### Related Formula textR_1text-CH=textCH-R_2 xrightarrow[textZn / H_2textO]textO_3 textR_1text-CHO + textR_2text-CHO$\text{R}_1\text{-CH}=\text{CH-R}_2 \xrightarrow[\text{Zn / H}_2\text{O}]{\text{O}_3} \text{R}_1\text{-CHO} + \text{R}_2\text{-CHO}$ ### Core Logic Complete oxidative cleavage of all double bonds via reductive ozonolysis breaks the molecule into smaller fragments: textCH_3text-CH=textCH-CH(textCH_3)text-CH=textCH-CH(textCH_3)text-CH=textCH-CH_3$\text{CH}_3\text{-CH}=\text{CH-CH}(\text{CH}_3)\text{-CH}=\text{CH-CH}(\text{CH}_3)\text{-CH}=\text{CH-CH}_3$ Let's trace the fragmentation logic mapping visually:

### Step 1: Tracking Fragment Structures The chemical reaction outputs two major molecular product types: 1. textCH3text-CHO$\text{CH}3\text{-CHO}$ (Acetaldehyde): Optically inactive as it lacks a chiral carbon. 2. textOHC-CH(textCH_3)text-CHO$\text{OHC-CH}(\text{CH}_3)\text{-CHO}$ (2-methylpropanedial): Let's inspect the substituted central carbon. It is bonded to: a hydrogen atom (-textH$-\text{H}$), a methyl group (-textCH_3$-\text{CH}_3$), and two identical formyl groups (-textCHO$-\text{CHO}$). Because two of the groups are identical (-textCHO$-\text{CHO}$), this molecule does not have a chiral center and is entirely optically inactive. ### Step 2: Total Summation Since every single product formed is achiral, the number of optically active products is zero. ### Pattern Recognition Symmetry check shortcut: When a symmetrical dialkene is cleaved, it yields symmetric fragments. The central carbon is attached to identical flanking aldehyde units post-cleavage, destroying any prior asymmetry and leaving 0 optically active compounds. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Hydrocarbons Class 11 Chemistry: Organic Chemistry - Some Basic Principles and Techniques