Suppose you want biphenyl: two benzene rings joined by a single bond between one carbon of each. Bromobenzene looks like the natural starting point. Its bromine sits on a ring carbon, and bromide is a good leaving group.
A carbon of a benzene ring is an aryl carbon, and a halogen bonded to one makes an aryl halide. Neither SN2 nor SN1 can replace that halogen. An SN2 needs the nucleophile to reach the back of the carbon that carries the bromine, and the back of a ring carbon faces into the ring. An SN1 needs the bromide to leave on its own, and the cation it would leave behind, an aryl cation, is far too unstable to form.
A phenyl Grignard, PhMgBr, runs into the same wall. (Ph is short for a phenyl group, C6H5–.) Its carbon is a strong nucleophile, but it cannot reach the carbon that carries the bromine.
A Gilman reagent, R2CuLi, is the one reagent so far that does couple with an aryl halide (Cuprates and conjugate addition). That coupling has limits, though. It works best when the R group it brings is a simple alkyl group, and it joins two aryl groups poorly. Every coupling uses up a full equivalent of the copper reagent, and the cuprate tolerates fewer functional groups elsewhere in the molecule.
One other route does replace the halogen, but only when strong electron-withdrawing groups sit on the ring. Aromatic Follow-Through covers it. Bromobenzene has no such groups.
The same problem comes up at a C=C. A carbon of a C=C is an alkenyl carbon, and a halogen bonded to one makes an alkenyl halide, also called a vinyl halide (Alkynes). Its carbon is sp2, like a ring carbon, and it does not undergo SN2 or SN1 either.
Cross-coupling solves the problem with a metal catalyst, almost always palladium. The palladium bonds to the two partners one at a time and then joins them. It makes the aryl halide reactive by bonding to it directly, so no nucleophile ever has to attack the ring carbon.
Cross-coupling is also far more general than the cuprate coupling. The halide can be aryl or alkenyl. The partner can bring an aryl, alkenyl or alkynyl group, or, with the right catalyst, an alkyl group. A small amount of palladium does the job, and most other functional groups in the molecule come through untouched.
Heck, Negishi and Suzuki shared the 2010 Nobel Prize in Chemistry for this work. Cross-coupling is now the most-used carbon–carbon bond-forming reaction in the pharmaceutical industry.
One catalytic cycle, three steps
Start with the simplest case. Stir bromobenzene with phenylboronic acid, PhB(OH)2, a palladium catalyst and sodium carbonate in hot water and toluene, and biphenyl forms. A boronic acid is a boron atom carrying one carbon group and two OH groups. This reaction is a Suzuki coupling.
The catalyst is usually Pd(PPh3)4, a palladium atom with four triphenylphosphine molecules bonded to it. Each phosphine gives the metal the lone pair on its phosphorus. A molecule bonded to a metal this way is a ligand (Lewis acids and bases). The ligands stop the palladium atoms from clumping together into bare metal, and one or two of them come off to make room for the molecules that react.
Palladium(II) acetate, Pd(OAc)2, with a phosphine added, is a common alternative. It is reduced to Pd(0) in the flask before the cycle starts.
The palladium does three things, one after another, and is then ready to start again. The figure follows it around the cycle.
| Step | What happens | Palladium goes |
|---|---|---|
| Oxidative addition | Pd inserts into the C–Br bond of bromobenzene. That bond breaks, and Pd bonds to both Ph and Br. | Pd(0) → Pd(II) |
| Transmetalation | The second phenyl moves from boron onto palladium, and bromide leaves. | stays Pd(II) |
| Reductive elimination | The two phenyls on palladium bond to each other and leave as biphenyl. | Pd(II) → Pd(0) |
Now generalize. Write any aryl group as Ar, so any aryl halide is Ar–X. Swap bromobenzene for Ar–X, and phenylboronic acid for any partner that carries a carbon group R, and the three steps stay the same. Ar–X goes in, and Ar–R comes out.
The Roman numeral after Pd is its oxidation state: the charge the palladium would carry if each group bonded to it took the shared electron pair away. In Ph–Pd–Br, count the phenyl as Ph− and the bromine as Br−. That leaves the palladium at +2, so it is Pd(II). A ligand such as PPh3 brought its own pair and would leave as a neutral molecule, so it does not change the count. That is why Pd(PPh3)4 is Pd(0). It is the same bookkeeping Oxidation levels in organic chemistry uses for carbon: each bond’s electrons go to the more electronegative atom.
The step names follow that number. Oxidative addition oxidizes the palladium by two, from 0 to +2. Reductive elimination reduces it by two and hands back the Pd(0) the cycle began with. Transmetalation only swaps one group for another, so the number stays at +2.
Because the palladium ends each turn exactly as it began, it is a catalyst, like the palladium in Catalytic hydrogenation or the acid in a Fischer esterification, which gets its H+ back at the end. Here, though, the catalyst returns through a change of oxidation state. A little of it does the whole job. Chemists give the amount in mol %: 2 mol % means two palladium atoms for every hundred molecules of aryl halide. At that loading, each palladium atom goes around the cycle about fifty times.
Oxidative addition is also the step that decides which halide reacts. Give 1-bromo-4-chlorobenzene one equivalent of phenylboronic acid, and the new bond forms where the bromine was. The chlorine stays on the ring.
The reason is the bond the metal has to break. A C–Cl bond is much stronger than a C–Br bond, and a C–I bond is weaker than either. So aryl iodides react fastest and bromides react well. Chlorides are slow unless the palladium carries a specially designed bulky ligand. If a question offers you a choice of halide on the same ring, it is asking about this step.
An aryl triflate, Ar–OTf, is the common alternative to a halide. OTf stands for –OSO2CF3. Alcohol reactions turned an alcohol’s OH into this group, R–OTf, to make it an excellent leaving group, one that leaves as easily as a halide. A phenol is converted into an aryl triflate the same way.
The named reactions, by what the partner is
The Suzuki is one of five named couplings, each named after the chemist who developed it. In all five, the aryl or alkenyl halide and the palladium play the same parts. What changes is the partner that brings the second carbon group.
| Reaction | Nucleophilic partner | Forms |
|---|---|---|
| Suzuki | Boronic acid, R–B(OH)2, with base | Biaryls, most often |
| Stille | Organotin compound, R–SnBu3 | The same, with no base needed |
| Negishi | Organozinc, R–ZnX | The same, and tolerates esters |
| Sonogashira | Terminal alkyne, with a copper cocatalyst and an amine base | An aryl alkyne |
| Heck | An alkene, with no organometallic at all | A substituted alkene |
The Suzuki is the one to know well. Boronic acids are stable in air and water, low in toxicity and easy to store. The Stille coupling uses an organotin compound (an organostannane), R–SnBu3, a tin atom carrying three butyl groups and the R group. It works at least as well chemically, but organotin compounds are seriously toxic. That is why the Suzuki took over in industry.
The Suzuki’s base is not optional. A neutral boronic acid hands its carbon group to palladium only slowly. Its boron has only six valence electrons and an empty p orbital, like the boron in BF3. Hydroxide, which the carbonate supplies in water, adds to that empty orbital and gives a borate, [PhB(OH)3]−. The borate gives up its phenyl to palladium far more readily, so the borate is what transmetalates.
The Negishi coupling uses an organozinc, R–ZnX. A C–Zn bond is less polar than a C–Mg bond (Why C–metal means nucleophilic carbon), so an organozinc is a milder nucleophile than a Grignard. It can carry an ester that a Grignard would attack.
The Sonogashira coupling joins an aryl halide to a terminal alkyne. Iodobenzene and phenylacetylene, Ph–C≡C–H, give diphenylacetylene, Ph–C≡C–Ph. The flask holds palladium, copper(I) iodide and triethylamine, Et3N, and each has its own job.
Palladium runs the same cycle as before. Copper supplies the partner, the alkynyl group Ph–C≡C–, but copper has to make that group first. Copper(I) bonds side-on to the triple bond, to its π electrons rather than to one carbon, and pulls electron density away from the terminal C–H. Without copper, removing that proton (pKa 25) takes a base as strong as NaNH2 (Alkynes). With copper bound, triethylamine is strong enough to remove it. The product is a copper acetylide, Ph–C≡C–Cu.
The copper acetylide is the partner that transmetalates. It passes its alkynyl group to Ph–Pd–I and takes the iodide back as CuI. So two cycles turn at once: a copper cycle that makes the acetylide, and the palladium cycle that uses it. The triethylamine ends up as its salt, Et3NH+ I−. It is often used in large excess, as the solvent as well as the base.
The Heck reaction is the odd one out. Iodobenzene and methyl acrylate, CH2=CH–CO2CH3, react with a palladium catalyst and triethylamine to give methyl trans-cinnamate, Ph–CH=CH–CO2CH3. The phenyl group has replaced a hydrogen on the CH2 end of the alkene.
The partner here is a plain alkene, with no metal to hand anything over, so there is no transmetalation. After oxidative addition, the alkene bonds side-on to the palladium. Then comes migratory insertion: the phenyl moves from the palladium onto one carbon of the C=C, and the palladium bonds to the other. Both arrive from the same side of the double bond. The phenyl goes to the CH2 carbon, the less crowded end.
Next comes β-hydride elimination. The carbon bonded to palladium is the α carbon, and its neighbor is the β carbon. The palladium takes a hydrogen from the β carbon as a hydride, H with its two electrons (Reducing carbonyls), and the C=C forms again. The H and the Pd must sit on the same side of the C–C bond, so the molecule first turns about that single bond to line them up.
The β carbon has two hydrogens, and either one could line up with the palladium. Losing one leaves the phenyl and the ester on opposite sides of the new C=C. Losing the other would put them on the same side, crowded together. The first path is easier, so the product is the trans alkene.
What is left on the metal is H–Pd–I, which is Pd(II), not the Pd(0) catalyst. The base, triethylamine here or a carbonate, removes HI from it and gives back Pd(0). So the Heck needs a full equivalent of base, even though the alkene is never deprotonated. Without that last step, the cycle would not close.
Why the halide is aryl or alkenyl, and not alkyl
Try the Suzuki with 1-bromobutane in place of bromobenzene. Oxidative addition into an alkyl C–Br bond is slower, but the bigger problem comes after it. The butyl–Pd–Br that forms has hydrogens on its β carbon, so the same β-hydride elimination that ends the Heck can happen here. It gives but-1-ene and H–Pd–Br, and with the simple palladium catalysts on this page it is usually faster than reductive elimination. So the butyl group never couples.
An aryl group on palladium has no such hydrogen to lose. The carbons next to the Pd-bonded carbon are sp2 ring carbons, and their hydrogens lie in the ring plane, pointing away from the metal. The step removes a hydrogen only from an sp3 β carbon. An alkenyl group is safe too, because its β carbon is usually the other sp2 carbon of the C=C. So an aryl or alkenyl group stays on the palladium until reductive elimination joins it to its partner. The restriction comes from β-hydride elimination, not from how good the halide is as a leaving group.
A methyl group has no β carbon at all, so a methyl on palladium cannot lose a hydrogen this way either. Newer ligands, and nickel catalysts, now couple many other alkyl groups as well. With the palladium catalysts on this page, though, the halide partner is aryl or alkenyl.
Step 1 — cut the bond you cannot otherwise make. The target is a biaryl: two benzene rings joined directly by a single bond. It carries an acetyl group, CH3C(=O)–, on one ring, para to the other ring. Cutting a bond in a target and asking which two pieces could form it is called a disconnection. Synthesis & Retrosynthesis makes a method of it; here one cut is enough. No earlier reaction joins two rings well, so cut the bond between them. The pieces are an aryl halide and an aryl boronic acid.
Step 2 — decide which half carries which. Put the bromine on the ring that has the ketone, which gives 4′-bromoacetophenone. Put the boron on plain benzene, which gives phenylboronic acid. The reverse assignment also works; this one uses the simplest boronic acid.
Step 3 — check the other functional groups. The molecule has a ketone. A phenyl Grignard would add to it and give a tertiary alcohol (Grignard reagents). The boronic acid leaves it alone. Its C–B bond is much less polar than a C–Mg bond, so its phenyl carbon is barely nucleophilic. The phenyl moves only to palladium, and once there its only way off is reductive elimination.
Step 4 — write the conditions. Pd(PPh3)4 (2 mol %), Na2CO3 (aq), toluene/ethanol, heat.
Step 5 — trace the cycle. Pd(0) inserts into the C–Br bond of 4′-bromoacetophenone and becomes Pd(II). The borate, [PhB(OH)3]−, hands its phenyl to the palladium, which stays Pd(II). The two rings join, and Pd(0) is released. The ketone takes no part at any stage.
Why this changed how molecules get made
Compare two routes to the same target. A linear route runs ten steps in a row, each in 85% yield. Yields multiply, so it ends with 0.8510, about 20%. A convergent route builds two halves separately, in five steps each, and then joins them in one more step. Each half comes through at 0.855, about 44%, and the joining step still leaves about 37%. That is nearly double, for one extra reaction.
Cross-coupling suits that joining step, because it runs late in a synthesis without disturbing the rest of the molecule. The two halves can also be varied independently. A medicinal chemistry team that wants fifty versions of one drug candidate can make fifty versions of one half and couple each to the other half, instead of running fifty separate syntheses.
What carries forward
A palladium catalyst joins an aryl or alkenyl halide to a carbon partner by oxidative addition, transmetalation and reductive elimination. The named reactions differ in the partner. The Suzuki, with a boronic acid, is the default. The Heck has no organometallic partner and ends with a β-hydride elimination instead. When you reach Synthesis & Retrosynthesis, a bond between two rings is a cut to spot at once, because this is the reaction that makes it.