Question 1

Why is the C–Cl bond in chlorobenzene shorter than an ordinary alkyl-halide C–Cl bond?

Accepted Answer

One chlorine lone pair participates in resonance with the benzene ring. This contribution gives the C–Cl bond partial double-bond character and makes it shorter and stronger than a typical single bond to chlorine.

Question 2

Are Kekulé and charge-separated drawings separate chlorobenzene molecules?

Accepted Answer

No. They are resonance contributors used to describe one delocalised hybrid. The real molecule does not physically switch between the drawings, and the negative charge in charge-separated contributors is not permanently localised on one carbon.

Question 3

Why are the ortho and para positions involved in chlorobenzene resonance?

Accepted Answer

The chlorine lone pair can overlap with the ring p-orbital system and distribute negative charge into positions directly connected with the conjugated pathway. The relevant charge-separated contributors place negative charge at the two ortho positions or at the para position, but not at a meta position.

Question 4

Why does chlorobenzene undergo nucleophilic substitution less readily than an alkyl halide?

Accepted Answer

The carbon bonded to chlorine is sp2-hybridised, which makes the required backside approach for SN2 substitution difficult. Resonance also stabilises the starting chlorobenzene, while the aryl carbocation required for a simple SN1 pathway is highly unstable.

Question 5

How can halogen be deactivating but ortho/para-directing?

Accepted Answer

The two effects have different origins. Chlorine donates electron density by resonance and directs electrophiles toward ortho and para positions, but its electronegativity withdraws electron density through sigma bonds. The inductive withdrawal dominates the overall activation, so the ring remains less reactive toward electrophilic substitution.

Chlorobenzene Resonance Lab: The Hidden Double Bond

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