An organic compound contains $69.77 \%$ carbon,$11.63 \%$ hydrogen and the rest is oxygen. The molecular mass of the compound is $86$. It does not reduce Tollens' reagent but forms an addition compound with sodium hydrogensulphite and gives a positive iodoform test. On vigorous oxidation,it gives ethanoic acid and propanoic acid. Write the possible structure of the compound.

(A) $1$. Calculate the empirical formula:
$C:H:O = \frac{69.77}{12} : \frac{11.63}{1} : \frac{18.6}{16} = 5.81 : 11.63 : 1.16 = 5 : 10 : 1$.
Empirical formula is $C_5H_{10}O$.
$2$. Determine molecular formula:
Empirical formula mass $= 5(12) + 10(1) + 16 = 86$.
Since molecular mass is $86$,the molecular formula is $C_5H_{10}O$.
$3$. Analyze chemical properties:
- Does not reduce Tollens' reagent: Not an aldehyde.
- Forms addition compound with $NaHSO_3$: Contains a carbonyl group.
- Positive iodoform test: Contains a $CH_3CO-$ group.
- Oxidation yields ethanoic acid $(CH_3COOH)$ and propanoic acid $(CH_3CH_2COOH)$: The cleavage occurs at the $C-C$ bond adjacent to the carbonyl group.
$4$. Conclusion:
The compound is $Pentan-2-one$ $(CH_3COCH_2CH_2CH_3)$.