Nucleophilic Aromatic Substitution
SNAr addition-elimination through the Meisenheimer complex, benzyne mechanism, and substituent effects
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🎯 Nucleophilic Aromatic Substitution (SNAr)
Aromatic rings can undergo nucleophilic substitution when (a) a strong electron-withdrawing group (NO₂, CN, C=O) is positioned ortho or para to the leaving group, or (b) very harsh conditions form a benzyne intermediate.
Addition–Elimination (SNAr) mechanism
- Nucleophile attacks the ring carbon bearing the leaving group → Meisenheimer (σ-complex)
- Resonance delocalizes negative charge onto the ortho/para EWGs
- Leaving group departs, restoring aromaticity
Reactivity order of leaving groups in SNAr: F > Cl > Br > I (opposite of SN2; rate-determining step is nucleophilic addition, not C–LG cleavage).
Benzyne (elimination-addition) mechanism — strong base (NaNH₂) on aryl halides without activating EWGs. Goes through a strained benzyne intermediate; nucleophile can add to either sp carbon, giving substitution at both positions.
📚 Practice Problems
1Problem 1medium
❓ Question:
(a) Rank these aryl chlorides toward NaOMe / MeOH: chlorobenzene, 4-chloronitrobenzene, 2,4-dinitrochlorobenzene, 2,4,6-trinitrochlorobenzene. (b) Explain why fluoride is a better leaving group than iodide in this reaction series.
💡 Show Solution
(a) Reactivity order (fastest → slowest): 2,4,6-trinitrochlorobenzene > 2,4-dinitrochlorobenzene > 4-chloronitrobenzene ≫ chlorobenzene
Each ortho/para nitro group accepts the negative charge of the Meisenheimer intermediate by resonance, lowering the activation barrier for nucleophilic addition.
(b) F vs I leaving group — In SNAr the rate-determining step is nucleophilic addition, not C–LG cleavage. Fluorine is the most electronegative leaving group, so it activates the ipso carbon most strongly toward attack. Loss of fluoride happens only in the (fast) second step. Therefore F > Cl > Br > I in SNAr, the opposite of SN2.
2Problem 2hard
❓ Question:
Treatment of 4-bromotoluene with NaNH₂ in liquid NH₃ at –33 °C gives a roughly 50:50 mixture of meta- and para-toluidine (3- and 4-aminotoluene). Explain mechanistically.
💡 Show Solution
Benzyne (E2-type elimination–addition) mechanism.
- NaNH₂ deprotonates ortho to the Br; loss of Br⁻ gives a benzyne (3,4-didehydrotoluene).
- Benzyne is symmetric across the original C–Br carbon, so NH₂⁻ can add to either sp carbon of the triple-bond-like intermediate.
- Addition at C4 → 4-aminotoluene (para); addition at C3 → 3-aminotoluene (meta).
Because the methyl group is mildly electron-donating, the two addition sites are nearly isoenergetic and the product ratio is close to 1:1.
⚠️ Common Mistakes: Nucleophilic Aromatic Substitution
Avoid these 3 frequent errors
🌍 Real-World Applications: Nucleophilic Aromatic Substitution
See how this math is used in the real world
📝 Worked Example: Stoichiometry — Limiting Reagent
mol of reacts with mol of . How many grams of water are produced? Which is the limiting reagent? ()
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