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 Tollens’ reagent but forms an addition compound with sodium hydrogensulphite and give positive iodoform test. On vigorous oxidation it gives ethanoic and propanoic acid. Write the possible structure of the compound.
Answers
Answer:
% 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:
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
Molecular mass of the compound = 86
Therefore, the molecular formula of the compound is given by C5H10O.
Since the given compound does not reduce Tollen's reagent, it is not an aldehyde. Again, the compound forms sodium hydrogen sulphate addition products and gives a positive iodoform test. Since the compound is not an aldehyde, it must be a methyl ketone.
The given compound also gives a mixture of ethanoic acid and propanoic acid.
Hence, the given compound is Pentan-2-one.
The given reactions can be explained by the following equations:
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Answer:
Pentan-2-one
[CH3-CO-(CH2)2-CH3]
Explanation:
Carbon : 69.77%
Hydrogen:11.63%
Oxygen:100-(69.77+18.6)
= 18.6%
C:H:O = 69.77/12 : 11.63/1 : 18.6/6
= 5.88 : 11.63 : 1.16
( Dividing all with 1.16)
= 5:10:1
Therefore it's compound will be C5H10O
Since the compound does not give tollens test it means it is not an aldehyde.it gives iodoform test so methyl ketone .on oxidation ,it gives ethanoic acid and propanoic acids.therefore the compound should be Pentan-2-one [CH3-CO--(CH2)2-CH3]