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

Everything you need is built from the parent's one idea: an electron-poor $E^{+}$ replaces an $H$ , and aromaticity is restored. Before we start, two things need plainly stating so nobody is lost:

Figure s01 below is the picture we lean on all page: it shows the arenium ion with its positive charge shared over exactly three carbons. Keep it in mind from L2.2 onward.

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

Figure s01 — The arenium ion. One carbon (orange) is now $s p^{3}$ and holds both $E$ and $H$ . The three magenta dots are the two ortho and one para carbons that share the $+$ charge; the meta carbons stay neutral.

Figure s04 — Primary vs secondary vs tertiary carbocation: stability rises left to right as more carbons surround the $+$ carbon.

Level 1 — Recognition

L1.1

Name the electrophile ( $E^{+}$ ) in each: (a) nitration, (b) bromination, (c) Friedel–Crafts acylation.

Recall Solution

(a) Nitronium ion $N O_{2}^{+}$ . (b) A polarized " $B r^{+}$ " — really $B r^{+} \dots F e B r_{4}^{-}$ , the bromine made electron-poor by the Lewis acid. (c) The acylium ion $R - C \equiv O^{+}$ . Why these and not the neutral reagent? Neutral $H N O_{3}$ , $B r_{2}$ , or $R C O C l$ are too weak/electron-rich to overcome benzene's stability. Each reaction's Step 1 manufactures a much hungrier cation.

L1.2

Which of these EAS reactions is reversible, and what does that let you do?

Recall Solution

Sulfonation is reversible. Heating an arenesulfonic acid with dilute acid removes the $- S O_{3} H$ group. This lets you use $- S O_{3} H$ as a temporary blocking group: park it on a position, run another EAS elsewhere, then strip it off.

Level 2 — Application

L2.1

Write the balanced generation of the nitronium ion and confirm the atom/charge bookkeeping. $H N O_{3} + 2 H_{2} S O_{4} \to N O_{2}^{+} + H_{3} O^{+} + 2 H S O_{4}^{-}$

Recall Solution

Nitrogen: 1 left, 1 right ✓. Sulfur: 2 left, 2 right ✓. Hydrogen: left $= 1 (H N O_{3}) + 2 \times 2 (H_{2} S O_{4}) = 5$ ; right $= 3 (H_{3} O^{+}) + 2 \times 1 (H S O_{4}^{-}) = 3 + 2 = 5$ ✓. Oxygen: left $= 3 (H N O_{3}) + 2 \times 4 (H_{2} S O_{4}) = 11$ ; right $= 2 (N O_{2}^{+}) + 1 (H_{3} O^{+}) + 2 \times 4 (H S O_{4}^{-}) = 2 + 1 + 8 = 11$ ✓. Charge: left $= 0$ ; right $= (+ 1) + (+ 1) + 2 (- 1) = 0$ ✓. Why two $H_{2} S O_{4}$ , not one? Follow the atoms. The first $H_{2} S O_{4}$ donates a proton to $H N O_{3}$ , giving protonated nitric acid $H_{2} N O_{3}^{+}$ ; this then sheds an $- O H$ to expose $N O_{2}^{+}$ . But a bare $- O H$ cannot just float off — it must be captured. So the second $H_{2} S O_{4}$ donates its proton to that $- O H$ , turning it into water ( $H_{2} O \to H_{3} O^{+}$ ). Each $H_{2} S O_{4}$ that gives up a proton becomes one $H S O_{4}^{-}$ , which is why you get two $H S O_{4}^{-}$ on the right. One acid protonates the nitric acid; the other mops up the leaving oxygen as water.

L2.2

A benzene ring gives its arenium ion. The parent note says the $+$ charge is delocalized over 3 carbons. Which 3 (relative to the attacked carbon), and how many resonance structures of the ion carry the charge? (See figure s01.)

Recall Solution

The three positions are the two ortho carbons and the one para carbon relative to the $s p^{3}$ carbon that grabbed $E$ — exactly the three magenta dots in figure s01. That gives 3 resonance structures (see Resonance and delocalization) — each places $+$ on one of those three carbons. The meta carbons never carry the charge. More sharing → lower energy → the reaction is possible at all.

L2.3

Product-naming: name the product of Friedel–Crafts acylation of benzene with $C H_{3} C O C l / A l C l_{3}$ .

Recall Solution

The acylium ion $C H_{3} - C \equiv O^{+}$ attaches; product is $C_{6} H_{5} C O C H_{3}$ = acetophenone (phenyl methyl ketone).

Level 3 — Analysis

L3.1

$1 -chloropropane + A l C l_{3}$ + benzene gives mostly isopropylbenzene (cumene), not $n$ -propylbenzene. Explain via the mechanism.

Recall Solution

$A l C l_{3}$ pulls $C l^{-}$ off, giving the primary (1°) propyl cation $C H_{3} C H_{2} C H_{2}^{+}$ (only 1 carbon attached to the $+$ centre — see figure s04). Primary cations are unstable, so before the ring can attack, a ==hydride ( $H^{-}$ ) shifts== from the middle carbon to the cationic carbon, converting the $1°$ cation into the more stable secondary (2°) cation $C H_{3} C + H C H_{3}$ (2 carbons on the $+$ centre; see Carbocation stability and rearrangements). The ring then attacks that, giving isopropylbenzene. Net: the alkyl skeleton rearranged.

L3.2

Rank the reactivity of these rings toward EAS: anisole ( $C_{6} H_{5} O C H_{3}$ ), benzene, nitrobenzene ( $C_{6} H_{5} N O_{2}$ ). Justify with electron-density language.

Recall Solution

anisole > benzene > nitrobenzene.

$- O C H_{3}$ donates a lone pair into the ring → more electron-rich → stabilizes the positive arenium ion → faster.
$- N O_{2}$ pulls electron density out → less electron-rich → destabilizes the arenium ion → much slower. This is exactly the activation/deactivation logic of Directing effects (ortho/para vs meta directors).

L3.3

Why does Friedel–Crafts acylation stop cleanly at one substitution, while alkylation over-substitutes?

Recall Solution

Acylation installs a $- C O R$ group. The $C = O$ is electron-withdrawing → the product ring is less reactive than the starting benzene → it won't react again. Alkylation installs a $- R$ group, which is electron-donating → product ring is more reactive → it keeps reacting (polyalkylation).

Level 4 — Synthesis

L4.1

You want 1-nitro-4-bromobenzene (para) as the major product from benzene. In which order do you nitrate and brominate? Explain using directing effects.

Recall Solution

Brominate first, then nitrate.

$B r$ is an ortho/para director (a weakly deactivating halogen). So bromobenzene, when nitrated, sends $N O_{2}$ mostly para → the 1,4 product you want.
If you nitrated first, $- N O_{2}$ is a meta director, so subsequent bromination would go meta (1,3), giving the wrong isomer. Order matters because the first group installed decides where the second goes. See Directing effects (ortho/para vs meta directors).

L4.2

Design a route to n-propylbenzene (a straight $- C H_{2} C H_{2} C H_{3}$ chain) from benzene, avoiding the rearrangement problem of L3.1.

Recall Solution

Use Friedel–Crafts acylation, then reduce:

$C_{6} H_{6} + C H_{3} C H_{2} C O C l / A l C l_{3} \to C_{6} H_{5} - C O - C H_{2} C H_{3}$ (propiophenone). The acylium ion does not rearrange, so the straight 3-carbon chain is preserved.
Reduce the $C = O$ to $C H_{2}$ (Clemmensen $Z n (H g) / H C l$ or Wolff–Kishner) → $C_{6} H_{5} C H_{2} C H_{2} C H_{3}$ = n-propylbenzene. This sidesteps the hydride shift that ruins direct alkylation.

L4.3

You must run a nitration that ends up on the position ortho to a $- C H_{3}$ but you want to block the para position. Sketch a blocking-group strategy.

Recall Solution

Use sulfonation as a reversible block:

Sulfonate toluene → $- S O_{3} H$ parks at the para position. Why para, mechanistically? Sulfonation is reversible and under thermodynamic (equilibrium) control. The bulky $- S O_{3} H$ group and the incoming $S O_{3}$ suffer steric strain if they sit ortho (right next to) the $- C H_{3}$ , so the ortho arenium ion and ortho product are both higher in energy. The equilibrium therefore drains toward the less strained para product, whose arenium ion is also well stabilized by the methyl's electron donation from the far side. Because the reaction is reversible, any ortho product that forms is undone and re-formed until the low-energy para isomer dominates. (This is different from an irreversible EAS, where the fastest-formed product would win.)
Nitrate → the para spot is occupied, so $N O_{2}$ is forced ortho to $C H_{3}$ .
Desulfonate by heating with dilute acid → removes $- S O_{3} H$ , revealing the free para position. The reversibility of sulfonation (L1.2) is exactly what makes this possible.

Level 5 — Mastery

L5.1

Nitrobenzene refuses Friedel–Crafts alkylation. Explain in two independent ways why, and state one general rule you can extract.

Recall Solution

Both reasons apply directly to nitrobenzene itself: Reason 1 (ground-state electron density): $- N O_{2}$ strongly withdraws electron density from the ring, so the $π$ cloud is too electron-poor to act as a nucleophile toward the (already weak) alkyl cation — the attack barely happens. Reason 2 (intermediate destabilization): even if the ring did attack, the resulting arenium ion places positive charge next to the electron-withdrawing $- N O_{2}$ group. Putting two positive centres side by side is very high energy, so the RDS transition state is prohibitively tall — the reaction stalls. These are independent: one is about the starting ring, the other about the intermediate. Both make the barrier too high. General rule: Friedel–Crafts needs a sufficiently electron-rich ring — it fails on rings bearing strong deactivators like $- N O_{2}$ .

L5.2

Full mechanism, all three steps, for the nitration of benzene — WHAT/WHY at each step, using the arrow-pushing diagram (figure s02).

Recall Solution

Follow the three labelled panels of figure s02 left to right (curved arrows show where electron pairs move): Panel 1 — make $E^{+}$ . $H_{2} S O_{4}$ protonates $H N O_{3}$ ; water leaves → $N O_{2}^{+}$ . Why: $N O_{2}^{+}$ is a far stronger electrophile than neutral $H N O_{3}$ . Panel 2 — $π$ attacks $N O_{2}^{+}$ . A ring $π$ -pair (curved arrow) reaches out to $N O_{2}^{+}$ , forming the arenium ion; the $+$ charge is delocalized over the two ortho + one para carbons (compare figure s01). This is the RDS because aromaticity is broken. Why: the electron-rich ring is the nucleophile; the ring "chooses" the most stabilized (3-position-delocalized) intermediate. Panel 3 — $H S O_{4}^{-}$ removes $H^{+}$ . The base grabs the proton on the $s p^{3}$ carbon; that C–H pair swings back into the ring → rearomatize → nitrobenzene, and $H_{2} S O_{4}$ is regenerated. Why: dropping $H^{+}$ (not adding a nucleophile) restores the aromatic sextet — the big thermodynamic prize.

Figure s02 — Arrow-pushing for nitration, in three labelled panels: Panel 1 generate $N O_{2}^{+}$ ; Panel 2 $π$ -attack forming the arenium ion (RDS); Panel 3 base removes $H^{+}$ to rearomatize. Each panel is titled and boxed on the figure.

L5.3

An energy-profile reasoning question. In EAS, why is forming the arenium ion the rate-determining step (RDS) and not the loss of $H^{+}$ ? Answer with the profile in figure s03.

Recall Solution

Forming the arenium ion destroys aromaticity — it climbs to a high-energy, non-aromatic intermediate, so its transition state is the tallest barrier (figure s03, first peak, explicitly labelled as the highest point). Losing $H^{+}$ afterward goes downhill to the rearomatized product; its barrier (the second peak) is drawn and annotated as clearly lower than the first. The overall rate is set by the tallest hill, which is arenium-ion formation. Hence RDS = arenium-ion formation.

Figure s03 — EAS energy profile. A dashed guide line marks the height of the first (tallest) transition state; the second transition state is annotated with its height so you can see it is markedly lower. The tallest hump (forming the arenium ion, aromaticity broken) is the RDS; loss of $H^{+}$ is a small downhill hump to the product.

Edge cases & when these reactions fail

Common mistake Comprehensive edge-case checklist — read before any exam

Every exercise above assumes "textbook" conditions. Here is where each reaction breaks down:

Over-sulfonation / poly-sulfonation: with excess $S O_{3}$ / very hot oleum, a second $- S O_{3} H$ can add (meta to the first, since $- S O_{3} H$ is itself a meta-directing deactivator). The L4.3 blocking trick assumes mono-sulfonation under controlled conditions.
Wrong-temperature sulfonation: because sulfonation is reversible, hot conditions favour removal, cooler conditions favour addition. Choosing the wrong temperature can strip your blocking group prematurely, or fail to install it.
Friedel–Crafts alkylation carbocation rearrangement (L3.1): any $1°$ or $2°$ alkyl halide risks a hydride/methyl shift to a more stable cation → wrong skeleton. Acylation avoids this.
Friedel–Crafts polyalkylation (L3.3): without limiting the alkyl halide, the activated product keeps reacting → mixtures.
Friedel–Crafts fails entirely on deactivated rings (nitrobenzene, L5.1) and on rings with basic $- N H_{2}$ (the lone pair binds $A l C l_{3}$ , deactivating the ring).
Over-nitration / over-halogenation: with forcing conditions, even the deactivated mono-product can pick up a second group (meta-directed), so "clean mono-product" needs mild, controlled conditions.

Recall One-line self-check before you leave

The whole page is one skeleton: make $E^{+}$ → $π$ attacks (slow, arenium ion) → lose $H^{+}$ (rearomatize). Directing, ordering, blocking, and reactivity are all just consequences of how electron-rich the ring is and how stable the arenium ion becomes.