Electrophilic substitution reactions

Why substitution, not addition. An alkene's $\pi$ bond is a single, localised pair of electrons with nothing to lose by becoming two $\sigma$ bonds. Benzene's six $\pi$ electrons are delocalised over all six carbons, giving it a resonance (aromatic) stabilisation energy that addition would sacrifice — one ring carbon would turn $sp^3$ and the delocalised loop would break permanently. Substitution avoids this: the ring carbon that is attacked briefly becomes $sp^3$, but the very next step throws out a proton and restores full conjugation, so the aromatic system is only ever borrowed, never destroyed.

The electrophile must be strong. Benzene's $\pi$ cloud is a much weaker nucleophile per encounter than an isolated alkene $\pi$ bond (spread over six carbons instead of two), so it will not react with mild electrophiles at all. Every one of the named reactions therefore begins with a dedicated step — a strong acid, an oleum, or a Lewis-acid catalyst — whose entire job is to manufacture a potent electrophile ($\text{NO}_2^+$, $\text{SO}_3$, $\text{Cl}^+$-like species, or a carbocation/acylium ion) before the ring ever gets involved.

One mechanism, four faces. Once a strong electrophile $E^+$ exists, every reaction below follows the identical two remaining stages: the ring attacks $E^+$ to give a positively charged, non-aromatic arenium ion (Wheland intermediate), and then a base removes the proton from the newly $sp^3$ carbon, regenerating the aromatic ring. Only the identity of $E^+$ and how it is generated differ from reaction to reaction:

where: $E^+$ is the electrophile specific to each named reaction ($\text{NO}_2^+$ for nitration, $\text{SO}_3$/$\text{HSO}_3^+$ for sulphonation, a halogen–Lewis-acid complex for halogenation, a carbocation or acylium ion for Friedel–Crafts), and $[\text{arenium ion}]^+$ is the resonance-stabilised, non-aromatic intermediate common to all four reactions.

The goal is to see why the three-step mechanism above is not arbitrary bookkeeping but a direct consequence of aromatic stability, and why it forces substitution as the outcome rather than addition.

Step 1. Benzene's delocalised $\pi$ system carries a real energetic bonus — the aromatic stabilisation energy — that comes only from having all six $p$-orbitals overlapping in one continuous ring. Any process that permanently converts a ring carbon to $sp^3$, as simple addition would, forfeits this bonus for good, which is why benzene resists the addition reactions that alkenes undergo so readily.

Step 2. When a strong electrophile does attack, the ring is only temporarily interrupted. The attacked carbon becomes $sp^3$, but the other five carbons still share a delocalised, resonance-stabilised positive charge (the arenium ion draws contributing structures with the charge on alternating carbons). This partial retention of delocalisation is what makes arenium-ion formation energetically feasible at all, even though it costs aromaticity temporarily.

Step 3. The arenium ion is a dead end unless it can regain full aromaticity, and it has an easy way out: losing the proton on the $sp^3$ carbon puts that carbon's electrons back into the ring, restoring the complete six-electron delocalised system. This step is strongly favoured thermodynamically — the system recovers the full aromatic stabilisation energy it briefly lost — which is why deprotonation, not any competing pathway, is essentially always what happens next.

Step 4. Put together, the ring never has an incentive to stay as a saturated, addition-type product: doing so would abandon aromaticity permanently. Losing H⁺ to regenerate the ring is always the lower-energy route, so substitution — not addition — is the universal outcome whenever benzene meets a sufficiently strong electrophile.

Problem: Write the complete mechanism for the nitration of benzene, showing the arenium-ion intermediate, and name the product.

Step 1. Generate the electrophile. Concentrated $\text{HNO}_3$ is protonated by concentrated $\text{H}_2\text{SO}_4$ and loses water to form the nitronium ion:

Step 2. Attack the electrophile. Benzene's delocalised $\pi$ electrons attack the electrophilic nitrogen of $\text{NO}_2^+$, forming a new C–N $\sigma$ bond. The attacked carbon becomes $sp^3$ (bonded to H, $\text{NO}_2$, and two ring carbons), and the positive charge is delocalised over the remaining five carbons as the resonance-stabilised arenium ion.

Step 3. Deprotonate. The base $\text{HSO}_4^-$ (generated in Step 1) removes the proton from the $sp^3$ carbon. Its electron pair returns to the ring, restoring full conjugation and the aromatic sextet.

Step 4. Read off the product. The product is nitrobenzene, $\text{C}_6\text{H}_5\text{NO}_2$, and $\text{H}_2\text{SO}_4$ is regenerated overall ($\text{H}_3\text{O}^+ + \text{HSO}_4^- \rightarrow \text{H}_2\text{SO}_4 + \text{H}_2\text{O}$), confirming its role as a catalyst rather than a consumed reagent.

Competitive stretch

Problem: Benzene reacts with acetyl chloride, $\text{CH}_3\text{COCl}$, in the presence of anhydrous $\text{AlCl}_3$. Identify the reaction type and product, and explain one advantage of acylation over Friedel–Crafts alkylation.

Step 1. Identify the reaction. This is Friedel–Crafts acylation: the Lewis acid $\text{AlCl}_3$ coordinates to the chlorine of the acid chloride and helps generate the electrophile, the acylium ion, $\text{CH}_3\text{CO}^+$, which is resonance-stabilised by the adjacent oxygen lone pair.

Step 2. Run the standard mechanism. The ring's $\pi$ electrons attack the acylium carbon to give the arenium-ion intermediate, which loses $\text{H}^+$ to restore aromaticity, regenerating the $\text{AlCl}_3$ catalyst.

Step 3. State the advantage. Unlike an alkyl carbocation, the acylium ion cannot rearrange by hydride or methyl shifts (its positive charge is already resonance-stabilised on the carbonyl carbon), so acylation gives one clean, unrearranged product. The product is acetophenone, $\text{C}_6\text{H}_5\text{COCH}_3$, and because the ketone product is deactivating, the reaction stops cleanly at mono-substitution — unlike alkylation, which over-reacts because the alkylbenzene product is more reactive than benzene itself.

• Drawing electrophilic aromatic substitution as a one-step process — examiners expect all three stages shown explicitly: electrophile generation, arenium-ion formation, and deprotonation.
• Forgetting the Lewis-acid catalyst for halogenation ($\text{FeCl}_3$ or $\text{AlCl}_3$) — without it, benzene does not react with $\text{Cl}_2$/$\text{Br}_2$, unlike alkenes, which add halogens directly with no catalyst.
• Confusing Friedel–Crafts alkylation (alkyl halide, alkyl carbocation, prone to rearrangement and over-alkylation) with acylation (acid chloride, resonance-stabilised acylium ion, self-limiting to mono-substitution).
• Drawing the arenium-ion intermediate as still fully aromatic — it is not: one ring carbon is $sp^3$, and the positive charge is delocalised over only the remaining five carbons until the proton is lost.

Ask the tutor: if the curved-arrow flow from ring to electrophile to arenium ion to product is confusing, ask the AI tutor to walk through it one arrow at a time for any single reaction above.