Aromatic Nucleophilic Substitution

Aryl halides can undergo nucleophilic substitution reactions (SNAr) but in a very different manner as compared to how alkyl halides do it.

The “normal” reaction of an aromatic compound is the SEAr because the ring is electron rich. However, should the aromatic compound bear one or more electron-attracting substituents, the nucleophilic substitution (SNAr) can be carried out.

Electrostatic potential maps of chlorobenzene and of its derivatives with one, two and three nitro groups: the ring turns from green to blue as it loses electron density — From left to right: chlorobenzene, 1-chloro-4-nitrobenzene, 1-chloro-2,4-dinitrobenzene and 1-chloro-2,4,6-trinitrobenzene.

The presence of an increasing number of strong electron-withdrawing nitro groups, makes the ring increasingly electron deficient ("bluer"). In this situation, a good leaving group, like chlorine, can be displaced by a strong nucleophile, like hydroxide:

Chlorobenzene reacts with hydroxide in very drastic conditions. In contrast, the same reaction on 2,4,6-trinitrochlorobenzene can be performed in much more smooth conditions, proving that a SNAr process is strongly facilitated if there are electron-withdrawing groups attached to the ring.

Hydrolysis of chlorobenzenes with NaOH: chlorobenzene needs 350 °C and high pressure; with one, two and three nitro groups 150, 100 and 30 °C are enough

Model of the attack of the hydroxide ion perpendicular to the ring plane of 1-chloro-4-nitrobenzene

The nucleophilic attack by (HO⁻) takes place along a trajectory perpendicular to the ring plane, where the p orbitals of the π cloud point.

The mechanism of the SNAr has nothing to do with the aliphatic substitution (either S_N1 or S_N2) on sp³ carbons.

Nevertheless, you will surely find a strong resemblance to the SEAr, the main difference being that in the SNAr a negative (instead of positive) charge is developed. This is reasonable because in the SNAr the attacking species (nucleophile) bears an electron excess.

The electron deficiency of the carbon bonded to the chlorine, increased by the resonance effect of the para-NO₂ group, makes it easy for the hydroxide to attack it. A negatively charged intermediate is thus formed, named after its discoverer, the German chemist Meisenheimer.

Addition-elimination mechanism: HO⁻ adds to 1-chloro-4-nitrobenzene to form the Meisenheimer intermediate, whose negative charge is delocalized onto the nitro group; loss of Cl⁻ gives 4-nitrophenol

Electrostatic potential map of the Meisenheimer intermediate: the negative charge concentrates at the ortho and para positions and on the nitro group

It can be seen in the molecular model how the negative charge concentrates in the ortho and para positions relative to the attack site. The resonance forms help us to understand that. The 4-nitro group assists in the stabilization of the developed negative charge.

Did you understand the mechanism? If you did you must then be able to answer the following question:
Why does 3-nitrochlorobenzene react less easily with hydroxide than 4-nitrochlorobenzene?

Benzyne mechanism

There is another mechanism that explains the following facts:

For example:

1) 4-Chlorotoluene reacts with the strong base sodium amide and produces m-toluidine (m-methylaniline), together with p-toluidine.

2) When the same reaction is carried out in the presence of a good diene, like 1,3-cyclohexadiene, the Diels-Alder adduct is formed.

4-Chlorotoluene with NaNH₂ in NH₃ gives 3-methylaniline (m-toluidine) and, in the presence of 1,3-cyclohexadiene, a Diels-Alder adduct

Benzyne must be involved because otherwise the formation of the Diels-Alder adduct cannot be understood:

Benzyne mechanism: the amide removes an H next to the Cl, NaCl is lost and benzyne forms, which adds NH₂⁻ or is trapped by cyclohexadiene

The structure of benzyne is not explained as a conventional triple bond in which there are two regular π clouds formed by the overlap of the corresponding p orbitals. In the case of benzyne, one of the π clouds is formed from the incomplete overlap of two vicinal sp² orbitals sharing two electrons. The overlap is incomplete because the two sp² orbitals are not parallel for obvious geometrical reasons.

Conventional triple bonds are reactive but very stable. Benzyne is very unstable and extremely reactive. That's why it has to be "captured" in situ with the diene.

Orbitals of benzyne: the second π bond, in the plane of the ring, comes from the weak overlap of two sp2 orbitals