Question

Identify the product C in the series.
C$_6$H$_5$NO$_2 \xrightarrow{\text{Fe/HCl}} A \xrightarrow{\text{NaNO}_2\text{+HCl, } 273\text{ K}} B \xrightarrow{\text{H}_2\text{O, } 283\text{ K}} C$

• A. C$_6$H$_5$OH
• B. C$_6$H$_5$CH$_2$OH
• C. C$_6$H$_5$CHO
• D. C$_6$H$_5$NH$_2$

Hint:

Key Exam Tip:
Remember the key transformations:
• Nitro group reduction to amine: e.g., R-NO$_2 \xrightarrow{\text{Fe/HCl or Sn/HCl}} R-NH_2$.
• Diazotization of primary aromatic amines: R-NH$_2 \xrightarrow{\text{NaNO}_2\text{+HCl, low temp}} R-N_2^+Cl^-$.
• Hydrolysis of diazonium salts: R-N$_2^+X^- \xrightarrow{\text{H}_2\text{O, heat}} R-OH + N_2 + HX$.

Answer 1

The Correct Option is A

This is a classic sequence of reactions involving nitrobenzene and its transformation into phenol. Let's break down the steps:

Step 1: Reduction of nitrobenzene to aniline C$_6$H$_5$NO$_2$ (Nitrobenzene) $\xrightarrow{\text{Fe/HCl}}$ A Nitrobenzene is reduced to aniline (aminobenzene) in the presence of a reducing agent like Fe/HCl. Therefore, A is C$_6$H$_5$NH$_2$ (Aniline).

Step 2: Diazotization of aniline A (C$_6$H$_5$NH$_2$) $\xrightarrow{\text{NaNO}_2\text{+HCl, } 273\text{ K}}$ B Aniline reacts with sodium nitrite (NaNO$_2$) and hydrochloric acid (HCl) at a low temperature (0-5 $^\circ$C, which is 273 K) to form a diazonium salt. This process is called diazotization. Therefore, B is C$_6$H$_5$N$_2^+$Cl$^-$ (Benzenediazonium chloride).

Step 3: Hydrolysis of the diazonium salt B (C$_6$H$_5$N$_2^+$Cl$^-$) $\xrightarrow{\text{H}_2\text{O, } 283\text{ K}}$ C Benzenediazonium chloride reacts with water at a slightly elevated temperature (around 10-15 $^\circ$C, which is 283 K) to undergo hydrolysis. The diazonium group (-N$_2^+$) is replaced by a hydroxyl group (-OH), forming phenol. Nitrogen gas (N$_2$) and HCl are released as byproducts. The reaction is: C$_6$H$_5$N$_2^+$Cl$^-$ + H$_2$O $\xrightarrow{\Delta}$ C$_6$H$_5$OH + N$_2$ + HCl Therefore, the product C is C$_6$H$_5$OH (Phenol). Let's check the options: (1) C$_6$H$_5$OH (Phenol) - This matches our product. (2) C$_6$H$_5$CH$_2$OH (Benzyl alcohol) - Incorrect. (3) C$_6$H$_5$CHO (Benzaldehyde) - Incorrect. (4) C$_6$H$_5$NH$_2$ (Aniline) - This is intermediate A, not final product C. Final Answer: \(\boxed{1}\)