Optically Active Products from Ozonolysis of Alkene

To find the number of optically active products obtained from the complete ozonolysis of the given compound, let us first identify its structure and analyze the products formed upon cleavage of all carbon-carbon double bonds ($\text{C}=\text{C}$).

1. Structure of the Given Compound

The given compound is: $\text{CH}_3-\text{CH}=\text{CH}-\underset{\underset{\text{H}}{\shortmid}}{\overset{\overset{\text{CH}_3}{\shortmid}}{\text{C}}}-\text{CH}=\text{CH}-\underset{\underset{\text{CH}_3}{\shortmid}}{\overset{\overset{\text{H}}{\shortmid}}{\text{C}}}-\text{CH}=\text{CH}-\text{CH}_3$

This molecule contains three double bonds at positions $\text{C2}=\text{C3}$, $\text{C5}=\text{C6}$, and $\text{C8}=\text{C9}$.

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2. Ozonolysis Reaction

Reductive ozonolysis breaks each carbon-carbon double bond ($\text{C}=\text{C}$) and converts each double-bonded carbon into a carbonyl group ($\text{C}=\text{O}$).

Cleaving the three double bonds gives the following products:

1. From the terminal double bonds ($\text{C2}=\text{C3}$ and $\text{C8}=\text{C9}$): $\text{CH}_3-\text{CHO} \quad (\text{Acetaldehyde})$

   • Two moles of acetaldehyde ($\text{CH}_3\text{CHO}$) are formed.

2. From the internal double bonds ($\text{C3}-\text{C4}-\text{C5}$ and $\text{C6}-\text{C7}-\text{C8}$): $\text{OHC}-\underset{\underset{\text{H}}{\shortmid}}{\overset{\overset{\text{CH}_3}{\shortmid}}{\text{C}}}-\text{CHO} \quad (\text{2-methylpropanedial})$

   • Two moles of 2-methylpropanedial ($\text{OHC}-\text{CH}(\text{CH}_3)-\text{CHO}$) are formed.

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3. Optical Activity of the Products

• Acetaldehyde ($\text{CH}_3\text{CHO}$):

  • Contains no chiral (asymmetric) carbon center.
  • It possesses a plane of symmetry and is optically inactive.

• 2-Methylpropanedial ($\text{OHC}-\text{CH}(\text{CH}_3)-\text{CHO}$):

  • The central carbon atom (C-2) is bonded to four groups:
    1. $-\text{H}$
    2. $-\text{CH}_3$
    3. $-\text{CHO}$
    4. $-\text{CHO}$
  • Since two of the four groups attached to the central carbon atom are identical (two $-\text{CHO}$ groups), C-2 is not a chiral center.
  • The molecule possesses a plane of symmetry ($\sigma$) and is achiral, making it optically inactive.

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Conclusion

None of the organic products formed from the complete ozonolysis are optically active. Therefore, the total number of optically active organic products is $0$.

Correct Answer: A