Aryl halides — low reactivity, addition-elimination (benzyne mechanism), nucleophilic aromatic substitution: Exercises

HOW to use this page. Read each problem, cover the solution, and try it yourself first. The answer lives inside a collapsible [!recall]- callout — click to reveal only after you have committed to an answer. Levels climb from simply recognising a mechanism up to mastering predictions across all cases. Every idea here is built in the parent note the parent topic.

Before we begin, three plain-word reminders (so no symbol arrives unearned):

A quick vocabulary anchor, all from the parent:

Nucleophile (Nu $^{-}$ ) = an electron-rich "attacker" that brings a lone pair, e.g. hydroxide $O H^{-}$ or amide $N H_{2}^{-}$ .
EWG = electron-withdrawing group, e.g. $- N O_{2}$ , that pulls electron density toward itself.
Ortho / meta / para = neighbouring / one-gap / opposite positions on the ring relative to the halogen. Picture below.

Figure — Aryl halides — low reactivity, addition-elimination (benzyne mechanism), nucleophilic aromatic substitution

Chlorobenzene ( $C_{6} H_{5} C l$ ) is heated with aqueous $N a O H$ . It refuses to react at $60^{\circ} C$ but does react at $350^{\circ} C$ under high pressure. In one sentence, why is plain chlorobenzene so unreactive toward substitution?

Recall Solution 1.1

Four reasons, straight from the parent:

Resonance — the chlorine lone pair overlaps into the ring, giving the C–Cl bond partial double-bond character (shorter, stronger, harder to break).
The carbon is $s p^{2}$ (more s-character → holds electrons tighter) than the $s p^{3}$ carbon of an alkyl halide.
$S_{N} 2$ is blocked because the flat ring physically blocks back-side attack.
$S_{N} 1$ is blocked because the aryl cation $C_{6} H_{5}^{+}$ is hugely unstable. The very high temperature ( $350^{\circ} C$ ) is needed to force the benzyne route since there is no EWG.

Problem 1.2

Match each intermediate to its mechanism: (a) Meisenheimer complex, (b) benzyne.

Recall Solution 1.2

(a) Meisenheimer complex → SNAr (addition–elimination). It is anionic, $s p^{3}$ , non-aromatic. (b) Benzyne → elimination–addition (benzyne mechanism). It is neutral, strained, with an extra in-plane π bond. See Reactive intermediates — carbanions, benzyne, Meisenheimer.

Level 2 — Application

Problem 2.1

For each substrate + reagent, decide which mechanism operates (SNAr or benzyne): (a) $p$ -nitrochlorobenzene + $N a O H$ , $160^{\circ} C$ (b) chlorobenzene + $N a N H_{2}$ (sodamide) in liquid $N H_{3}$ (c) 2,4-dinitrochlorobenzene + $C H_{3} O^{-}$ , room temperature

Recall Solution 2.1

(a) SNAr. A single para- $N O_{2}$ EWG stabilises the Meisenheimer charge, so the moderate base $O H^{-}$ suffices. (b) Benzyne. No EWG, but $N a N H_{2}$ is a very strong base — it deprotonates the ortho H and forces elimination. (c) SNAr, very fast. Two EWGs (2,4-dinitro) at ortho and para stabilise the charge strongly, so the reaction runs at room temperature.

Problem 2.2

Rank the leaving-group reactivity of $p$ -nitrofluorobenzene, $p$ -nitrochlorobenzene, $p$ -nitrobromobenzene, $p$ -nitroiodobenzene toward $O H^{-}$ (SNAr). Give the order and the reason.

Recall Solution 2.2

Order: F > Cl > Br > I — the opposite of $S_{N} 2$ . Why: in SNAr the addition step (nucleophile joining the ring) is rate-determining. The most electronegative halogen (F) polarises the ring carbon most strongly, making it most electrophilic and speeding that addition. Bond-breaking of C–X happens in the fast second step, so how easily the halide leaves barely matters.

Level 3 — Analysis

Problem 3.1

Draw/describe why a meta- $N O_{2}$ group fails to accelerate SNAr, while ortho or para works. Use the idea of "where the negative charge can park."

Recall Solution 3.1

When the nucleophile adds, the ring's negative charge spreads (by resonance) onto the carbons ortho and para to the attack site. If the EWG sits at one of those carbons, its resonance can accept the charge onto the nitro oxygens — huge stabilisation. A meta- $N O_{2}$ sits on a carbon that does not carry any of the delocalised negative charge, so it can offer no resonance stabilisation. Result: no acceleration. Look at the shaded carbons in the figure — meta is never shaded.

Problem 3.2

$^{14} C$ -labelled chlorobenzene (label on the C–Cl carbon) is treated with $N a N H_{2}$ . Predict the percentage of the aniline product that has the $- N H_{2}$ group on the labelled carbon versus on the neighbouring carbon. Explain.

Recall Solution 3.2

50% labelled : 50% neighbour. The base removes the ortho H; elimination of $C l^{-}$ builds benzyne — a symmetric intermediate whose extra bond lies between the labelled carbon and its neighbour. The nucleophile $N H_{2}^{-}$ can add to either end of that bond with equal probability. Since the two carbons are equivalent by symmetry, the product is a $50 : 50$ mixture. This scrambling is the classic proof that benzyne exists.

Level 4 — Synthesis

Problem 4.1

You want to convert chlorobenzene → phenol ( $C_{6} H_{5} O H$ ). Two industrial/lab routes exist. (a) Name the harsh classic conditions and mechanism. (b) If you first put a para- $N O_{2}$ on the ring, why does the reaction become much milder?

Recall Solution 4.1

(a) The Dow process (see Phenols — preparation from chlorobenzene (Dow process)): chlorobenzene + $N a O H$ at $\sim 350^{\circ} C$ and high pressure. With no EWG, it must go through benzyne (elimination–addition), which needs brutal conditions. (b) With a para- $N O_{2}$ , the pathway switches to SNAr. The nitro group stabilises the Meisenheimer intermediate, dropping the required temperature from $\sim 350^{\circ} C$ to about $160^{\circ} C$ . Two nitro groups (2,4-dinitro) drop it further to near room temperature.

Problem 4.2

$p$ -chlorotoluene ( $C H_{3}$ at para, $C l$ where we substitute) is treated with $N a N H_{2}$ . Two benzyne carbons are possible attack sites. Predict the structure(s) of the amino products and roughly their ratio.

Recall Solution 4.2

Benzyne forms between the C–Cl carbon and its ortho neighbour (the base removes the ortho H, not the para- $C H_{3}$ carbon's H because $C H_{3}$ is para, not adjacent to Cl). The strained bond spans C1 (was C–Cl) and C2. $N H_{2}^{-}$ adds to either C1 or C2:

Attack at C1 → $- N H_{2}$ para to $C H_{3}$ = $p$ -toluidine (4-methylaniline).
Attack at C2 → $- N H_{2}$ meta to $C H_{3}$ = $m$ -toluidine (3-methylaniline). Roughly $50 : 50$ (the small directing influence of the distant methyl only lightly biases it). So expect a mixture of $p$ - and $m$ -toluidine.

Level 5 — Mastery

Problem 5.1

Given four reactions, predict the major mechanism, the intermediate, and whether position is conserved: (a) 2,4,6-trinitrochlorobenzene + $N H_{3}$ (b) bromobenzene + $K N H_{2}$ in liquid $N H_{3}$ (c) $p$ -nitrofluorobenzene + $C H_{3} O^{-}$ (d) chlorobenzene + weak base ( $N a H C O_{3}$ ), warm

Recall Solution 5.1

(a) SNAr (three EWGs → extremely fast, runs mild). Intermediate: Meisenheimer complex. Position conserved. (b) Benzyne. Intermediate: benzyne. Position not conserved (scrambles between the two carbons). (c) SNAr. F is the best SNAr leaving group (fastest addition). Intermediate: Meisenheimer. Position conserved. (d) No reaction. No EWG (rules out SNAr) and the base is far too weak to deprotonate an aryl ortho-H (rules out benzyne). Chlorobenzene just sits there.

Problem 5.2

Explain, using rate-determining steps, why in SNAr the order is $F > C l > B r > I$ but in the benzyne route the halide identity mostly reflects the acidity of the ortho hydrogen instead.

Recall Solution 5.2

SNAr: the rate-determining step is addition of the nucleophile, before C–X breaks. The most electronegative halogen (F) makes the ring carbon most electrophilic → fastest addition. Hence $F > C l > B r > I$ .
Benzyne: the rate-determining step is usually deprotonation of the ortho H to form the carbanion. A more electronegative halogen next door makes that ortho H more acidic (easier to remove), so the key factor is now proton acidity, not leaving-group ability. The two routes therefore rank halogens by completely different logic because their slow steps are different.

Recall One-screen recap

SNAr: needs EWG ortho/para, moderate base, Meisenheimer intermediate, position conserved, LG order F > Cl > Br > I.
Benzyne: no EWG, very strong base, benzyne intermediate, position scrambles (50:50), $^{14} C$ labelling proves it.
No EWG + weak base = no reaction.

Exercises — Aryl halides — low reactivity, addition-elimination (benzyne mechanism), nucleophilic aromatic substitution

Level 1 — Recognition

Problem 1.1