Difficulty: Advanced | Prerequisites: EAS Core Concepts (Doc 1) and Friedel–Crafts / Synthesis (Doc 2) from this series. Familiarity with nucleophilic substitution (SN1/SN2) from earlier chapters.
Most of Chapter 18 focuses on electrophilic aromatic substitution, where the ring is the nucleophile. But aromatic rings can also act as electrophiles when they carry strong electron-withdrawing groups and a good leaving group, opening up nucleophilic aromatic substitution (SNAr) and the benzyne pathway. These reactions round out the chapter and appear on exams as mechanism questions or product-prediction problems. This doc also covers the trickier end of directing-effect problems: compounds with multiple substituents, ranking rings by reactivity, and predicting which position reacts on complex molecules.
Nucleophilic aromatic substitution requires electron-withdrawing groups (especially ortho/para to the leaving group) to stabilise the anionic intermediate. The benzyne mechanism operates with strong bases and no electron-withdrawing groups. For complex molecules with multiple substituents, the strongest activator controls regiochemistry, and steric effects break ties.
Meisenheimer complex
The anionic, resonance-stabilised intermediate (also called a sigma complex for NAS) formed when a nucleophile adds to an aromatic ring bearing electron-withdrawing groups. Think of it as: the NAS equivalent of the arenium ion in EAS, except it carries a negative charge rather than a positive one.
Benzyne (aryne)
A highly reactive, strained intermediate with a formal triple bond in a six-membered ring. Formed by elimination of HX from an aryl halide under strongly basic conditions. In simple terms, it is a fleeting species where two adjacent carbons share an extra bond that is immediately attacked by a nucleophile.
Addition–elimination (SNAr)
The two-step mechanism for nucleophilic aromatic substitution: the nucleophile adds to the ring (forming the Meisenheimer complex), then the leaving group departs. Think of it as: the reverse order of EAS, where EAS is addition of E+ then loss of H+, and SNAr is addition of Nu− then loss of X−.
Elimination–addition (benzyne pathway)
The two-step mechanism where a strong base first eliminates HX to form benzyne, then a nucleophile adds across the triple bond. In simple terms, the leaving group and a neighbouring H leave first, creating the reactive benzyne, and then the nucleophile attacks.
A good leaving group on the ring (typically F, Cl, Br, or I; fluorine is often the best leaving group in SNAr because the rate-determining step is nucleophilic addition, not loss of the leaving group, and C–F is the most polar bond, making the carbon most electrophilic).
One or more strong electron-withdrawing groups (EWGs) positioned ortho and/or para to the leaving group. These stabilise the Meisenheimer complex by delocalising the negative charge. The more EWGs, the faster the reaction.
A nucleophile (OH−, OR−, NH2−, NR2−, etc.).
Addition (rate-determining): The nucleophile attacks the carbon bearing the leaving group. The ring temporarily loses aromaticity, forming the Meisenheimer complex. The negative charge is stabilised by resonance with the EWGs at ortho/para positions.
Elimination: The leaving group departs, restoring aromaticity. The nucleophile has replaced the leaving group.
EWGs must be ortho or para to the leaving group to stabilise the Meisenheimer complex. A nitro group meta to the leaving group cannot stabilise the intermediate by resonance (the negative charge never reaches the meta carbon in the resonance structures).
Problem 18.72 from the problem set illustrates this: ortho-bromonitrobenzene reacts with NaOH via SNAr because –NO2 is ortho to –Br. Meta-bromonitrobenzene would not undergo this reaction because –NO2 is meta to –Br and cannot stabilise the intermediate. Para-bromonitrobenzene would react (–NO2 para to –Br).
18.35a: Bromobenzene + NaOH at 350°C → phenol. At this extreme temperature, the mechanism is benzyne (no EWGs present), not SNAr.
18.35b: Bromination of a ring already carrying a substituent, proceeding via standard EAS.
18.35c: ortho-Nitroiodobenzene + NaNH2 → ortho-nitroaniline. The –NO2 group is ortho to –I, stabilising the Meisenheimer complex. This is SNAr (addition–elimination).
Benzyne forms when an aryl halide is treated with a very strong base (NaNH2, NaH) at elevated temperature, and the ring does not carry EWGs ortho/para to the leaving group (so SNAr is not possible).
Elimination: The strong base removes a proton from the carbon adjacent to the carbon bearing the halogen. The halide departs simultaneously, forming benzyne, a species with a strained "triple bond" in the ring (actually two overlapping sp2 orbitals in the plane of the ring, not a true alkyne triple bond).
Addition: A nucleophile (often the conjugate base of the base used, e.g. NH2−) attacks one of the two carbons of the benzyne. The other carbon picks up a proton from the solvent.
Because benzyne is symmetrical with respect to the two carbons that formed the "triple bond," the nucleophile can attack either carbon. This means two products can form, with the nucleophile bonded to the carbon that originally bore the leaving group, or to the adjacent carbon. With para-bromotoluene and NaNH2, two amine products are obtained.
Para-bromotoluene treated with NaNH2 gives two products: para-methylaniline (NH2 where Br was) and meta-methylaniline (NH2 on the adjacent carbon). Drawing the benzyne intermediate makes the two products obvious.
The more electron-rich a ring, the faster it reacts. The order from most to least reactive:
Aniline (–NH2) > phenol (–OH) > anisole (–OCH3) > toluene (–CH3) > benzene > halobenzenes (–F, –Cl, –Br) > benzoic acid (–COOH) > nitrobenzene (–NO2)
This is tested directly in Problem 18.40 (which compound is most/least activated).
Step 1: Classify each substituent (activating or deactivating, ortho/para or meta director).
Step 2: Identify the positions each group directs to.
Step 3: If both groups agree on a position, that position reacts. If they disagree, the stronger activator wins.
Step 4: Among the "winning" positions, rule out any that are sterically blocked (flanked by bulky groups).
For each compound in this problem, you identify the most reactive position by:
Drawing the ring and numbering the carbons.
Marking the directing effect of each substituent with arrows.
Finding the position(s) where both arrows agree, or where the stronger group's arrows point.
Eliminating positions with heavy steric crowding.
For example, in a compound with –CH3 (ortho/para director) and two –NO2 groups (meta directors), the –CH3 group is the strongest activator (the –NO2 groups are deactivators). The electrophile enters ortho or para to –CH3, but only at positions that are not between two nitro groups.
When a molecule has multiple aromatic rings with different substituents, rank them by electron density. The ring with the strongest activating group reacts first. A ring attached to nitrogen (as in an amine) is more activated than one attached to an ester oxygen. The ring with the strongest deactivation (e.g. bearing two carbonyls) reacts last or not at all under mild conditions.
This kind of selectivity is exactly how pharmaceutical chemists modify drug molecules: they exploit differences in electron density between rings to target one position selectively. The same logic guides dye chemistry, polymer synthesis, and materials science wherever aromatic functionalisation is needed.
"SNAr and SN2 are the same thing on a ring." They are not. SN2 is a one-step, backside-attack mechanism that does not occur on sp2 carbons. SNAr is a two-step addition–elimination through the Meisenheimer complex. The geometry of the aromatic ring prevents SN2.
"Benzyne is a true triple bond like an alkyne." Benzyne's "third bond" consists of two sp2 orbitals overlapping in the plane of the ring, not the perpendicular p-orbital overlap of a normal triple bond. It is much weaker and much more reactive.
"If a ring has any EWG, NAS will occur." The EWG must be ortho or para to the leaving group. A meta EWG cannot stabilise the Meisenheimer complex and NAS does not proceed.
⚠️ Exam questions on NAS typically ask you to draw the Meisenheimer complex and explain why meta-substituted EWGs do not work.
⚠️ Benzyne questions ask you to draw two products and explain why both form.
⚠️ "Rank these compounds by reactivity" is a standard multiple-choice or short-answer format.
⚠️ Multi-ring molecules (Problem 18.44, 18.74) require you to identify which ring reacts and at which position.
Q: Which of the following best describes the effect of an –OH group on EAS? (a) EDG, meta director (b) EDG, ortho/para director (c) EWG, meta director (d) EWG, ortho/para director
A: (b). The –OH group donates electron density by resonance (lone pairs on oxygen feed into the ring), making it an electron-donating group and an ortho/para director. This is Problem 18.76.
Q: Predict the major product when the compound shown (a toluene derivative with –NO2 at the para position) is treated with HNO3/H2SO4, then Cl2/FeCl3 (Problem 18.77).
A: The first step nitrates toluene to give para-nitrotoluene. The –CH3 group is an ortho/para director (weak activator), and the –NO2 group is a meta director (strong deactivator). The –CH3 controls regiochemistry because it is the activator. Chlorination puts –Cl ortho to –CH3 (and meta to –NO2), giving 2-chloro-4-nitrotoluene as the major product.
Q: When para-chloronitrobenzene is treated with NaOH, then H3O+, what is the major product? (Problem 18.78)
A: The –NO2 group is para to –Cl, so SNAr is operative. OH− attacks the carbon bearing Cl, forming the Meisenheimer complex stabilised by the para-NO2. Cl− departs. The product is para-nitrophenol. Answer is (d): HO–C6H4–NO2.
Q: ortho-Bromonitrobenzene is treated with NaOH at elevated temperature. Draw the product and identify the mechanism (Problem 18.72).
A: SNAr (addition–elimination). –NO2 is ortho to –Br, so it stabilises the Meisenheimer complex. The product is ortho-nitrophenol. Meta-bromonitrobenzene would not react because –NO2 cannot stabilise the intermediate from the meta position. Para-bromonitrobenzene would react (–NO2 para to –Br stabilises the intermediate).
Q: Rank these from most to least reactive toward EAS: nitrobenzene, phenol, bromobenzene, anisole (–OCH3).
A: Phenol > anisole > bromobenzene > nitrobenzene. Phenol and anisole are both strongly activated (–OH and –OCH3 donate by resonance); phenol is slightly more activated because OH is a better donor than OCH3. Bromobenzene is weakly deactivated. Nitrobenzene is strongly deactivated.
SNAr connects to nucleophilic substitution (Ch. 7) by contrasting how aromatic sp2 carbons behave differently from aliphatic sp3 carbons. The benzyne mechanism links to elimination reactions (E2, Ch. 8), as it involves base-induced elimination. Ranking reactivity ties into thermodynamics and kinetics (Hammond postulate): the more stable sigma complex leads to a lower-energy transition state and a faster reaction. These topics also set the stage for diazonium chemistry and aromatic side-chain reactions in later chapters.
nucleophilic aromatic substitution, SNAr, addition elimination aromatic, Meisenheimer complex, benzyne, aryne mechanism, elimination addition, strong base aromatic, NaNH2 aromatic, directing effects multiple substituents, competing directors, reactivity ranking aromatic, activated ring, deactivated ring, electron donating group, electron withdrawing group, fluorine leaving group SNAr, ortho-bromonitrobenzene NaOH, picric acid nitration, organic chemistry chapter 18, NYU chemistry