Take acetone and remove any one of its six α hydrogens. You get the same enolate every time, because the two α carbons are alike. Now take butan-2-one, CH3COCH2CH3. Its α hydrogens come in two sets: three on the CH3 and two on the CH2. Each set gives a different enolate.
So for a ketone like butan-2-one, “form the enolate” leaves a question open: which one? The reaction conditions answer it; the ketone itself does not. This section works through one ketone, 2-methylcyclohexanone, because its two α carbons differ so clearly.
Two α carbons, two enolates
Number the ring of 2-methylcyclohexanone. The carbonyl carbon is C1. The carbons on either side of it, C2 and C6, are both α carbons.
- C6 is a CH2. Its two hydrogens stand in the open, with nothing crowding them.
- C2 carries the methyl group and one hydrogen. That hydrogen sits right beside the methyl.
Removing a C6 hydrogen gives an enolate whose C=C joins C1 and C6. Removing the C2 hydrogen gives an enolate whose C=C joins C1 and C2. These are two different compounds.
Now count the carbons attached to each C=C. The C1=C6 bond carries two: C2 on one end and C5 on the other. So this C=C is disubstituted. The C1=C2 bond carries three: C6 on C1, and both C3 and the methyl on C2. So this C=C is trisubstituted. (The count leaves out the oxygen. Some books count it too, and call these enolates tri- and tetrasubstituted.)
A C=C that carries more carbons is more stable. The alkyl groups feed electron density into the π bond by hyperconjugation, as Alkenes & Alkynes showed. Each extra carbon is worth roughly 1–1.5 kcal/mol. It is the same count that picks the Zaitsev product of an elimination in Substitution & Elimination. So the C1=C2 enolate is the more stable one.
The two sides pull in opposite directions. A base removes a C6 hydrogen faster: there are two of them, and nothing blocks the way. But the C2 hydrogen gives the more stable enolate. The enolate that forms faster is called the kinetic enolate. The more stable one is the thermodynamic enolate. This is the kinetic-versus-thermodynamic control you met in Conjugation & Pericyclic Reactions, now applied to a deprotonation.
The kinetic enolate: get there first and stay there
Keeping the C6 enolate takes two things. The base must take a C6 hydrogen. Then the flask must hold nothing that could turn that enolate into the other one. The standard base is LDA (lithium diisopropylamide), which you met in Alpha hydrogens & enolates. Its nitrogen takes the C6 hydrogen in a single step.
Each condition in the recipe does one job:
- LDA. Its conjugate acid, diisopropylamine, has a pKa near 36. The ketone’s α hydrogen has a pKa near 20. So the deprotonation is complete, and it does not run back.
- Bulky. The two isopropyl groups make LDA choosy. It reaches the open C6 hydrogens far more easily than the crowded C2 one. A small base of the same strength would not discriminate this well.
- One equivalent, meaning one mole of LDA per mole of ketone, or a slight excess. Every ketone molecule becomes enolate, and none is left over.
- −78 °C. The cold sharpens LDA’s preference for the open hydrogen. It also slows any proton transfer that a stray acid could still manage.
- THF (tetrahydrofuran, a cyclic ether) as the solvent. It dissolves the lithium enolate and does not destroy the base.
- Ketone added to the base, called inverse addition. If you add the base to the ketone instead, unreacted ketone stays in the flask until the last of the base goes in. Free ketone passes protons between the enolates, as the next part shows. So the recipe leaves no free ketone behind.
The thermodynamic enolate: let it equilibrate
To get the C1=C2 enolate, the two enolates must be able to interconvert, or turn into each other. Then the mixture can settle into the more stable one. For that, a proton must travel from one α carbon to the other, and something in the flask has to carry it.
The easiest carrier to picture is free ketone. Leave some 2-methylcyclohexanone beside the kinetic enolate. The enolate’s C6 takes the C2 hydrogen from a ketone molecule. The enolate becomes ketone again, and that ketone molecule becomes the C1=C2 enolate.
Each hand-over can also run backward, so after many of them the enolates collect in the more stable C1=C2 form. Two kinds of carrier are common:
- The base’s conjugate acid, or any protic solvent (one with an O–H). Sodium ethoxide in ethanol is the simplest example, and there ethanol is both. Its pKa is comparable to the ketone’s, so only a small amount of enolate exists at any moment, and ethanol hands a proton back to it constantly.
- Free ketone, as in the figure. This is how sodium hydride, NaH, works. Its deprotonation does not reverse: its conjugate acid is H2, which leaves the flask as a gas. But NaH is a solid and reacts slowly, so a large pool of un-ionized ketone sits in the flask throughout. That pool equilibrates the two enolates.
Diisopropylamine, the amine LDA leaves behind, is not a carrier. As an acid it is about 16 pKa units weaker than the ketone, far too weak to protonate an enolate. So a lithium enolate made with a slight excess of LDA, with no free ketone, keeps its regiochemistry, the position of its C=C. The enolate keeps that position even if the flask warms up, as long as nothing else can carry a proton.
Even at equilibrium the more stable enolate does not win outright. The mixture runs about 90:10 in favor of C1=C2, which fits a gap of about one substituent’s worth of stability. The kinetic recipe runs about 99:1 the other way. Thermodynamic control is a preference, not a switch.
| Kinetic | Thermodynamic | |
|---|---|---|
| Base | LDA, 1 equiv | NaOEt (reverses) or NaH (leaves free ketone) |
| Temperature | −78 °C | Room temperature or warmer |
| Addition | Ketone into base, so no free ketone survives | Irrelevant — it equilibrates anyway |
| Can the enolates interconvert? | No route left open | Yes, and that is the point |
| Deprotonates | Less hindered α carbon | More substituted α carbon |
| Enolate C=C | Less substituted | More substituted |
Substrate: 2-methylcyclohexanone. Electrophile: CH3I, added once the enolate has formed.
Route A — LDA (1 equiv), THF, −78 °C, ketone added to the base; then CH3I.
Step 1: LDA takes a hydrogen from C6, the open CH2. That gives the kinetic enolate, C1=C6.
Step 2: nothing in the flask can carry a proton, so it is still the C6 enolate when the CH3I arrives.
Step 3: C6 attacks the carbon of CH3I in an SN2 step, and iodide leaves. The new methyl lands on C6. Product: 2,6-dimethylcyclohexanone. C6 becomes a new stereocenter, so this product forms as a mixture of cis and trans isomers; this section leaves that aside.
Route B — NaOEt in EtOH, room temperature; then CH3I.
Step 1: a small amount of enolate forms and reverts constantly, so the enolate collects mostly as C1=C2.
Step 2: that enolate attacks CH3I through C2, the carbon that already carries a methyl. The main product is 2,2-dimethylcyclohexanone. Its C2 is now bonded to four carbons, which makes it a quaternary carbon.
Route B is not clean, though. Three things spoil it:
(1) About a tenth of the enolate is the C1=C6 one, so some 2,6-dimethylcyclohexanone forms too. The product ratio need not even match the enolate ratio: the two enolates keep interconverting while they react, and they react with CH3I at different rates.
(2) The methylated product still has α hydrogens. Ethoxide removes them too, so a second and third methyl can go on. This is polyalkylation.
(3) Enolate and unreacted ketone sit side by side, so some enolate attacks a ketone C=O instead. That gives aldol products (see Aldol reactions, earlier in this chapter).
Same ketone, same electrophile, two different main products, constitutional isomers of each other. Nothing about the substrate chose between them; the conditions did. Route A gives mainly 2,6-dimethylcyclohexanone; a little is over-alkylated, as “A second alkylation” explains. Route B gives 2,2-dimethylcyclohexanone only as the largest part of a mixture, and a clean route to it takes extra steps beyond this section.
The same switch, one step earlier
You met this kind of control in Conjugation & Pericyclic Reactions. HBr adds to buta-1,3-diene through an allylic cation, which carries its positive charge at C2 and at C4. In the cold, bromide adds at C2 of the cation, the faster route, giving 3-bromobut-1-ene (the same carbon is C3 once the product is numbered). When the flask is warm, the C–Br bond can break again, so the mixture settles into the more stable product, 1-bromobut-2-ene, with the more substituted C=C.
Here the branch point is the deprotonation. Reversing it means a proton moving back to carbon, not a C–Br bond breaking.
One difference matters. In the diene case the two outcomes were products. Here they are intermediates. You choose an enolate to control where the next reaction happens, so the choice is made before any electrophile is added.
The third option: don't use an enolate
A ketone and a secondary amine, such as pyrrolidine, give an enamine: a C=C with a nitrogen on one end. Carbonyl Chemistry built enamines and ran the Stork sequence with them. That sequence forms the enamine, alkylates or acylates its α carbon, then hydrolyzes back to the ketone.
Two more points from that chapter matter here. A secondary amine’s iminium ion has no N–H to lose, so it loses an α hydrogen instead and gives the neutral C=C–N. And the Stork sequence needs a reactive halide. What is new here is where the enamine puts its C=C.
An enamine is nucleophilic at its α carbon for the same reason an enolate is. The nitrogen lone pair spreads into the C=C and piles electron density onto the far carbon. So an enamine is a neutral relative of an enolate, with nitrogen doing the job of the O−.
Make the pyrrolidine enamine of 2-methylcyclohexanone, and its C=C forms toward the less substituted side: C1=C6.
The reason is a clash. The nitrogen can donate only while its lone pair stays lined up with the C=C π bond. That holds the groups on nitrogen in the plane of the C=C. In the C1=C2 enamine, the methyl on C2 lies in that same plane, and it runs into the ring CH2 next to nitrogen.
This kind of crowding is called allylic strain: a group on one end of a C=C bumps a group on the atom attached to the other end. To escape it, the nitrogen twists, and its lone pair falls out of line with the C=C. The C1=C6 enamine avoids the problem. There C2 is sp3, so its methyl can turn out of the plane.
So an enamine puts its C=C on the same, less substituted side as the kinetic enolate. Unlike the kinetic enolate, this enamine is also the more stable isomer, so it forms with no LDA, no cold and no anion. The three routes line up as a choice:
- LDA in the cold: the kinetic enolate, less substituted side.
- A weak base at equilibrium: the thermodynamic enolate, more substituted side.
- An enamine: the less substituted side, alkylated exactly once under neutral conditions, provided the halide is reactive enough.
The limit nobody gets around
Alkylating a lithium enolate is an SN2 (Substitution & Elimination), and an enolate is a base as well as a nucleophile. Methyl, primary, allylic and benzylic halides work well. Secondary halides give substantial elimination. Tertiary halides give elimination and nothing else. The malonic ester synthesis, in the next section, meets the identical limit. Judge this by the halide you plan to use, not by the product you want.
A second alkylation
Even Route A can put on a second methyl. Its product, 2,6-dimethylcyclohexanone, still has an α hydrogen on each side. While the reaction runs, enolate that has not reacted yet can take one of those hydrogens. It is the same hand-over drawn under “The thermodynamic enolate”, with the product as the donor. The new enolate then picks up a second methyl.
A reactive halide such as CH3I keeps this small, because the alkylation then outruns the proton exchange. This is the over-alkylation that the Stork enamine route avoids.
Why the halide goes to carbon
Most of an enolate’s negative charge sits on oxygen, as Alpha hydrogens & enolates showed. Yet CH3I usually bonds to the α carbon. Attack through carbon gives a ketone. Attack through oxygen would give an enol ether: a C=C carrying an OR group.
Three things favor carbon when the electrophile is an alkyl halide:
- The product. Attack through carbon ends with a C=O; attack through oxygen ends with a C=C instead. A C=O is worth far more than a C=C; the same gap keeps the keto tautomer on top.
- The metal. In a lithium enolate, the lithium sits on the oxygen and shields it.
- The electrophile. A soft electrophile is large and polarizable. It reacts where orbital overlap is best, which for an enolate is the carbon. A hard electrophile is small or highly charged, and it reacts where the charge is, at oxygen. Alkyl iodides are soft, so they go to carbon almost exclusively.
The balance can shift. Hard electrophiles push some of the product toward O-alkylation. So do very polar solvents, which pull the metal away from the oxygen. That is why these alkylations are run in THF or DME (1,2-dimethoxyethane) rather than in DMSO (dimethyl sulfoxide).
What carries forward
Read the conditions before drawing the enolate. LDA at −78 °C means the less hindered side. A weak base and warmth mean the more substituted side, usually as the main part of a mixture. A secondary amine means an enamine instead. Then check the halide.