Step 1: Percentage composition.
Given:
\[
%C = 69.77, \quad %H = 11.63, \quad %O = 100 - (69.77 + 11.63) = 18.60
\]
Step 2: Calculate empirical formula.
\[
\text{For C: } \frac{69.77}{12} \approx 5.81
\]
\[
\text{For H: } \frac{11.63}{1} \approx 11.63
\]
\[
\text{For O: } \frac{18.60}{16} \approx 1.16
\]
Dividing all by 1.16 (smallest):
\[
C : H : O = 5.81/1.16 : 11.63/1.16 : 1.16/1.16 \approx 5 : 10 : 1
\]
So, empirical formula = \(C_5H_{10}O\).
Step 3: Molecular formula.
Empirical formula mass = \(5 \times 12 + 10 \times 1 + 16 = 86\).
Since molecular mass = 86, the molecular formula is the same:
\[
C_5H_{10}O
\]
Step 4: Functional group tests.
\begin{itemize}
\item Compound does not reduce Tollen’s reagent → Not an aldehyde.
\item Forms addition compound with NaHSO$_3$ → Presence of carbonyl group.
\item Positive iodoform test → Presence of methyl ketone group (\(-COCH_3\)).
\end{itemize}
Thus, compound must be a ketone with structure \(R-COCH_3\).
Step 5: Oxidation products.
On strong oxidation, compound gives ethanoic acid and propanoic acid. This indicates the structure is pentan-2-one.
\[
CH_3-CO-CH_2-CH_2-CH_3 \quad (Pentan\text{-}2\text{-}one)
\]
Step 6: Reaction equations.
(i) NaHSO$_3$ addition:
\[
CH_3COCH_2CH_2CH_3 + NaHSO_3 \;\longrightarrow\; \text{Addition compound}
\]
(ii) Iodoform test:
\[
CH_3COCH_2CH_2CH_3 + 3I_2 + 4NaOH \;\longrightarrow\; CHI_3 \downarrow + CH_3CH_2COONa + 3NaI + 3H_2O
\]
(iii) Oxidation:
\[
CH_3COCH_2CH_2CH_3 \;\xrightarrow{[O]}\; CH_3COOH + CH_3CH_2COOH
\]
Conclusion:
The possible structure of the compound is:
\[
\boxed{CH_3-CO-CH_2-CH_2-CH_3 \quad \text{(Pentan-2-one)}}
\]