Question

Which of the following reagents will bring out the following transformation ?

Choose the correct answer from the options given below:

• A. (A) and (C) only
• B. (A), (B) and (D) only
• C. (A), (B), (C) and (D)
• D. (B), (C) and (D) only

Hint:

Aniline to Chlorobenzene:
1. \(\text{NaNO}_2 + \text{HCl} \xrightarrow{0\text{--}5^\circ\text{C}} \text{Ar-N}_2^+\text{Cl}^-\) (Diazotization).
2. Followed by either Sandmeyer (\(\text{Cu}_2\text{Cl}_2/\text{HCl}\)) or Gattermann (\(\text{Cu}/\text{HCl}\)).

Answer 1

The Correct Option is A

Concept:
The primary route for converting aniline into aryl halides involves converting aniline into a benzenediazonium salt (diazotization) followed by displacement of the diazonium group by a chloride nucleophile using copper-based reagents.

Step 1: The Diazotization Requirement:
In the initial step, primary aromatic amines must be treated with nitrous acid (\(\text{HNO}_2\)), generated in situ from sodium nitrite and hydrochloric acid (\(\text{NaNO}_2 + \text{HCl}\)) at low temperature (\(273\text{--}278\text{ K}\)):
\[ \text{C}_6\text{H}_5\text{NH}_2 + \text{NaNO}_2 + 2\text{HCl} \xrightarrow{0\text{--}5^\circ\text{C}} \text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- + \text{NaCl} + 2\text{H}_2\text{O} \] This reaction requires an acidic medium.
Using \(\text{NaNO}_2/\text{NaOH}\) (as in options B and D) provides basic conditions that fail to produce nitrous acid, preventing diazonium salt formation.

Step 2: Sandmeyer Reaction via Reagent (C):
When the freshly prepared benzenediazonium chloride solution is treated with cuprous chloride dissolved in hydrochloric acid (\(\text{Cu}_2\text{Cl}_2/\text{HCl}\)), chlorobenzene is formed:
\[ \text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- \xrightarrow{\text{Cu}_2\text{Cl}_2/\text{HCl}} \text{C}_6\text{H}_5\text{Cl} + \text{N}_2\uparrow \] This process is known as the Sandmeyer reaction. Hence, set (C) carries out the transformation.

Step 3: Gattermann Reaction via Reagent (A):
Alternatively, treating the benzenediazonium salt with copper powder in the presence of hydrochloric acid (\(\text{Cu}/\text{HCl}\)) also produces chlorobenzene:
\[ \text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- \xrightarrow{\text{Cu}/\text{HCl}} \text{C}_6\text{H}_5\text{Cl} + \text{N}_2\uparrow + \text{CuCl} \] This variation is known as the Gattermann reaction. Hence, set (A) also successfully carries out the transformation.
Final Answer:
The transformation can be accomplished using (A) and (C) only, corresponding to option (A).