Stir two equivalents of methyllithium with one of copper(I) iodide in cold ether, and both methyl groups move onto copper. The product is lithium dimethylcuprate, (CH3)2CuLi: a copper atom holding two methyl groups and a negative charge, with a lithium ion beside it.
Any reagent of this kind, R2CuLi, is a Gilman reagent, or lithium dialkylcuprate. A cuprate is a negative ion built around copper. Every Gilman reagent is made the same way:
2 R–Li + CuI → R2CuLi + LiI
Making the cuprate changes the metal on carbon from lithium to copper. Carbon's electronegativity is 2.55, lithium's is 0.98 and copper's is 1.90. Copper's value is far closer to carbon's, so the C–Cu bond is much less polar than the C–Li bond it came from. Less negative charge sits on carbon, and the reagent is much less reactive.
That sounds like a loss, but it is the point. A cuprate does three things a Grignard cannot, and it can do each of them because it is the weaker reagent. The next three headings take them in turn.
One: conjugate addition
Take cyclohex-2-en-1-one, a six-membered ring with a C=O and a C=C right next to it. It is an enone, and as Carbonyl Chemistry showed, a nucleophile can attack it at two carbons: the carbonyl carbon, and the β carbon at the far end of the C=C.
Number the four atoms of this system from the oxygen. The oxygen is atom 1, the carbonyl carbon atom 2, the α carbon atom 3 and the β carbon atom 4. A nucleophile that bonds to atom 2 makes a 1,2-addition. One that bonds to atom 4 makes a 1,4-addition, also called conjugate addition. The numbers count along the conjugated system, as they did for dienes in Conjugation, and are not the locants in the product's name.
Add CH3MgBr to cyclohexenone, then aqueous acid, and the methyl bonds to atom 2. The product is 1-methylcyclohex-2-en-1-ol, an allylic alcohol: its OH carbon sits right next to a C=C. Add (CH3)2CuLi instead and the methyl bonds to atom 4. After workup the product is 3-methylcyclohexan-1-one, a ketone with the new methyl on the old β carbon.
| Nucleophile | Adds | Product |
|---|---|---|
| RMgX or RLi | 1,2 — at the carbonyl carbon (atom 2) | Allylic alcohol |
| R2CuLi | 1,4 — at the β carbon (atom 4) | Ketone with a new alkyl group at β |
Two properties of the enone decide which carbon wins. The first is charge. Oxygen pulls electron density out of the C=O, so atom 2 carries a partial positive charge. A resonance form of the enone also puts a positive charge on atom 4, as the left half of the next figure shows. Of the two, atom 2 carries the larger positive charge.
The second is the LUMO, the lowest-energy empty orbital, which you met in Conjugation. A nucleophile's electron pair goes into it. The enone's LUMO spreads over all four atoms, but not evenly: its lobe is largest on atom 4 and next largest on atom 2. The size of an orbital's lobe on one atom is that atom's orbital coefficient. A large coefficient means the atom holds a large share of the orbital, so an incoming nucleophile overlaps with the orbital best there.
Which of the two a nucleophile follows depends on whether it is hard or soft, a sorting you met in Lewis acids/bases. A hard nucleophile packs its negative charge onto one small atom. It goes to the largest positive charge, and chemists call that charge control. A soft nucleophile holds its electrons loosely and spread out, so they are easily distorted. Chemists call such a nucleophile polarizable. It goes to the atom with the largest orbital coefficient in the LUMO, where its electrons overlap best, and that is called orbital control.
In a Grignard or an organolithium, the very polar bond to the metal concentrates charge on carbon. That makes the reagent hard, so it adds at atom 2. In a cuprate, the carbon is bonded to a large copper atom by a much less polar bond. That makes the reagent soft, so it adds at atom 4.
Follow the electrons of the 1,4-addition in the next figure. The pair in one C–Cu bond goes to atom 4 and becomes the new C–C bond. The C=C pair moves over to the bond between atoms 3 and 2, and the C=O pair moves onto oxygen. The other methyl stays on copper and ends up in CH3Cu, which is discarded.
So what sits in the flask is not yet a ketone. It is an enolate: an anion with its negative charge on oxygen and a C=C between the α carbon and the old carbonyl carbon. A second resonance form puts the charge on the α carbon instead. Enolate Chemistry is built on this anion.
Workup with aqueous acid, H3O+, adds a proton. If it lands on the α carbon, the ketone forms at once. If it lands on oxygen, it gives the enol, which then tautomerizes to the ketone (the same step that turns an enol into a ketone in Alkynes). Either way the C=O comes back, now with the new methyl on the β carbon.
Step 1 — locate the new bond. The methyl has to end up on C3 of the ring. Count from the carbonyl and C3 is the β carbon of the enone, atom 4 in the 1,4 count. A bond to the β carbon means conjugate addition. Nothing that attacks the carbonyl carbon can get there.
Step 2 — reject the obvious reagent, and say why. CH3MgBr is hard, so it adds 1,2 and gives 1-methylcyclohex-2-en-1-ol. That is a tertiary allylic alcohol with the methyl on the wrong carbon. Oxidation cannot rescue it. The carbon that carries the OH, the carbinol carbon, has no hydrogen, so PCC cannot turn that C–OH into a C=O (Oxidizing alcohols). The methyl would stay on the carbinol carbon anyway, not on the β carbon, and the ring would keep its C=C.
Step 3 — make the soft reagent. 2 CH3Li + CuI → (CH3)2CuLi + LiI.
Step 4 — add. The cuprate delivers one of its two methyls to the β carbon. What forms first is the enolate, not the ketone: the negative charge is on oxygen, the C=C now lies between the α carbon and the old carbonyl carbon, and the new methyl sits on the β carbon.
Step 5 — work up. H3O+ protonates the enolate, either on the α carbon directly or on oxygen to give the enol, which tautomerizes. The C=O comes back, and the product is 3-methylcyclohexan-1-one. It takes one reagent choice and one step.
What the question is really testing is whether you see that the C=O comes back even though the nucleophile never touched the carbonyl carbon.
Two: coupling with halides
A Grignard and an alkyl halide in the same flask do not reliably couple (Why C–metal means nucleophilic carbon shows why). A Gilman reagent and a halide do.
Mix (E)-1-bromoprop-1-ene with lithium dibutylcuprate, (C4H9)2CuLi. One butyl group takes the place of the bromine, and the product is (E)-hept-2-ene. The new C–C bond sits exactly where the C–Br bond was.
In general, a Gilman reagent joins one of its R groups to the carbon that held the halogen:
R2CuLi + R′–X → R–R′ + RCu + LiX
Only one of the two R groups is transferred. The other leaves in RCu and is wasted. This coupling is called the Corey–House synthesis.
The halide R′–X should be methyl, primary, vinyl or aryl. A vinyl halide has its halogen on a C=C carbon, like the bromopropene above. An aryl halide has it on a carbon of a benzene ring, like bromobenzene.
The vinyl and aryl cases are why the reaction matters. Neither kind of halide does SN2 at all, as Substitution & Elimination showed. The carbon is sp2, its C–X bond is short and strong, and the path an SN2 nucleophile needs, straight in from behind the C–X bond, is screened by the π system. In bromobenzene that path would run through the middle of the ring. SN1 fails too, because a vinyl or aryl cation is far too unstable to form.
At a vinyl or aryl carbon a cuprate does not need that path. The copper bonds to the carbon first and then joins one of its R groups to it. The halogen is simply replaced, so the double bond keeps its geometry: E stays E, and Z stays Z. No substitution reaction can offer that at a vinyl carbon.
Three: stopping at the ketone
Treat butanoyl chloride with (CH3)2CuLi at −78 °C and one methyl replaces the chlorine. The product is pentan-2-one, a ketone, and the reaction stops there. Treat the same acyl chloride with excess CH3MgBr and two methyls add; after workup the product is the tertiary alcohol 2-methylpentan-2-ol.
The Grignard adds twice because the ketone it makes is still reactive enough to be attacked as fast as it forms, so using only one equivalent does not help, as Grignard reagents showed. A cuprate is not reactive enough to attack that ketone. It adds once and stops, which makes it a reliable way to turn an acyl chloride into a ketone.
Opening an epoxide
A cuprate also opens epoxides. A Grignard can do this too, so it is not one of the three abilities above. Lithium dimethylcuprate opens 2-methyloxirane at its CH2 carbon, the less hindered end of the ring. It attacks that sp3 carbon from the back, opposite the C–O bond that breaks, in the same SN2 a Grignard uses. After workup the product is butan-2-ol.
Selectivity as a general principle
Because a cuprate is mild, it tolerates groups a Grignard would attack: esters, nitriles and, in many cases, a ketone elsewhere in the molecule. So a cuprate can open an epoxide in a molecule that also holds an ester, where a Grignard would attack the ester too. When a synthesis needs one site changed in a molecule that has several, the answer is usually a less reactive reagent, not cleverer conditions.
What carries forward
A Gilman reagent is the selective carbon nucleophile. It adds 1,4 to an enone where a Grignard adds 1,2. It couples with vinyl and aryl halides, which SN2 cannot touch. And it adds once to an acyl chloride where a Grignard adds twice. The next section, Palladium cross-coupling, joins vinyl and aryl partners more generally, with a catalyst instead of a full equivalent of copper. Whenever a question gives you an enone and asks for the product, check first which metal the nucleophile carries.