An alcohol has the molecular formula C4H7OH. Oxidation of this alcohol gives a ketone.
The production of a ketone on oxidation shows that the alcohol is a secondary alcohol. It must be unsaturated since it does not have enough hydrogen atoms to be saturated unless there is also a ring. In that case it would not react with ozone.
Since ozonolysis of the alcohol gives methanal as one of the products the C=C double bond must be at the end of the molecule. The only structure consistent with all of this evidence is CH2=CHCH(OH)CH3. Ozonolysis of this molecule will give CH2O and CHOCH(OH)CH3
The alcohol molecule will be optically active since it is chiral, the asterisked carbon being the chiral centre: CH2=CH*CH(OH)CH3. The necessary and sufficient condition for its being chiral is that it is non-superimposable on its mirror image.
Problem 4 Organic problems contents Home page