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On ozonolysis, an alkene (x) forms two c...

On ozonolysis, an alkene (x) forms two carbonyl compounds namely butan - 2 - one and methanal. The total number of methyl groups present in alkene (x) is

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To solve the problem, we need to determine the structure of the alkene (x) that, upon ozonolysis, yields butan-2-one and methanal. We will analyze the products to deduce the structure of the starting alkene and count the number of methyl groups present. ### Step-by-Step Solution: 1. **Identify the Products**: The products of ozonolysis are butan-2-one (C4H8O) and methanal (HCHO). - Butan-2-one has the structure: CH3-CO-CH2-CH3 - Methanal has the structure: HCHO 2. **Determine the Structure of the Alkene**: - Ozonolysis cleaves the double bond of the alkene and forms carbonyl compounds. - The presence of butan-2-one suggests that the alkene must have contributed a 4-carbon chain. - The methanal indicates that one of the fragments must be a 1-carbon fragment. 3. **Construct the Alkene**: - To form butan-2-one and methanal, we can deduce that the alkene must have a structure that, when cleaved, gives these two products. - A possible structure for the alkene is 2-methylbut-2-ene (C5H10), which can be represented as: ``` CH3 | CH3- C = C - CH3 | H ``` - Upon ozonolysis, this alkene will yield: - Butan-2-one from the chain (CH3-CO-CH2-CH3) - Methanal from the terminal carbon (HCHO) 4. **Count the Methyl Groups**: - In the structure of 2-methylbut-2-ene, we can identify the methyl groups: - There are two methyl groups attached to the second carbon. - There is one methyl group on the first carbon. - Total methyl groups = 3 (from the two CH3 groups and one from the chain). 5. **Conclusion**: The total number of methyl groups present in alkene (x) is 3. ### Final Answer: The total number of methyl groups present in alkene (x) is **3**.
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