To solve this problem, we need to understand the reaction of compound X with ozone ($$O$_3$) followed by reduction with Zn/H$_2$O. This process is known as ozonolysis, which cleaves double bonds in alkenes to form carbonyl compounds.$\\$Given that the products are formaldehyde ($$HCHO$$) and 2-methylpropanal ($$C$_4$H$_8$O$$), we need to determine the structure of compound X.$\\$Step 1: Identify the carbonyl products.$\\$• Formaldehyde is $$HCHO$,$ which indicates that one of the carbon atoms involved in the double bond is a terminal carbon ($$CH$_2$ group).$\\$• 2-Methylpropanal is $$(CH$_3$)_2CHCHO$,$ which suggests the presence of a branched alkyl group adjacent to the carbonyl group.$\\$Step 2: Analyze the possible structures of compound X.$\\$• Compound X must have a double bond that, when cleaved, results in the given carbonyl compounds.$\\$Step 3: Consider the options.$\\$Option 1: 3-Methylbut-1-ene$\\$• Structure: CH$_2$=C(CH$_3$)CH$_2$CH$_3 \\$• Ozonolysis of this compound would cleave the double bond between the terminal CH$_2$ and the adjacent carbon, forming formaldehyde ($$HCHO$$) and 2-methylpropanal ($$(CH$_3$)_2CHCHO).} \\ Option 2: 2-Methylbut-1-ene$\\$• Structure: CH$_2$=CHCH(CH$_3$)CH$_3 \\$• Ozonolysis would not produce formaldehyde, as the double bond is not terminal.$\\$Option 3: 2-Methylbut-2-ene$\\$• Structure: (CH$_3$)_2C=CHCH$_3 \\$• Ozonolysis would not produce formaldehyde, as the double bond is internal.$\\$Option 4: Pent-2-ene$\\$• Structure: CH$_3$CH=CHCH$_2$CH$_3 \\$• Ozonolysis would not produce formaldehyde, as the double bond is internal.$\\$Conclusion: The only compound that can produce both formaldehyde and 2-methylpropanal upon ozonolysis is 3-Methylbut-1-ene.$\\$Therefore, the correct option is Option 1: 3-Methylbut-1-ene.