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Topic: L-D Configurations: Threose and Erythrose [Question Corrected]  (Read 6084 times)

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Offline Biocheistry_97

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L-D Configurations: Threose and Erythrose [Question Corrected]
« on: October 26, 2015, 10:30:26 AM »
I am a little confused about the concept of L and D configurations to distinguish between optical isomers after seeing these diagrams. The image below shows the L and D configurations of Threose and Erythrose as Fischer Projections.





Please could somebody explain why, in a Fischer projection, D-erythrose has both -OH groups on the right-hand side and E-erythrose has both -OH groups on the left-hand side whereas D-threose has only one of the -OH groups on the right-hand side, despite both carbons which are bonded to -OH groups being chiral.

Thanks

« Last Edit: October 26, 2015, 10:56:56 AM by Biocheistry_97 »

Offline Dan

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Re: L-D Configurations: Threose and Erythrose [Question Corrected]
« Reply #1 on: October 26, 2015, 11:40:01 AM »
Your questions is a little unclear, do you mean:

For threose, why is "left, right" D and and "right, left" L (rather than the other way around)?

If this is the question, the answer is that the D/L descriptor is assigned on the basis of the configuration of the last stereocentre (the stereocentre closest to the bottom of the Fisher projection). In the case of erythrose/threose, this is C3. For a pentose, you would assign D/L based on C4, for a hexose C5 and so on.

If the last stereocentre points right in the Fisher projection, it's a D-sugar, if it points left, it's an L-sugar. It doesn't matter what the other OHs are doing, only the last stereocentre is used to assign D/L. This is simply the definition of D/L, it is an arbitrary (but consistent) convention.
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Offline Biocheistry_97

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Re: L-D Configurations: Threose and Erythrose [Question Corrected]
« Reply #2 on: October 26, 2015, 11:46:07 AM »
Your questions is a little unclear, do you mean:

For threose, why is "left, right" D and and "right, left" L (rather than the other way around)?

If this is the question, the answer is that the D/L descriptor is assigned on the basis of the configuration of the last stereocentre (the stereocentre closest to the bottom of the Fisher projection). In the case of erythrose/threose, this is C3. For a pentose, you would assign D/L based on C4, for a hexose C5 and so on.

If the last stereocentre points right in the Fisher projection, it's a D-sugar, if it points left, it's an L-sugar. It doesn't matter what the other OHs are doing, only the last stereocentre is used to assign D/L. This is simply the definition of D/L, it is an arbitrary (but consistent) convention.

Thank you for your reply Dan. The position of the -OH facing to the left on both C2 and C3, like in the Fischer projection, for L-Erythrose (and both to the right for D-Erythrose) whereas C2 OH is on the opposite side of C3 OH for the other.

Offline Dan

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Re: L-D Configurations: Threose and Erythrose [Question Corrected]
« Reply #3 on: October 26, 2015, 11:51:07 AM »
I meant that the position of the -OH facing to the left on both C2 and C3 like in the Fischer projection for L-Erythrose (and both to the right for D-Erythrose) whereas C2 OH is on the opposite side of C3 OH for the other.

I don't understand the question. Both pointing in the same direction in the Fischer projection is called erythro and pointing opposite directions is called threo - that is just the definition of the erythro/threo stereodescriptors.
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Offline Yggdrasil

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Re: L-D Configurations: Threose and Erythrose [Question Corrected]
« Reply #4 on: October 26, 2015, 11:53:04 AM »
Remember that L-erythrose and D-erythrose are enantiomers, meaning that they are mirror images.  Thus, the stereochemistry at every single stereocenter will be flipped.  If you flip the configuration only at one sterocenter, you get a different molecule, not an enantiomer.

Offline Biocheistry_97

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Re: L-D Configurations: Threose and Erythrose [Question Corrected]
« Reply #5 on: October 26, 2015, 12:03:34 PM »
Remember that L-erythrose and D-erythrose are enantiomers, meaning that they are mirror images.  Thus, the stereochemistry at every single stereocenter will be flipped.  If you flip the configuration only at one sterocenter, you get a different molecule, not an enantiomer.
I understand now. Thank you kindly.

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