What is the value of ' $n$ ' in ' $Z$ ' of the following sequence?

Lauryl alcohol $\xrightarrow{\mathrm{H}_{2} \mathrm{SO}_{4}}$

( $X$ )

Lauryl hydrogen sulphate $\xrightarrow{\mathrm{NaOH}(a q)}$

( $Y$ )

$ \mathrm{CH}_{3}-\left(\mathrm{CH}_{2}\right)_{n}-\mathrm{CH}_{2} \mathrm{OSO}_{3} \mathrm{Na} $

(Z)

Sodium lauryl sulphate

Match the following

The correct answer is

The bromides formed by the cleavage of ethers $A$ and $B$ with HBr respectively are

The major product ' $Y^{\prime}$ in the given sequence of reactions is

$$ \mathrm{C}_3 \mathrm{H}_7 \mathrm{OH} \xrightarrow[443 \mathrm{~K}]{\text { Conc. } \mathrm{H}_2 \mathrm{SO}_4} X \xrightarrow[\left(\mathrm{C}_6 \mathrm{H}_5 \mathrm{CO}\right)_2 \mathrm{O}_2]{\mathrm{HBr}} Y $$