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  4. One mole of a symmetrical alkene on ozonolysis gives two moles of an aldehyde having a molecular mass of 44 u. The alkene is

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Q. One mole of a symmetrical alkene on ozonolysis gives two moles of an aldehyde having a molecular mass of $44 \,u$. The alkene is

Hydrocarbons

Solution:

$\underset{\text{(Symm.alkene)}}{R C H=C H R}\xrightarrow[{(ii)Zn,H_2O}] {(i)O_3}2RCHO$
Molecular mass of $R CHO =44$
$\Rightarrow R+12+1+16=44$
Mol. mass of $R=44-29=15$
This is possible, only when $R$ is $- CH _{3}$ group.
The aldehyde is $CH _{3} CHO$ and the symmetrical alkene is
$CH _{3} CH = CHCH _{3}$
$\underset{\text { 2-Butene}}{ CH _{3} CH }= CHCH _{3} \xrightarrow[(ii) Zn, H_2O]{(i) O_3} \underset{\text {Acetaldehyde}}{ 2CH _{3} CHO}$