Electrophilic Substitution in Haloarenes and Reactions with Metals

In the previous topic, you saw that haloarenes put up a fierce resistance against nucleophiles. But what about electrophiles? Since the benzene ring in a haloarene still carries a pi-electron cloud, electrophiles can and do attack it. The twist is that the halogen atom plays a double game: it makes the ring slightly sluggish (deactivating), yet it steers every incoming electrophile to very specific positions. Understanding how one atom can slow a reaction down while simultaneously controlling where it happens is one of the most satisfying ideas in aromatic chemistry.

The Halogen’s Double Role: Deactivating yet Ortho, Para-Directing

Haloarenes undergo all the classic electrophilic substitution reactions that benzene does: halogenation, nitration, sulphonation, and Friedel-Crafts alkylation and acylation. The key difference is that the halogen substituent influences both the speed and the position of the attack, and it does so through two competing electronic effects.

How Resonance Steers the Electrophile

A halogen atom bonded to a benzene ring possesses lone pairs that can overlap with the ring’s pi system. Drawing the resonance structures of halobenzene makes this clear:

In these structures, the halogen donates a lone pair into the ring, picking up a partial positive charge itself. The extra electron density lands on the ortho and para carbons (not the meta carbons). When an electrophile comes looking for electron-rich spots to attack, it naturally heads for these ortho and para positions. So resonance controls the orientation of the substitution.

How the Inductive Effect Slows Things Down

At the same time, the halogen is highly electronegative. Through the sigma bond, it pulls electron density away from the ring. This is the inductive effect ( $-I$ ), and it operates in the opposite direction to resonance. Since the inductive withdrawal is stronger overall than the resonance donation, the net effect is that the ring loses some electron density compared to plain benzene.

An electrophile finds the ring of a haloarene less inviting than unsubstituted benzene. The reaction still takes place, but it needs more vigorous conditions: higher temperatures, stronger catalysts, or longer reaction times. So the inductive effect controls reactivity, making haloarenes slower to react than benzene.

Two Effects, Two Jobs

Here is the elegant summary:

Resonance ( $+R$ effect) increases electron density at ortho and para positions, directing the electrophile there. It opposes the inductive withdrawal at these positions, making deactivation less severe at ortho and para compared to meta.
Inductive effect ( $-I$ effect) withdraws electrons from the entire ring through the sigma framework, causing net deactivation.

The result: halogen substituents are the unusual case of groups that are ortho, para-directing yet deactivating.

The Four Classic Electrophilic Substitution Reactions

Each of these reactions follows the same pattern: the electrophile enters at the ortho and para positions relative to the halogen. The para product is typically the major product, and the ortho product is the minor one (meta is negligible).

Halogenation

When chlorobenzene is treated with $Cl_2$ in the presence of anhydrous $FeCl_3$ (a Lewis acid catalyst), the second chlorine enters at the ortho or para position:

$C_6H_5Cl + Cl_2 \xrightarrow{\text{anhyd. } FeCl_3} \text{1,4-Dichlorobenzene (major)} + \text{1,2-Dichlorobenzene (minor)}$

Nitration

Chlorobenzene reacts with a nitrating mixture of concentrated $HNO_3$ and concentrated $H_2SO_4$ to give a nitro group at the para (major) or ortho (minor) position:

$C_6H_5Cl + HNO_3 \xrightarrow{\text{conc. } H_2SO_4} \text{1-Chloro-4-nitrobenzene (major)} + \text{1-Chloro-2-nitrobenzene (minor)}$

Sulphonation

Heating chlorobenzene with concentrated $H_2SO_4$ introduces a sulphonic acid group ( $-SO_3H$ ) at the para (major) or ortho (minor) position:

$C_6H_5Cl + H_2SO_4 \xrightarrow{\Delta} \text{4-Chlorobenzenesulphonic acid (major)} + \text{2-Chlorobenzenesulphonic acid (minor)}$

Friedel-Crafts Reactions

Both alkylation and acylation follow the same ortho, para pattern. Anhydrous $AlCl_3$ serves as the Lewis acid catalyst.

Friedel-Crafts alkylation with methyl chloride:

$C_6H_5Cl + CH_3Cl \xrightarrow{\text{anhyd. } AlCl_3} \text{1-Chloro-4-methylbenzene (major)} + \text{1-Chloro-2-methylbenzene (minor)}$

Friedel-Crafts acylation with acetyl chloride:

$C_6H_5Cl + CH_3COCl \xrightarrow{\text{anhyd. } AlCl_3} \text{4-Chloroacetophenone (major)} + \text{2-Chloroacetophenone (minor)}$

In every case, the pattern is the same: para product dominates, ortho product is secondary, meta product is essentially absent.

Worked Example: Why is Chlorine Ortho, Para-Directing Despite Being Electron-Withdrawing?

This is one of the most commonly asked conceptual questions in organic chemistry, and it ties together everything above.

The question: Chlorine is an electron-withdrawing group. So why does it direct incoming electrophiles to the ortho and para positions rather than meta?

The answer, step by step:

Chlorine influences the ring through two different pathways, and each has a distinct effect:

Through the inductive effect ( $-I$ ): Chlorine withdraws electrons through the sigma bond. This withdrawal destabilises the carbocation intermediate that forms during electrophilic substitution, making the overall reaction slower compared to unsubstituted benzene.

Through resonance ( $+R$ ): Chlorine releases electrons by donating a lone pair into the ring. When the electrophile attacks at the ortho or para position, the resonance structures show that the positive charge in the intermediate can be placed directly on the carbon bearing the chlorine. The chlorine’s lone pair then stabilises this charge through direct conjugation.

When the electrophile attacks at the meta position, the positive charge in the intermediate never lands on the carbon bonded to chlorine, so the chlorine’s lone pair cannot stabilise it through resonance.

Pulling it all together:

The inductive effect is stronger than resonance in absolute terms, so the net result is electron withdrawal and ring deactivation. This controls the overall reactivity (haloarenes react more slowly than benzene).
The resonance effect opposes the inductive effect specifically at the ortho and para positions. Deactivation at these positions is less severe than at the meta position. This controls the orientation (electrophile goes to ortho and para).

In short: reactivity is governed by the stronger inductive effect; orientation is governed by the resonance effect.

Reactions with Metals: Coupling Aryl and Alkyl Groups

Haloarenes also participate in important coupling reactions when treated with sodium metal in dry ether. These reactions build new carbon-carbon bonds and are named after the chemists who discovered them.

The Wurtz-Fittig Reaction

When a mixture of an aryl halide and an alkyl halide is heated with sodium in dry ether, sodium strips away the halogen atoms and joins the aryl group to the alkyl group. The product is an alkylarene (an arene with an alkyl side chain).

$Ar{-}X + R{-}X + 2Na \xrightarrow{\text{dry ether}} Ar{-}R + 2NaX$

For example, bromobenzene and methyl bromide with sodium in dry ether produce toluene:

$C_6H_5Br + CH_3Br + 2Na \xrightarrow{\text{dry ether}} C_6H_5CH_3 + 2NaBr$

This is an extension of the Wurtz reaction (which couples two alkyl halides). The “Fittig” part of the name acknowledges that one of the coupling partners is an aryl halide.

The Fittig Reaction

When two aryl halides (rather than a mix of aryl and alkyl) are treated with sodium in dry ether, the two aryl groups join together. The product is a biaryl compound (two aromatic rings linked directly).

$2Ar{-}X + 2Na \xrightarrow{\text{dry ether}} Ar{-}Ar + 2NaX$

For example, two molecules of bromobenzene give biphenyl (also called diphenyl):

$2C_6H_5Br + 2Na \xrightarrow{\text{dry ether}} C_6H_5{-}C_6H_5 + 2NaBr$

Quick Comparison: Wurtz vs Wurtz-Fittig vs Fittig

Reaction	Reactants	Metal / Solvent	Product
Wurtz	Alkyl halide + alkyl halide	$Na$ / dry ether	Alkane ( $R{-}R'$ )
Wurtz-Fittig	Aryl halide + alkyl halide	$Na$ / dry ether	Alkylarene ( $Ar{-}R$ )
Fittig	Aryl halide + aryl halide	$Na$ / dry ether	Biaryl ( $Ar{-}Ar$ )

All three reactions use sodium metal in dry ether and work by removing halogen atoms to form new $C{-}C$ bonds. The only difference is which types of halides are being coupled.