Stir 1-bromobutane with lithium metal in a dry hydrocarbon or ether solvent, and you get butyllithium, CH3CH2CH2CH2–Li. Chemists write it n-BuLi: Bu for butyl, and n for “normal”, the unbranched chain. It is an organolithium, a carbon bonded directly to lithium.
R–Br + 2 Li → R–Li + LiBr
This is the same recipe as a Grignard reagent, a halide plus a metal, with one difference in the count. Magnesium gives two electrons, but each lithium atom gives only one. So the reaction takes two lithium atoms: one ends up bonded to carbon, and the other leaves with the bromide as LiBr.
Everything a Grignard does, an organolithium also does. The reason to reach for one is electronegativity. Carbon’s electronegativity is 2.55. Lithium’s, 0.98, is even lower than magnesium’s, 1.31, so the carbon of a C–Li bond carries more negative charge. That makes an organolithium both a stronger nucleophile and a stronger base. Sometimes that extra reactivity is exactly what you need, but more often it is something you have to manage.
Where RLi succeeds and RMgX does not
RLi is shorthand for any organolithium, and RMgX for any Grignard. Two cases come up again and again where RLi works and RMgX does not.
A hindered ketone. Take diisopropyl ketone, 2,4-dimethylpentan-3-one, whose C=O sits between two bulky isopropyl groups. Isopropylmagnesium bromide, a bulky Grignard, struggles to reach that carbonyl carbon, so it does one of two other things.
It can act as a base. The carbon next to a C=O is called the α carbon, and a hydrogen on it is an α hydrogen. The Grignard pulls off an α hydrogen, and the anion left behind shares its charge between that carbon and the oxygen. That anion is an enolate, the subject of Enolate Chemistry later in the course. Here it simply sits until workup, and workup protonates it back to the ketone you started with. This path is called enolization.
It can also act as a reducing agent. The Grignard’s magnesium binds to the carbonyl oxygen, which brings a hydrogen on the Grignard’s β carbon (the carbon next to the one bonded to Mg) close to the carbonyl carbon. That hydrogen moves across with its pair of electrons, as a hydride, through a ring of six atoms. The Grignard’s own carbons leave as propene, and workup gives 2,4-dimethylpentan-3-ol, a secondary alcohol with no new carbon at all.
Isopropyllithium is smaller and more reactive, and it adds to the carbonyl carbon that the Grignard could not reach. Workup then gives the tertiary alcohol with three isopropyl groups on one carbon.
A carboxylic acid straight to a ketone. Treat butanoic acid with two equivalents of methyllithium, then work up with aqueous acid, and you isolate pentan-2-one. One equivalent is one mole of reagent for each mole of starting material.
At first glance this looks impossible. A carboxylic acid has an O–H, so the first equivalent of CH3Li is spent taking that proton, giving methane and the lithium carboxylate. A carboxylate is a very poor electrophile, because the group already carries a full negative charge.
An organolithium is reactive enough to add to it anyway. The second equivalent attacks the carbonyl carbon and gives a dianion: one carbon carrying two O− groups. A Grignard is not reactive enough to make this addition.
The dianion cannot collapse. To re-form a C=O it would have to push out O2−, which is no leaving group at all. So it waits in the flask. Acidic workup protonates both oxygens to give the hydrate, a carbon with two OH groups, and the hydrate loses water to give the ketone.
Deprotonation as the goal
An organolithium is an extremely strong base, so some of the ones you will meet are used to remove a proton rather than to form a C–C bond. The table lists them, along with LDA, a still more useful base made from n-BuLi:
| Reagent | Used for |
|---|---|
| n-BuLi | The general-purpose strong base. It deprotonates terminal alkynes, and some rings that contain oxygen or sulfur, such as furan, a flat ring of four carbons and one oxygen |
| t-BuLi | An even stronger and more reactive base; also the standard reagent for lithium–halogen exchange (below) |
| LDA (from n-BuLi and diisopropylamine) | The classic base for making enolates: very strong, but too bulky to add to a carbonyl (below) |
LDA is not itself an organolithium. Its lithium sits on nitrogen, in an N–Li bond, so it is a lithium amide, but it is made from an organolithium.
Bromobenzene, a benzene ring carrying a bromine, reacts with n-BuLi at −78 °C to give phenyllithium and 1-bromobutane. The lithium and the bromine simply trade places. This is lithium–halogen exchange, the second way to make an organolithium. It works on aryl and vinyl bromides and iodides, where the halogen sits on a benzene-ring carbon or on a C=C carbon. In general, Ar–Br + n-BuLi → Ar–Li + n-BuBr, where Ar stands for an aryl group: a benzene ring bonded through one of its carbons.
The exchange runs toward the more stable carbanion. The ring carbon is sp2, so its orbital has more s character than that of the sp3 carbon in a butyl group, and it holds a lone pair more tightly. The phenyl anion is therefore the weaker base, and the lithium ends up on the ring.
The exchange is over in seconds at −78 °C. Because it is so fast and so cold, chemists use it to put a lithium onto a molecule far too delicate to stir with lithium metal.
With t-BuLi, two equivalents are used. The first does the exchange and makes t-BuBr. The second removes that t-BuBr by E2, giving 2-methylpropene, so it cannot react with the new organolithium.
LDA needs a closer look. Its full name is lithium diisopropylamide, and it is made by using n-BuLi as a base on diisopropylamine, an amine whose nitrogen carries two isopropyl groups. The N–H has a pKa of about 36 and butane about 50, so the proton transfer goes all the way.
LDA combines two things. It is strong enough to remove an α hydrogen from a ketone completely, yet its two isopropyl groups keep it from attacking the C=O itself. The anion it makes is an enolate, the same kind of anion the Grignard made in the hindered-ketone case above. That pairing of high basicity with low nucleophilicity is something no Grignard can offer, and it is what makes controlled enolate chemistry possible when you reach Enolate Chemistry.
Acetylides, the cheapest carbon nucleophile
A terminal alkyne, such as propyne, has a C–H with a pKa of about 25. That is very acidic for a C–H. When the proton leaves, the lone pair stays in an sp orbital, which is half s in character, so the pair sits close to the carbon nucleus and the anion is stabilized. Alkynes made this argument in full.
NaNH2 or n-BuLi removes that proton to give an acetylide. It is a genuine carbon nucleophile.
An acetylide does three things, and the first is the one that matters most:
- SN2 on a methyl or primary halide, which joins two chains with a new carbon–carbon bond.
- Addition to an aldehyde or ketone, which gives a propargylic alcohol: an OH on the carbon next to the triple bond.
- Opening an epoxide at the less hindered carbon. This puts the epoxide’s two carbons between the triple bond and the OH, giving a homopropargylic alcohol. HC≡C− and ethylene oxide, then H3O+, give HC≡C–CH2CH2OH, but-3-yn-1-ol.
The substitution is the surprising one, because a Grignard or an organolithium does not reliably do it (see Why C–metal means nucleophilic carbon).
The metal is not what sets the acetylide apart: it is itself a sodium, lithium or magnesium salt. What sets it apart is that it is an sp carbanion, and so a far weaker base. Its conjugate acid has a pKa of 25, while the conjugate acid of an sp3 carbanion has a pKa of about 50. Being twenty-five orders of magnitude less basic is a large part of what lets the acetylide act as a nucleophile on a primary halide.
Step 1 — count. The target has five carbons and propyne has three, so the electrophile has to bring two.
Step 2 — make the nucleophile. NaNH2 in liquid ammonia takes the terminal C–H, whose pKa is 25. The conjugate acid of NH2− is NH3, with a pKa of 38. Ammonia is thirteen orders of magnitude weaker as an acid than the alkyne, so the deprotonation goes to completion. That gives CH3C≡C− Na+. n-BuLi would do the same job.
Step 3 — choose the electrophile. Bromoethane brings two carbons, and it is primary. Nothing more substituted will do. With 2-bromopropane, the acetylide acts as a base instead: E2 gives propene, and you recover your alkyne unchanged.
Step 4 — alkylate. CH3C≡C− + CH3CH2Br → CH3C≡CCH2CH3, pent-2-yne. The acetylide attacks the CH2 from the side opposite the bromine.
Step 5 — set the geometry. H2 over Lindlar catalyst adds both hydrogens to the same face of the triple bond, so it gives cis-pent-2-ene. Sodium in liquid ammonia would give the trans isomer instead. One alkyne, two sets of conditions, two answers; Catalytic hydrogenation explains why.
Trap to notice. The order cannot be reversed. If you reduce the alkyne first, no acidic C–H is left to deprotonate, so there is no nucleophile to alkylate with.
What carries forward
An organolithium is a Grignard with the volume turned up: the same reactions, more reactivity and less selectivity. Use one where a Grignard stalls on a hindered ketone, or where you want a ketone from a carboxylic acid. Recognize it, too, in its other role, as the base behind LDA and the acetylides. The next section goes the opposite way and turns the reactivity down, and that turns out to be more useful.