Aryl Halides and Why They Resist Substitution | JEE Notes

Put the halogen on a ring and most of what you learned about alkyl halides stops working. Here is why, and what brings the reactions back.

Four reasons for low reactivity

Resonance shortens and strengthens the C-Cl bond.

Resonance. Chlorine's lone pair enters the ring, giving C-Cl partial double-bond character.
Hybridisation. An $s p^{2}$ carbon holds the bond pair more tightly. NCERT gives C-Cl as 169 pm in chlorobenzene and 177 pm in a haloalkane.
Unstable cation. The phenyl cation cannot use resonance, because its empty orbital lies in the plane of the ring.
Repulsion and blocking. The electron-rich ring repels a nucleophile, and backside attack would go through the ring.

Watch the animation: Four reasons chlorobenzene resists substitution

As a result, C-Cl in chlorobenzene is less polar than in chlorocyclohexane, and it gives no precipitate with alcoholic $A g N O_{3}$ .

Forcing conditions, and how nitro groups help

Substrate	Conditions to replace Cl by OH
Chlorobenzene	NaOH, 623 K, 300 atm, then acid
4-Nitrochlorobenzene	NaOH, 443 K, then acid
2,4-Dinitrochlorobenzene	NaOH, 368 K, then acid
2,4,6-Trinitrochlorobenzene	Warm water, 323 K

Addition-elimination

The nucleophile adds to the carbon holding X and breaks the ring's aromaticity. The negative charge goes round the ring to the carbons ortho and para to the attack. A nitro group on one of those carbons takes the charge onto its oxygens. Then X leaves and aromaticity returns.

Addition then elimination on 4-nitrochlorobenzene

Hydroxide adds to the carbon holding chlorine. The C-Cl bond is still intact, and this addition is the slow step.

Carbon 1 is now $s p^{3}$ and aromaticity is lost. The negative charge sits on C-2. It can also sit on C-6, on the other side.

The charge can also sit on C-4. Note the pattern: 2, 4 and 6 are the ortho and para positions to the attack. A meta nitro group never gets the charge.

Here is why nitro helps. From C-4 the pair moves onto a nitro oxygen, which is the best place for it.

The ring pushes chloride out and aromaticity returns. In base the product is the phenoxide, and acid then gives 4-nitrophenol.

Watch the animation: Nitro groups rescue the ring

The Meisenheimer intermediate: an ortho or para

N O_{2}

can hold the charge, a meta one cannot.

The halogen order reverses here. The first step is slow and the C-X bond is still intact in it. So the bond strength does not matter much, and strong electron withdrawal does. For this reason, activated aryl fluorides react fastest: $F ≫ C l \approx B r > I$ , which is the reverse of the $S_{N} 1$ and $S_{N} 2$ order.

Electrophilic substitution

Halogens withdraw electrons by induction, so the ring reacts more slowly than benzene. Their lone pairs donate by resonance only when the electrophile adds ortho or para, so those positions win. Rate comes from $- I$ , and direction comes from $+ R$ .

Why chlorine directs para: nitration of chlorobenzene

The ring attacks the electrophile at C-4. Chlorine has already made the ring poorer in electrons by induction, so this is slower than with benzene.

The ring now carries a positive charge. One form puts it on C-3.

Another form puts it on C-1, directly under chlorine. This looks bad at first, because chlorine pulls electrons.

But chlorine can share a lone pair, and now every atom has an octet. Only ortho and para attack allow this extra form. Meta attack never places the charge on C-1, so it gets no help.

A proton leaves and aromaticity returns. Para is the main product, because ortho is more crowded.

Watch the animation: Slower, yet ortho and para

Chlorobenzene undergoes halogenation, nitration, sulphonation and Friedel-Crafts reactions. The para isomer is major, because ortho is crowded.

Reactions with metals

Fittig coupling gives biaryls, and Wurtz-Fittig gives alkylbenzenes. Aryl bromides also form Grignard reagents.

Worked example: rank for hydrolysis

Rank by ease of replacing Cl with OH: (a) chlorobenzene, (b) 3-nitrochlorobenzene, (c) 4-nitrochlorobenzene, (d) 2,4-dinitrochlorobenzene.

Show the solution

Count nitro groups at ortho and para. (d) has two, (c) has one, and (b) has none, because a meta nitro cannot take the charge. It still pulls a little by induction, so (b) sits just above (a).

So (d) > (c) > (b) > (a).

Beyond JEE: benzyne

With a very strong base such as $N a N H_{2}$ , an unactivated aryl halide can lose HX to form benzyne, which then adds the nucleophile. The JEE Advanced syllabus excludes the benzyne mechanism and cine substitution, and NCERT does not cover it.

Why aryl halides resist, and how to make them react

In 60 seconds