An alcohol $X\left(\mathrm{C}_4 \mathrm{H}_{10} \mathrm{O}\right)$ on dehydration gave alkene $\left(\mathrm{C}_4 \mathrm{H}_8\right)$ as major product, which on bromination followed by treatment with $Y$ gave alkyne $\mathrm{C}_4 \mathrm{H}_6$. Alkyne $\mathrm{C}_4 \mathrm{H}_6$, does not react with sodium metal. What are $X$ and $Y$ ?

The correct statement regarding $X$ and $Y$ formed in the following reaction is

$$ \left(\mathrm{CH}_3\right)_3 \mathrm{COC}_2 \mathrm{H}_5 \xrightarrow[\Delta]{\mathrm{HI}} \text { halide }(X)+\text { alcohol }(Y) $$

The correct order of boiling points of the compounds given below is

(A) methoxy ethane

(B) propan-1-ol

(C) propanal

(D) propanone

The number of primary $\left(1^{\circ}\right)$, secondary $\left(2^{\circ}\right)$ and tertiary $\left(3^{\circ}\right)$ alcohols possible for the formula $\mathrm{C}_5 \mathrm{H}_{12} \mathrm{O}$ respectively are