(a) Give chemical tests to distinguish between compounds in the following pairs of substances ,
(i) Ethanal and Propanal
(ii) Benzoic acid and Ethyl benzoate
(b) An organic compound contains `69.77%` carbon , `11.63%` hydrogen and rest 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 positive iodoform test. On vigorous oxidation, it gives ethanoic and propanoic acids . Derive the structure of the compound.

Explanation

Step 1. Calculation of molecular formula of the compound
%C=69.77%:% H=11.63%
%O=100-(69.77 +11.63) = 18.6%
`C:H,O = (69.77)/12 : (11.63)/1 : (18.6)/16 = 5.88 : 11.63 : 1.16 = 5 : 10 : 1`
The EF of the given compound is `C_5H_10O`
EF mass =` 5 xx 12 + 10 xx 1 + 1 xx 16 = 86`
Mol. mass = 86 (given)
`M.F. = C_5H_10 O xx (86)/(86) = C_5H_10O`
Thus, molecular formula of the given compound =`C_5H_10O`
Step 2. Determination of the structure of the compound
(i) Since the given compound forms an addition product with sodium hydrogensulfite, it must be an aldehyde or ketone.
(ii) Since the compound does not reduce Tollens. reagent, it cannot be an aldehyde and thus must be a ketone.
(iii) Since the compound gives positive iodoform test, it must be a methyl ketone.
(iv) Since, the given compound on vigorous oxidation gives a mixture of ethanoic acid and propanoic acid, the methyl ketone is pentan-2-one, i.e.,

On the basis of this structure, the reactions taking place can be explained as follows: