Problem 105
Question
In the reaction given below, the product \(Y\) is \(\mathrm{C}_{6} \mathrm{H}_{6} \stackrel{\mathrm{HNO}_{3}}{\mathrm{H}_{2} \mathrm{SO}_{4}}-\mathrm{X} \frac{\mathrm{Cl}_{2}}{\mathrm{FeCl}_{3}}-\mathrm{Y}\) (a) 4 -nitrochlorobenzene (b) 3 -nitrochlorobenzene (c) 1 -nitrochlorobenzene (d) none of the above
Step-by-Step Solution
Verified Answer
The product \(Y\) is 1-nitrochlorobenzene (c).
1Step 1: Understand the Reaction
The reaction is a two-step electrophilic aromatic substitution involving benzene. The first step involves nitration using concentrated \(\text{HNO}_3\) in the presence of \(\text{H}_2\text{SO}_4\) to form nitrobenzene.
2Step 2: Write the First Reaction
The benzene reacts with \(\text{HNO}_3\) and \(\text{H}_2\text{SO}_4\). The \(\text{NO}_2\) group (nitro group) is introduced to the benzene ring, resulting in nitrobenzene, \(\text{C}_6\text{H}_5\text{NO}_2\).
Key Concepts
Nitration of BenzeneFriedel-Crafts AlkylationOrganic Reaction Mechanisms
Nitration of Benzene
Nitration of benzene is a significant reaction in organic chemistry, specifically an electrophilic aromatic substitution. This process introduces a nitro group (\(\text{NO}_2\)) into the benzene ring. The reaction involves the interaction of benzene with concentrated nitric acid (\(\text{HNO}_3\)) usually in the presence of sulfuric acid (\(\text{H}_2\text{SO}_4\)) as a catalyst.
Here's a simple breakdown of the nitration process:
Here's a simple breakdown of the nitration process:
- The \(\text{H}_2\text{SO}_4\) enhances the nitronium ion (\(\text{NO}_2^+\)) formation from \(\text{HNO}_3\), which is crucial for the nitration since it is a strong electrophile.
- The nitronium ion then attacks the electron-rich benzene ring, forming a sigma complex where the \(\text{NO}_2\) group is added to the aromatic ring.
- Following this, the intermediate complex undergoes deprotonation (a hydrogen ion is removed), restoring the aromaticity of the benzene ring, resulting in nitrobenzene.
Friedel-Crafts Alkylation
Friedel-Crafts Alkylation is another classic example of electrophilic aromatic substitution, where an alkyl group is attached to an aromatic ring. In many laboratory settings, it often involves benzene reacting with an alkyl halide in the presence of a Lewis acid catalyst like aluminum chloride (\(\text{AlCl}_3\)).
Let's delve into the Friedel-Crafts Alkylation mechanism:
Let's delve into the Friedel-Crafts Alkylation mechanism:
- The \(\text{AlCl}_3\), a Lewis acid, first helps form a stable carbocation from the alkyl halide, which serves as the powerful electrophile needed for the reaction.
- This carbocation then approaches the aromatic benzene ring, allowing the alkyl group to attach itself to one of the carbon atoms of benzene by substituting a hydrogen atom.
- Finally, a proton is removed from the intermediate, allowing the ring to regain its aromatic nature.
Organic Reaction Mechanisms
Organic reaction mechanisms are the step-by-step sequence of elementary reactions by which overall chemical change occurs. Understanding these mechanisms allows chemists to predict and control the outcome of reactions. In electrophilic aromatic substitutions like nitration and Friedel-Crafts Alkylation, the principal mechanism involves:
Moreover, environmental factors such as temperature, solvent, and catalyst presence critically influence these reactions. Gaining a solid understanding of these mechanisms provides a foundational toolset for designing new reactions and applications in synthesis.
- Formation of high-energy intermediates such as carbocations or sigma complexes, which are stabilized by resonance.
- Regaining aromaticity, which drives the reaction to completion.
Moreover, environmental factors such as temperature, solvent, and catalyst presence critically influence these reactions. Gaining a solid understanding of these mechanisms provides a foundational toolset for designing new reactions and applications in synthesis.
Other exercises in this chapter
Problem 99
Among the following, the strongest base is (a) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}\) (b) \(\mathrm{p}-\mathrm{NO}_{2}-\mathrm{C}_{6} \mathrm{H}_{4}
View solution Problem 103
When nitrobenzene is treated with \(\mathrm{Br}_{2}\) in presence of \(\mathrm{FeBr}_{3}\), the major product formed is \(\mathrm{m}\)-bromonitrobenzene. The st
View solution Problem 106
p-nitrotoluene on further nitration gives
View solution Problem 112
\(\mathrm{p}\) - Nitroaniline is obtained by (a) 1,4 dinitrobenzene \(\stackrel{\mathrm{NH}_{4} \mathrm{HS}}{\longrightarrow}\) (b) Benzene sulphonic acid \(\st
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