What is the product formed when phenol reacts with $Br_2$ in $CCl_4$ at low temperature?

The correct statement$(s)$ about the following reaction sequence is(are):
Cumene $(C_9H_{12})$ $\xrightarrow[ii) H_3O^+]{i) O_2}$ $P$ $\xrightarrow{CHCl_3/NaOH}$ $Q$ (major) + $R$ (minor)
$Q$ $\xrightarrow[PhCH_2Br]{NaOH}$ $S$
$(A)$ $R$ is steam volatile
$(B)$ $Q$ gives dark violet coloration with $1 \%$ aqueous $FeCl_3$ solution
$(C)$ $S$ gives yellow precipitate with $2, 4-dinitrophenylhydrazine$
$(D)$ $S$ gives dark violet coloration with $1 \%$ aqueous $FeCl_3$ solution

The correct acidity order of phenol $(I)$, $4-$hydroxybenzaldehyde $(II)$ and $3-$hydroxybenzaldehyde $(III)$ is:

The major product $(B)$ of this reaction is:

Phenol $\xrightarrow{Br_2 / H_2O}$ Product formed is

Cumene on oxidation in air gives a compound,$X$. This on reaction with dilute acid gives $Y$ and $Z$. $Y$ reacts with sodium metal and not $Z$. What is $Z$ ?