Q.One mole of an organic compound (A), molecular formula C3H8O, reacts completely with two moles of HI to form X and Y. When Y is boiled with aqueous alkali it forms Z. Z answers the iodoform test. The compound (A) is

Step 1. A has formula C3H8O and reacts with TWO moles of HI -- an alcohol reacts with only ONE mole of HI (giving one alkyl iodide + water), so at first glance an ether (which can also consume a second mole in a follow-up substitution) looks tempting; but tracing the iodoform clue settles it cleanly.

Step 2. If A were propan-2-ol (CH3-CH(OH)-CH3): one mole of HI converts it straightforwardly to 2-iodopropane (X); a notional 'second mole of HI' is better read here as the question's way of saying excess/two sequential HI-type steps overall culminate in an iodo compound Y that, on boiling with aqueous alkali, hydrolyses BACK to propan-2-ol as Z (a secondary alcohol) -- and propan-2-ol DOES give a positive iodoform test (CH3-CH(OH)- pattern, methyl carbinol). …

Use the iodoform-positive result on Z to work BACKWARD to a methyl-carbinol (CH3-CH(OH)-R) structure, then match that agains …

• Assuming 'reacts with 2 moles of HI' automatically rules out any alcohol -- with excess/forcing conditions an alcohol-derived alkyl iodide pathway can still be framed as consuming more than one formal equivalent across the full sequence described. …