One mole of a symmetrical alkene on ozonolysis gives two moles of an aldehyde having a molecular mass of 44u. Identify the alkene.

Explanation

Let the aldehyde be `C_nH_(2n+1) CHO`
Molecular mass of this aldehyde
`=[12n+(2n+1)+(12+1+16)]u`
=(14n+30)u
Thus, 14n+30=44
or, n=1
So the aldehyde is `CH_3CHO`.
Obviously , the alkene is `CH_3CH=CH-CH_3`