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(D) The molecular formula $C_3H_8O$ corresponds to either an alcohol or an ether. Since it reacts with $HI$ to form two products ($X$ and $Y$),it must

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Methoxy ethane

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(D) The molecular formula $C_3H_8O$ corresponds to either an alcohol or an ether. Since it reacts with $HI$ to form two products ($X$ and $Y$),it must be an ether. The reaction is: $CH_3-CH_2-O-CH_3 + 2HI \rightarrow CH_3-I (X) + CH_3-CH_2-I (Y) + H_2O$.When ethyl iodide $(Y)$ is boiled with aqueous alkali $(NaOH)$,it forms ethanol $(CH_3CH_2OH)$.However,the iodoform test is given by compounds containing the $CH_3CO-$ group or $CH_3CH(OH)-$ group. Ethanol $(CH_3CH_2OH)$ gives a positive iodoform test. Thus,$(A)$ is methoxyethane.

One mole of an organic compound $(A)$ with the formula $C_3H_8O$ 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

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