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Topic: Optical activity of stereoisomers  (Read 2505 times)

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Offline cseil

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Optical activity of stereoisomers
« on: November 09, 2014, 05:16:46 AM »
I did an easy exercise of organic chemistry about chlorination of alkanes.
It says to rapresent the stereoisomers of the products obtained by chlorination of (2S)-2-chlorobutanoic acid considering the monosubstitution on C-3 and to establish which one is optical active or inactive.

The products are:
(2S,3S)-2,3-dichlorobutanoic acid and (2S,3R)-2,3,dichlorobutanoic acid.

There are 2 chiral centers in each molecule. Which one is active?
I said that the 2S,3S is active, because every center rotate the plane of the polarized light in the same verse, and the 2S,3R is inactive.
Is it right?

Offline Archer

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Re: Optical activity of stereoisomers
« Reply #1 on: November 10, 2014, 01:08:02 AM »
I did an easy exercise of organic chemistry about chlorination of alkanes.
It says to rapresent the stereoisomers of the products obtained by chlorination of (2S)-2-chlorobutanoic acid considering the monosubstitution on C-3 and to establish which one is optical active or inactive.

The products are:
(2S,3S)-2,3-dichlorobutanoic acid and (2S,3R)-2,3,dichlorobutanoic acid.

There are 2 chiral centers in each molecule. Which one is active?
I said that the 2S,3S is active, because every center rotate the plane of the polarized light in the same verse, and the 2S,3R is inactive.
Is it right?

I can't rationalise why both diastereoisomers wouldn't be optically active unless each centre rotates light by an equal and opposite amount. I am not familiar with this compound but I have never experienced this phenomenon.

Draw both diastereoisomers above with it's corresponding enantiomer (i.e. 2R,3R and 2R,3S respectively), if they can't be superimposed irrespective of which rotational conformation you draw, then they should polarise light to some extent.

I think perhaps the question is relating to the stereo specificity of the reaction. If there is an equal mixture of both diastereoisomers in the product then one centre is not optically active because it would be racemic. Maybe you are being asked to identify which chiral centre is active in your mixed product of diastereoisomers.
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Offline discodermolide

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Re: Optical activity of stereoisomers
« Reply #2 on: November 10, 2014, 06:02:40 AM »
One stereocenter is fixed as (S), adding another forms a mixture of diastereoisomers. Now unless the original chiral center has been racemised the two diastereoisomers with rotate the plane of polarised light.
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Offline Babcock_Hall

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Re: Optical activity of stereoisomers
« Reply #3 on: November 10, 2014, 01:19:20 PM »
I have to wonder whether or not there is a typographical error in the problem.  If the problem specified 2-Chlorobutane (not ...butanoic acid) as the starting material, it would make more sense.

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