L or D Sugar from Ring?

[justadream]

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When a sugar is provided in ring form (as a pyranose or furanose), how do you tell if it is L or D?

I know L or D comes from the last chiral carbon. This is easy to determine in Fischer projection. But what about when it's in a ring?

 

[Czarcasm]

When a sugar is provided in ring form (as a pyranose or furanose), how do you tell if it is L or D?

I know L or D comes from the last chiral carbon. This is easy to determine in Fischer projection. But what about when it's in a ring?

For the Haworth projection:

If the -CH2OH substituent is RIGHT of the anomeric carbon, it's D.
If the -CH2OH substituent is LEFT of the anomeric carbon, it's L.

This can be confusing, because it's with reference to the way it's pictorialized on paper/screen, NOT your left or right.
This is not something commonly tested so I wouldn't worry about it too much. Fischer is more important.

For the Fischer projection:

If the last chiral center points to YOUR RIGHT, it's D.
If the last chiral center points to YOUR LEFT, it's L.

For the anomer:

If the anomeric carbon's -OH points TRANS (opposite) to the -CH2OH substituent, it's ALPHA.
If the anomeric carbon's -OH points CIS (same) to the -CH2OH substituent, it's BETA.

Hope this helps.

 

[justadream]

@Czarcasm

With regard to
"If the -CH2OH substituent is RIGHT of the anomeric carbon, it's D.
If the -CH2OH substituent is LEFT of the anomeric carbon, it's L."


How does that apply here?
I think the anomeric carbon is the one on the right-most position (for each of those structures) and the CH2OH would be left of it. But these are all D.

 

[Czarcasm]

@Czarcasm

With regard to
"If the -CH2OH substituent is RIGHT of the anomeric carbon, it's D.
If the -CH2OH substituent is LEFT of the anomeric carbon, it's L."


How does that apply here?
I think the anomeric carbon is the one on the right-most position (for each of those structures) and the CH2OH would be left of it. But these are all D.

It's still no different. Remember, right or left is with respect to the -CH2OH group to the anomer itself, not your right or left. If it's right of the anomer, it's D. If it's left of the anomer, it's L. In all four scenarios above, -CH2OH is right of the anomer. Like I said, this is confusing because typically right/left is oriented with respect to the reader's right or left side. In this case, it's respect to the actual orientation on paper/screen. Unfortunately, that's the way it was taught to me. Maybe other's know an easier way, or you could just recognize that typically the -CH2OH group for D glucose points up and ignore everything else.

 

[justadream]

@Czarcasm

"In all four scenarios above, -CH2OH is right of the anomer."

Excuse my ignorance but how exactly are you viewing the structure to see that the CH2OH is to the right of the anomer? Are you flipping the molecule in your head (or rotating or something)?

 

[Czarcasm]

@Czarcasm

The depiction I'm using is confusing, but that's how it was taught to me. Something that's probably more intuitive instead, is rather than focusing on where the -CH2OH group is appearing within the molecule, instead just look at the anomer itself:

If the anomer is on the right side, it's D.
If the anomer is on the left side, it's L.

I think that will clear things up a lot more and be less confusing. These conventions are probably more familiar because they are referring to your right/left side rather than the molecules right/left side. Either way you look at it, it represents the same exact thing.

The convention I explained before though, is no different than the anatomical depiction given to patients: the right and left side of the heart (labeled on a diagram) for instance are that of the patient (not yours). Likewise, the convention I gave earlier was for the molecules right/left side (again, not yours):

 

[pepocho]

.

 

[Czarcasm]

Hey czarcasm, I have been checking other sources, and some places say that D and L depends on whether the -CH2OH is above or below the plane of the ring

http://science.marshall.edu/castella/chm204/projection.pdf

Yes, that's also something I learned in review books, but if you look at my earlier example, some people don't follow that convention. I'm not sure if it's because it's newer or what, but I suppose knowing both might be useful. The problem is, using either convention would yield different configurations.

Take a look at this image for D-Glucose:

For L-Glucose, everything would have to flip over, causing inversion of all the chiral centers, but also it would shift the anomer to the left side, according to the way this is depicted in some textbooks.

 

[pepocho]

.

 

[Czarcasm]

Yeah youre right. I think I can push my neurotic mind away from this one. If there is more than a couple ways to do it, I doubt it'll be a problem. And even if it was a question on the MCAT, i bet it won't be a discrete. Thanks for everything Czarcasm

Yeah, it's honestly not worth the added stress. I've never encountered a single question in practice asking me to specifically determine whether a haworth projection was D or L (only understanding D or L for fischer is of importance).

 

[hannahrenee]

To determine L or D confirmation you can look at the position of the non-ring carbon. A D monosaccharide has the non-ring carbon attached to C-4 "up" relative to the hydrogen on C-4. In an L monosaccharide this same non-ring carbon is down.

 

[theonlytycrane]

Given a sugar in ring form, uncyclize the ring form and redraw the sugar in a fischer projection. If the -OH group on the last chiral carbon is on the right it's D. If it's on the left it's L.

 

[deleted801953]

Are we supposed to know the position of the other -OH and -H in the ring structure for compounds like glucose, mannose or galactose?

 

[BerkReviewTeach]

Are we supposed to know the position of the other -OH and -H in the ring structure for compounds like glucose, mannose or galactose?

Yes!

 

[aldol16]

I really wouldn't worry about this. It's a really low-yield concept.

 

[Swagster]

I really wouldn't worry about this. It's a really low-yield concept.

I would beg to differ, based on my experience not too long ago. I can't talk about specifics, but suffice to say it was a very interesting passage on a useful case of an L-sugar.

The point being, there really is no such thing as high yield or low yield anymore.

 

[aldol16]

I would beg to differ, based on my experience not too long ago. I can't talk about specifics, but suffice to say it was a very interesting passage on a useful case of an L-sugar.

The point being, there really is no such thing as high yield or low yield anymore.

Where your performance on the passage hinged on your identifying a sugar as either L or D based on the ring structure alone? There certainly is high and low yield material. High yield material will come up more often and will have many passages based on it. Low yield material might come up only once or not at all. Unless your exam had multiple questions where you had to identify L vs. D from structure alone, then it's low yield material. Everything in your education from here on out will have high and low yield information and the only way you stay sane is to identify what's high- or low-yielding.

 

[Swagster]

I can't talk about specifics, but if I were to ever study again for this exam, I would definitely take some time to familiarize myself with common L-sugars found in nature, specifically in the human body. And I would have looked more carefully at whether a ring structure was D or L.

I don't mean to make this an argument of semantics, but high frequency is not the same thing as high yield. Every exam is different and there will be only one (maybe two) questions on any highly specific topic. There may be two pH questions or two gas exchange questions, but each exam is designed to spread the questions out. So using your perspective for a given exam, all questions are technically low frequency on that exam. If we had a bigger sample size than two MCATs (one being old) for me and however many you took, we could reach a better conclusion about high versus low frequency of this particular material. But given that you and I have not taken every MCAT, neither of us can argue whether a topic is high or low frequency. We can speculate, but never say with confidence.

But I can definitely say that being able to distinguish D versus L for a ring-structure is a skill that has value.

 

[aldol16]

I don't mean to make this an argument of semantics, but high frequency is not the same thing as high yield. Every exam is different and there will be only one (maybe two) questions on any highly specific topic. There may be two pH questions or two gas exchange questions, but each exam is designed to spread the questions out. So using your perspective for a given exam, all questions are technically low frequency on that exam. If we had a bigger sample size than two MCATs (one being old) for me and however many you took, we could reach a better conclusion about high versus low frequency of this particular material. But given that you and I have not taken every MCAT, neither of us can argue whether a topic is high or low frequency. We can speculate, but never say with confidence.

You can go into it studying every "highly specific topic" for the exam or advise others to. That's not the best way to spend one's time studying for the exam. Concepts like amino acids (knowing the names, three- and one-letter codes, properties) will net you quite a few points on the exam. There will be concepts that appear multiple times on any given exam because they're central concepts. Glycolysis, for example, in some shape or form. It's like an introductory chemistry course. Knowing the equation for particle-in-a-box might net you a point or two on an exam because it might not come up. But understanding the basic principles of quantum mechanics (high level overview) will certainly net you more points because it's going to come up.

 

[PeaPodPreMed]

D (equitorial C6) and L (axial C6) forms of glucose. Alpha or Beta on anomeric carbon (C1) is determined when compared to C6. Cis=beta; trans=alpha when comparing C1 to C6. Alpha-D-glucose and beta-D-glucose are anomers (hint: change in orientation of -OR).

 

[vahidb]

For the Haworth projection:

Hope this helps.

i've never saw someone show L and D like this(and my major is food engineering!) i'm not saying this is wrong (actually i am 🙂 )but there is a problem. we know that after the formation of the ring(only in Aldopyranoses), the O of H-C=O group is attached to 5th carbon and what is attached to other side of that O? the 1st Carbon. so in any Chemical structure for Glucopyranose we should have an Oxygen between 1st and 5th Carbon and that is not the case in this chemical structure. thats why i think it is wrong.
the L and D are vertical mirror images of each other but not in the Haworth (ring) projection. imagine that you have two Fisher formations that are vertical mirror images of each other now roll them and turn them to Haworth projections. now you can see that "right" is "up" and "left" is "down" so that property of being mirror images now will present itself horizontally. in D formation the CH2OH group of the last chiral carbon is Up. in L the CH2OH of last chiral carbon is down.
and every OH which is down in L will be up in D and vice versa.

 

[BerkReviewTeach]

I believe what Czarcasm was emphasizing is the D and L sugars are enantiomers of one another, so showing them as mirror images emphasizes that visually. You are correct that L-sugars usually drawn using the same Haworth skeleton that is used for D-sugars, the difference being that the last carbon is found below the ring (in the L-configuration) as opposed to above the ring (in the D-configuartion.)