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The number of optically active compound(s) obtained upon complete ozonolysis of the following optically active compound is

(1E,3R,4Z,6S,7E,10S,11Z,13S)-3,6,10,13-tetramethylcyclotetradeca-1,4,7,11-tetraene

In the question, I broke all the double bonds and added a $\ce{-CHO}$ group according to the question (ozonolysis). Checking for the optically active compounds I found two of them:

products

But the solution has given only one compound! Can you please help me with this?

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2 Answers 2

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Ozonolysis of the given compound gives $\ce{I}$ , $\ce{II}$ , $\ce{III}$ and $\ce{IV}$.

  • $\ce{I}$ and $\ce{II}$ are Homomers and have same configuration "S" and therefore identical (as shown in the figure).
  • $\ce{III}$ and $\ce{IV}$ are achiral . Therefore , the total number of optically active compounds formed after ozonolysis is $\ce{= 1}$.

enter image description here

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The given answer is correct. The product scheme you drew is correct as well. However, as I marked in your scheme (see below), products A and B are essentially the same enanthiomer (both have (2S)-configuration). The compounds you have drown in right-hand side are also an identical compound, which is not optically active. Therefore, ozonolysis has given only one optically active compound.

enter image description here

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