Organic chemistry..

1. The compound contains 66.97% C , 11.63% H and 21.4% O. The atomic mass of C is 12, O is 16 and H is1.  The molecualr formul of teh compound is 86.Therefore, the formula of the compound is C₅H₁₀O.

1. The compound reacts with sodium bisulphite to give an addition product. Therefore, it is a carbonyl compound.

2. The compound does not reduce tollen's reagent. Therefore, it is a ketone.

3. To identify the position of carbonyl group it is given that the compound gives a positive iodoform test. So, it is a methyl ketone. Also, it is given that on vigorous oxidation , it gives ethanoic acid and propanoic acid.

 Therefore, the formula of the compound is CH₃CH₂CH₂COCH₃

_{Teh reactions are:}

CH₃CH₂CH₂COCH₃ + NaHSO₃  CH₃CH₂CH₂C(OH)(OSO₂Na)CH₃

                                        vig oxidation

CH₃CH₂CH₂COCH₃           CH₃COOH +  CH₃CH₂CH₂COOH