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Chapter 20: Problem 88

The reaction least likely to oecur is (1) \(\mathrm{C}_{6} \mathrm{H}_{6}+\mathrm{HNO}_{3} \stackrel{\mathrm{II}, \mathrm{so}}{\longrightarrow} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NO}_{2}\) (2) \(\mathrm{C}_{6} \mathrm{H}_{6}+\mathrm{H}_{2} \mathrm{SO}_{4} \stackrel{\text { Heat }}{\longrightarrow} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{SO}_{3} \mathrm{H}\) (3) \(\mathrm{C}_{6} \mathrm{H}_{6}+\mathrm{Cl}_{2} \stackrel{\text { UY }}{\longrightarrow} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{Cl}\) (4) \(\mathrm{C}_{6} \mathrm{H}_{6}+\mathrm{Br}_{2} \longrightarrow \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{Br}\)

Short Answer

Expert verified
4. \text{C}_6\text{H}_6 + \text{Br}_2 \rightarrow \text{C}_6\text{H}_5\text{Br} is least likely to occur without a catalyst.

Step by step solution

01

Analyze Reaction 1

Check the reaction \(\text{C}_6\text{H}_6 + \text{HNO}_3 \rightarrow \text{C}_6\text{H}_5\text{NO}_2\) using \(\text{H}_2\text{SO}_4\). This is the nitration of benzene, a common reaction for forming nitrobenzene. This reaction is quite feasible.
02

Analyze Reaction 2

Check the reaction \(\text{C}_6\text{H}_6 + \text{H}_2\text{SO}_4 \rightarrow \text{C}_6\text{H}_5\text{SO}_3\text{H}\) with heat. This is the sulfonation of benzene, a typical reaction used to form benzene sulfonic acid. It is also feasible.
03

Analyze Reaction 3

Check the reaction \(\text{C}_6\text{H}_6 + \text{Cl}_2 \rightarrow \text{C}_6\text{H}_5\text{Cl}\) using UV light. This is a chlorination reaction, which is feasibly carried out under UV light.
04

Analyze Reaction 4

Check the reaction \(\text{C}_6\text{H}_6 + \text{Br}_2 \rightarrow \text{C}_6\text{H}_5\text{Br}\). This looks like a bromination reaction, but benzene typically requires a Lewis acid catalyst, such as \(\text{FeBr}_3\), for bromination. Without the catalyst, this reaction is unlikely.

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Key Concepts

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

Nitration of Benzene
The nitration of benzene is a well-known and important chemical reaction.
Benzene reacts with concentrated nitric acid (HNO鈧�) in the presence of sulfuric acid (H鈧係O鈧�) as a catalyst to form nitrobenzene (C鈧咹鈧匩O鈧�).
This process involves an electrophilic aromatic substitution.

Here鈥檚 a detailed breakdown of what happens:
  • Sulfuric acid protonates nitric acid, turning it into a nitronium ion (NO鈧傗伜), a very strong electrophile.
  • The nitronium ion attacks the benzene ring, creating a resonance-stabilized intermediate.
  • Finally, a proton (H鈦�) is abstracted, restoring the aromaticity of the benzene ring and forming nitrobenzene.

This reaction is highly feasible and widely used in the production of nitrobenzene, which is a precursor for many dyes, drugs, and explosives.
Sulfonation of Benzene
Sulfonation of benzene is another electrophilic aromatic substitution reaction.
This reaction occurs when benzene reacts with sulfuric acid (H鈧係O鈧�) under heat to form benzene sulfonic acid (C鈧咹鈧匰O鈧僅).
Sulfonation involves the introduction of a sulfonyl group (SO鈧僅) into the benzene ring.

Here鈥檚 how it happens:
  • Under heat, sulfuric acid generates sulfur trioxide (SO鈧�), a strong electrophile.
  • Sulfur trioxide then attacks the benzene ring, forming a resonance-stabilized intermediate.
  • This intermediate then loses a proton (H鈦�), resulting in the formation of benzene sulfonic acid.

Sulfonation is a practical reaction used to manufacture detergents, dyes, and pharmaceuticals.
Chlorination of Benzene
Chlorination of benzene is a common reaction where benzene reacts with chlorine (Cl鈧�) in the presence of ultraviolet (UV) light to produce chlorobenzene (C鈧咹鈧匔l).
This is also an electrophilic aromatic substitution reaction.

The steps include:
  • UV light breaks down chlorine molecules (Cl鈧�) into chlorine radicals (Cl鈥�).
  • One of these chlorine radicals attacks the benzene ring, forming a resonance-stabilized intermediate.
  • The intermediate then loses a proton (H鈦�), leading to the formation of chlorobenzene.

This reaction requires UV light to generate the chlorine radicals and is widely used in organic synthesis and the production of pesticides, dyes, and pharmaceuticals.
Bromination of Benzene
Bromination of benzene is a reaction where benzene reacts with bromine (Br鈧�) to form bromobenzene (C鈧咹鈧匓r).
However, unlike chlorination, bromination requires the presence of a Lewis acid catalyst such as iron(III) bromide (FeBr鈧�).
This reaction is also an electrophilic aromatic substitution.

Here鈥檚 the reaction mechanism:
  • FeBr鈧� polarizes the Br鈧� molecule, forming a more electrophilic bromine species (Br鈦�).
  • This more electrophilic bromine attacks the benzene ring, creating a resonance-stabilized intermediate.
  • The intermediate then loses a proton (H鈦�), resulting in the formation of bromobenzene.

Without the catalyst, the bromination of benzene is highly unlikely to proceed, making the presence of FeBr鈧� crucial for the reaction's success. Brominated compounds are frequently used in fire retardants and pharmaceuticals.

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Most popular questions from this chapter

Among the following the wrong statcment is (1) aromatic hydrocarbons are the derivatives of benzene (2) benzenc contains 9 sigma and 3 pi bonds (3) aromaticity of benzene is due to delocalization of \(\pi\) -clectrons (4) all carbon atoms in benzene are involved in sp \(^{2}\) hybridisation

Arrange toluene (I), benzyl chloride (II), benzalchloride (III) and benzotrichloride (IV) in order of the inductive cffect of the group attached to the benzene nucleus. (1) \(\mid<][

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

Toluene can be converted into benzaldehyde by oxidation with (1) \(\mathrm{KMnO}_{4}\) /alkali (2) \(\mathrm{CrO}_{2} \mathrm{Cl}_{2}\) (3) \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7} / \mathrm{H}_{2} \mathrm{SO}_{4}\) (4) \(\mathrm{O}_{2} / \mathrm{V}_{2} \mathrm{O}_{5}\)

The wrong statement in the following is (1) Sulphonation of benzenc takes place only with hot concentrated sulphuric acid. (2) In the nitration mixture concentrated sulphuric acid is uscd for the formation of nitronium ion. (3) Bccause of unsaturation benzene casily undergoes addition rcactions. (4) Benzene burns with a sooty flame.
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