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

• A

  $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}$
• Correct answer
  B

  $\mathrm{CH}_{3} \mathrm{CHOHCH}_{2} \mathrm{CH}_{3}$
• C

  $\left(\mathrm{CH}_{3}\right)_{3} \mathrm{COH}$
• D

  $\mathrm{CH}_{3} \mathrm{CH}_{2}-\mathrm{O}-\mathrm{CH}_{2} \mathrm{CH}_{3}$

Solution

$$ \mathrm{C}_{4} \mathrm{H}_{8} \mathrm{O}\left(\mathrm{R}-\mathrm{COCH}_{3}\right) $$ 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 $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CHOHCH}_{3}$

Hence, the correct answer is option: 'B'.