Reactions of Amines

One Lone Pair, Two Jobs: Base and Nucleophile

Everything an amine does traces back to a single feature you already know: the lone pair sitting on nitrogen. That lone pair is what makes an amine a base — it reaches out and grabs a proton — and it is also what makes the amine a nucleophile, reaching out to grab a partly positive carbon. Base and nucleophile are not two different talents; they are the same lone pair pointed at two different targets. Aim it at H+ and you call it basicity; aim it at a carbon and you call it nucleophilicity. The whole of amine chemistry is just bookkeeping on where that lone pair goes.

As a base, a typical alkylamine has a conjugate acid (R-NH3+) with a pKa around 10-11 — so the amine itself is a decent base, far stronger than aniline, whose lone pair is partly soaked into the ring. This basicity is not just a number on a chart; it has a practical sting. Protonate an amine and you switch off its nucleophile, because R-NH3+ has no free lone pair left to attack with. That single fact — a protonated amine is a dead nucleophile — is the hidden hinge behind half the reactions in this guide, the same trade-off you saw govern the bell-shaped pH curve of imine formation.

Acylation: Amines into Amides and Sulfonamides

Point that nucleophilic nitrogen at the carbon of an acyl chloride and you run a textbook nucleophilic acyl substitution: the amine's lone pair attacks the carbonyl carbon, the tetrahedral intermediate collapses, and chloride leaves. The product is an [[amide|amide]] — a C(=O)-N linkage, the same bond that stitches every protein together. Because the amine is consumed as it reacts (it becomes an amide, no longer basic and nucleophilic), you generally add a second base, like pyridine or a little extra amine, to mop up the HCl that is released so it does not just protonate your starting amine into uselessness.

Swap the carbon-based acyl chloride for a sulfur-based one — a sulfonyl chloride, R-SO2-Cl — and the very same logic plays out. The nitrogen attacks the electrophilic sulfur, chloride leaves, and you get a [[sulfonamide|sulfonamide]], an R-SO2-N linkage. This is not a curiosity: the original 'sulfa drugs' that launched the antibiotic era are sulfonamides, and the S-N bond they contain is exactly the bond formed in this reaction. The pattern to carry away is that amines acylate readily at any hungry electrophilic centre that carries a good leaving group — whether that centre is a carbonyl carbon or a sulfonyl sulfur.

Written as plain relations, the two acylations are twins: 'R-NH2 + R'-C(=O)-Cl -> R'-C(=O)-NH-R + HCl' for the amide, and 'R-NH2 + R'-SO2-Cl -> R'-SO2-NH-R + HCl' for the sulfonamide. In both, the nitrogen's lone pair attacks the electrophilic centre (the carbonyl carbon or the sulfonyl sulfur), chloride departs as the leaving group, and a proton is lost to a spare base. One move, two flavours of product — and as the next section shows, that same acylation becomes a clever diagnostic tool.

The Hinsberg Test: Telling 1°, 2°, and 3° Amines Apart

Sulfonamide formation does more than make drugs — it gives chemists a clever way to tell a primary, secondary, and tertiary amine apart in the absence of a spectrometer, called the [[hinsberg-test|Hinsberg test]]. The trick rests on a simple count: how many N-H bonds does the amine bring? Treat your unknown amine with benzenesulfonyl chloride in aqueous base (NaOH) and read the result by eye. The logic is the same nucleophilic substitution you just learned, watched three times with three different amines.

1. Primary amine (R-NH2, two N-H). It reacts to give a sulfonamide that still keeps ONE N-H. That remaining N-H sits next to the electron-pulling SO2 group, so it is acidic enough for the NaOH to deprotonate. The resulting anion is charged and water-soluble: the mixture stays a clear solution. Add acid afterward and a solid sulfonamide crashes out.
2. Secondary amine (R2N-H, one N-H). It also reacts and forms a sulfonamide — but now the nitrogen has NO N-H left after substitution (its one H went into making the bond... actually both R groups stay, so the product N carries no acidic H). With no N-H to deprotonate, the sulfonamide is neutral and insoluble: a solid precipitate appears directly, even in base.
3. Tertiary amine (R3N, zero N-H). Its nitrogen can still attack the sulfonyl chloride, but the intermediate it would form has a positively charged nitrogen with no N-H to lose, so it simply falls back apart — no stable sulfonamide forms at all. In base the tertiary amine just sits there unreacted: no precipitate while basic, and on acidifying it dissolves (as its ammonium salt).

So you read three different stories: a clear solution that precipitates only on acidifying (primary), a precipitate that forms straight away in base (secondary), and an oily layer that stays unreacted in base but dissolves on acidifying (tertiary). Be honest about the test's limits, though — the classic descriptions are idealized, real amines can blur the boundaries, and modern labs reach for IR and NMR instead. The Hinsberg test earns its place not as a workhorse but as a beautiful illustration that counting N-H bonds predicts behaviour: the same count that decided imine versus enamine now decides solution versus precipitate.

Hofmann Elimination: When the Less Substituted Alkene Wins

A neutral amine refuses to leave a molecule — an amide ion, R2N-, is a terrible leaving group, far too unstable. But there is a way to bribe the nitrogen into leaving: keep methylating it until it carries four carbon groups and a positive charge, a quaternary ammonium ion (exhaustive methylation with excess CH3I does this). Now the nitrogen leaves as a neutral trimethylamine, R3N, which is a perfectly respectable leaving group. Convert the salt to its hydroxide form, heat it, and hydroxide acts as a base to pull off a beta hydrogen, triggering an E2 elimination that ejects the amine and forms an alkene. This is the [[hofmann-elimination|Hofmann elimination]].

Here is the twist that makes this reaction famous. Back in the elimination rung you learned Zaitsev's rule: ordinary eliminations favour the MORE substituted, more stable alkene. Hofmann elimination breaks that rule — it gives the LESS substituted alkene instead, exactly what Hofmann's rule predicts. The reason is honest and physical: the leaving ammonium group is enormous and bulky. To reach the proton next to the more substituted (more crowded) carbon, the base would have to push into a thicket of alkyl groups around that fat positive nitrogen. It is far easier for the base to grab a proton on the least hindered carbon at the rim — usually a terminal CH3 — so the double bond forms there, giving the terminal alkene.

Cope Elimination: A Quieter Route to the Same Place

The [[cope-elimination|Cope elimination]] reaches that same less-substituted alkene, but it skips the strong base and the harsh conditions of the Hofmann route. First you oxidize a tertiary amine to an amine oxide (R3N+-O-) using hydrogen peroxide — nitrogen picks up an oxygen and a formal positive charge while the oxygen carries the negative. Then you just heat it gently. No external base is needed at all, because the oxide's own oxygen reaches over and plucks a neighbouring beta hydrogen from within the same molecule.

Because the abstracting oxygen and the departing nitrogen are tethered together, the whole thing happens in one smooth concerted step through a flat, five-membered ring of atoms — the oxygen reaching for the beta H, the C-H and C-N bonds breaking, and the new C=C forming all at once. This is called a SYN elimination: unlike E2, which strictly demands the H and leaving group be anti-periplanar (on opposite sides), the Cope requires the H and the amine oxide to be on the SAME side, because they must meet inside that little ring. The result is the same Hofmann-style terminal alkene, made under mild, base-free heating — handy when a sensitive molecule could not survive the strong base or high temperatures of a true Hofmann elimination.

The Enamine: A Nitrogen-Stabilized Nucleophile

There is one more reaction where the amine does not stay as a spectator base but becomes the engine of carbon-carbon bond-making: the [[enamine-formation|enamine]]. You met enamines in the carbonyl rung as the product of a secondary amine plus a carbonyl (C=C-N). What matters here is WHY they are so useful. The nitrogen's lone pair can push into the neighbouring C=C, spilling electron density all the way out to the far carbon. That far carbon becomes electron-rich and genuinely nucleophilic — a soft, neutral cousin of the enolate you would otherwise have to make with a strong base from keto-enol chemistry.

   N:               +N
    \               ||
     C = C   <-->   C - C(-)     the lone pair pushes the
                                  far carbon electron-rich

use:  ketone -> enamine -> alkylate far C -> hydrolyze -> alpha-substituted ketone

This is the heart of the Stork enamine synthesis. Take a ketone, condense it with a secondary amine to form an enamine, and now its alpha carbon is nucleophilic without any strong base in the flask. Let that carbon attack an electrophile — an alkyl halide, or the C=C of a Michael acceptor — and you forge a new carbon-carbon bond right at the alpha position. A final splash of water hydrolyzes the C=N-bearing iminium back to a carbonyl (remember, that condensation was reversible), handing you an alpha-substituted ketone and giving the amine back unchanged. The amine acted as a temporary, removable activator: it lent the carbon its nucleophilic punch, then quietly left.

Step back and the whole guide rhymes. Whether the amine grabs a proton as a base, attacks a carbonyl or sulfonyl as a nucleophile, leaves as a bulky ammonium in a Hofmann elimination, or lends its lone pair to an enamine's distant carbon — every move is the same lone pair, traded into a different role. That is the deep payoff of amine chemistry: not a list of named reactions to memorize, but one electron pair you can follow, faithfully, from base to nucleophile to leaving group to the quiet architect of a new carbon-carbon bond.