The reaction,
$C_6H_5OH + CHCl_3 \xrightarrow{NaOH/\Delta} \text{o-hydroxybenzaldehyde} + H^+$
is known as:

Compound $I$ is heated with Conc. $HI$ to give a hydroxy compound $A$ which is further heated with $Zn$ dust to give compound $B$. Identify $A$ and $B$.

Which of the following reagents is used to convert phenol to benzene?

When phenol reacts with dilute nitric acid at room temperature,the major product obtained is

Phenol $\xrightarrow[(ii) CO_2 / 140^{\circ}C]{(i) NaOH} A$ $\xrightarrow{H^{+} / H_2O} B$ $\xrightarrow[CH_3COOH, \Delta]{Al_2O_3} C$
In this reaction,the end product $C$ is :

Given below are two statements:
Statement $I$: Bromination of phenol in solvent with low polarity such as $CHCl_3$ or $CS_2$ requires Lewis acid catalyst.
Statement $II$: The Lewis acid catalyst polarises the bromine to generate $Br^{+}$.
In the light of the above statements,choose the correct answer from the options given below: