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Why is ch3cooh optically inactive it doesnt even have a plane of symetry and also its image do not superimpose it

Why is ch3cooh optically inactive it doesnt even have a plane of symetry and also its image do not superimpose it

Grade:12

1 Answers

dolly bhatia
54 Points
7 years ago
Why is CH3COOH optically inactive?
A molecule has a center of symmetry when for any atom in the molecule, an identical atom can be found when it moves in a straight line through this center at equal distance on other side.
A plane of symmetry is any plane cutting through the molecule such that one side is a perfect reflection of the other. So, it does not have a plane of symmetry.
To be optically active, a molecule needs to be not identical to its own mirror image. Chiral centers are an easy way to achieve this but are not sufficient and it’s possible to have structures without a single chiral center which are nonetheless not identical to their mirror images.
A compound is optically active if it rotates plane polarized light. It has nothing to do with chirality of compound (whether it’s chiral or achiral). Symmetrical compound (having a plane of symmetry) cannot rotate plane of symmetry. Hence, CH3COOH is optically inactive.
A carbon atom is chiral if it has a carbon atom to which different groups are attached. Since different groups are attached so it cannot be symmetrical but that’s not a criterion for optical activity.
All optically active compounds are unsymmetrical but not all unsymmetrical compounds are optically active.

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