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 $C_{n} H_{2 n} O$

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

So, the molecular formula will be $C_{2} H_{4} O$

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

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

Hence, option (d) is correct.

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Similar questions

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:

An alkene ‘A’ contains three C – C, eight C – H σ bonds and one C – C π bond. ‘A’ on ozonolysis gives two moles of an aldehyde of molar mass 44 u. Write IUPAC name of ‘A’.

Q. An alkene

A

contains three

C - C

, eight

C - H

σ

-bonds and one

C - C π

bond.

A

on ozonolysis gives two moles of an aldehyde of molar mass

44 u

. IUPAC name of

A

is:

Ozonolysis of alkenes gives ozonide which gives aldehydes and/or ketones. When a stream of $O_{3}$ is passed through a solution of alkene in an inert solvent at low temperature, the molecule adds up to form ozonide. This ozonide is unstable and easily decomposes either by reduction or hydrolysis and forms compound having the carbonyl group $> C = O$ . This is known as ozonolysis. 2-Methyl-but-2-ene on ozonolysis followed by reductive hydrolysis gives: