This reaction involves the ozonolysis of an alkene followed by reductive workup (Zn/H₂O). Ozonolysis cleaves the double bond of an alkene and converts each carbon of the double bond into a carbonyl group. Reductive workup (Zn/H₂O) ensures aldehydes are produced instead of carboxylic acids.

The given reactant is a cyclic alkene with one double bond. Ozonolysis will break the ring at the double bond, opening it to form a dicarbonyl compound. Since it is a symmetric molecule, the product will have two identical carbonyl groups.

Step 1: Ozonolysis cleaves the double bond. Each carbon of the double bond becomes a carbonyl carbon.

Step 2: With reductive workup (Zn/H₂O), both carbonyls are aldehydes.

The product is a dialdehyde: OHC-CH₂-CH₂-CH₂-CHO.

This corresponds to the structure: $O = C H - C H_{2} - C H_{2} - C H_{2} - C H = O$

Thus, the correct product (X) is the first option:

Related Topics & Formulae

Ozonolysis Reaction: Ozonolysis is used to cleave alkenes or alkynes at the multiple bond. For alkenes, general reaction is:

Alkene + O₃ → Ozonide → (with workup) Carbonyl compounds

Workup can be:

Reductive (e.g., Zn/H₂O): Produces aldehydes (or ketones).
Oxidative (e.g., H₂O₂): Produces carboxylic acids (or ketones).

Key point: Each carbon of the double bond becomes a carbonyl carbon. The substituents on the carbon remain attached.