Question

Which statement(s) is/are correct :

• A. Reduced product of $P$ and $Q$ will be metamers to each other.
• B. By dry distillation of hydrolysed products of $P$ with $Ca(OH)_2$, gives benzophenone.
• C. Hydrolysed product of $Q$, reacts with $NaNO_2$ + $HCl$ followed by reaction with phenol, gives orange red dye.
• D. Electrophile involved in the formation of $Q$ is dichlorocarbene.

Answer 1

Answer: (B), (C), (D)

B, C, D

Answer 2

The starting material is benzamide ($\text{C}_6\text{H}_5\text{CONH}_2$).

Synthesis of P: Benzamide is treated with $\text{P}_2\text{O}_5$ and heat. $\text{P}_2\text{O}_5$ is a dehydrating agent. Dehydration of benzamide gives benzonitrile. $\text{C}_6\text{H}_5\text{CONH}_2 \xrightarrow{\text{P}_2\text{O}_5, \Delta} \text{C}_6\text{H}_5\text{CN}$ So, P is benzonitrile.

Synthesis of Q: Benzamide is treated with (i) $\text{Br}_2 + \text{KOH} + \Delta$ and (ii) $\text{CHCl}_3 + \text{OH}^- + \Delta$. Step (i) is Hoffmann bromamide degradation, which converts benzamide to aniline. $\text{C}_6\text{H}_5\text{CONH}_2 \xrightarrow{\text{Br}_2, \text{KOH}, \Delta} \text{C}_6\text{H}_5\text{NH}_2$ Step (ii) is Carbylamine reaction (Hofmann isocyanide synthesis), where a primary amine (aniline) reacts with chloroform and a strong base to form an isocyanide. $\text{C}_6\text{H}_5\text{NH}_2 + \text{CHCl}_3 + 3\text{OH}^- \xrightarrow{\Delta} \text{C}_6\text{H}_5\text{NC} + 3\text{Cl}^- + 3\text{H}_2\text{O}$ So, Q is phenyl isocyanide.

Now, let's evaluate the statements:

Statement A: Reduced product of P and Q will be metamers to each other. Reduced product of P (benzonitrile): $\text{C}_6\text{H}_5\text{CN} \xrightarrow{\text{Reduction}} \text{C}_6\text{H}_5\text{CH}_2\text{NH}_2$ (Benzylamine, a primary amine). Reduced product of Q (phenyl isocyanide): $\text{C}_6\text{H}_5\text{NC} \xrightarrow{\text{Reduction}} \text{C}_6\text{H}_5\text{NHCH}_3$ (N-methylaniline, a secondary amine). Benzylamine and N-methylaniline have the same molecular formula $\text{C}_7\text{H}_9\text{N}$. However, they are functional isomers (primary amine vs secondary amine), not metamers. Metamers are isomers with the same molecular formula and the same functional group but different alkyl groups around the functional group. Thus, statement A is incorrect.

Statement B: By dry distillation of hydrolysed products of P with $\text{Ca(OH)}_2$, gives benzophenone. Hydrolysed product of P (benzonitrile) is benzoic acid ($\text{C}_6\text{H}_5\text{COOH}$). Dry distillation of calcium salt of benzoic acid (calcium benzoate) gives benzophenone. $2\text{C}_6\text{H}_5\text{COOH} + \text{Ca(OH)}_2 \rightarrow (\text{C}_6\text{H}_5\text{COO})_2\text{Ca} + 2\text{H}_2\text{O}$ $(\text{C}_6\text{H}_5\text{COO})_2\text{Ca} \xrightarrow{\text{Dry distillation}} (\text{C}_6\text{H}_5)_2\text{CO} + \text{CaCO}_3$ So, dry distillation of hydrolysed product of P with $\text{Ca(OH)}_2$ (implying formation of calcium salt and then dry distillation) gives benzophenone. Thus, statement B is correct.

Statement C: Hydrolysed product of Q, reacts with $\text{NaNO}_2 + \text{HCl}$ followed by reaction with phenol, gives orange red dye. Hydrolysis of Q (phenyl isocyanide) gives aniline. $\text{C}_6\text{H}_5\text{NC} + 2\text{H}_2\text{O} \xrightarrow{\text{H}^+} \text{C}_6\text{H}_5\text{NH}_2 + \text{HCOOH}$ Aniline reacts with $\text{NaNO}_2 + \text{HCl}$ (nitrous acid) at 0-5 $^\circ\text{C}$ to form benzenediazonium chloride (diazotization). $\text{C}_6\text{H}_5\text{NH}_2 + \text{NaNO}_2 + 2\text{HCl} \xrightarrow{0-5 ^\circ\text{C}} \text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- + \text{NaCl} + 2\text{H}_2\text{O}$ Benzenediazonium chloride undergoes azo coupling with phenol in alkaline medium to form p-hydroxyazobenzene, which is an orange-red dye. $\text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- + \text{C}_6\text{H}_5\text{OH} \xrightarrow{\text{OH}^-} \text{C}_6\text{H}_5\text{-N=N-}\text{C}_6\text{H}_4\text{-OH (para)} + \text{HCl}$ Thus, statement C is correct.

Statement D: Electrophile involved in the formation of Q is dichlorocarbene. Q is formed in the Carbylamine reaction. The Carbylamine reaction involves the generation of dichlorocarbene (:$\text{CCl}_2$) from chloroform and a strong base, which acts as the electrophile that attacks the primary amine. $\text{CHCl}_3 + \text{OH}^- \rightleftharpoons \text{CCl}_3^- + \text{H}_2\text{O}$ $\text{CCl}_3^- \rightarrow :\text{CCl}_2 + \text{Cl}^-$ Thus, the electrophile involved in the formation of Q is dichlorocarbene. Thus, statement D is correct.

Statements B, C, and D are correct.