Q.Why do alkenes prefer to undergo electrophilic addition reaction while arenes prefer electrophilic substitution reactions? Explain.

Concept: Benzene Aromatic Stability — The key difference lies in the exceptional stability of the benzene ring due to delocalisation of $\pi$ electrons (aromaticity).

1. Alkenes have a localised $\pi$ bond. An electrophilic addition reaction breaks the $\pi$ bond to form two new $\sigma$ bonds, which is energetically favourable because the product is more stable than the reactant (no special stabilisation is lost).

2. Arenes (benzene) have a delocalised $\pi$ electron cloud (aromatic sextet). If an electrophilic addition occurred, the aromatic ring would be destroyed, losing its resonance stabilisation energy (≈ 150 kJ/mol). This is energetically very unfavourable. …

The question gets at the very heart of what makes aromatic compounds special. An alkene and an arene both contain $\pi$ bonds, but those $\pi$ bonds are fundamentally different in character. The alkene’s $\pi$ bond is localised between two carbons; the arene’s $\pi$ electrons are delocalised over the entire ring. That delocalisation confers a huge thermodynamic stabilisation — the resonance energy of benzene is about $150\ \text{kJ mol}^{-1}$.

When an electrophile attacks an alkene, the product is a saturated compound. One $\pi$ bond is lost, but the molecule gains two new $\sigma$ bonds. The energy released by forming those $\sigma$ bonds more than compensates for breaking the $\pi$ bond, so the reaction is exothermic and proceeds readily. There is no special stabilisation to lose — the alkene has none.

Now consider benzene. If an electrophile added across the ring to give a non-aromatic cyclohexadiene product, the delocalised $\pi$ system would be destroyed. That costs roughly $150\ \text{kJ mol}^{-1}$ of resonance stabilisation. The $\sigma$ bonds formed do not release enough energy to offset that loss, so direct addition is thermodynamically unfavourable.

Instead, benzene uses a different pathway. The electrophile attacks the ring to form a sigma complex (also called an arenium ion or Wheland intermediate). In this intermediate, the ring is no longer fully aromatic — it has four $\pi$ electrons delocalised over five carbons, while the sixth carbon is $sp^3$ hybridised. The stabilisation is less than that of benzene, but the intermediate is still partially stabilised by delocalisation. Crucially, the next step is deprotonation: a base removes the proton from the $sp^3$ carbon, and the $\pi$ electrons return to restore the full aromatic sextet. The net result is substitution — one hydrogen is replaced by the electrophile — and the aromatic stabilisation is regained.

Let us walk through the two reaction pathways side by side.

1. Alkene + electrophile (addition)

   The electrophile $E^+$ attacks the $\pi$ bond, forming a carbocation. A nucleophile (often the counterion) then attacks the carbocation. The product is saturated. The driving force is the formation of two strong $\sigma$ bonds at the cost of one $\pi$ bond. There is no hidden stabilisation to lose.

2. Benzene + electrophile (substitution)

   • Step 1: $E^+$ attacks the ring, generating the sigma complex. This step is endothermic because the aromatic stabilisation is partially lost. …