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Topic: Optically Active?  (Read 9978 times)

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Offline aldoxime_amine

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Optically Active?
« on: March 21, 2009, 08:52:00 AM »
Consider compounds 1 and 2. Compound 3 is a generalized case.

Offline azmanam

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Re: Optically Active?
« Reply #1 on: March 21, 2009, 10:44:46 AM »
I've redrawn (plus an enantiomer).  R = Cl, R1 = OH, R2 = H.  What do you think?
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Offline aldoxime_amine

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Re: Optically Active?
« Reply #2 on: March 21, 2009, 04:00:59 PM »
Ah! I see, that is no "enantiomer", the 2 compounds are identical!
But does this mean optically inactive? ???
Can we say that if a compound does not have an enantiomer, then it is optically inactive?

Or can we say, the molecule is symmetric, then it is optically inactive?
More precisely, plane of symmetry or axis of symmetry exist, then it is optically inactive?

For carbons greater than or equal to 4, I don't see why a plane of symmetry must always exist...

Just an observation, why this particular example? Conformer 2 predominating in concentration over 1 because of H-bonding?

Offline azmanam

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Re: Optically Active?
« Reply #3 on: March 21, 2009, 04:47:23 PM »
there's a word for 'enantiomers' that are really the same compound.  The molecule has a C2 axis of symmetry (horizontal through the middle of the fisher projection).

remember the definition of an enantiomer is a non-superimposable mirror image.
There's a similar word that has as it's definition molecules with a plane of symmetry.

Yes, it is optically inactive, though.

As for the example - completely arbitrary.  No logic whatsoever.  I just don't like looking at things as fisher projections, they don't give me enough information in one glance.  I much rather prefer looking at the typical straight chain orientation.  So I showed my thought process in rotating from fisher projection to sideways look to straight chain orientation.
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Offline aldoxime_amine

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Re: Optically Active?
« Reply #4 on: March 23, 2009, 03:19:52 PM »

For carbons greater than or equal to 4, I don't see why a plane of symmetry must always exist...


Can you answer this question with respect to the original question?

With respect to your question, the are called meso compounds...

Thank you for the time.

Offline azmanam

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Re: Optically Active?
« Reply #5 on: March 23, 2009, 03:27:37 PM »
yup.  all of these types of compounds, regardless of chain length, will be meso and therefore optically inactive.  for chains with odd numbers of carbon atoms, the C2 axis of symmetry will go through the middle carbon.  for chains with even numbers of carbon atoms, the C2 axis of symmetry will go between the two middle carbons.

ps, my mnemonic for meso compouds is 'enantiomers that are really the same compound.'  for any compound of the type drawn of any chain length, you can draw it's mirror image, then rotate 180 degrees and it's the same compound.
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Offline Squirmy

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Re: Optically Active?
« Reply #6 on: March 26, 2009, 01:04:47 PM »
the C2 axis of symmetry will go between the two middle carbons.
Not to nitpick...it's an internal mirror plane, not a C2 axis of symmetry.

Chiral compounds can (and often do) have C2 symmetry, but can't have an internal mirror plane.

Offline azmanam

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Re: Optically Active?
« Reply #7 on: March 26, 2009, 02:19:18 PM »
fair enough.
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