A Grignard reagent is made by stirring an alkyl, vinyl or aryl halide with magnesium turnings in dry ether. The metal inserts into the carbon–halogen bond:
R–Br + Mg → R–MgBr
Nothing is added and nothing leaves. The same carbon that was δ+ in the halide comes out δ− in the product, which is the polarity reversal the previous section described.
Why the solvent has to be ether
Diethyl ether or THF is not incidental. The magnesium of a Grignard is electron-poor, and the two lone pairs on the ether oxygen coordinate to it, which is what keeps the reagent in solution and stable enough to use. Try the same reaction in a hydrocarbon and it does not work; try it in anything with an O–H and the reagent destroys the solvent.
What a Grignard attacks
The mechanism is always the same: the nucleophilic carbon attacks an electrophilic carbon, and an acidic workup protonates whatever anion is left. What differs is the electrophile, and the products are worth learning as a set, because the carbon count changes differently in each case.
| Electrophile | Product after H3O+ | Carbons gained |
|---|---|---|
| Formaldehyde, HCHO | Primary alcohol | one |
| Any other aldehyde | Secondary alcohol | two or more |
| Ketone | Tertiary alcohol | three or more |
| Carbon dioxide | Carboxylic acid | one |
| Epoxide (at the less hindered carbon) | Alcohol, further along the chain | two, for ethylene oxide |
| Nitrile, then hydrolysis | Ketone | one plus the R group |
| Ester or acyl chloride | Tertiary alcohol, two R groups added | see below |
The one-carbon extensions are the three worth memorizing together, because each ends somewhere different: formaldehyde gives a primary alcohol, CO2 gives a carboxylic acid, and a nitrile gives a ketone. Asked to lengthen a chain by exactly one carbon, you pick from that list according to what functional group you want at the end.
The ester problem
An ester looks like a way to make a ketone, and it is not. The Grignard adds once, the tetrahedral intermediate collapses by expelling the alkoxide, and what is left is a ketone that is more electrophilic than the ester was. The second equivalent of Grignard attacks it immediately.
So the product is a tertiary alcohol carrying two identical R groups from the Grignard, and stopping at the ketone by using one equivalent does not work — the ketone is consumed faster than it forms. An acyl chloride ends up in the same place by a different route. It is the most electrophilic acyl derivative, so the ester argument does not transfer — the ketone it gives is less reactive than the acyl chloride was. It is still reactive enough to be attacked as fast as it forms, which is enough to spoil the selectivity.
Grignards as bases, which is usually the problem
Every one of the reactions above assumes the substrate has no acidic hydrogen. Put an alcohol, a carboxylic acid, an amine, a thiol or a terminal alkyne anywhere in the molecule and the Grignard removes that proton instead, giving R–H and a magnesium salt. The reagent is consumed and the carbonyl is untouched.
Three routes around it, in the order you should consider them:
- Reorder. Do the Grignard step before the acidic group is installed. Free, and usually possible.
- Use extra reagent. One equivalent to kill the acidic proton and a second to do the chemistry. Wasteful but sometimes simplest.
- Protect. A silyl ether on the alcohol, removed with fluoride afterwards. Two extra steps, so it is the last resort rather than the first.
What carries forward
A Grignard is nucleophilic carbon that adds once to an aldehyde or ketone and twice to an ester, gives a primary, secondary or tertiary alcohol depending on what it hit, and is destroyed by any acidic proton in the flask. When a route needs a C–C bond next to an oxygen, this is almost always the reaction — and when a route on paper does not work in practice, an acidic hydrogen is almost always the reason.