Which of the following compounds will show optical isomerism?

A student has been given a compound " $x$ " of molecular formula- $\mathrm{C}_{6} \mathrm{H}_{7} \mathrm{~N}$. ' $x$ ' is sparingly soluble in water. However, on addition of dilute mineral acid, ' $x$ ' becomes soluble in water. ' $x$ ' when treated with $\mathrm{CHCl}_{3}$ and $\mathrm{KOH}($ alc $)$, ' $y$ ' is produced. ' y ' has a specific unpleasant smell. On treatment with benzenesulphonyl chloride, ' $x$ ' gives a compound ' $z$ ' which is soluble in alkali. The number of different " $H$ " atoms present in ' $z$ ' is:-

Given below are two statements : Statement I: Compound $(X)$, shown below, dissolves in $\mathrm{NaHCO}_{3}$ solution and has two chiral carbon atoms  Statement II: Compound $(\mathrm{Y})$, shown below, has two carbons with $\mathrm{sp}^{3}$ hybridization, one carbon with $\mathrm{sp}^{2}$ and one carbon with sp hybridization  In the light of the above statements, choose the correct answer from the options given below :

$4.7$ g of phenol is heated with Zn to give product X. If this reaction goes to $60\%$ completion then the number of moles of compound X formed will be _______ $\times 10^{-2}$. (Nearest Integer) (Given molar mass in g mol$^{-1}$: H:1, C:12, O:16)

The mass of benzanilide obtained from the benzoylation reaction of 5.8 g of aniline, if yield of product is $82 \%$, is $\_\_\_\_$ g (nearest integer). (Given molar mass in $\mathrm{g} \mathrm{mol}^{-1} \mathrm{H}: 1, \mathrm{C}: 12, \mathrm{~N}: 14, \mathrm{O}: 16$)

Consider the following sequence of reactions.  The percentage of nitrogen in the yellow product (X) formed is _______%. (Nearest Integer) (Given Molar mass in g mol$^{-1}$ H:1, C:12, N:14)