Kinetic & thermodynamic enolates — Enolate Chemistry

Every reaction in this chapter has begun with "form the enolate." For acetone or cyclohexanone that is a complete instruction, since every α hydrogen is equivalent. For an unsymmetrical ketone it is not an instruction at all — it is a question.

Two α carbons, two enolates

Take 2-methylcyclohexanone. The carbonyl is C1, with an α carbon on each side: C2 carries the methyl and a single hydrogen, while C6 is an unsubstituted CH₂ with two hydrogens and nothing crowding it. Either can be deprotonated, and the two enolates are different compounds:

Deprotonating C6 gives the less substituted C=C — disubstituted.
Deprotonating C2 gives the more substituted C=C — trisubstituted, because the methyl already sits on one of the two carbons now sharing the double bond. (Counting carbon substituents only; some texts count the enolate oxygen as well and would call these tri- and tetrasubstituted.)

So the more accessible proton and the more stable anion are on opposite sides of the molecule. That conflict is settled by conditions, not by the ketone.

The kinetic enolate: get there first and stay there

Removing a C6 proton is faster: there are two of them rather than one, and nothing blocks a bulky base from reaching them. Isolating that outcome means preventing the system from ever reconsidering, and each condition does one job:

LDA, lithium diisopropylamide. Its conjugate acid has a pK_a near 36 against a ketone's 20, so deprotonation is complete and irreversible.
Bulky. The two isopropyl groups are what make LDA discriminate between the crowded proton and the open one. A small base of the same strength would not.
One equivalent, −78 °C, THF. Cold enough that nothing equilibrates once formed, in a solvent that dissolves the lithium enolate without destroying the base.
Ketone added to the base — inverse addition. Done the other way around, the last of the base sits in a flask full of unreacted ketone, and ketone shuttles protons from one enolate to the other. Never leave free ketone present.

The thermodynamic enolate: let it equilibrate

The thermodynamic enolate needs the opposite: the two enolates must be able to interconvert, so that the mixture can find the more stable one. A proton has to be able to travel from one α carbon to the other, and there is more than one way for that to happen.

The deprotonation itself reverses. Sodium ethoxide in ethanol is the clean case: its conjugate acid is an alcohol of comparable pK_a, so only a small equilibrium amount of enolate exists and it is handed back constantly.
Free ketone shuttles it. Un-ionized ketone is an acid, and any enolate in the flask can take a proton from it. This is the route that operates with sodium hydride — whose own deprotonation does not reverse, since the conjugate acid is H₂ and it leaves the flask. NaH is slow and heterogeneous, so a large pool of un-ionized ketone sits there throughout, and that pool equilibrates the two enolates.
The amine shuttles it. Deprotonating with LDA generates a full equivalent of diisopropylamine, which is a proton donor too. It is why warming a lithium enolate erodes its regiochemistry even though the base is long gone.

Whichever route is open, the mixture accumulates as the more stable enolate: the trisubstituted one from C2. The alkene-substitution rule that ranks Zaitsev products ranks these.

	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

The same switch, one step earlier

Nothing about the ketone settles which enolate you get, which is why the conditions are written out in full rather than abbreviated to a reagent name. Each item in the kinetic column blocks one route back: the base is strong enough not to reverse, bulky enough to pick the open proton, cold enough not to equilibrate, and added first so no free ketone is left to shuttle protons.The test is whether the two enolates can trade a proton, and there are three ways they can. The deprotonation reverses — ethoxide. Free ketone shuttles it — which is why NaH, whose deprotonation is as irreversible as LDA’s, is a thermodynamic base: it is slow enough that un-ionized ketone is always present. Or the amine the base generated shuttles it, which is what erodes a lithium enolate on warming. Each item in the kinetic column closes one of the three, which is why missing any single one is enough to lose the regiochemistry.

This is the control problem from the conjugation chapter, applied to a different step. There, an allylic cation could be captured at two carbons, and warming let bromide leave again so the mixture settled into the more substituted alkene. Here the branch point is the deprotonation, and reversing it means a proton moving back from enolate to ketone rather than a C–Br bond breaking.

One difference matters. In the diene case the two options 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.

So the question to ask is not “is the base strong?” and not “is it cold?” but can the two enolates trade a proton? A base strong enough to deprotonate completely closes only one of the three routes above — its own reversal — which is exactly why NaH, whose deprotonation is as irreversible as LDA's, still delivers the thermodynamic enolate. Every item in the kinetic recipe is there to close a different route: LDA for the reversal, inverse addition for the free ketone, and the cold for the amine. Miss any one and the regiochemistry drifts.

The third option: don't use an enolate

A ketone and a secondary amine give an enamine, nucleophilic at the α carbon for the reasons set out in the imines and enamines section. It reaches the same regiochemistry by a different route: the double bond forms toward the less substituted side, because the substituted alternative twists the nitrogen lone pair out of alignment with the C=C.

It is also neutral rather than anionic, so no strong base is needed, and it stops cleanly after one alkylation where a lithium enolate can go twice. The reason is structural rather than a matter of reactivity: alkylating an enamine converts it into an iminium salt, which has no nucleophilic carbon at all, and there is no base present to turn it back into an enamine. A second alkylation is not slow, it is impossible. Compare the lithium enolate, where the monoalkylated ketone is simply deprotonated by enolate that has not yet reacted. Alkylate, then hydrolyze back to the ketone — the Stork enamine sequence.

The limit nobody gets around

Alkylating a lithium enolate is an S_N2, and an enolate is a base as well as a nucleophile. Methyl, primary, allylic and benzylic halides work well, secondary halides give substantial elimination, and tertiary halides give elimination and nothing else — the identical constraint that limits the malonic ester synthesis. Check it on the halide, never on the target.

What carries forward

Read the conditions before drawing the enolate. LDA at −78 °C means the less hindered side; a weak base and warmth means the more substituted side; a secondary amine means an enamine instead. Then check the halide.