Hint:  The molecular formula of the compound is C₅H₁₀O

Step 1:

First calculate the molecular formula of the compound as follows:

% of carbon = 69.77 %

% of hydrogen = 11.63 %

% of oxygen = {100 − (69.77 + 11.63)}%

= 18.6 %

Thus, the ratio of the number of carbon, hydrogen, and oxygen atoms in the organic compound can be given as:

$\mathrm{C} : \mathrm{H} : \mathrm{O}=\frac{69.77}{12}:\frac{11.63}{1}:\frac{18.6}{16}$
$=5.81 : 11.63 : 1.16$
$=5 : 10 : 1$

Therefore, the empirical formula of the compound is C₅H₁₀O.

Now, the empirical formula mass of the compound can be given as:

5 × 12 + 10 ×1 + 1 × 16 = 86

The molecular mass of the compound = 86

Therefore, the molecular formula of the compound is given by C₅H₁₀O.

Step 2:

Since the given compound does not reduce Tollen’s reagent, it is not an aldehyde. Again, the compound forms sodium hydrogen sulfate addition products and gives a positive iodoform test. Since the compound is not an aldehyde, it must be a methyl ketone.

Step 3:

The given compound also gives a mixture of ethanoic acid and propanoic acid.
Hence, the given compound is pentan−2−ol. 

The given reactions can be explained by the following equations: