An organic compound (P) on treatment with aqueous ammonia under hot condition forms compound (Q) which on heating with Br₂ and KOH forms compound (R) having molecular formula C₆H₇N. Names of P, Q and R respectively are.
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Recognizing named reactions is a superpower in organic chemistry. Hofmann Bromamide Degradation (Amide → Amine with one less carbon) is very frequently tested. When a molecular formula is given for an aromatic compound, try to fit it to common structures like benzene, aniline, phenol, etc.
Step 1: Understanding the Question:
This question presents a two-step reaction sequence and asks to identify the reactant (P), intermediate (Q), and final product (R). The key is to recognize the named reactions involved. It's often easiest to work backwards from the final product whose formula is given. Step 2: Detailed Explanation: Analyze the final product R (C₆H₇N):
\begin{itemize}
\item The molecular formula C₆H₇N strongly suggests an aromatic amine. C₆H₅-NH₂ (Aniline) has the formula: (6 x 12.01) + (7 x 1.01) + (1 x 14.01) and contains a benzene ring. The structure is C₆H₅NH₂, which adds up to C₆H₇N. So, R is Aniline.
\end{itemize}
Analyze the reaction Q → R:
\begin{itemize}
\item The reaction is heating with Bromine (Br₂) and Potassium Hydroxide (KOH). This is the Hofmann Bromamide Degradation reaction.
\item This reaction converts a primary amide (R'-CONH₂) into a primary amine (R'-NH₂) with one less carbon atom.
\item Since R is Aniline (C₆H₅NH₂), the reactant Q must be an amide with one more carbon atom. The amide must be Benzamide (C₆H₅CONH₂).
\item Reaction: \( C_6H_5CONH_2 + Br_2 + 4KOH \rightarrow C_6H_5NH_2 + K_2CO_3 + 2KBr + 2H_2O \)
\item So, Q is Benzamide.
\end{itemize}
Analyze the reaction P → Q:
\begin{itemize}
\item Compound Q (Benzamide) is formed from compound P by treatment with aqueous ammonia under hot conditions.
\item Amides are formed from carboxylic acids upon reaction with ammonia, which first forms an ammonium salt that upon heating dehydrates to form the amide.
\item Therefore, P must be the corresponding carboxylic acid, which is Benzoic Acid (C₆H₅COOH).
\item Reaction: \( C_6H_5COOH + NH_3 \rightarrow [C_6H_5COO^-NH_4^+] \xrightarrow{\Delta} C_6H_5CONH_2 + H_2O \)
\item So, P is Benzoic Acid.
\end{itemize}
Step 3: Final Answer:
The compounds are:
\begin{itemize}
\item P: Benzoic acid
\item Q: Benzamide
\item R: Aniline
\end{itemize}
This combination matches option (C).
