[ksr985]

presence of a chiral carbon atom is a sufficient but not a necessary condition for optical activity in non allene systems.

is the above statement correct?

how can one tell if a compound is optically active in terms of simple or alternating axes of symmetry. does the presence of either of these axes make a compound optically inactive?

help.

[Mitch]

The only 100% way to know if a molecule is chiral is to draw the mirror image and see if its superimposable. Whether it has a chiral atom or a certain axis is an inferior method for determining chirality than the above statement.

[BelarusGirl]

By the way, ksr985,  did you know that:
substance may have hiral atom, but it may be not optically active if it has symmetry plane - it is called meso-form.

[drewauerbach]

the statement is false.  Here are two counter-examples:
1.  Just as ksr985 said, a compound can be meso, which would mean that the net optical activity is zero.  (Really?  What if you have (R*,S*)-2-bromo-3-chloro-butane? Hmm)
2.  You can also have a funky situation where two carbons form a cage compound.  Think of butane.  Now, think of carbons two and three.  Imagine us splitting the bond between the two apart and inserting tow carbons such that the resulting figure is a square where carbons two and three are at entirely opposite ends.  Now add two didfferent substituents to carbons one and four, and you have carbon two and three being chiral, but the overall compound can have no optical activity.  At least I think ...

[ksr985]

mitch:
given that checking for the superimposability of the mirror image is the only way to tell for sure if the compound iis optically active, why are the simple and alternating axes of symmetry defined? do they give any kind of reliable clue about optical activity? the intention behind my original question was to figure this out, really.

once again, please help.

[movies]

I think the take-home message is that the presence of one of these symmetry elements is not sufficient to guarantee optical activity.  Things such as a chiral carbon center at one point in a molecule do not preclude the presence of another chiral carbon atom at another point in the molecule.  The combination of multiple individual chiral centers can lead to a molecule with other symmetry elements that destroy the optical activity (such as mirror planes and inversion centers).

What Mitch is getting at is that you should draw out the entire molecule and it's mirror image so that you account for all of the symmetry elements.  It's easy to miss them sometimes, especially the strange ones like inversion centers.

One class of compounds that are yet more difficult to deal with are conformational enantiomers, where two conformations of one molecule are enantiomeric.  Examples of this are the ring flips of cis-1,2-dimethylcyclohexane and gauche-butane.  In most cases, it is practically impossible to obtain pure samples of a single enantiomer of these compounds because they racemize via relatively low energy conformational changes.