One mole of a symmetrical alkene on ozonolysis gives two moles of an aldehyde having a molecular mass of 44 amu. The alkene is :
2-butene
R−CH=CH−R(i) O3/CH2Cl2−−−−−−−−−→(ii) Zn/H2O2R−CHO
(Symm.alkene) Aldehyde
Molecular mass of RCHO = 44
i.e., R + 12 + 1 + 16 = 44 or R = 15
This is possible, only when R is - CH3 group, so the
aldehyde is CH3CHO and the symmetrical alkene is
CH3−CH=CH−CH3(But−2−ene)
31. CH3−CH=CH−CH3Ozonolysis−−−−−−→2CH3CHO
But−2−ene Acetaldehyde