Question

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

• A. Ethene
• B. Propene
• C. But-1-ene
• D. But-2-ene

Answer 1

The correct option is D But-2-ene

Molecular formula of an aldehyde is $C_n{H}_{2n}O$

$n\times 12+2n\times 1+16=44\Rightarrow n=2$

So, the molecular formula will be $C_2{H}_{4}O$

The aldehyde with molecular mass $44u$ is $C{H}_{3}CHO$ (acetaldehyde).
Therefore, the symmetrical alkene is but-2-ene.

$C{H}_{3}CH=CHC{H}_3\underset{\left(ii\right)Zn/H_{2}O}{\overset{\left(i\right)O_3}{\to }}2C{H}_{3}CHO$

Hence, option (d) is correct.