Question

An aromatic organic compound A on treatment with aqueous ammonia and heating forms compound B, which on heating with Br$_2$ and KOH forms a compound C of molecular formula C$_6$H$_7$N. Write the structures and names of compounds A, B and C.

Hint:

Hoffmann bromamide reaction always converts an amide (R–CONH$_2$) into a primary amine (R–NH$_2$) with one carbon less.

Answer 1

Step 1: Identify compound A.
Since the product (C) has molecular formula C$_6$H$_7$N, it must be an aromatic amine (aniline). The intermediate step suggests that compound B is amide, which upon Hoffmann bromamide reaction gives amine. Hence, A must be an aromatic acid derivative.
Thus, compound A is
benzamide (C$_6$H$_5$CONH$_2$).
Step 2: Reaction with aqueous ammonia.
An aromatic acid chloride (benzoyl chloride, C$_6$H$_5$COCl) reacts with aqueous ammonia to form
benzamide (B).
\[ C_6H_5COCl + NH_3 \rightarrow C_6H_5CONH_2 + HCl \] Step 3: Hoffmann bromamide reaction.
When benzamide (B) is treated with Br$_2$ and KOH, Hoffmann bromamide reaction occurs. The –CONH$_2$ group is converted into –NH$_2$. Thus, the product is
aniline (C$_6$H$_5$NH$_2$).
\[ C_6H_5CONH_2 + Br_2 + 4KOH \rightarrow C_6H_5NH_2 + K_2CO_3 + 2KBr + 2H_2O \] Step 4: Structures of compounds.
\begin{itemize} \item Compound A: Benzoyl chloride (C$_6$H$_5$COCl) \item Compound B: Benzamide (C$_6$H$_5$CONH$_2$) \item Compound C: Aniline (C$_6$H$_5$NH$_2$) \end{itemize} Conclusion:
- A = Benzoyl chloride (C$_6$H$_5$COCl)
- B = Benzamide (C$_6$H$_5$CONH$_2$)
- C = Aniline (C$_6$H$_5$NH$_2$)