Question

An organic compound ‘A’, molecular formula C\(_6\)H\(_6\)O oxidises with CrO\(_3\) to form a compound ‘B’. Compound ‘B’ on warming with iodine and aqueous solution of NaOH gives a yellow precipitate of compound ‘C’. When compound ‘A’ is heated with conc. H\(_2\)SO\(_4\) at 413 K gives a compound ‘D’, which on reaction with excess HI gives compound ‘E’. Identify compounds ‘A’, ‘B’, ‘C’, ‘D’ and ‘E’ and write chemical equations involved.

Hint:

These reactions show common organic transformations such as oxidation, iodination, and reaction with acids to introduce functional groups.

Answer 1

- Compound ‘A’ is phenol (C\(_6\)H\(_6\)O), which reacts with CrO\(_3\) to form compound ‘B’, which is quinone (C\(_6\)H\(_4\)O\(_2\)). - Compound ‘B’ reacts with iodine and aqueous NaOH to form compound ‘C’, which is iodoquinone (C\(_6\)H\(_3\)I\(_2\)O\(_2\)). - Compound ‘A’ (phenol) when heated with conc. H\(_2\)SO\(_4\) at 413 K gives compound ‘D’, which is bisphenol (C\(_6\)H\(_4\)OH), through sulfonation. - Compound ‘D’ reacts with excess HI to form compound ‘E’, which is 2-iodophenol (C\(_6\)H\(_4\)IOH).
\text{The chemical equations involved are:}
\[ \text{Phenol (C\(_6\)H\(_6\)O) + CrO\(_3\) } \rightarrow \text{ Quinone (C\(_6\)H\(_4\)O\(_2\)) } \]
\[ \text{Quinone (C\(_6\)H\(_4\)O\(_2\)) + I\(_2\) + NaOH } \rightarrow \text{ Iodoquinone (C\(_6\)H\(_3\)I\(_2\)O\(_2\)) } \]
\[ \text{Phenol (C\(_6\)H\(_6\)O) + H\(_2\)SO\(_4\) } \xrightarrow{413K} \text{ Bisphenol (C\(_6\)H\(_4\)OH) } \]
\[ \text{Bisphenol (C\(_6\)H\(_4\)OH) + HI} \rightarrow \text{ 2-Iodophenol (C\(_6\)H\(_4\)IOH) } \]