Stir acetophenone, C₆H₅COCH₃, with bromine in acetic acid. One hydrogen on the methyl group is replaced by a bromine, and the product is C₆H₅COCH₂Br. (C₆H₅ is a benzene ring, often written Ph.)
Now stir the same ketone with bromine and excess sodium hydroxide. All three methyl hydrogens are replaced. Then the molecule splits in two, and one carbon leaves as CHBr₃.
In both flasks the bromine goes onto the α carbon, the carbon bonded directly to the carbonyl carbon. Putting a halogen there is called α-halogenation. This page explains why acid stops after one halogen, why base keeps going, and how chemists use the runaway.
You met both nucleophiles in Alpha hydrogens & enolates, earlier in this chapter: the enol (C=C–OH) in acid and the enolate (its anion) in base. Both are electron-rich at the α carbon. The electrophile is a halogen molecule: Br₂, Cl₂ or I₂. Its X–X bond breaks, one X bonds to the α carbon, and the other leaves as halide, X⁻.
Acid: through the enol, and it stops
Start with acetophenone and Br₂ in acid. The ketone first forms its enol, by the same two steps as in Alpha hydrogens & enolates. H₃O⁺ puts a proton on the carbonyl oxygen. Then water takes an α hydrogen, and the C=C forms.
Next the enol attacks Br₂ with its C=C. The near Br bonds to the α carbon. The Br–Br electrons leave with the far Br as bromide. A lone pair on the enol oxygen moves in to make the C=O, so that oxygen is briefly positive. Water takes the proton off it, and the product is α-bromoacetophenone, C₆H₅COCH₂Br.
Making the enol is the rate-determining step: the slowest step, the one that sets the rate of the whole reaction. Two observations show this. The rate does not change when you change the halogen’s concentration, and Br₂, Cl₂ and I₂ all react at the same rate. So the halogen takes no part in the slow step; it catches each enol as soon as the enol forms.
The product is slower to react a second time. To react again, it must form an enol again, and that starts with a proton on the carbonyl oxygen. The new Br is strongly electron-withdrawing: it pulls electron density toward itself and away from that oxygen. So the oxygen is a weaker base and harder to protonate. The second enol forms more slowly than the first, so the reaction can be stopped after one halogen.
Base: through the enolate, and it runs away
Now use excess NaOH. Hydroxide takes an α hydrogen and gives the enolate. The enolate attacks Br₂ with the same three arrows the enol used. The only difference is the oxygen: it starts with a negative charge instead of an H.
The product still has two α hydrogens, on the carbon that now carries the Br. Those two are more acidic than the ones you started with. The Br pulls electron density toward itself, and that helps hold the negative charge of the next enolate. So the second deprotonation is faster than the first, and the third is faster again. The reaction does not stop cleanly at one.
The two routes also differ in what happens to the acid or base. In acid, the H₃O⁺ used in the first step comes back when the enol forms, so the acid is a true catalyst (the reaction also makes HBr). In base, every round turns one HO⁻ into water, and nothing gives it back. The base is consumed, so this is called base-promoted halogenation rather than base-catalyzed. That is why haloform conditions call for excess hydroxide as well as excess halogen.
What stops the base route in the end is running out of α hydrogens, not running out of base. Isopropyl phenyl ketone, C₆H₅COCH(CH₃)₂, has only one α carbon that carries hydrogen, and it carries just one. Under base it takes one bromine however much bromine you add, because after that there is no α hydrogen left to remove.
Pentan-3-one, CH₃CH₂COCH₂CH₃, has two α carbons, C2 and C4. Each is a CH₂, and neither is a methyl.
(a) Br₂, CH₃CO₂H. The enol forms and attacks Br₂, giving 2-bromopentan-3-one. Its carbonyl oxygen is now harder to protonate, so the next enol is slow to form. Stop here: one bromine.
(b) Excess Br₂, excess NaOH. The first Br goes on C2. The last H on C2 is now the most acidic α hydrogen in the molecule, so the second Br goes there too: 2,2-dibromopentan-3-one. With more reagent, C4 is brominated in turn. No carbon ever carries three bromines, so the molecule never splits. The product is a polybrominated ketone.
The habit: read the conditions before you read the substrate. The same ketone gives different products in acid and in base.
Which α carbon takes the halogen?
2-Methylcyclohexanone has two different α carbons. C2 carries the methyl and one H. C6 is a CH₂. With Br₂ in acid, the Br goes on C2, and the product is 2-bromo-2-methylcyclohexanone.
The Br goes where the enol’s C=C points. The enol with its C=C toward C2 is the more substituted enol: its C=C carries more carbon groups. A more substituted C=C is more stable (Alkenes & Alkynes). This enol also forms faster, because the step that removes the α hydrogen is already building its C=C. So in acid the halogen lands on the more substituted α carbon.
Under base, the first halogen tends to go to the less substituted α carbon instead, C6 here. Its hydrogens are less crowded, and an alkyl group on a carbon makes a negative charge there slightly less stable. Why each base picks the enolate it does is the subject of Kinetic & thermodynamic enolates, the next section. With excess halogen and hydroxide, the question matters less, because the reaction does not stop at one.
The haloform reaction
Chemists put the runaway to use. Go back to acetophenone with excess Br₂ and excess NaOH. Acetophenone is a methyl ketone: a ketone with a CH₃ bonded directly to the carbonyl carbon, CH₃COR. Base brominates that methyl three times, and each round is faster than the one before. The methyl ends as a CBr₃ group.
Now hydroxide adds to the C=O and gives a tetrahedral intermediate, a carbon with four single bonds, as in Carboxylic Acids & Derivatives. The C=O then re-forms, and the C–CBr₃ bond breaks. Its two electrons leave with the CBr₃ group as the anion ⁻CBr₃. This is nucleophilic acyl substitution with an unusual leaving group.
A carbanion is normally far too strong a base to leave. ⁻CBr₃ can leave because its three bromines pull electron density away from the carbon and help hold the negative charge. That makes it a much weaker base than an ordinary carbanion such as ⁻CH₃.
One fast step remains. ⁻CBr₃ is still basic enough to take the proton from benzoic acid, the other product. So the flask holds benzoate and bromoform, CHBr₃. Adding acid at the end, the acidic workup, turns benzoate into benzoic acid.
In general, a methyl ketone gives a carboxylate (the carboxylic acid after acidic workup) and CHX₃, a haloform. Chlorine gives chloroform, CHCl₃; bromine gives bromoform, CHBr₃; iodine gives iodoform, CHI₃.
With I₂, the iodoform comes out as a yellow precipitate with a distinctive medicinal smell, which makes the reaction a classical test. A positive iodoform test means a methyl ketone. Acetaldehyde, CH₃CHO, is also positive. So is any alcohol the reagent first oxidizes to one of these: ethanol, and any secondary alcohol CH₃–CH(OH)–R. The oxidant is hypoiodite, IO⁻, which I₂ forms in base.
The test reports a CH₃CO (or CH₃CH(OH)) bonded to H or to carbon, not a methyl group anywhere in the molecule. 2-Methylcyclohexanone has a methyl and is a ketone, yet it is negative. Its methyl sits on C2, one bond out from the carbonyl carbon, so neither of its α carbons is a CH₃. It still brominates, but it never forms a CX₃ to expel.
Hell–Volhard–Zelinsky: halogenating an acid
Stir butanoic acid with Br₂, and essentially nothing happens. A carboxylic acid has almost no enol to work with, because its OH oxygen already donates a lone pair into the C=O. The fix is to turn the acid into a derivative that does form an enol.
Add a catalytic amount of PBr₃, or red phosphorus, which reacts with Br₂ to make PBr₃ in the flask. PBr₃ turns some of the acid into butanoyl bromide, an acyl bromide: the bromine version of an acid chloride, with Br in place of the OH. Br shares its lone pairs with the C=O far less than OH does, so the acyl bromide forms its enol readily. That enol attacks Br₂ at the α carbon, just as a ketone’s enol does, giving 2-bromobutanoyl bromide.
That α-bromo acyl bromide then trades groups with a fresh molecule of butanoic acid: its Br moves to the new acid, and the new acid’s OH moves to it. Out come 2-bromobutanoic acid, the α-bromo acid, and a new butanoyl bromide that goes round again. The whole sequence is the Hell–Volhard–Zelinsky (HVZ) reaction.
The product matters because an α-bromo acid is an SN2 substrate (Substitution & Elimination). Displace the bromide with NH₃, used in excess, and you have an α-amino acid: an acid with an NH₂ on its α carbon, the building block of proteins. Displace it with hydroxide or water and you have an α-hydroxy acid.
It is one of three classical laboratory routes to the α-amino acids; the other two, the Strecker synthesis and the Gabriel–malonic ester sequence, come later in the course. Because the enol is flat, Br₂ adds to either face equally. So the α-bromo acid comes out racemic, a 50:50 mix of both enantiomers, and so does the amino acid made from it.
The α-halo ketone itself
An α-halo ketone is a good electrophile in its own right. A carbon next to a carbonyl reacts unusually fast in SN2, so compounds like C₆H₅COCH₂Br are strong alkylating agents: they transfer their carbon group to a nucleophile.
Heated with a weak base such as pyridine, an α-halo ketone can instead lose HX by E2 elimination. 2-Bromocyclohexanone loses the Br from C2 and an H from C3, and a new C=C forms between them. A strong base such as hydroxide is avoided, because it sends 2-bromocyclohexanone down other paths, such as a rearrangement that shrinks the ring. The product, cyclohex-2-en-1-one, is an α,β-unsaturated ketone: its C=C sits between the α and β carbons, next to the C=O. It is the same kind of enone the aldol condensation reaches by another road.