Ozonolysis of one mole of a symmetrical alken | vedclass

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(D) The general reaction for the ozonolysis of a symmetrical alkene is: $R-CH=CH-R + O_3 \rightarrow 2R-CHO$. Given that the product is an aldehyde wi

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$2$-Butene

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(D) The general reaction for the ozonolysis of a symmetrical alkene is: $R-CH=CH-R + O_3 \rightarrow 2R-CHO$. Given that the product is an aldehyde with a molar mass of $44 \ u$. The general formula for an aldehyde is $R-CHO$. The molar mass of $CHO$ is $12 + 1 + 16 = 29 \ u$. Therefore,the mass of the alkyl group $R$ is $44 - 29 = 15 \ u$. An alkyl group with a mass of $15 \ u$ is a methyl group $(-CH_3)$. Thus,the aldehyde is acetaldehyde $(CH_3CHO)$. Substituting this back into the reaction: $CH_3-CH=CH-CH_3 + O_3 \rightarrow 2CH_3CHO$. The reactant is $2$-butene.

Ozonolysis of one mole of a symmetrical alkene yields two moles of an aldehyde with a molar mass of $44 \ u$. Identify the alkene.

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