Chapter 26: Problem 86
An organic compound \(A, \mathrm{C}_{8} \mathrm{H}_{6}\), on treatment with dilute sulphuric acid containing mercuric sulphate gives a compound \(B\), which can also be obtained from a reaction of benzene with an acid chloride in the presence of anhydrous aluminium chloride. The compound \(B\), when treated with iodine in aqueous \(\mathrm{KOH}\), yields \(C\) and a yellow compound \(D .\) Identify \(A, B, C\) and \(D\) with justification. Show how \(B\) is formed from \(A\).
Short Answer
Step by step solution
Identify Compound B with Friedel-Crafts Acylation
Identify Compound A Through Reaction Conditions
Mechanism of Formation from A to B
Reaction of B to form C and D
Identify Yellow Compound D
Complete the Identification
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Friedel-Crafts Acylation
The process involves the reaction of an aromatic compound with an acid chloride in the presence of a strong Lewis acid, typically aluminum chloride (AlCl\(_3\)). In the typical scenario for this process, acetyl chloride (CH\(_3\)COCl) serves as the acylating agent, while benzene acts as the aromatic compound.
- First, AlCl\(_3\) coordinates with the chlorine atom in acetyl chloride, creating a highly reactive electrophile.
- This electrophile is capable of attacking the aromatic ring of benzene, leading to the formation of an intermediate.
- Finally, after re-aromatization and the removal of HCl, acetophenone is formed as the final product.
Alkyne Reactions
When terminal alkynes, such as phenylacetylene (C\(_6\)H\(_5\)C\equivCH), are subjected to hydration under specific conditions, they convert into ketones. The hydration of phenylacetylene using dilute sulfuric acid (H\(_2\)SO\(_4\)) and mercuric sulfate (HgSO\(_4\)) is a classic example of this transformation.
- Initially, the alkyne reacts with the sulfuric acid and HgSO\(_4\), facilitating the formation of an enol.
- This unstable enol immediately tautomerizes to form a stable ketone.
- In this particular case, acetophenone forms, which matches the compound B identified in the exercise.
Haloform Reaction
In the described exercise, compound B, acetophenone, undergoes the haloform reaction in the presence of iodine (I\(_2\)) and aqueous potassium hydroxide (KOH). This reaction serves to highlight the unique characteristics of the haloform process.
- Initially, the methyl group adjacent to the carbonyl group in acetophenone is halogenated to form a tetrahalomethyl group.
- Subsequently, this group is cleaved in the presence of base, resulting in the formation of iodoform (yellow compound D, CHI\(_3\)) and a carboxylate anion.
- Upon acidification, the carboxylate anion ultimately becomes a carboxylic acid, such as benzoic acid (compound C).