Is This Compound Optically Active?

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    Isomerism Optical
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Discussion Overview

The discussion revolves around the optical activity of a specific compound, focusing on whether it is optically active or inactive. Participants explore concepts related to symmetry and chirality, as well as structural considerations of the compound.

Discussion Character

  • Homework-related, Conceptual clarification, Debate/contested

Main Points Raised

  • One participant initially claims the compound is optically inactive due to a perceived lack of a plane of symmetry.
  • Another participant questions the assertion that a lack of symmetry implies optical inactivity, suggesting a need for clarification on this point.
  • A later reply corrects the initial claim, stating that the compound should be optically active instead.
  • Further discussion raises the question of where the chiral atom is located within the compound.
  • Participants are reminded that the molecule's actual three-dimensional structure may differ from its two-dimensional representation, which could affect chirality considerations.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the optical activity of the compound, with competing views on the implications of symmetry and chirality remaining unresolved.

Contextual Notes

There are limitations regarding the assumptions made about symmetry and chirality, as well as the structural representation of the compound, which may affect the conclusions drawn.

nil1996
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Homework Statement


Is this compound optically active or inactive??



Homework Equations

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The Attempt at a Solution


This compound doesn't have the plane of symmetry.So according to me it should be optically inactive.is it right??
 
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nil1996 said:
This compound doesn't have the plane of symmetry.

It doesn't?

So according to me it should be optically inactive.is it right??

What you wrote suggests lack of plane of symmetry makes the compound optically inactive. Are you sure about it?
 
sorry i mean it should be optically active.
 
So, where is the chiral atom?

I guess you should start finding out how to deal with rings...

Also remember that while the molecule is drawn flat, in fact it is not flat at all - which is especially important for carbons substituted with methyls and Cl.
 

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