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(D) The molecular formula $C_{3}H_{8}O$ corresponds to the general formula $C_{n}H_{2n+2}O$,which suggests that the compound $A$ is either an alcohol

Vedclass · Accepted Answer

methoxyethane

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(D) The molecular formula $C_{3}H_{8}O$ corresponds to the general formula $C_{n}H_{2n+2}O$,which suggests that the compound $A$ is either an alcohol or an ether.Since the compound reacts with $HI$ to form two different products ($X$ and $Y$),it must be an unsymmetrical ether. The reaction is: $CH_{3}OC_{2}H_{5} + HI \longrightarrow CH_{3}I + C_{2}H_{5}OH$.Here,$X$ is $CH_{3}I$ and $Y$ is $C_{2}H_{5}OH$ (ethanol).When ethanol $(Y)$ is treated with aqueous alkali (like $NaOH$) and $I_{2}$,it undergoes the iodoform test because it can be oxidized to acetaldehyde,which contains the $CH_{3}CO-$ group.The reaction for the iodoform test is: $C_{2}H_{5}OH + 4I_{2} + 6NaOH \longrightarrow CHI_{3} + HCOONa + 5NaI + 5H_{2}O$.Thus,the compound $A$ is methoxyethane $(CH_{3}OC_{2}H_{5})$.

One mole of an organic compound $A$ with the formula $C_{3}H_{8}O$ 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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