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 Tollen’s reagent but forms an addition compound with sodium hydrogen sulphite and gives a positive iodoform test. On vigorous oxidation it gives ethanoic and propanoic acids. Deduce the possible structure of the organic compound.

Calculation of emperical formula:
We are given that the percentage of carbon and hydrogen in the given compound is 69.77 % 11.63 % respectively.
Therefore, the percentage of oxygen = (100 - 69.77 - 11.63) % = 18.60 %.
So, 100g of the sample will contain 69.77g of carbon, 11.63g of hydrogen and 18.6g of oxygen.

Number of moles in 69.77g of carbon = mass of carbon / molar mass of carbon = 69.77 / 12 = 5.814

11.63g of hydrogen = 11.63 / 1 = 11.63

18.6g of oxygen = 18.6 / 16 = 1.163

To obtain the simplest ratio of element, we will divide the number of moles of each element by the smallest value that is 1.163. So, we get

C : H : O = C5H10O

The mass from the empirical formula comes out to be 86.
We are given that the molar mass of the compound is 86g. So, the molecular formula is the same as the empirical formula that is C5H10O. Now, we are given that the compound gives sodium bisulphite test. This means that it contains a carbonyl group. So, it could be either an aldehyde or a ketone. Also, this compound does not give Tollen's test. Tollen's reagent is a mild reducing agent which gives positive test with aldehydes and not with ketones. So, the given compound is a ketone.

Since the compound gives Iodoform test, therefore it contains the CH3CO group. Also, the compound gives ethanoic acid and propanoic acid on vigorous oxidation. Thus the compound is pentan-2-one which has the following structure

CH3COCH2CH2CH3

An organic compound contains 69.77% carbon, 11.63% hydrogen
percentage of oxygen =18.6%
ratio of C,H and O in the compound
percentage of carbon =12x/86 =.6977,x=5
percentage of hydrogen = y/86 =.1163 =10
for oxygen =16z/86 =.186 ,z=1
therefore the formula is C5H10O
It gives a positive idoform test .Therefore the structure is CH3CHO-
the formula is CH3COCH2CH2CH3