Hey all, From the way I have always learned in Orgo 2 and from nearly any MCAT prep book, it is pretty easy to determine the chirality of 5 carbon sugars. But this one example I am thinking of is confusing now to me and I keep thinking of it.
Imagine we have a 5 carbon sugar oxidized (or reduced if you like that better) on each end. Also, when I say "right" i mean the OH is on the right of the fischer projection and "left" means just the opposite. Almost any prep book will tell you that a 5 carbon sugar that is oxidized at each end and has the OH's from Carbon's 2-4 in an order of left, right, left or left, left, left or right, right, right or right, left, right is meso. But, they say that anytime where carbon 2 and carbon 4 differ (say carbon 2 is left, carbon 4 is right), the molecule is chiral (not meso).
However, if we look at carbon 3, that carbon cannot be chiral as it has identical substituents above it and below it (its not asymmetric). Now, one might say that the molecule is indeed chiral, but only about the other two carbons (carbon 2 and carbon 4). BUT, you could say the same thing about the previously described meso compounds. I mean, even in the meso compounds, at least carbon 2 and carbon 4 are still chiral. I don't know why this is confusing to me now. Never seemed to confuse me before.
Edit - Wow never mind I got it. Yes, carbon 3 is indeed achiral in the example, but the other two carbons still retain chiraltiy. In the meso compounds, those two carbons, although chiral, are meso and thus would not rotate plane polarized light. I guess I was thinking about this too much.
Imagine we have a 5 carbon sugar oxidized (or reduced if you like that better) on each end. Also, when I say "right" i mean the OH is on the right of the fischer projection and "left" means just the opposite. Almost any prep book will tell you that a 5 carbon sugar that is oxidized at each end and has the OH's from Carbon's 2-4 in an order of left, right, left or left, left, left or right, right, right or right, left, right is meso. But, they say that anytime where carbon 2 and carbon 4 differ (say carbon 2 is left, carbon 4 is right), the molecule is chiral (not meso).
However, if we look at carbon 3, that carbon cannot be chiral as it has identical substituents above it and below it (its not asymmetric). Now, one might say that the molecule is indeed chiral, but only about the other two carbons (carbon 2 and carbon 4). BUT, you could say the same thing about the previously described meso compounds. I mean, even in the meso compounds, at least carbon 2 and carbon 4 are still chiral. I don't know why this is confusing to me now. Never seemed to confuse me before.
Edit - Wow never mind I got it. Yes, carbon 3 is indeed achiral in the example, but the other two carbons still retain chiraltiy. In the meso compounds, those two carbons, although chiral, are meso and thus would not rotate plane polarized light. I guess I was thinking about this too much.
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