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One mole of an organic compound A with the formula $\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}$ reacts completely with two moles of HI to form $\mathrm{X}$ and $\mathrm{Y}$. When $\mathrm{Y}$ is boiled with aqueous alkali it forms Z. Z answers the iodoform test. The compound A is
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The correct answer is:
methoxyethane
Molecular formula $\quad \mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}\left(\mathrm{C}_{\mathrm{n}} \mathrm{H}_{2 \mathrm{n}}+2 \mathrm{O}\right)$
suggests that the organic compound is either alcohol or ether.
Since, the compound on reaction with HI gives two different compounds, it must be an unsymmetrical ether, and its formula must be $\mathrm{CH}_{3} \mathrm{OC}_{2} \mathrm{H}_{5}$ (methoxyethane).
The reactions are as follows.
$\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}+\underset{\text { (aqueous) }}{\mathrm{NaOH}}+\mathrm{I}_{2} \longrightarrow \underset{\text { iodoform }}{\mathrm{CHI}_{3}}+\mathrm{HCOONa}+\mathrm{H}_{2} \mathrm{O}+\mathrm{NaI}$
suggests that the organic compound is either alcohol or ether.
Since, the compound on reaction with HI gives two different compounds, it must be an unsymmetrical ether, and its formula must be $\mathrm{CH}_{3} \mathrm{OC}_{2} \mathrm{H}_{5}$ (methoxyethane).
The reactions are as follows.
$\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}+\underset{\text { (aqueous) }}{\mathrm{NaOH}}+\mathrm{I}_{2} \longrightarrow \underset{\text { iodoform }}{\mathrm{CHI}_{3}}+\mathrm{HCOONa}+\mathrm{H}_{2} \mathrm{O}+\mathrm{NaI}$
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