Every substitution on a benzene ring so far has run one way. The ring is electron-rich, so it attacks an electrophile, then loses H+ and becomes aromatic again. That is Electrophilic aromatic substitution.
Now try the reverse on chlorobenzene. Warm it with methoxide, CH3O−, a nucleophile that easily displaces chloride from chloromethane. Nothing happens. Chlorobenzene is an aryl halide: a halogen bonded directly to a carbon of an aromatic ring. Neither of the substitutions you know works on it.
SN2 fails for a geometric reason. The nucleophile has to attack C1 from the side opposite the C–Cl bond, and on a ring that side points into the middle of the ring. To get there, the nucleophile would have to pass through the ring. C1 would also have to turn inside out, and a carbon held flat in a ring cannot.
SN1 fails because the cation it needs is very unstable. If chloride left on its own, C1 would keep an empty sp² orbital lying in the plane of the ring. The ring’s π electrons are in p orbitals at right angles to that plane, so they cannot spread into the empty orbital, and nothing stabilizes the positive charge.
Two classic routes can still put a nucleophile on an aromatic ring. The other groups on the ring decide which one runs. A third, newer route is palladium cross-coupling. A metal catalyst joins a carbon group to an unactivated aryl halide, with no nitro group and no strong base.
Route one: SNAr, when the ring is electron-poor
Start with 1-chloro-2,4-dinitrobenzene: chlorine on C1, and nitro groups (NO2) on C2 and C4. Warm it with methoxide, and OCH3 replaces the chlorine to give 1-methoxy-2,4-dinitrobenzene. The two nitro groups make the difference. They are strong electron-withdrawing groups (EWGs): groups that pull electron density out of the ring, as you saw in Ortho/meta/para directing effects. They leave the ring electron-poor.
This reaction is SNAr, which stands for nucleophilic substitution at an aromatic carbon. It takes two steps. First, methoxide adds to C1, the carbon that carries the chlorine. C1 now holds both OCH3 and Cl, so it is sp³, and the ring is no longer aromatic. The electrons of the C1=C2 double bond move onto C2 as a lone pair, which gives C2 a negative charge. This negatively charged intermediate is a Meisenheimer complex.
Second, the lone pair on C2 moves back in to re-form the C1=C2 double bond. Chloride leaves, and the ring is aromatic again.
The first step is the slow one, because it costs the ring its aromaticity. The nitro groups pay part of that cost. Follow the lone pair around the ring: it can sit on C2, C4 or C6, the carbons ortho and para to C1. It never sits on C3 or C5.
In 1-chloro-2,4-dinitrobenzene, both nitro groups sit on carbons that carry the charge. From there the charge moves onto a nitro oxygen, and an electronegative oxygen holds a negative charge far better than carbon does. Spreading the charge this way lowers the complex’s energy enough for the reaction to go.
Count the electrons in the complex. Because C1 is sp³, the ring of p orbitals is broken, and the other five carbons share six π electrons: two C=C bonds and the lone pair. That makes a cyclohexadienyl anion: a six-membered ring with two C=C bonds and a lone pair, whose negative charge is shared by C2, C4 and C6. It has the same shape as the arenium ion of electrophilic substitution, with the opposite charge. EAS runs through a cation that electron donors stabilize, and SNAr runs through an anion that withdrawing groups stabilize.
The mechanism predicts four things you can test.
The withdrawing group must be ortho or para to the leaving group. Compare 1-chloro-4-nitrobenzene with 1-chloro-3-nitrobenzene. In the first, the nitro group is on C4, a carbon the charge visits, so the charge can reach its oxygens. In the second, it is on C3, which the charge never visits. A meta nitro group still pulls electrons through the σ bonds (induction), and that speeds the reaction up measurably. But induction is a small effect, far too small to make SNAr practical. The reaction needs resonance.
More withdrawing groups make it faster. One ortho or para nitro group is enough with a good nucleophile and heat. Two make it an easy reaction. Three, as in 2-chloro-1,3,5-trinitrobenzene (picryl chloride), make it nearly instant. Exam questions usually use two, which is why 1-chloro-2,4-dinitrobenzene keeps coming up. For SNAr, nitro is the strongest common activating group; cyano (C≡N) and carbonyl groups also work, less strongly. These are the same groups that deactivate the ring toward EAS. Both effects come from pulling electrons out of the ring: that slows attack by an electrophile and speeds attack by a nucleophile.
Fluorine is the best leaving group here. 1-Fluoro-2,4-dinitrobenzene reacts far faster than its chloro partner. The order usually given is F > Cl > Br > I, the reverse of SN1 and SN2. The reason lies in which step is slow. The slow step, which sets the rate, is the nucleophile adding (see Energy diagrams), and the C–X bond does not break in it. So bond strength barely matters. What matters is how hard the halogen pulls electrons away from C1. Fluorine, the most electronegative, pulls hardest: it makes C1 the most electrophilic, and it helps hold the negative charge that the attack creates.
No rearrangement. The nucleophile arrives exactly where the leaving group was, so SNAr gives one product.
Step 1. Read the ring. The nitro groups are on C2 and C4, one ortho and one para to the fluorine. Both can take charge from the carbon under attack, so SNAr can run.
Step 2. Check the leaving group. Fluorine is fine here. It is the best leaving group in SNAr, because the C–F bond does not break in the slow step.
Step 3. Add. The lone pair on the nitrogen of methylamine attacks C1. C1 becomes sp³ and holds both F and N, so the ring is no longer aromatic. The nitrogen now has four bonds and a positive charge, and the ring carries the negative charge.
Step 4. Place the negative charge. It sits on C2, C4 and C6. Two of those carry a nitro group, so the charge spreads onto the nitro oxygens.
Step 5. Eliminate. The lone pair comes back into the ring, fluoride leaves, and the ring is aromatic again.
Step 6. Lose H+. The nitrogen still carries two hydrogens and a positive charge. A second molecule of methylamine removes one of those hydrogens as H+.
Product: N-methyl-2,4-dinitroaniline, with the nitrogen exactly where the fluorine was. One product, no rearrangement.
1-Fluoro-2,4-dinitrobenzene is Sanger’s reagent. A preview of Peptides and proteins: it reacts the same way with the free NH2 at one end of a protein chain (the N-terminus). The new bond from that nitrogen to the ring survives the acid that later breaks the chain into amino acids, so the label stays on the end amino acid and identifies it.
Route two: benzyne, when the ring has no withdrawing group
Chlorobenzene has no nitro groups, so SNAr is out. Treat it with sodium amide, NaNH2, in liquid ammonia (ammonia cooled below its boiling point, −33 °C), and it does react. The product is aniline, C6H5NH2. The amide ion, NH2−, is a very strong base. Despite the name, it has nothing to do with the C=O amide group of Esters & amides.
This mechanism starts with the base. Amide removes the hydrogen on C2, the carbon next to the chlorine. That leaves an aryl anion: a lone pair and a negative charge on a ring carbon, in the sp² orbital the C–H bond used. The lone pair then pushes chloride off C1. What remains is benzyne, a benzene ring with an extra bond between C1 and C2, drawn as a triple bond.
Benzyne does not last long. Amide adds to one end of the extra bond, which leaves a lone pair on the other end. That aryl anion takes H+ from an ammonia molecule, and the product is aniline.
The extra bond is a poor one. Each of its two carbons supplies an sp² orbital that lies in the plane of the ring and points outward, away from the other. The two orbitals overlap only weakly, side on, so benzyne is strained, extremely reactive, and lasts only until something attacks it.
The extra bond is also not part of the aromatic system. The ring’s six π electrons stay in p orbitals above and below the plane, at right angles to the new bond, so benzyne is still aromatic. Its reactivity comes from the weak in-plane bond alone.
Each end of the extra bond is a carbon, and in the benzyne from chlorobenzene the two are equivalent. Amide can add to either one. That gives the reaction its signature: two products, one with the nucleophile where the leaving group was and one with it on the neighboring carbon.
This scrambling is how benzyne was shown to exist. In 1953, Roberts made chlorobenzene with a 14C atom at C1. (14C is a radioactive isotope of carbon, so it can be tracked into the product.) He treated it with potassium amide, KNH2, in liquid ammonia. In about half the aniline, NH2 sat on the labeled carbon. In the other half it sat on the carbon next door. The split was within a few percent of even. No direct displacement can put the nucleophile on a carbon that never carried the leaving group.
Step 1. Find the hydrogens next to the halogen. Toluene is methylbenzene, so 2-chlorotoluene has CH3 on C1 and Cl on C2. Of the two carbons beside C2, C1 carries the methyl, so only C3 has a hydrogen to lose.
Step 2. Place the benzyne. Amide removes the hydrogen from C3, chloride leaves from C2, and the extra bond forms between C2 and C3.
Step 3. Add at both ends. Amide adding at C2 puts NH2 where the chlorine was and gives 2-methylaniline. Adding at C3 gives 3-methylaniline. These methylanilines also go by the common names o-toluidine and m-toluidine.
The two products need not form in equal amounts. Once the ring carries a substituent, adding to one end of the benzyne leaves a carbanion, a carbon with a negative charge, on the other end. The substituent can favor one of those two positions for the charge.
A methyl group barely chooses. 4-Chlorotoluene gives a benzyne between C3 and C4, and amide adds at both ends, so the product is a real mixture of 4-methylaniline and 3-methylaniline.
An oxygen does choose. Anisole is methoxybenzene, so 3-bromoanisole has OCH3 on C1 and Br on C3. The base removes the hydrogen on C2, between the two groups, because both pull electrons from it and make it the most acidic. The benzyne forms between C2 and C3. Amide then adds to C3, the end farther from the oxygen. That leaves the negative charge on C2, right beside the carbon that carries the oxygen, where the oxygen’s inductive pull stabilizes it. The main product is 3-methoxyaniline (m-anisidine). The same holds for any inductively withdrawing group: the charge ends up on the carbon next to it.
The oxygen’s lone pairs play no part here, even though OCH3 donates electrons to the ring by resonance in EAS. The carbanion’s lone pair sits in an sp² orbital in the ring plane, at right angles to the π system, so resonance cannot reach it. Only the inductive pull counts. Calculations describe the same result in a second way: the oxygen bends the benzyne slightly, which leaves C3 more open to attack. Both pictures predict the same product.
The same question, where a carbanion is most stable, comes back in Birch reduction later in this chapter.
Benzyne has one more use, as a dienophile: the two-carbon partner in a Diels–Alder reaction. Its weak in-plane bond reacts far more readily than an ordinary alkene C=C. Benzyne made in the presence of furan gives the bicyclic adduct at room temperature, with no heating and no withdrawing group to activate it. The furan oxygen ends up as a one-atom bridge across the new ring. When a question makes benzyne with a diene present, this is the reaction it is testing.
Telling them apart
Two questions sort any aryl halide. Ask them in this order.
| SNAr | Benzyne | |
|---|---|---|
| Ring needs | Electron-withdrawing group ortho or para | No EWG needed; an H next to the halogen |
| Conditions | Good nucleophile (CH3O−, an amine), often warm | Very strong base, NaNH2 |
| Order | Add, then eliminate | Eliminate, then add |
| Intermediate | Meisenheimer complex | Benzyne |
| Best halide | F, by a wide margin | Cl, Br and I work; F barely reacts with NaNH2 and needs a stronger base, such as an organolithium. |
| Product | One, at the original position | Two, at neighboring positions |
What carries forward
There are two ways to put a nucleophile on a ring, and they are opposites. SNAr is addition then elimination through a stabilized anion. It needs withdrawing groups in the right places and gives one product. Benzyne is elimination then addition through a strained intermediate. It needs no activation, only a very strong base, and gives two products.