Mix butanoic acid with diethylamine and you do not get an amide. You get a salt, because the amine simply takes the acid’s proton. Yet the same amide forms cleanly in two steps: turn the acid into an acid chloride first, then add the amine. This page is about that first step, and about what acid chlorides and anhydrides do once you have them.
An acyl group is the R–C=O unit, and an acylating agent is a compound that hands its acyl group to a nucleophile. An acid chloride (also called an acyl chloride), RCOCl, carries a chlorine on the carbonyl carbon. An anhydride, RCO–O–COR, is two acyl groups joined through one oxygen. These two are the most reactive acylating agents on the ladder from Esters & amides: acid chloride > anhydride > ester > amide.
You can turn a derivative into any derivative below it, but never into one above it. The reason is in the tetrahedral intermediate from Nucleophilic acyl substitution: it expels the better leaving group of the two it holds. Going down the ladder, that is the group you want to lose. Going up would mean expelling the worse one, so it does not happen. A synthesis that starts from a carboxylic acid usually climbs to the top of the ladder first, and then runs down from there.
Why the acid itself is a poor acylating agent
Put methylamine on acetic acid and watch where the nitrogen’s lone pair goes. It does not attack the carbonyl carbon. It takes the O–H proton instead, and the products are acetate and methylammonium. The nucleophile is now protonated and has no lone pair left. The electrophile is now an anion, which a nucleophile has no reason to approach. Both halves of the reaction are switched off by one proton transfer, and that transfer is faster than anything else in the flask.
The general rule follows. A carboxylic acid carries a proton with a pKa of 4–5, and good nucleophiles are usually also bases. Any nucleophile basic enough to take that proton will take it first, and the ammonium carboxylate left behind is a dead end.
On carbonyl reactivity alone, the acid would not be so badly placed. Its OH donates into the carbonyl a little less than an ester’s OR does, because the OH lacks the electron-releasing alkyl group that pushes the ester oxygen’s lone pair toward the carbonyl. That puts the neutral acid just above an ester. In practice the acid behaves far worse than that ranking suggests, and the reason is its proton, not its carbonyl.
A few reactions do run directly on the acid, among them Fischer esterification and LiAlH4 reduction. Anything that needs a reactive acylating agent does not.
Making an acid chloride
Warm butanoic acid with thionyl chloride, SOCl2, and you get butanoyl chloride, with SO2 and HCl bubbling out of the flask. The name follows the acid: -oic acid becomes -oyl chloride. Acetic acid, which keeps its common name, gives acetyl chloride.
Three kinds of reagent do this job, and the choice between them is practical rather than mechanistic.
- SOCl2 (thionyl chloride) → acid chloride + SO2 + HCl
- PCl3 or PCl5 → acid chloride, plus H3PO3 from PCl3, or POCl3 and HCl from PCl5
- (COCl)2 (oxalyl chloride) with catalytic DMF → acid chloride + CO + CO2 + HCl
DMF is dimethylformamide, a small amide. “Catalytic” means only a small amount is added, and it is not used up.
Now look at the by-products. Thionyl chloride and oxalyl chloride give only gases, which leave the flask as they form. Losing a product that way pulls the equilibrium forward and drives the reaction to completion. It also makes the workup an evaporation rather than a separation. The phosphorus reagents leave phosphorus acids and oxychlorides behind, which you must then remove. They still work; they are just less convenient.
The mechanism has four steps. In step 1, the OH oxygen of the acid uses a lone pair to bond to sulfur, and a chloride leaves from sulfur. In step 2, that chloride takes the proton off the now positive oxygen. HCl forms, and what is left is an acyl chlorosulfite, RCO–O–S(=O)Cl.
In step 3, chloride adds to the carbonyl carbon, as in any acyl substitution, and a tetrahedral intermediate forms. In step 4, the C=O re-forms and the chlorosulfite group leaves. As it leaves, it falls apart into SO2 gas and chloride. Nothing is left in the flask that could put the OH back.
What an acid chloride then does
Add water to acetyl chloride and you get acetic acid and HCl. Add ethanol and you get ethyl acetate and HCl. Each time the nucleophile adds to the carbonyl carbon, chloride leaves, and one HCl is released. The table lists the nucleophiles you will meet most.
| Nucleophile | Product | Note |
|---|---|---|
| H2O | carboxylic acid | often just moisture in the air |
| R′OH | ester | irreversible, unlike Fischer |
| R′NH2 | amide | 2 equivalents, or 1 in excess pyridine |
| R′COO− | anhydride | the carboxylate, not the acid |
| LiAlH(OtBu)3, −78 °C | aldehyde | a deliberately weakened hydride |
| R′2CuLi | ketone | preview: a weakened carbon nucleophile, taught later in the course |
The amine row hides a favorite exam trap: it takes two equivalents of amine, not one. Each acylation releases one HCl, and the amine is the most basic thing in the flask. So the amine takes that proton and becomes an ammonium ion, with no lone pair left to attack. With one equivalent, half the amine is used up as a base and only half becomes amide. The second equivalent is there only to be used up as that base. Chemists call it a sacrificial base.
Pyridine is the usual replacement for that second equivalent. You met it in Leaving groups as a mild base: a benzene-like ring with a nitrogen in place of one CH. It is normally used in excess, often as the solvent, so there is always enough base to take up the HCl.
Pyridine is a much weaker base than an ordinary amine. Its conjugate acid, pyridinium, has a pKa of 5.2, against about 10.7 for an alkylammonium ion. The reason is hybridization. Pyridine’s lone pair sits in an sp2 orbital in the plane of the ring. That orbital has more s character than the sp3 orbital holding an amine’s lone pair, and more s character holds the pair closer to the nucleus. That makes pyridine strong enough to take up HCl, yet too weak to pull off protons it should leave alone.
Pyridine is also a good nucleophile, and it attacks the acid chloride before the amine does. It gives an N-acylpyridinium ion, an acyl group bonded to the ring nitrogen, which now carries a positive charge. That costs nothing, for two reasons. The acylpyridinium ion is itself a potent acylating agent, so it passes the acyl group straight on to the amine, often faster than the direct route. And pyridine has no N–H, so it can never end up in the product the way the amine does. It simply leaves again and picks up a proton.
This relay is the whole basis of DMAP catalysis. DMAP is 4-dimethylaminopyridine, a pyridine with an NMe2 group opposite the ring nitrogen. It is a better nucleophile and a better relay, because the NMe2 lone pair shares the positive charge of the acylpyridinium ion and stabilizes it. A small amount of DMAP speeds acylations up by thousands of times.
The LiAlH(OtBu)3 row works because the acid chloride is so reactive. A deliberately weakened reagent still reacts with it, and that weakened reagent is then too feeble to touch the aldehyde it has just made. So the reaction stops one step early. LiAlH(OtBu)3 is LiAlH4 with three of its hydrogens replaced by bulky tert-butoxy groups, and −78 °C is the temperature of a dry-ice/acetone bath. This is the same “crippled, not rationed” idea that Reducing carbonyls uses for DIBAL-H on esters.
Preview. The Gilman reagent in the last row, R′2CuLi, is a copper-based carbon nucleophile that stops at the ketone for the same reason. Nothing on this page depends on it; it is taught in Organolithiums, Cuprates & Cross-Coupling.
From butanoic acid, make (a) isopropyl butanoate and (b) N,N-diethylbutanamide.
Both start the same way. Butanoic acid’s leaving group would be hydroxide, and its proton (pKa 4.8) is taken by any basic nucleophile. So step one for both targets is SOCl2. That gives butanoyl chloride.
(a) Add propan-2-ol. The alcohol attacks, chloride leaves, a proton is lost, and you have isopropyl butanoate. One equivalent of alcohol is enough if a base such as pyridine is there to take up the HCl. Without a base, use the alcohol in excess.
(b) Add diethylamine: two equivalents, or one in excess pyridine. The first equivalent is acylated. The second takes the HCl and ends up as diethylammonium chloride. With only one equivalent, the yield stops near 50%.
Why not go direct? Fischer esterification would reach (a) eventually, but only as an equilibrium that has to be driven. Nothing met so far in this course reaches (b) directly: butanoic acid and diethylamine give the ammonium carboxylate salt and stop.
Anhydrides
Acetic anhydride, (CH3CO)2O, is two acetyl groups joined through one oxygen. Add methanol and the alcohol oxygen attacks one of the two carbonyl carbons. The tetrahedral intermediate then collapses, and the other half of the molecule leaves as acetate. The products are methyl acetate and acetic acid.
So an anhydride’s leaving group is a carboxylate. A carboxylate is resonance-stabilized, so it leaves willingly, but not as willingly as chloride, because HCl is a far stronger acid than a carboxylic acid. That places anhydrides exactly where the ladder puts them: below acid chlorides and above esters. They are reactive enough to acylate alcohols and amines directly, yet stable enough to weigh out in air.
Acetic anhydride is the standard acetylating agent, the reagent chemists reach for to attach an acetyl group. Its best-known use is aspirin. Salicylic acid carries two OH groups. One belongs to a carboxylic acid; the other belongs to a phenol, which is an OH bonded directly to a benzene ring. Acetic anhydride acetylates the phenol OH, giving an ester there, and releases acetic acid.
Why the phenol and not the carboxyl? Acylating the carboxylic acid would give another anhydride, a mixed one, which sits on the same rung of the ladder as the reagent and so gains nothing. Acylating the phenol gives an ester, one rung down. The reaction goes the way the ladder allows, so aspirin is still a carboxylic acid.
The acetic acid by-product is the real cost of the method. Half of the anhydride is spent as a leaving group, which hurts the atom economy, the share of the reagents’ atoms that end up in the product. You buy convenience with atom economy.
Anhydrides are named from their acid: replace “acid” with “anhydride”. Ethanoic acid gives ethanoic (acetic) anhydride, and benzoic acid gives benzoic anhydride. When two different acyl groups share the oxygen, the compound is a mixed anhydride, and its name lists both acids, as in acetic benzoic anhydride.
To make an anhydride from an acid chloride, the carboxylate is the nucleophile that works well. Sodium acetate and acetyl chloride, for example, give acetic anhydride and NaCl.
A neutral carboxylic acid is a much weaker nucleophile than its anion. Stir an acid chloride with a carboxylic acid and no base, and very little happens at room temperature. Heating does force the reaction, slowly, with HCl given off. A base makes it fast: it removes the acid’s proton, and the carboxylate that forms attacks the acid chloride.
Triethylamine, Et3N, is a common choice. Its three ethyl groups crowd the nitrogen, so it acts mainly as a base here rather than as a nucleophile, and it has no N–H, so it cannot end up as an amide. One caution applies. When the acid chloride has a hydrogen on the carbon next to the C=O, triethylamine can remove HCl from it in an elimination. That gives a ketene, R2C=C=O, a reactive side product.
Cyclic anhydrides
Heat succinic acid, a four-carbon diacid, and it loses one water from within the same molecule. The product is succinic anhydride, a five-membered ring. Phthalic acid closes a five-membered anhydride the same way, and glutaric acid closes a six-membered one. Each closure is a condensation: two groups join and a small molecule, here water, is lost. Succinic anhydride’s ring is the four chain carbons plus one oxygen. Five- and six-membered rings like this are almost free of strain, which is why these diacids close readily. Diacids that would have to close a larger ring do not.
Add ammonia or an amine and the ring opens at one carbonyl. The product is an amic acid, which has an amide and a carboxylic acid on the same molecule. Both groups stay on one molecule because the leaving carboxylate is still tied to the other end of the chain.
Heat the amic acid and it loses water again, closing the ring onto nitrogen. The product is an imide, a nitrogen bonded to two carbonyls, as in succinimide and phthalimide. That closure is a condensation too. Phthalimide comes back in Amines.
The other use: Friedel–Crafts
Preview. Everything above forms a carbon–heteroatom bond. The acid chloride’s other major use forms a carbon–carbon bond. With aluminum chloride, AlCl3, it loses its chloride and gives an acylium ion, R–C≡O+, which acylates a benzene ring. The product is an aryl ketone, a ketone with a benzene ring on its carbonyl carbon. Aromatic Chemistry teaches the reaction in full. What matters here is the link: the SOCl2 step is the usual way to reach an aryl ketone from a carboxylic acid.
What carries forward
Almost every multistep problem that involves a carboxylic acid opens with the same move: activate with SOCl2, then run down the ladder. Recognize that move and half the planning is done before you start.