Electrophilic aromatic substitution (EAS) — nitration, halogenation, sulfonation, Friedel-Crafts alkylation - acylation;: Worked examples

The scenario matrix

Think of EAS the way you'd think of a checklist. There are only a handful of case classes, and if you can do one worked example from each row, nothing on an exam is new.

#	Case class	What is being tested	Filled by
A	Standard electrophile ( $E^{+}$ ) generation	Can you make each of the 5 electrophiles?	Ex 1 (nitration), Ex 2 (halogenation)
B	Reversible reaction	Sulfonation forward + backward (blocking group)	Ex 3
C	Degenerate / "no reaction" input	Missing catalyst, ring too electron-poor	Ex 4
D	Rearrangement trap	$1° \to 2°$ carbocation shift in alkylation	Ex 5
E	Over-substitution vs clean	Alkylation multiplies, acylation stops at one	Ex 6
F	Real-world multi-step synthesis	Order-of-operations word problem	Ex 7
G	Exam-style twist	Acylation-then-reduce to beat rearrangement	Ex 8

Every "sign/quadrant" analogue in chemistry is: which electrophile, is the ring rich or poor, and is the step forward or reverse. Those are our axes.

Cell A — Standard electrophile ( $E^{+}$ ) generation

Figure s01 — Nitration in three snapshots. Left: benzene's cyan hexagon with its central π cloud. Middle: after the amber $N O_{2}^{+}$ attacks, the top carbon goes $s p^{3}$ (labelled "E,H (sp3)") and three amber "+" signs show where the charge is delocalised in the arenium ion. Right: nitrobenzene after $H^{+}$ leaves, with $N O_{2}$ drawn on the ring. Follow the white arrows left→right; the amber arrow is $N O_{2}^{+}$ approaching the ring.

Figure — Electrophilic aromatic substitution (EAS) — nitration, halogenation, sulfonation, Friedel-Crafts alkylation - acylation;

Build $E^{+}$ . $H_{2} S O_{4}$ protonates the $- O H$ of $H N O_{3}$ ; water leaves, giving the ==nitronium ion $N O_{2}^{+}$ ==. $H N O_{3} + 2 H_{2} S O_{4} \to N O_{2}^{+} + H_{3} O^{+} + 2 H S O_{4}^{-}$ Why this step? Neutral $H N O_{3}$ is a weak electrophile; the ring needs a strongly electron-poor target. Stripping off water makes a full $+$ charge on nitrogen — a hungry electrophile. $H_{2} S O_{4}$ is the proton donor that pays for water's exit.
Ring attacks (this is the rate-determining step, RDS — the slowest step, defined above). Two of benzene's π electrons form a bond to $N O_{2}^{+}$ ; one carbon goes $s p^{3}$ and holds both $H$ and $N O_{2}$ . The $+$ charge spreads over the other 3 carbons (look at the amber charges in the figure). The species you now have is the arenium ion. Why this step? The electron-rich cloud is the nucleophile. Breaking aromaticity here is why this is the slow, rate-determining step.
Lose $H^{+}$ . $H S O_{4}^{-}$ plucks the proton off the $s p^{3}$ carbon → nitrobenzene, and $H_{2} S O_{4}$ is regenerated. Why this step? Kicking out $H^{+}$ (not adding a nucleophile) is the only way to get the aromatic ring back — the thermodynamic prize.

Verify: Balance atoms. Left of overall reaction: $C_{6} H_{6} + H N O_{3} \to C_{6} H_{5} N O_{2} + H_{2} O$ . Carbons $6 = 6$ , hydrogens $6 + 1 = 5 + 2$ ✓, nitrogens $1 = 1$ , oxygens $3 = 2 + 1$ ✓. The catalyst $H_{2} S O_{4}$ appears on both sides of the $E^{+}$ -generation step (regenerated), so it is truly catalytic.

Polarize $B r_{2}$ . The Lewis acid $F e B r_{3}$ has an empty orbital; it accepts a lone pair from $B r_{2}$ , dragging electron density onto itself: $B r_{2} + F e B r_{3} \to B r^{+} \dots F e B r_{4}^{-}$ Why this step? Neutral $B r_{2}$ is not electrophilic enough for an aromatic ring (alkenes are richer and can do it, benzene cannot). The Lewis acid manufactures a $B r^{+}$ -like end. Without $F e B r_{3}$ , no reaction — this is exactly the "benzene ≠ alkene" mistake.
Ring attacks $B r^{+}$ → the arenium ion (the $s p^{3}$ -carbon, charge-spread intermediate defined in the opening box; identical geometry to Ex 1's figure). Why this step? This is the same rate-determining step as Ex 1: the ring's electron-rich π cloud is the nucleophile, and it donates two π electrons into the polarized $B r^{+}$ end. Aromaticity breaks here, so this is the slow step that sets the reaction's speed.
Lose $H^{+}$ to $F e B r_{4}^{-}$ → bromobenzene + $H B r$ + regenerated $F e B r_{3}$ . Why this step? The arenium ion throws off the proton on the $s p^{3}$ carbon (rather than adding a nucleophile) because that is the only path back to the stable aromatic ring. $F e B r_{4}^{-}$ acts as the base that removes $H^{+}$ , and in doing so it releases $B r^{-}$ to rebuild the $F e B r_{3}$ catalyst — proving the catalyst is regenerated.

Verify: $C_{6} H_{6} + B r_{2} \to C_{6} H_{5} B r + H B r$ . Carbons $6 = 6$ , hydrogens $6 = 5 + 1$ ✓, bromines $2 = 1 + 1$ ✓. Catalyst regenerated. Forecast answer: no, plain $B r_{2}$ does nothing — you need the catalyst.

Cell B — Reversible reaction (sulfonation)

Forward: attach $- S O_{3} H$ . The electrophile is neutral but very electron-poor $S O_{3}$ (sulfur is starved by three oxygens). Ring attacks → arenium ion → lose $H^{+}$ → benzenesulfonic acid $C_{6} H_{5} S O_{3} H$ . Why this step? $S O_{3}$ 's sulfur carries a strong partial $+$ ; it acts as the $E^{+}$ even without a formal charge.
Reverse: strip $- S O_{3} H$ . With dilute aqueous acid and heat, water re-protonates the ring carbon and $S O_{3} H$ leaves. Sulfonation is the one EAS that runs both ways. Why this step? Concentrated/fuming conditions favour forward (little water); dilute + hot favours reverse (lots of water, Le Chatelier pushes back).
Use as a blocking group. Put $- S O_{3} H$ where you want to forbid substitution, do your real reaction elsewhere, then boil it off. Reversibility is the feature.

Verify: Forward: $C_{6} H_{6} + S O_{3} \to C_{6} H_{5} S O_{3} H$ . Carbons $6 = 6$ , hydrogens $6 = 5 + 1$ ✓, sulfur $1 = 1$ , oxygen $3 = 3$ ✓. Reverse is the exact mirror ( $+ H_{2} O$ pushes back). Forecast answer: out (reverse).

Cell C — Degenerate / "no reaction" inputs

(i) No catalyst = no electrophile. $C H_{3} C l$ alone won't ionize; there is no $C H_{3}^{+}$ . Result: no reaction. Why this step? Step 1 of every EAS is "make $E^{+}$ ." Remove the Lewis acid and Step 1 fails, so nothing downstream can happen. This is the degenerate/zero-input case.
(ii) Ring too electron-poor. The $- N O_{2}$ group is strongly deactivating — it pulls electron density out of the ring (see directing effects). The π cloud is now too poor to attack even a good acylium ion. Why this step? EAS needs a sufficiently rich ring to act as nucleophile. A deactivated ring is like a customer with no money — the electrophile can't do business. Friedel–Crafts is the most demanding EAS and fails first.
Verdict: neither reacts.

Verify (logic): Rate $\propto$ ring electron density $\times$ electrophile strength. (i) sets electrophile strength $= 0$ (none formed). (ii) sets ring density far below benzene's threshold. In both, one factor collapses → overall rate $\to 0$ . Forecast answer: neither.

Cell D — The rearrangement trap

Figure s02 — The hydride shift, drawn out. On the left, the primary cation $C H_{3} - C H_{2} - C H_{2}^{+}$ has its amber "+" on an end carbon (labelled "primary (unstable)"). A curved cyan arrow shows an $H^{-}$ (hydride) hopping from the middle carbon to that end. On the right, the resulting secondary cation $C H_{3} - C H^{+} - C H_{3}$ carries its amber "+" on the middle carbon (labelled "secondary (stable)"), and an amber note reads "→ attacks ring → cumene."

Form the carbocation. $A l C l_{3}$ pulls $C l^{-}$ off, giving the primary cation $C H_{3} C H_{2} C H_{2}^{+}$ . Why this step? Standard alkylation Step 1. But a $1°$ cation is unstable.
Hydride shift. A hydrogen (with its bonding electrons) hops from the middle carbon to the empty $+$ carbon, converting the unstable $1°$ cation into a stable secondary cation $C H_{3} C + H C H_{3}$ (isopropyl cation). Why this step? Carbocation stability rises $1° < 2° < 3°$ because more alkyl groups spread the $+$ charge. Nature takes the more stable cation before attacking the ring. This is the trap the parent note warned about.
Ring attacks the rearranged cation → lose $H^{+}$ → isopropylbenzene (cumene), not $n$ -propylbenzene. Why this step? By the time the ring's π cloud reaches an electrophile, the electrophile is already the stable $2°$ isopropyl cation (the rearrangement in step 2 happened first, because it lowers energy). So the ring bonds to that cation in the usual rate-determining attack, forming the arenium ion, and then throws off $H^{+}$ to rearomatize — exactly the standard two-step skeleton. The only twist is which cation the ring meets, which is why the product is the branched cumene rather than the straight chain.

Verify: Count carbons in the added group: $C_{3} H_{7}$ before and after — a hydride shift only moves an $H$ , it doesn't change the formula. Product $C_{6} H_{5} - C_{3} H_{7}$ : total $C_{9} H_{12}$ . Carbons $6 + 3 = 9$ ✓, hydrogens $5 + 7 = 12$ ✓. Forecast answer: rearranged (cumene) dominates.

Cell E — Over-substitution vs clean single addition

(a) Methyl is activating. Once one $- C H_{3}$ is on, it donates electron density (weak $+ I$ effect) → the ring is now richer than plain benzene → the next $C H_{3}^{+}$ attacks even faster → di-, tri-methylbenzenes. Why this step? An activating group raises ring density, raising rate; the product out-competes the reactant for more electrophile. Runaway.
(b) Acetyl is deactivating. The introduced $- C O C H_{3}$ has a $C = O$ that withdraws electrons → the product ring is poorer than benzene → it does not react again. You cleanly stop at acetophenone. Why this step? A deactivating group lowers ring density, lowering rate below the reactant's — self-limiting. This is why acylation is preferred for putting a carbon group on a ring.
Bonus: the acylium ion $C H_{3} - C \equiv O^{+}$ is resonance-locked and cannot rearrange — so acylation dodges both alkylation problems (Cell D and Cell E) at once.

Verify (logic): Let benzene reactivity $= 1$ . Toluene ring $> 1$ (activated) → 2nd substitution favoured. Acetophenone ring $< 1$ (deactivated) → 2nd substitution suppressed. So (a) over-substitutes, (b) stops at one. Forecast answer: the methylated ring is more reactive.

Cell F — Real-world synthesis word problem

Rule out nitrate-first. If you nitrate benzene first, you get nitrobenzene, whose ring is strongly deactivated — and from Ex 4 (ii) we proved Friedel–Crafts fails on nitrobenzene. Dead end. Why this step? Sequencing must respect Cell C: never try Friedel–Crafts on a poor ring.
So acylate first. Benzene + $C H_{3} C O C l / A l C l_{3}$ → acetophenone. The $- C O C H_{3}$ is deactivating and a meta-director. Why this step? Now the demanding step (Friedel–Crafts) is done while the ring is still rich enough (plain benzene).
Then nitrate. Nitrate acetophenone. Because $- C O C H_{3}$ directs meta, the $N O_{2}$ enters the meta position, giving $m$ -nitroacetophenone (a $1, 3$ arrangement). The ring is deactivated but nitration (with strong $N O_{2}^{+}$ ) still works, just slower. Why this step? The already-present group decides where the second one lands — this is directing effects in action.

Verify: Both routes are checked for feasibility, not just position. Nitrate-first route contains an impossible step (F–C on nitrobenzene) → invalid. Acylate-first route has every step allowed → valid, product is $m$ -nitroacetophenone. Forecast answer: acylate first.

Cell G — Exam-style twist

Acylate, don't alkylate. Use propanoyl chloride $C H_{3} C H_{2} C O C l + A l C l_{3}$ . The electrophile is the acylium ion $C H_{3} C H_{2} C \equiv O^{+}$ , which is resonance-locked → cannot rearrange (contrast Ex 5). Why this step? Acylation preserves the straight carbon skeleton because acylium ions don't do hydride shifts. Product: propiophenone $C_{6} H_{5} C O C H_{2} C H_{3}$ .
Reduce the $C = O$ to $C H_{2}$ . Apply Clemmensen ( $Z n (H g) / H C l$ ) or Wolff–Kishner reduction to turn the carbonyl into a methylene. This converts $- C O C H_{2} C H_{3}$ into $- C H_{2} C H_{2} C H_{3}$ . Why this step? Reduction removes only the oxygen (as water), leaving the intact straight three-carbon chain — exactly the skeleton alkylation could never build cleanly.
Read off the final product. The ring now carries $- C H_{2} C H_{2} C H_{3}$ : this is $n$ -propylbenzene $C_{6} H_{5} C H_{2} C H_{2} C H_{3}$ , the unbranched target, with no rearrangement anywhere in the route. Why this step? The whole point of the twist is that the straight chain survives because the branch-inducing carbocation was never formed — the acylium route sidestepped it, and reduction only stripped the oxygen. This "acylate → reduce" sequence is the classic exam workaround for the alkylation rearrangement problem.

Verify: Track carbons. Propanoyl group $C_{3} H_{5} O$ on ring → propiophenone $C_{9} H_{10} O$ : $C = 6 + 3 = 9$ ✓. Reduction $C = O \to C H_{2}$ adds 2 H, removes 1 O → $C_{9} H_{12}$ ( $n$ -propylbenzene): $C = 9$ ✓, $H = 5 + 7 = 12$ ✓, $O = 0$ ✓. Forecast answer: no single alkylation works (it would rearrange to cumene); acylate then reduce gives clean $n$ -propylbenzene.

Recall Quick self-test across the whole matrix

Missing catalyst outcome? ::: No $E^{+}$ forms → no reaction (Cell C). Why does nitrobenzene fail Friedel–Crafts? ::: $- N O_{2}$ deactivates the ring too much (Cell C). 1-chloropropane + benzene/ $A l C l_{3}$ major product? ::: Isopropylbenzene (cumene), via $1° \to 2°$ hydride shift (Cell D). Why does acylation stop at one substitution? ::: $- C O R$ is deactivating; product ring poorer than benzene (Cell E). To make both acetyl and nitro on one ring, which first? ::: Acylate first, then nitrate (nitrate-first kills F–C) (Cell F). How to get straight n-propylbenzene? ::: Acylate then reduce the carbonyl (Cell G). Which EAS is reversible? ::: Sulfonation — heat + dilute acid removes $- S O_{3} H$ (Cell B).