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A substance $\mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O}$ yields on oxidation a compound, $\mathrm{C}_{4} \mathrm{H}_{8} \mathrm{O}$ which gives an oxime and a positive iodoform test. The original substance on treatment with conc. $\mathrm{H}_{2} \mathrm{SO}_{4}$ gives $\mathrm{C}_{4} \mathrm{H}_{8} .$ The structure of the compound is
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Verified Answer
The correct answer is:
$\mathrm{CH}_{3} \mathrm{CHOHCH}_{2} \mathrm{CH}_{3}$
$$
\mathrm{C}_{4} \mathrm{H}_{8} \mathrm{O}\left(\mathrm{R}-\mathrm{COCH}_{3}\right)
$$
Thus $\mathrm{C}_{4} \mathrm{H}_{8} \mathrm{O}$ should be $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COCH}_{3}$,
hence $\mathrm{C}_{4} \mathrm{H}_{10} \mathrm{O}$ should be $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CHOHCH}_{3}$
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