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

Ethene

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Propene

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But-1-ene

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But-2-ene

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Solution

The correct option is D But-2-ene
Molecular formula of an aldehyde is $C_{n} H_{2 n} O$
$n \times 12 + 2 n \times 1 + 16 = 44 \Rightarrow n = 2$
$C_{2} H_{4} O$

The aldehyde with molecular mass $44 u$ is $C H_{3} C H O$ (acetaldehyde).

Therefore the symmetrical alkene is but-2-ene.

$C H_{3} C H = C H C H_{3} (i) O_{3} - ------ \to (i i) Z n / H_{2} O 2 C H_{3} C H O$

but-2-ene acetaldehyde

Option D is correct.

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

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