Question

An aromatic compound ‘A’ on treatment with aqueous ammonia and heating forms compound B which on heating with $B{{r}_{2}}$ and KOH forms a compound C of molecular formula${{C}_{6}}{{H}_{7}}N$. Write the structures and IUPAC names of compounds A, B and C.

Answer 1

Hint:To answer this type of question we should follow the leads given in the question. ‘A’ is an aromatic compound and the reaction with $B{{r}_{2}}$ and KOH is Hoffmann Bromamide degradation. This means that ‘A’ is an aromatic amine.

Complete step by step solution:Let’s see the answer of the give question:
First we will look at the facts given in the question,
‘A’ is an aromatic compound.
‘A’ on reaction with aqueous ammonia and on heating gives B. B on heating with $B{{r}_{2}}$ and KOH gives C. This is a Hoffmann Bromamide degradation reaction.
This means that B is an amide and C is an aromatic amine.
The only aromatic amine with the formula ${{C}_{6}}{{H}_{7}}N$is aniline.
So, B is benzamide and A is benzoic acid.
Hence, benzoic acid reacts with aqueous ammonia and on heating gives Benzamide.
Benzamide on heating with $B{{r}_{2}}$ and KOH gives aniline. This reaction is known as Hoffmann bromamide degradation.
${{C}_{6}}{{H}_{5}}COOH\,\xrightarrow[heat]{aqueous\,\,ammonia}\,{{C}_{6}}{{H}_{5}}CON{{H}_{2}}\,\xrightarrow[KOH]{bromine gas}\,{{C}_{6}}{{H}_{5}}N{{H}_{2}}$
A= Benzoic acid (${{C}_{6}}{{H}_{5}}COOH$)
B= Benzamide (${{C}_{6}}{{H}_{5}}CON{{H}_{2}}$)
C= Aninile (${{C}_{6}}{{H}_{5}}N{{H}_{2}}$)

Note: When an amide is reacted with bromine in an aqueous solution containing a strong base to give a primary amine is known as Hoffmann Bromamide Degradation. One of the characteristics of this reaction is that it leads to the reduction in the number of carbon atoms present originally in the compound.