Question

The compound from which chlorobenzene cannot be prepared easily is

• A. Aniline
• B. Benzene
• C. Phenol
• D. Benzene diazonium chloride

Hint:

The reactivity of the C-X bond (where X is a heteroatom like O, N, or a halogen) attached to a benzene ring is very different from its aliphatic counterpart. Resonance makes the bond stronger and less susceptible to nucleophilic substitution. This is why phenols are not easily converted to haloarenes.

Answer 1

The Correct Option is C

Step 1: Evaluate the synthesis of chlorobenzene from each starting material.
(A) Aniline (C\(_6\)H\(_5\)NH\(_2\)): Aniline can be easily converted to chlorobenzene via the Sandmeyer reaction. First, aniline is treated with NaNO\(_2\)/HCl at 0-5\(^\circ\)C to form benzene diazonium chloride. This is then treated with CuCl/HCl to yield chlorobenzene. This is a standard and easy preparation.
\[ \text{C}_6\text{H}_5\text{NH}_2 \xrightarrow{\text{NaNO}_2, \text{HCl}} \text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- \xrightarrow{\text{CuCl/HCl}} \text{C}_6\text{H}_5\text{Cl} \] (B) Benzene (C\(_6\)H\(_6\)): Chlorobenzene is prepared commercially by the direct halogenation of benzene. This is an electrophilic aromatic substitution reaction carried out by reacting benzene with chlorine in the presence of a Lewis acid catalyst like FeCl\(_3\) or AlCl\(_3\). This is an easy and direct method.
\[ \text{C}_6\text{H}_6 + \text{Cl}_2 \xrightarrow{\text{FeCl}_3} \text{C}_6\text{H}_5\text{Cl} + \text{HCl} \] (C) Phenol (C\(_6\)H\(_5\)OH): The C-O bond in phenol is very strong and difficult to break. This is because the lone pair of electrons on the oxygen atom participates in resonance with the benzene ring, giving the C-O bond a partial double bond character. Simple reagents used to convert aliphatic alcohols to alkyl chlorides (like HCl, SOCl\(_2\), PCl\(_5\)) do not work effectively with phenol to produce chlorobenzene. The reaction is not easy and requires harsh conditions with very low yield.
(D) Benzene diazonium chloride (C\(_6\)H\(_5\)N\(_2\)\(^+\)Cl\(^-\)): This is an excellent starting material for preparing chlorobenzene. As mentioned in part (A), treatment with cuprous chloride (CuCl) in HCl (Sandmeyer reaction) or with copper powder in HCl (Gattermann reaction) easily converts it to chlorobenzene.
Step 2: Final Answer.
Due to the partial double bond character of the C-O bond, it is very difficult to prepare chlorobenzene from phenol. Therefore, this is the compound from which chlorobenzene cannot be prepared easily.