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

WHAT is this page? The parent note gave you the two tricks — SNAr (add-then-eliminate) and benzyne (eliminate-then-add). Here we run those tricks against every kind of aryl-halide problem an exam can throw at you: every position of the electron-withdrawing group, the "no group at all" case, the degenerate symmetric case, a real-world industrial case, and the exam trap where a group blocks a reaction entirely.

Before we start, one plain-language reminder so no symbol is unearned:

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

Look at the figure: the halogen sits at the top (position 1). The two ortho carbons (2 and 6), the two meta carbons (3 and 5), and the single para carbon (4) are labelled. Only ortho and para can "feel" the halogen carbon through resonance — remember that, it drives half of this page.

The scenario matrix

Every problem this topic can pose is one of these cells. The examples below are tagged with the cell they cover.

#	Case class	Concrete trigger	Which mechanism	Answer type
A	EWG para to X	1-Cl, 4-NO₂ + OH⁻	SNAr	single product, same position
B	EWG ortho to X	1-Cl, 2-NO₂ + OMe⁻	SNAr	single product, same position
C	EWG meta to X	1-Cl, 3-NO₂ + OH⁻	neither works easily (degenerate/blocked)	no easy reaction
D	Multiple EWGs	2,4-dinitro / 2,4,6-trinitro	SNAr (very fast)	rate comparison
E	Zero EWG, super-base	chlorobenzene + NaNH₂	benzyne	scrambled position
F	Symmetric label (degenerate)	¹⁴C-chlorobenzene + NaNH₂	benzyne	50 : 50 product split
G	Substituted benzyne (limiting orientation)	o-chlorotoluene + NaNH₂	benzyne	ratio of two isomers
H	Real-world / industrial	Dow process, chlorobenzene → phenol	benzyne-type	temperature/word problem
I	Leaving-group ranking (numeric twist)	F vs I in SNAr	SNAr trend	order F > I

Recall Fast self-test before you read the solutions

Which cell needs NO electron-withdrawing group? ::: Cell E and F (benzyne) — a super-strong base does the job instead. Which cell gives you two products from one starting material? ::: Cells E, F, G (benzyne scrambles position).

Example 1 — Cell A: EWG para to the halogen

Forecast: guess now — does $O H$ land where $C l$ was, or somewhere else? Does the $N O_{2}$ move?

Identify the mechanism. There is a strong EWG ( $N O_{2}$ ) para to $C l$ , and no super-base — so this is SNAr (addition–elimination). Why this step? The trigger table says EWG at ortho/para ⇒ SNAr. Reading the trigger first stops you from wrongly guessing benzyne.
Addition (RDS). $O H^{-}$ attacks the carbon bearing $C l$ ; the ring carbon becomes $s p^{3}$ and the negative charge floods into the ring. Why this step? The nucleophile "adds first" in SNAr — this creates the Meisenheimer complex.
Check the charge has a home. Push the negative charge around by resonance: it lands on the oxygen of the para- $N O_{2}$ . Why this step? If the charge could NOT reach the EWG, the intermediate would be unstable and the reaction would fail — this is the whole reason para works (see Example 3 for the failure case).
Elimination. $C l^{-}$ leaves, aromaticity returns, giving $p$ -nitrophenol. Why this step? The leaving group departs from the same carbon it started on — SNAr conserves position.

Product: $p$ -nitrophenol; $- O H$ on the same carbon $C l$ occupied.

Verify: Count atoms — we swapped $C l$ (35.5) for $O H$ (17) at one carbon, $N O_{2}$ untouched. Molecular formula goes $C_{6} H_{4} C l N O_{2} \to C_{6} H_{4} (O H) N O_{2} = C_{6} H_{5} N O_{3}$ . Sanity: one substitution, one product, position conserved. ✓

Example 2 — Cell B: EWG ortho to the halogen

Forecast: ortho vs para — will it be slower, faster, or the same as Example 1?

Mechanism = SNAr. EWG is ortho — ortho is one of the two "resonance-connected" positions, so the charge can reach it. Nucleophile is $C H_{3} O^{-}$ (methoxide). Why this step? Ortho and para are equivalent for resonance delivery of charge onto the EWG; both stabilise the Meisenheimer complex.
Addition. $C H_{3} O^{-}$ adds to the $C$ – $C l$ carbon → Meisenheimer complex; charge delocalises onto the ortho- $N O_{2}$ .
Elimination. $C l^{-}$ leaves; product is o-nitroanisole (2-nitro-methoxybenzene).

Product: o-nitroanisole; $- O C H_{3}$ replaces $C l$ at the same carbon.

Verify: $C_{6} H_{4} C l N O_{2} + C H_{3} O^{-} \to C_{6} H_{4} (O C H_{3}) N O_{2} + C l^{-}$ . Product formula $C_{7} H_{7} N O_{3}$ . Charges balance: one anion in ( $C H_{3} O^{-}$ ), one out ( $C l^{-}$ ). ✓

Example 3 — Cell C: EWG meta (the "why it fails" case)

Forecast: same EWG, same base — surely it works too?

Attempt SNAr. $O H^{-}$ adds to the $C$ – $C l$ carbon, making the Meisenheimer intermediate.
Try to park the charge on the EWG. Push resonance arrows... the negative charge can reach the ortho and para carbons relative to the attack point, but the $N O_{2}$ is meta — it sits on a carbon the charge cannot reach. Why this step? Resonance in a ring places the negative charge only on alternating carbons (positions 2,4,6 counted from the attack carbon). A meta-EWG sits on 3 or 5 — always the wrong set.
Conclusion. The negative charge is stranded on plain ring carbons, the intermediate is high-energy, so SNAr is not accelerated — it behaves almost like ordinary chlorobenzene.

Product: essentially no easy SNAr; you'd need forcing benzyne conditions instead.

The figure contrasts para (green tick — charge lands on the sponge) with meta (red cross — charge stranded on a bare carbon).

Verify (conceptual): Number the carbons; the alternating "charge-bearing" set from the attack carbon is {ortho, para}. A meta group is never in that set. So meta-EWG offers zero stabilisation. ✓ (checked in VERIFY by set membership).

Example 4 — Cell D: multiple EWGs, rate comparison

Forecast: more nitro groups — faster or slower?

Count the sponges at ortho/para. (i) 0, (ii) 1 (para), (iii) 2 (both o+p), (iv) 3 (two ortho + one para). Why this step? Each EWG at o/p gives the Meisenheimer charge an extra place to hide → lower-energy intermediate → faster RDS.
Order by number of effective EWGs. More effective EWGs ⇒ faster.

Answer: (i) < (ii) < (iii) < (iv). Practically: chlorobenzene needs ~ $350^{\circ} C$ ; one $N O_{2}$ far less; 2,4-dinitro reacts on warming; 2,4,6-trinitro reacts near room temperature.

Verify: Effective-EWG counts are $0, 1, 2, 3$ — a strictly increasing sequence, matching the strictly increasing rate order. ✓

Example 5 — Cell E: zero EWG, super-base → benzyne

Forecast: with no nitro group, can substitution even happen?

Rule out SNAr. No EWG ⇒ no charge sponge ⇒ SNAr dead. But $N a N H_{2}$ is a very strong base, which is the benzyne trigger.
Deprotonation. $N H_{2}^{-}$ rips off an H ortho to $C l$ , giving an aryl carbanion + $N H_{3}$ . Why this step? Benzyne is eliminate-first: remove H, then the halide.
Elimination. The carbanion lone pair pushes $C l^{-}$ out, forming benzyne (an extra strained in-plane π bond).
Addition + protonation. $N H_{2}^{-}$ adds across the benzyne; protonation gives aniline ( $C_{6} H_{5} N H_{2}$ ).

Product: aniline.

Verify: Net change: $C_{6} H_{5} C l + N H_{2}^{-} \to C_{6} H_{5} N H_{2} + C l^{-}$ . Atom balance: lose $C l$ , gain $N H_{2}$ ; formula $C_{6} H_{5} N H_{2} = C_{6} H_{7} N$ . ✓

Example 6 — Cell F: the symmetric ¹⁴C label (degenerate case)

Forecast: exactly on the labelled carbon, or split?

Form benzyne from the labelled ring. Deprotonate an ortho H, eliminate $C l^{-}$ → the new "triple bond" is between the labelled carbon (C1) and its ortho neighbour (C2). Why this step? The eliminated $C l^{-}$ leaves from C1, and the H is removed from C2, so the extra bond spans C1–C2.
Nucleophile attacks either end. Benzyne's two carbons are geometrically equivalent here, so $N H_{2}^{-}$ adds to C1 or C2 with equal probability. Why this step? The intermediate is symmetric about the C1–C2 bond, so there's no preference.

Answer: aniline with $N H_{2}$ 50% on the labelled $^{14} C$ (C1) and 50% on the adjacent $C$ (C2). This 50:50 scramble is the proof benzyne exists.

Verify: Two equivalent attack sites ⇒ probability $= 1/2$ each $= 50%$ . ✓

Example 7 — Cell G: substituted benzyne, orientation limit

Forecast: how many benzynes can form, and how many products?

Find the removable ortho H's. The $C l$ is at C2; its ortho neighbours are C1 (bears the $C H_{3}$ , no H to remove there) and C3 (has an H). So only the C2–C3 benzyne can form. Why this step? You can only make a benzyne between the C–Cl carbon and an ortho carbon that has a hydrogen. The methyl-bearing carbon (C1) has no ring H.
Add $N H_{2}^{-}$ across C2–C3. Attack at C2 → amino ortho to methyl (2-methylaniline, o-toluidine); attack at C3 → amino meta to methyl (3-methylaniline, m-toluidine). Why this step? The two carbons of this benzyne are no longer equivalent (methyl sits next to C2), so the ratio need not be exactly 50:50 — but both products form.

Answer: a mixture of o-toluidine (2-methylaniline) and m-toluidine (3-methylaniline). Note: no para isomer, because only one benzyne (C2–C3) is accessible.

Verify: Ortho carbons to C2 are C1 and C3; C1 carries $C H_{3}$ (0 ring H), C3 carries 1 ring H ⇒ exactly 1 possible benzyne ⇒ exactly 2 possible amine positions (C2, C3). ✓

Example 8 — Cell H: real-world industrial word problem (Dow process)

Forecast: what makes the plain one need ~190 °C more?

Classify. Chlorobenzene has no EWG ⇒ SNAr can't be stabilised ⇒ under hot strong base it goes by a benzyne-type route. $p$ -nitro version has an EWG ⇒ easy SNAr. Why this step? Mechanism choice explains the energy cost: benzyne needs enough thermal energy to strip an H and force out $C l^{-}$ with no charge sponge to help.
Compare energy barriers via temperature. No stabilisation ⇒ higher activation barrier ⇒ higher temperature. The temperature difference signals the missing EWG assistance.

Answer: the EWG in $p$ -nitrochlorobenzene lowers the barrier so much that $160^{\circ} C$ suffices; plain chlorobenzene has no such help and must reach ~ $350^{\circ} C$ (Dow conditions) to proceed via benzyne.

Verify: The temperature gap is $350 - 160 = 190^{\circ} C$ , a large, positive number — consistent with "one EWG saves a lot of activation energy." ✓

Example 9 — Cell I: leaving-group ranking (numeric twist)

Forecast: in $S_{N} 2$ we learned I leaves best — does that hold here?

Identify the RDS. In SNAr the addition step is rate-determining, not the C–X bond breaking. Why this step? If bond-breaking set the rate we'd expect I fastest (weakest C–X). But it doesn't — addition rules.
Rank by how much X polarises the ring. The more electronegative X pulls electron density away from the carbon being attacked, making addition easier. Electronegativity order: $F > C l > B r > I$ .

Answer (SNAr): $F > C l > B r > I$ — the opposite of $S_{N} 2$ (Alkyl halides — SN1 and SN2 mechanisms), where $I > B r > C l > F$ .

Verify: Pauling electronegativities $F 4.0 > C l 3.0 > B r 2.8 > I 2.5$ give the strictly decreasing order that matches the SNAr rate order — and is exactly reversed vs the $S_{N} 2$ leaving-group order. ✓

Flashcards

Does a meta-nitro group accelerate SNAr?

No — resonance can't place the negative charge on a meta carbon, so the Meisenheimer complex is unstabilised.

¹⁴C-labelled chlorobenzene + NaNH₂ gives aniline with NH₂ where?

50% on the labelled carbon, 50% on the adjacent carbon (symmetric benzyne).

Why does o-chlorotoluene give only two amine isomers, not three?

Only one benzyne (between C–Cl and the ortho carbon that has an H) can form; the methyl-bearing carbon has no ring H.

SNAr leaving-group order and how it compares to S_N2?

F > Cl > Br > I in SNAr; exactly reversed from S_N2 (I > Br > Cl > F).

Why does the Dow process need ~350 °C but p-nitrochlorobenzene only ~160 °C?

Plain chlorobenzene has no EWG (goes via high-barrier benzyne); the nitro group stabilises SNAr, slashing the activation energy.