Is This Compound Optically Active?

  • Thread starter Thread starter nil1996
  • Start date Start date
  • Tags Tags
    Isomerism Optical
Click For Summary
SUMMARY

The compound in question is determined to be optically active due to the absence of a plane of symmetry, which indicates chirality. The discussion highlights the importance of identifying chiral centers, particularly in cyclic compounds. Participants emphasize that even if a molecule appears flat in its representation, its three-dimensional structure can affect optical activity, especially when substituents like methyl and chlorine are involved.

PREREQUISITES
  • Understanding of chirality and chiral centers in organic chemistry
  • Familiarity with optical activity and its significance in stereochemistry
  • Knowledge of molecular geometry and three-dimensional structures
  • Basic concepts of cyclic compounds and their properties
NEXT STEPS
  • Study the concept of chiral centers in organic molecules
  • Learn about the relationship between molecular symmetry and optical activity
  • Explore the three-dimensional representations of cyclic compounds
  • Investigate the effects of substituents on optical activity in organic compounds
USEFUL FOR

Chemistry students, organic chemists, and anyone studying stereochemistry and optical activity in compounds.

nil1996
Messages
301
Reaction score
7

Homework Statement


Is this compound optically active or inactive??



Homework Equations

Untitled.jpg




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??
 
Physics news on Phys.org
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.
 

Similar threads

Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 9 ·
Replies
9
Views
6K
  • · Replies 3 ·
Replies
3
Views
3K