JEE Advance - Chemistry (1987 - No. 17)
An unknown compound of carbon, hydrogen and oxygen
contains 69.77% carbon and 11.63% hydrogen and has
a molecular weight of 86. It does not reduce Fehling
solution, but forms a bisulphite addition compound and
gives a positive iodoform test. What are the possible
structures?
CH3CH2CH2COCH3
CH3CH2COCH2CH3
CH3CH(CH3)COCH3
CH3CHO
Explanation
| Element | % of element | Relative number of atoms | Relative Ratio | Simplest Ratio |
|---|---|---|---|---|
| C | 69.77 | $${{69.77} \over {12}} = 5.81$$ | $${{5.81} \over {1.16}} = 5$$ | 5 |
| H | 11.63 | $${{11.63} \over 1} = 11.63$$ | $${{11.63} \over {1.16}} = 10$$ | 10 |
| O | 18.60 | $${{18.60} \over {16}} = 1.16$$ | $${{1.16} \over {1.16}} = 1$$ | 1 |
$$\therefore$$ Empirical formula = C5H10O
Empirical formula weight
= 5 $$\times$$ 12 + 10 $$\times$$ 1 + 1 $$\times$$ 16
= 86
Molecular weight = 86
$$\therefore$$ $$n = {{86} \over {86}} = 1$$
$$\therefore$$ Molecular formula = C5H10O
Since the compound forms bisulphite addition compound so
it has a carbonyl group (i.e. it is an aldehyde or a ketone)
since it does not reduce Fehling’s solution so it is a ketone,
since it gives positive iodoform test so it has 
grouping.
From the above we find that the compound is
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