Chemical Reactions of Haloarenes

Why the C–X bond itself resists nucleophilic attack. In haloarenes the halogen's lone pair delocalises into the aromatic ring through resonance, giving the C–X bond partial double-bond character. This makes the bond shorter and stronger than a haloalkane C–X bond, and it also pushes extra electron density onto the ring — which then repels an incoming nucleophile (itself electron-rich). Plain chlorobenzene therefore needs forcing conditions (high temperature and pressure) for nucleophilic substitution to occur at all, unlike a haloalkane, which reacts readily by $S_N1$ or $S_N2$. Substitution becomes easy only if strongly electron-withdrawing groups such as $-NO_2$ sit at the ortho or para positions relative to the halogen, because they stabilise the negatively charged (Meisenheimer-type) intermediate that forms during substitution.

Electrophilic aromatic substitution — the halogen's split personality. This is the syllabus's central point. On the benzene ring itself, a halogen substituent is: An ortho/para director — it sends incoming electrophiles predominantly to the ortho and para positions relative to itself, not meta. Yet net deactivating — the substituted ring reacts more slowly with electrophiles than benzene does, not faster. Two effects, pulling in different directions on different properties, explain both halves at once.

The resonance ($+M$) effect decides position. The halogen's lone pair can donate into the ring. When the arenium-ion intermediate forms during ortho or para attack, one of its resonance structures places the positive charge directly on the carbon bearing the halogen — and the halogen's lone pair then neutralises that charge, giving an extra, unusually stable resonance form. No such structure is available for meta attack, so the meta intermediate is comparatively higher in energy. Lower-energy intermediate means a faster pathway, so ortho/para attack wins.

The inductive ($-I$) effect decides rate. Halogens are strongly electronegative and pull electron density out of the ring through the $\sigma$-bond framework, at every ring position, not just ortho/para. This lowers the ring's overall electron density compared with unsubstituted benzene, making it harder for any electrophile to approach anywhere on the ring. Because this inductive withdrawal outweighs the resonance donation, the net effect on rate is deactivation.

where: the left relation (an inequality, not a numeric formula) records that the inductive effect dominates the resonance effect in determining overall reactivity, while the right relation records that resonance still governs which positions are attacked once a reaction does occur — the two conclusions come from the same halogen but answer different questions.

Typical electrophilic substitution reactions of chlorobenzene.

where: the reaction is halogenation, catalysed by iron (which generates $FeCl_3$ in situ); para product usually predominates over ortho because it suffers less steric hindrance near the bulky chlorine already on the ring.

where: the nitrating mixture generates the electrophile $NO_2^+$; sulphonation with fuming $H_2SO_4$ and Friedel–Crafts alkylation/acylation follow the same ortho/para pattern, though Friedel–Crafts can be sluggish on a deactivated ring.

This topic is purely descriptive — a mechanistic argument about resonance and induction rather than a numeric formula — so instead of a derivation, here is the reasoning chain that explains both halves of the halogen's behaviour end to end.

Start from the ground state. In chlorobenzene, chlorine's lone pair already conjugates into the ring even before any electrophile arrives; this is the same resonance that shortens the C–Cl bond and makes it resist nucleophilic substitution. That conjugation places some negative character on the ring carbons ortho and para to chlorine specifically (those are the positions the resonance structures reach).

Bring in the electrophile. When an electrophile such as $NO_2^+$ or $Br^+$ attacks, it forms a positively charged arenium-ion (sigma-complex) intermediate. The ring redistributes that positive charge over three resonance structures for ortho or para attack, and over three different resonance structures for meta attack — but only in the ortho/para case does one resonance structure land the positive charge on the very carbon bearing chlorine. Chlorine's lone pair is right there to donate into that carbon and quench the charge, so that particular resonance structure is unusually low in energy.

Compare stabilities. A resonance hybrid is stabilised by every contributing structure, but especially by its lowest-energy contributor. Because ortho/para intermediates get that extra chlorine-stabilised structure and the meta intermediate does not, the ortho/para intermediates are lower in overall energy. Lower-energy intermediate means a lower activation energy pathway (by Hammond-postulate-style reasoning already familiar from mechanism study), so the reaction preferentially proceeds through ortho/para attack. This is the entire origin of "ortho/para directing" — it is a statement about relative intermediate stability, not about electron density in the starting material.

Now ask about overall speed, a separate question. Whether or not ortho/para attack is preferred, the reaction is still slower on chlorobenzene than on benzene at every position, ortho/para included. That is because chlorine's electronegativity withdraws electron density inductively through the $\sigma$ framework — a withdrawal that is present regardless of resonance and does not discriminate between ring positions the way resonance does. It raises the activation energy of attack everywhere relative to benzene, and this inductive penalty is numerically larger than the resonance bonus at any single position. So even the most favoured ortho/para intermediate for chlorobenzene is less stabilised, and the reaction less facile, than the analogous intermediate for benzene itself.

The conclusion follows without contradiction. Resonance and induction are not competing to answer the same question; they are answering two different questions about the same molecule — resonance discriminates among ring positions (ortho/para preferred over meta), while induction discriminates between molecules (chlorobenzene slower than benzene, uniformly). A substituent can therefore be simultaneously a director (a positional preference) and a deactivator (an overall rate effect), which is exactly the halogen case.

Problem. Chlorobenzene is treated with $Br_2$ in the presence of $Fe$/$FeBr_3$. Predict the major products, state whether the reaction is faster or slower than bromination of benzene, and justify both using resonance and induction.

Step 1. Identify the electrophile. $Fe$ reacts with $Br_2$ to generate $FeBr_3$, a Lewis acid that polarises $Br_2$ into an electrophilic bromine, effectively $Br^+$, the species that attacks the ring.

Step 2. Decide the rate relative to benzene using the inductive effect. Chlorine withdraws electron density from the ring through the $\sigma$ framework at every position, lowering the ring's ability to stabilise the approaching $Br^+$ compared with benzene. So bromination of chlorobenzene is slower than bromination of benzene — chlorine is a deactivating substituent.

Step 3. Decide the position using the resonance effect. Despite the overall slowdown, the arenium-ion intermediate formed by attack ortho or para to chlorine gains an extra resonance structure in which chlorine's lone pair neutralises the ring's positive charge; no such structure exists for meta attack. So among the positions still available, ortho and para are strongly preferred.

Step 4. State the products and steric note. The major products are ortho-bromochlorobenzene and para-bromochlorobenzene, with the para isomer typically predominating because it suffers less steric crowding than the ortho isomer, whose bromine sits next to the existing chlorine. Answer: predominantly o- and p-bromochlorobenzene (para favoured), formed more slowly overall than bromination of benzene, because resonance directs the position while induction controls the rate.

• Assuming "ortho/para director" always means "activating." For halogens this breaks: they direct ortho/para via resonance while still deactivating the ring overall via a dominant inductive effect — unlike $-OH$ or $-NH_2$, which are both activating and ortho/para directing.
• Expecting easy nucleophilic substitution on plain haloarenes. Chlorobenzene needs forcing conditions, or electron-withdrawing groups at ortho/para, to undergo nucleophilic substitution at all — it is nothing like a haloalkane's routine $S_N1/S_N2$ chemistry.
• Mixing up which effect answers which question. Inductive effect explains the rate (why the ring is slower than benzene); resonance effect explains the position (why ortho/para beats meta). Keep them as separate arguments, not one blended explanation.
• Ignoring sterics when predicting the ortho to para ratio. Both positions are electronically equally favoured by resonance, but a bulky incoming or existing substituent usually shifts the balance toward the less crowded para product.

Ask the tutor: Want to work through why a specific haloarene reaction gives the product distribution it does, or practise separating resonance from inductive arguments on a new example?