. Direct tribromination gives 1,2,4-tribromobenzene. The 1,3,5 pattern requires meta directing, and no halogen directs meta.

So install a group that will hold the three bromines in the right places, then delete it.

${\mathrm{C}}_6{\mathrm{H}}_6\stackrel{\mathrm{HN}{\mathrm{O}}_3/{\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4}{\to }{\mathrm{C}}_6{\mathrm{H}}_5\mathrm{N}{\mathrm{O}}_2\stackrel{Sn/HCl}{\to }{\mathrm{C}}_6{\mathrm{H}}_5\mathrm{N}{\mathrm{H}}_2$

${\mathrm{C}}_6{\mathrm{H}}_5\mathrm{N}{\mathrm{H}}_2\stackrel{\mathrm{B}{\mathrm{r}}_2\mathrm{water}}{\to }\text{2,4,6-tribromoaniline}\stackrel{\mathrm{NaN}{\mathrm{O}}_2/HCl,273\mathrm{K}}{\to }\text{diazonium salt}\stackrel{{\mathrm{H}}_3\mathrm{P}{\mathrm{O}}_2}{\to }\text{1,3,5-tribromobenzene}$

The amino group did exactly one job: it activated the ring so powerfully that all three of its ortho and para positions brominated at once, in bromine water with no catalyst. Those three positions are 2, 4 and 6, which become 1, 3 and 5 the moment the nitrogen leaves.

A group installed to be deleted is a legitimate synthetic move, and hypophosphorous acid is what makes it available. Any question asking for a substitution pattern that directing rules forbid is asking for this.

Illustration 10

Why does the KI conversion need no copper catalyst when the chloride and bromide conversions do?

Sandmeyer and Gattermann both need copper because the reaction is a radical process: copper(I) transfers an electron to the diazonium ion, which loses nitrogen to give an aryl radical, and the halide is delivered from the copper.

Chloride and bromide cannot start that electron transfer themselves. Iodide can, because it is the most easily oxidised of the halides, its outer electrons being furthest out and least tightly held.

$\mathrm{Ar}{\mathrm{N}}_2^{+}+{\mathrm{I}}^{-}\to \mathrm{Ar}\cdot +{\mathrm{N}}_2+\mathrm{I}\cdot$

Iodide performs the copper's job unaided, so potassium iodide alone suffices.

This is the same property met in the halogen chapters, where iodide is the best nucleophile in water and the strongest reducing agent of the halides, all for the identical reason: large, polarisable and holding its electrons loosely.

Coupling reactions

Diazonium salts couple with electron-rich aromatics to give intensely coloured azo compounds, which is the basis of the dye industry.

Illustration 11

Coupling with phenol is run at pH 9 to 10 and coupling with aniline at pH 4 to 5. Both are narrow windows, and getting either wrong stops the reaction. Explain both.

The diazonium ion is a weak electrophile, so it can only attack a strongly activated ring, and it is also fragile.

With phenol, raise the pH. Mild alkali converts phenol to phenoxide, which carries a full negative charge and is far more nucleophilic than phenol itself. Go too alkaline, though, and the diazonium ion is converted to a diazotate, which is not an electrophile at all.

With aniline, lower the pH, but only a little. The medium must be acidic enough to keep the diazonium salt stable, and yet if it becomes properly acidic the amine is protonated to anilinium, whose lone pair is gone and whose ring is deactivated. pH 4 to 5 keeps most of the aniline unprotonated while the salt survives.

$\text{too alkaline: no electrophile}\text{too acidic: no nucleophile}$

Both windows are the same problem seen from opposite sides: one reagent needs to be electron-rich and the other needs to survive, and pH moves both at once in opposite directions.

The colour has a structural cause. Coupling joins two aromatic rings through an $\mathrm{N}=\mathrm{N}$ double bond, creating one continuous conjugated system. A larger conjugated system absorbs longer wavelengths, and once that absorption enters the visible range the compound is coloured. Adding donating groups at one end and withdrawing groups at the other extends the effect and shifts the shade, which is how a dye chemist tunes a colour.

Uses

Aliphatic amines are intermediates in pharmaceutical manufacture, and quaternary ammonium salts are the active ingredients in fabric softeners and many disinfectants. Aniline is the feedstock for dyes, rubber processing chemicals and paracetamol. Azo compounds account for the majority of synthetic dyes and pigments, and methyl orange, made exactly this way, is a standard acid-base indicator.

Summary

Everything here is a question about whether nitrogen's lone pair is available. Available, it is a base, a nucleophile and a ring activator; delocalised or blocked, it is none of those.

An amide protonates on oxygen, not nitrogen, because that preserves the delocalisation, and it is about ten billion times less basic than an amine. The same delocalisation shortens the amide C-N bond to 132 pm and locks it planar, which is why every peptide bond in every protein is flat.

Basicity runs aliphatic amine, aniline, amide, at $\mathrm{p}{K}_b$ near 3.3, 9.4 and 14.5, and substituents shift aniline in the direction their electronic effect predicts.

Carbon count names the preparation: one more carbon means the nitrile route, one fewer means Hofmann bromamide degradation, and the same number means Gabriel or a direct reduction. Gabriel fails for aryl amines because its alkylation step is ${\mathrm{S}}_{\mathrm{N}}2$.

In water three effects compete. Induction favours more alkyl groups; solvation and sterics favour fewer. The gas-phase order is purely inductive, and the aqueous order differs between methyl and ethyl series only because ethyl donates better.

Carbylamine detects primary amines alone, Hinsberg separates all three in one operation, and an amide responds to neither because its lone pair is unavailable. That same unavailability is exploited deliberately in the acetanilide strategy.

Aniline nitrates partly meta because the acid protonates it to the deactivating, meta-directing anilinium ion, so acetylate first. Friedel-Crafts fails on aniline because the lone pair binds the catalyst.

Diazonium salts are the hub of aromatic synthesis, since the group leaves as nitrogen gas. They install iodine and fluorine, which electrophilic substitution cannot, and hypophosphorous acid deletes the group entirely, which is how 1,3,5-tribromobenzene is made. Iodide needs no copper because it can transfer the electron itself.

Azo coupling needs pH 9 to 10 with phenol to form the phenoxide and pH 4 to 5 with aniline to avoid protonating it, and the colour comes from the extended conjugation across both rings.