Stir bromoethane with magnesium turnings in dry diethyl ether, and the magnesium slowly dissolves. The product is ethylmagnesium bromide, CH3CH2MgBr. The metal has inserted itself into the carbon–bromine bond.
A compound R–MgX made this way from a halide R–X is a Grignard reagent. R is the carbon group and X is the halogen. In general:
R–X + Mg → R–MgX
No atom is lost: the magnesium simply slots in between the carbon and the halogen. The carbon that was δ+ in the halide is δ− in the Grignard reagent, so it now behaves as a carbon nucleophile. This is the polarity flip that Why C–metal means nucleophilic carbon explained.
Which halides work
The halogen matters. Iodides and bromides react with magnesium easily. Chlorides are sluggish and usually need heat and THF (tetrahydrofuran, a ring-shaped ether). Fluorides do not react at all, because the C–F bond is far too strong.
The kind of carbon matters much less. Bromobenzene, a bromine on a benzene ring, gives phenylmagnesium bromide, PhMgBr (Ph is short for phenyl, a benzene ring used as a group). A halogen on a benzene ring makes an aryl halide, and a halogen on a C=C carbon makes a vinyl halide. Neither undergoes SN2 or SN1, yet both form Grignard reagents. Forming a Grignard reagent is not a substitution at carbon, so the rules that block SN2 and SN1 do not apply. Vinyl halides and aryl chlorides usually need THF rather than diethyl ether.
Nothing else in the halide may react with the reagent. A bromide that also carries an O–H, an N–H, a carboxyl group or a C=O cannot be turned into a Grignard reagent. The first molecules of reagent to form react with that group on the starting material still in the flask. Why C–metal means nucleophilic carbon set out this test, and the section on Grignards as bases, below, shows how to get around it.
Why the solvent has to be ether
Look at the magnesium in CH3CH2MgBr. It forms only two bonds, one to carbon and one to bromine, so it has just four valence electrons around it. That makes it electron-poor. An ether oxygen has two lone pairs and can share one of them with the magnesium.
Two ether molecules do this, each giving one lone pair, and that brings the magnesium to eight electrons. This coordination keeps the reagent dissolved and stable enough to use.
Diethyl ether and THF both work, and THF binds the magnesium more strongly, as Ether chemistry noted. A hydrocarbon solvent has no oxygen to offer, and in it the reaction does not work. A solvent with an O–H, such as ethanol or water, protonates the reagent as fast as it forms.
What a Grignard attacks
Methylmagnesium bromide adds to propanal, and acidic workup gives butan-2-ol. The mechanism takes two arrows and then a proton. The carbonyl carbon is δ+, an electrophile, because oxygen pulls the C=O electrons toward itself. The nucleophilic carbon attacks it, and the C=O π bond moves onto oxygen, leaving a magnesium alkoxide. H3O+, added afterward in a separate step called the workup, protonates the alkoxide to the alcohol. Carbonyl Chemistry, the next chapter, builds every reaction of aldehydes and ketones on this same attack.
Every reaction in this section follows that pattern. What changes is the electrophile, and with it the product and the number of carbons the chain gains. The table lists the set.
| 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, opened at the less hindered carbon | Alcohol, with the OH on the second new carbon, next to the one the reagent bonded to | two, for ethylene oxide |
| Nitrile, then hydrolysis | Ketone | the nitrile carbon plus its R′ group |
| Ester or acid chloride (2 equiv or excess) | Tertiary alcohol, two R groups added | see the next section |
The first three rows follow one counting rule. Count the carbon groups on the carbonyl carbon, then add one for the Grignard reagent. The total is the number of carbon groups on the carbinol carbon, the carbon that carries the OH, and that number is the class of the alcohol.
Two rows add exactly one carbon, and they are worth learning as a pair because they give different functional groups. Formaldehyde gives a primary alcohol. CO2 gives a carboxylic acid: the reagent adds to one C=O of CO2, and workup protonates the carboxylate salt that forms (Carboxylic acids, later, draws it in full). When a question asks you to lengthen a chain by exactly one carbon, choose between these two by the functional group you want at the end.
The nitrile row is the one people miscount. A nitrile R′C≡N hands over its own carbon and the R′ group attached to it. So R–MgX plus CH3C≡N gives the ketone R–CO–CH3, which is two carbons longer than R, not one. Nitriles, later in the course, shows why this reaction stops at the ketone. HCN cannot serve as a one-carbon nitrile here. It is an acid (pKa about 9), so it simply protonates the Grignard reagent.
The ester problem
Ethyl propanoate looks like a route to a ketone. Add CH3MgBr once, and it attacks the C=O just as it attacks a ketone’s. The alkoxide that forms then pushes its electrons back down to re-form the C=O, and ethoxide, CH3CH2O−, leaves. So butan-2-one should be left. It is not. The butan-2-one reacts with more CH3MgBr, and the product after workup is 2-methylbutan-2-ol.
The reason is that the ketone is more electrophilic than the ester. In the ester, the OEt oxygen donates a lone pair into the C=O, and that makes its carbonyl carbon less δ+. Once OEt has gone, nothing donates, so the ketone takes the next molecule of reagent first. (Add, then lose a group to re-form the C=O: Nucleophilic acyl substitution, later in the course, shows this is how every acid derivative reacts.)
The reaction therefore takes two equivalents of Grignard reagent, and in practice an excess. The product is a tertiary alcohol carrying two identical R groups from the reagent, and those two groups are how you spot this case in a question. Using only one equivalent does not stop it at the ketone, because the ketone is consumed faster than it forms.
An acid chloride, RCOCl (also called an acyl chloride), ends up in the same place for a different reason. Its chlorine barely donates to the C=O, so it is even more electrophilic than a ketone, and the ketone it forms is less reactive than the acid chloride was. That ketone is still reactive enough to be attacked as fast as it forms, though. So an acid chloride and a Grignard reagent also give the tertiary alcohol, not the ketone.
The Weinreb amide of propanoic acid is CH3CH2–CO–N(CH3)OCH3. It is an amide whose nitrogen carries a methyl group and a methoxy group, OCH3. CH3MgBr adds to its C=O once, giving the usual magnesium alkoxide. Its carbon now carries four groups, so this species is called a tetrahedral intermediate.
Next, the methoxy oxygen binds the same magnesium by sharing one of its lone pairs, so the magnesium is gripped by two oxygens of one molecule. Together they close a five-membered ring of carbon, oxygen, magnesium, oxygen and nitrogen. A metal held by two donor atoms of the same molecule like this is chelated.
The chelate holds together while the reaction runs. It cannot collapse to a ketone while reagent is still in the flask, so no second addition is possible. Water breaks the ring open during workup, and only then does butan-2-one form.
Grignards as bases: the usual problem
Try to add CH3MgBr to the ketone of 4-hydroxybutan-2-one. The reagent never reaches the ketone. It takes the proton from the O–H instead and leaves as methane. Why C–metal means nucleophilic carbon explained why the proton wins.
Every reaction above assumes the substrate has no acidic hydrogen. An alcohol, a carboxylic acid, an amine with an N–H (primary or secondary), a thiol (R–SH) or a terminal alkyne anywhere in the molecule gives up its proton to the Grignard reagent. The products are R–H and a magnesium salt. The reagent is used up and the carbonyl is untouched.
Two routes get around the problem. Consider them in this order:
- Reorder. Run the Grignard step before the acidic group is put in or unmasked. It costs no extra steps and is usually possible. For an N–H, an S–H, a carboxylic acid or a terminal alkyne, reordering is usually the answer.
- Protect. When the alcohol has to be present during the Grignard step, cap it as a silyl ether: an ether in which a silicon group, SiR3, replaces the hydrogen of the O–H. To put the cap on, treat the alcohol with a silyl chloride such as TBSCl (tert-butyldimethylsilyl chloride) and imidazole, a mild base that takes up the HCl released. The O–H becomes O–SiR3, and with no O–H left, nothing quenches the reagent. After the Grignard step, fluoride, usually as TBAF (tetrabutylammonium fluoride), takes the cap off and gives back the O–H. The cost is two extra steps, one on and one off. Protecting groups, later in the course, returns to silyl ethers alongside the other protecting groups. A C=O in the halide can be masked in the same spirit, as an acetal (Acetals & hemiacetals, in the next chapter).
Step 1 — count carbons. The halide has three and the target has five. Two carbons have to arrive with the electrophile.
Step 2 — decide where the new OH must go. The target is a primary alcohol, and its OH sits at the far end of the two new carbons. Only the epoxide row of the table does that. Ethylene oxide adds two carbons and leaves the oxygen on the end one.
Step 3 — check the alternatives and say why they fail. Formaldehyde adds one carbon and gives butan-1-ol. Acetaldehyde adds two, but it puts the OH on a carbon that already carries a methyl, so it gives pentan-2-ol, a secondary alcohol. CO2 adds one carbon and gives a carboxylic acid. Only the epoxide puts the OH in the right place.
Step 4 — write the conditions in order. (i) Mg, dry diethyl ether; (ii) ethylene oxide; (iii) H3O+. Step (iii) matters. Step (ii) leaves a magnesium alkoxide, and the alcohol exists only once that is protonated.
Step 5 — check where the ring opens. No acid is present when the reagent meets the epoxide, so nothing protonates the ring oxygen and no carbocation character develops. The opening is a plain SN2 at the less hindered carbon, as in Epoxides. Ethylene oxide is symmetrical, so here it makes no difference. With 2-methyloxirane it decides the answer: the reagent takes the CH2 end, and the product is hexan-2-ol. Under acid, which a Grignard reagent never meets, the rule changes: the protonated epoxide opens at the carbon best able to carry the positive charge.
Common trap. Students often swap steps (ii) and (iii), or write them as one step. Acid or water added before the epoxide destroys the reagent, so nothing is left to open the ring. (Adding the halide, the magnesium and the electrophile together, with the acid kept until the end, is a real variant called the Barbier reaction. The standard order is still to make the reagent first and then add the electrophile to it, slowly and cold.)
What carries forward
A Grignard reagent is a carbon nucleophile. It adds once to an aldehyde or ketone and twice to an ester, and gives a primary, secondary or tertiary alcohol depending on what it hit. Any acidic proton in the flask destroys it. When a route needs a new C–C bond next to an oxygen, this is usually the reaction to reach for. When a route that works on paper fails in the flask, an acidic hydrogen is usually the reason. Carbonyl Chemistry comes next and puts this addition to work across every aldehyde and ketone. A more reactive relative, the organolithium, waits until Organolithiums and acetylides, after the acid derivatives.