aroamtic 4

Why nucleophilic substitution fails on benzene

• Benzene is electron‑rich, flat (sp²), and aromatic

• Nucleophiles are electron‑rich too

• Electron‑rich vs electron‑rich → strong repulsion

• No favourable approach angle → Sₙ2 impossible

• Phenyl cation too unstable → Sₙ1 impossible

what do we need for an sn2 reaction and what happens in benzene?

• requires backside attack on tetrahedral (sp³) carbon with an accessible leaving group

• Benzene carbons are sp², planar, part of delocalized π system → no backside available

• Nucleophile cannot approach at correct angle

what do we need for sn1 and why dosent it happen with benzene?

• requires formation of a carbocation intermediate

• Leaving a bromide on benzene forms a phenyl cation

• Phenyl cation is extremely unstable, anti-aromatic, and not resonance-stabilized

What conditions allow nucleophilic aromatic substitution (SNAr) to occur?

• Must have a strong electron‑withdrawing group (EWG)
  e.g., NO₂

• EWG must be ortho or para to the leaving group

• Leaving group usually F > Cl > Br > I (F is best because it stabilises the Meisenheimer intermediate)

• Reaction proceeds by addition → elimination

When does nucleophilic substitution NOT occur on an aromatic ring?

• No strong electron‑withdrawing group on the ring

• EWG in the wrong position (meta)

• Leaving group without stabilisation of intermediate

• No way to stabilise the Meisenheimer complex

Why is fluoride (F⁻) a poor leaving group in typical sn1 and sn2 > reactions?

• Strong C–F bond: difficult to break

• Poorly stabilized anion: F⁻ is small, highly basic, and not stabilized in solution

• Result: fluoride rarely leaves

Why can fluoride act as a good leaving group in nucleophilic aromatic substitution (NAS)?

1. Electron-withdrawing groups (e.g., –NO₂) activate the aromatic ring

2. Nucleophile attacks the ring → forms Meisenheimer intermediate

3. Negative charge delocalized over the ring stabilizes the intermediate

4. Fluoride leaves easily → becomes a good leaving group on the activated ring

luorine as the leaving group in nucleophilic aromatic substitution

• Fluorine isn’t a good leaving group in general → reactions shown are rarely used

• Reason it works on aromatic rings: high electronegativity of F activates the ring

• Effect: increases reaction rate, lowers activation energy of the rate-determining step

• Second step: re-establishing aromaticity is fast and facile