To determine the structure of the alkene that produces methanal (formaldehyde) during reductive ozonolysis, we need to understand the process and products of ozonolysis.
Ozonolysis is a reaction where an alkene reacts with ozone (O3) to form ozonides, followed by reduction to yield aldehydes, ketones, or carboxylic acids. In reductive ozonolysis, the ozonide is cleaved into aldehydes or ketones.
When methanal (HCHO) is one of the products, it indicates that one carbon of the double-bonded carbons (C=C) must be directly bonded to two hydrogens. Therefore, the alkene must have a terminal =CH2 group.
Let's evaluate the given options:
\(CH_3CH_2CH_3\) - This is propane, an alkane, and will not undergo ozonolysis as it lacks a double bond. Hence, this option is incorrect.
\(CH_2=CH-CH_3\) - This is propene. On ozonolysis, the double bond is cleaved, yielding methanal (HCHO) and ethanal (CH3CHO), which matches our condition. Therefore, this option is correct.
\(CH_2CH_3\) - The structure is incomplete, and cannot form by itself; hence, it's not an alkene.
\(CH_3-CH=CH_2\) - This is 1-propene or propene, mentioned previously. On ozonolysis, it gives methanal, making this also a correct answer if considering alignment issues with options.
Given the structure \(CH_2=CH-CH_2\), when ozonized, breaks to produce methanal, fulfilling the conditions of the question.
Therefore, the correct answer is the structure of propene: \(CH_2=CH-CH_2\).
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