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The sulphonation of chlorobenzene gives exclusively (1) an o-product (2) an m-produet (3) a p-product (4) an o-, \(\mathrm{p}\) -disubstituted product

Short Answer

Expert verified
(4) an o-, \text{p}-disubstituted product

Step by step solution

01

Identify the Substituent Effect

Chlorobenzene has a chlorine substituent which is electron-withdrawing through induction but electron-donating through resonance. This makes chlorine an ortho-para directing group.
02

Determine the Directing Influence

Because chlorine directs incoming groups to the ortho and para positions relative to itself, the sulphonation will predominantly occur at these positions.
03

Understand the Reaction Site Preference in Sulphonation

For sulphonation, both ortho and para positions are possible, but there are generally more para products because of steric hindrance at the ortho positions.
04

Conclusion Based on Influence

Given the directing influence of chlorine and the steric factors, sulphonation of chlorobenzene will give mainly a para product, but ortho substitution is also observed.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

ortho-para directing effect
In the context of aromatic substitution reactions, some substituents can influence where new groups attach to a benzene ring. Chlorine is a good example here. Although it's an electron-withdrawing group through inductive effects, it can actually donate electrons through resonance and stabilize the benzene ring. This dual behavior makes chlorine an 'ortho-para director'. This means that, in a reaction, new groups prefer to add at the ortho (adjacent) or para (opposite) positions relative to the chlorine. As a result, when chlorobenzene undergoes sulphonation, the sulfonic acid group (SO₃H) will predominantly attach to these favored positions.
electron-withdrawing group
Chlorine in chlorobenzene acts as an electron-withdrawing group (EWG) through its inductive effect. This means it pulls electron density away from the benzene ring due to its higher electronegativity. However, chlorine also has lone pairs that can participate in resonance. Through this resonance, chlorine donates electron density back into the ring, which compensates for some of the electron withdrawal. The net effect leads to a decrease in overall electron density, yet enough remains to stabilize any incoming groups at ortho and para positions.
steric hindrance
Steric hindrance is about the physical space and the crowding around the reactive sites on a molecule. In other words, larger groups can block other reactions from happening close by. In the case of chlorobenzene, the ortho positions are close to the chlorine atom, making them more crowded. During sulphonation, this means that it's harder for the sulfonic acid group to fit into these ortho positions without causing strain or overlap with the chlorine. As a result, the para position, which is further away, generally becomes a more favorable site for the sulphonation product since there's less crowding and more room.
reaction site preference
When considering the sulphonation of chlorobenzene, we need to think about where the new sulfonic acid group will prefer to attach. Ortho and para directing effects, along with steric hindrance, influence this decision. Because chlorine directs new groups to the ortho and para positions and due to steric hindrance at the ortho positions, the para position becomes particularly preferred. This results in most of the sulphonation product being in the para form. However, some ortho substitution does still occur, but it's less common than the para substitution.

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Most popular questions from this chapter

Which of the following is the most reactive towards ring nitration? (1) Benzene (2) Toluene (3) m-Xylene (4) Mesitylene

The stability of the free radicals allyl, benzyl, \(3^{\circ}, 2^{\circ}\), \(1^{\circ}\) and \(\mathrm{CH}_{3}\) is of the order (1) allyl=benzyl \(>3^{\circ}>2^{\circ}>1^{\circ}>\mathrm{CH}_{3}\) (2) allyl \(>\) benzyl \(>3^{\circ}>2^{\circ}>1^{\circ}>\mathrm{CH}_{3}\) (3) \(3^{\circ}>2^{\circ}>1^{\circ}>\mathrm{CH}_{3}>\) allyl \(>\) benzyl (4) \(3^{\circ}>2^{\circ}>1^{\circ}>\mathrm{CH}_{3}>\) allyl=benzyl

Arrange the following groups in order of decreasing clectron attracting capacity. (1) \(\mathrm{COOII}>\mathrm{NO}_{2}>\mathrm{OH}>\mathrm{Cl}\) (2) \(\mathrm{NO}_{2}>\mathrm{COOH}>\mathrm{Cl}>\mathrm{OII}\) (3) \(\mathrm{OH}>\mathrm{Cl}>\mathrm{COOH}>\mathrm{NO}_{2}\) (4) \(\mathrm{NO}_{2}>\mathrm{COOH}>\mathrm{OH}>\mathrm{Cl}\)

The replacement of chlorine from chlorobenzene to give phenol requires a drastic condition, but the chlorine of 2,4 -dichloro-nitro-benzene is readily replaced since (1) nitro groups make the aromatic ring electron rich at ortho/para-positions (2) nitro groups withdraw electrons from the metaposition of the aromatic ring (3) nitro groups donate electrons at meta-position (4) nitro groups withdraw electrons from orthol para-positions of the aromatic ring

When nitrobenzene is treated with \(\mathrm{Br}_{2}\) in the presence of \(\mathrm{FeBr}_{3}\), the major product formed is \(\mathrm{m}\) -bromonitrobenzene. Statement which is related to obtain the m-isomer is (1) the clectron density on meta-carbon is more than that on ortho- and para- positions (2) loss of aromaticity when \(\mathrm{Br}^{+}\) attacks at the orthoand para- positions and not at meta-position (3) casier loss of II" to regain aromaticity from the meta-position than from ortho- and parapositions (4) None of the above

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