Thus, Empirical formula of A is ${\mathrm{C}}_3{\mathrm{H}}_5$.

$\therefore$ Empirical formula mass$=36+5=41.$

$\mathrm{n}=\frac{\mathrm{Molecular} \mathrm{mass}}{\mathrm{Empirical} \mathrm{formula} \mathrm{mass}}=\frac{83.1}{41}=2.02=2$

Molecular mass is double of empirical formula mass.

$\therefore$ Molecular formula is ${\mathrm{C}}_6{\mathrm{H}}_{10}$

To determine the structure of compounds (A) and (B)

$\underset{\left(\mathrm{A}\right)}{{\mathrm{C}}_6{\mathrm{H}}_{10}}\to 2-\mathrm{methyl} \mathrm{pentane}$

Hence, hydrogenation of hydrocarbon (A) requires 2 moles of hydrogen to form 2-methylpentane. Therefore, hydrocarbon (A) is an alkyne having five carbon atoms in a straight chain and a methyl substituent at position 2. Thus, the possible structures for the alkyne (A) are I and II.

Since, addition of H₂O to alkyne (A) in presence of Hg²⁺, give a ketone which gives positive iodoform test, therefore, ketone (B) must be a methyl ketone, i.e., it must contain a COCH₃ group.

Now addition of H₂O to alkyne (II) should give a mixture of two ketones in which 2-methylpentan-3 one (minor) and 4-methylpentan-2-one ketone (B) (which shows +ve iodoform test) predominates.

In contrast, addition of H₂O to alkyne (I) will give only one ketone, i.e., 4-methylpentan-2-one which gives iodoform test.

Thus, hydrocarbon ${\mathrm{C}}_{\mathrm{x}}{\mathrm{H}}_{\mathrm{y}}\left(\mathrm{A}\right)$ is 4-methylpent-1-yne. 4-methylpentan-2 one (gives +ve iodoform test)