Aromatic Follow-Through · Section 107 of 116

SNAr and benzyne

Practice this — interactive lesson

Everything you know about substitution on a benzene ring so far runs one way: the ring is electron-rich, so it attacks an electrophile. That is electrophilic aromatic substitution, and it is the whole of the aromatic chapter up to here.

The reverse — a nucleophile displacing a leaving group on the ring — looks impossible at first, and for a plain aryl halide it is. Chlorobenzene does not do SN2, because the backside of that carbon is behind the ring and the carbon is sp². It does not do SN1 either, because an aryl cation is badly unstable.

But there are two ways round it, and which one runs depends entirely on what else is on the ring.

Route one: SNAr, when the ring is electron-poor

Put strong electron-withdrawing groups on the ring, ortho or para to the leaving group, and the picture changes. The nucleophile adds to the carbon bearing the leaving group, giving a negatively charged intermediate called a Meisenheimer complex, and then the leaving group departs.

addition first, elimination second — the opposite order from every substitution you have seen

Aromaticity is lost in the intermediate and restored at the end. That is the cost, and it is why the reaction needs help: the nitro groups pay for it by delocalizing the negative charge onto their own oxygens.

Three consequences fall out of that mechanism, and all three are testable:

Route two: benzyne, when the ring is not activated

Take chlorobenzene with no activating groups and force it with a very strong base — NaNH2 in liquid ammonia. Something quite different happens. The base removes a hydrogen ortho to the chlorine, chloride leaves, and what remains is benzyne: a benzene ring with an extra bond made from two sp² orbitals in the plane of the ring.

That bond is a terrible one. The two orbitals point away from each other rather than overlapping face to face, so benzyne is strained, extremely reactive, and lasts only long enough to be attacked.

The mechanism is elimination first, addition second — the mirror image of SNAr. And because the triple bond is symmetric, the nucleophile can attack either of its two carbons, which gives the reaction its signature: a mixture of two products, one with the nucleophile where the leaving group was and one with it on the neighboring carbon.

That scrambling is how benzyne was proved to exist. Label the carbon bearing the chlorine with 14C, run the reaction, and the label comes out split roughly evenly between two positions — which no direct displacement could produce.

Telling them apart

SₙAr — needs an EWG ortho or paraAr–Xadd the nucleophileMeisenheimer anionONE productBenzyne — needs only an ortho HAr–Xeliminate HXstrained benzyneTWO productsadd first, then eliminateeliminate first, then addThe two steps are the same two steps, run in opposite orders — and that is whatdecides whether the nucleophile can land anywhere except where the halide was.
The two routes to a nucleophile on a ring, which are mechanistic opposites. SₙAr adds first, through an anion the withdrawing groups stabilize, so the nucleophile arrives exactly where the halide was. Benzyne eliminates first, and because the strained bond it forms is symmetric, the nucleophile can land on either of two carbons.Read the substrate before choosing. Withdrawing groups ortho or para to the halide mean SₙAr — a meta group helps only inductively, which is measurable and far too little, because only ortho and para reach the charge by resonance. A bare ring plus NaNH₂ means benzyne. A ring with neither activation nor an ortho hydrogen runs neither, which is the answer people skip past.
SNArBenzyne
Ring needsEWG ortho or paraNothing; works on plain rings
ConditionsMild nucleophile, warmVery strong base, NaNH2
OrderAdd, then eliminateEliminate, then add
IntermediateMeisenheimer complexBenzyne
Best halideF, by a wide marginAll work; rates differ, and an ortho H is the real requirement
ProductOne, at the original positionTwo, at adjacent positions

Read the substrate first. Nitro groups ortho or para to a halide mean SNAr; a bare aryl halide plus NaNH2 means benzyne; and a substrate with no hydrogen ortho to the halide cannot go through benzyne at all, because there is nothing to eliminate.

What carries forward

Two ways to put a nucleophile on a ring, and they are opposites. SNAr is addition–elimination through a stabilized anion, needs withdrawing groups in the right places, and gives one product. Benzyne is elimination–addition through a strained intermediate, needs no activation and a brutal base, and gives two.