Start with bromomethane, CH3Br. Its carbon is an electrophile: in an SN2 reaction, hydroxide attacks that carbon and bromide leaves. Now stir bromomethane with magnesium metal in dry ether. The magnesium slips in between the carbon and the bromine, and you get CH3MgBr, methylmagnesium bromide. That same carbon now does the attacking. It adds to the carbonyl carbon of a ketone, as the next section, Grignard reagents, shows.
A compound with a bond between carbon and a metal is an organometallic compound. CH3MgBr belongs to the most widely used family, the Grignard reagents, written R–MgX in general, where R is any carbon group and X is Cl, Br or I. This section explains why the metal turns carbon from an electrophile into a nucleophile. The next section puts that nucleophile to work, and the carbonyl chapters after it use it on almost every page.
The switch matters because a nucleophile supplies the atom that forms the new bond. Hydroxide, alkoxides, amines and halide ions all attach an oxygen, a nitrogen or a halogen. To join two carbon pieces, one of the two carbons has to be the nucleophile. Before the Grignard reagent, the carbon nucleophiles you met were narrow. Cyanide brings exactly one carbon, as a nitrile, and an acetylide ion has to come from a terminal alkyne. An organometallic reagent can bring almost any carbon group.
Bond a carbon to a metal and the polarity flips
The flip comes from electronegativity, the pull an atom exerts on the electrons it shares. On the Pauling scale, carbon is 2.55 and bromine is 2.96. Bromine pulls the shared pair of the C–Br bond toward itself, so bromine carries a partial negative charge (δ−) and carbon a partial positive charge (δ+). That δ+ carbon is the one a nucleophile attacks.
Magnesium is 1.31, far below carbon. In a C–Mg bond carbon is the more electronegative partner, so the shared pair sits closer to carbon. Now carbon is δ− and magnesium is δ+. Lithium, at 0.98, is lower still, so a C–Li bond is polarized the same way and more strongly.
Reversing the usual polarity of an atom on purpose is called umpolung, German for “polarity reversal”. Turning an alkyl halide into a Grignard reagent is the simplest case of it.
The C–Mg bond is polar covalent, not ionic. A Grignard reagent in ether is not a free carbanion, a carbon with a lone pair and a full negative charge. Still, it reacts as though it were one in every reaction in this course. So it is fair to treat its carbon as R−, and a curved arrow that shows it attacking starts at the C–Mg bond, where those electrons are.
The family, in order of reactivity
Other metals do the same job, some more strongly than magnesium and some less. The table lists the five reagents you will meet (boron, strictly, is a metalloid rather than a metal), from the element farthest below carbon to the one closest to it. The number after each metal is its electronegativity; carbon’s is 2.55.
| Reagent | Metal | Character |
|---|---|---|
| R–Li | Li (0.98) | Organolithium. More reactive and more basic than a Grignard reagent. |
| R–MgX | Mg (1.31) | Grignard reagent. Strongly nucleophilic and strongly basic. The workhorse. |
| R–ZnX | Zn (1.65) | Organozinc. Mild enough to leave an ester in the same molecule alone. |
| R2CuLi | Cu (1.90) | Gilman reagent, also called a cuprate. Much milder, and a soft nucleophile (see Lewis acids): its reactive electrons are spread over a large, easily distorted copper atom rather than packed onto one small carbon. It adds to the far carbon of a C=C joined to a C=O rather than to the C=O itself, and it couples with alkyl halides. Both come later in the course, in Cuprates and conjugate addition. |
| R–B(OH)2 | B (2.04) | Boronic acid. Barely nucleophilic on its own, so it needs a palladium catalyst (Palladium cross-coupling). |
The order follows one trend. The farther the metal sits below carbon, the more negative charge the carbon carries and the more it behaves like a carbanion. More carbanion character makes a reagent more reactive and less selective, that is, less able to pick out one group in a molecule and leave the others alone. Lithium sits farthest below carbon, so R–Li is the most aggressive. Copper and boron sit much closer to carbon, so their reagents are gentle and can be aimed at one site in a molecule that has several. The ranking is rough at the gentle end, but the direction holds.
The one thing that destroys all of them
Add a drop of water to CH3MgBr and you get methane, CH4, and Mg(OH)Br. The carbon has taken a proton from the water, and the reagent is gone.
The reason is basicity. A carbanion is the conjugate base of an alkane, and alkanes have pKa values around 50. That makes R− one of the strongest bases in organic chemistry, strong enough to take a proton from almost anything with an O–H, N–H or S–H bond, and from a terminal alkyne as well.
So an organometallic reagent and an acidic hydrogen cannot share a flask. Each of these destroys the reagent:
- Water (pKa 15.7) does it instantly, so the glassware and the solvent must be dry.
- Alcohols (pKa about 16) and carboxylic acids (about 5) do the same.
- Amines with an N–H do too, even though ammonia’s pKa is 38 and a dialkylamine’s about 36.
- A terminal alkyne (pKa 25) is deprotonated rather than attacked.
The same test applies to the halide you make the reagent from, and that check is easy to forget. Take 4-bromobutan-1-ol. As soon as the first molecules of Grignard reagent form, the O–H of a neighboring molecule protonates them. An N–H or a CO2H in the halide fails the same way. 5-Bromopentan-2-one fails in a different way: the new reagent attacks the ketone of a neighboring molecule. “Make the Grignard reagent from this halide” only works when nothing else in the molecule reacts with the reagent.
An acidic hydrogen wins even when the molecule also has a carbonyl group. Proton transfer to a base this strong is extremely fast, faster than addition to a C=O. So if your substrate carries an O–H anywhere, the reagent reacts with that proton first. Nothing adds to the carbonyl until every acidic hydrogen has been used up.
This is the most common way a synthesis that works on paper fails in the flask. One fix is to run the organometallic step before the acidic group is in the molecule. The other is to protect the group: convert it for a while into a group with no acidic hydrogen, and convert it back afterward. The Protecting groups section, later in the course, shows how.
You add 1.0 equivalent of CH3MgBr to 4-hydroxybutan-2-one, HOCH2CH2COCH3. One equivalent means one molecule of reagent for each molecule of ketone. After aqueous workup you isolate your starting ketone. Why, and what fixes it? The figure after this example draws each stage.
Step 1 — list the reactive sites. There are two: the O–H at the end of the chain, with a pKa near 16, and the ketone C=O on carbon 2.
Step 2 — rank them. CH3− is the conjugate base of methane, pKa about 50. It takes the O–H proton faster than it adds to the ketone, so the acid–base reaction comes first.
Step 3 — write what happens. CH3MgBr + HOCH2CH2COCH3 → CH4 + −OCH2CH2COCH3 +MgBr. Methane is a gas and leaves the flask. All the reagent has been spent, and the alkoxide that remains does not attack the ketone.
Step 4 — work up. Water protonates the alkoxide back to the alcohol, so you isolate the compound you started with. Nothing looks as if it went wrong; there is simply no product.
Step 5 — fix it. Use 2.0 equivalents. The first is spent on the O–H and the second adds to the ketone, so workup gives HOCH2CH2C(CH3)2OH, 3-methylbutane-1,3-diol. This fix can work, but it wastes a whole equivalent of reagent, and the yield is often poor. The cleaner route is the one named above: run the Grignard step before the OH is in the molecule. Grignard reagents, the next section, shows that route and a way to protect the OH.
Exams usually draw the molecule and ask for “the product”, and the OH at the far end is easy to overlook. The trap is that “starting material returned” is a legitimate answer, and it is the right one whenever the equivalents do not cover the acidic protons.
Sometimes the proton transfer is the point. Deuterium, D, is hydrogen with an extra neutron, and D2O is water made with it. To quench a reagent is to destroy it on purpose with a proton source. Quench CH3CH2MgBr with D2O and you get CH3CH2D. The deuterium sits exactly where the magnesium was, which is how chemists label one chosen position in a molecule.
Why a Grignard reagent and an alkyl halide do not simply couple
Suppose you want pentane from CH3MgBr and 1-bromobutane. On paper it is an SN2 reaction: the reagent’s carbon attacks C1, the carbon that carries the bromine, and bromide leaves. In practice it gives little pentane.
The substitution is slow. A Grignard reagent is a very strong base but a poor SN2 nucleophile toward an alkyl halide, so side reactions have time to take over. One of them is an E2 elimination: the reagent takes a hydrogen from C2, a C=C forms between C1 and C2, and bromide leaves. The products are methane and but-1-ene.
The fix is a less reactive reagent, not a more reactive one. A Gilman reagent couples cleanly with an alkyl halide precisely because it is milder, as Cuprates and conjugate addition shows. This is the reactivity–selectivity trade from earlier on this page.
What carries forward
Carbon bonded to a metal is nucleophilic carbon, and that is what makes carbon skeletons buildable. The course uses it in four ways: Grignard reagents, next, and organolithiums, which attack carbonyls directly; cuprates, which add where a Grignard reagent will not; and palladium chemistry, which joins two pieces that would not react with each other on their own. The last three come after the acid derivatives, in Organolithiums, Cuprates & Cross-Coupling.