Meso Sugars

[justadream]

Here is a diol created from a meso sugar:

It has an internal plane of symmetry.

What are the R/S configurations of C2 and C3?

Aren't they both "R" since the OH's are both on the right side.

However, I thought meso compounds needed to have chiral centers with OPPOSITE R/S configuration?

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[justadream]

To illustrate the source of my confusion, see TBR OChem II page 130 #45.

The answer explanation talks about how carbon 2 and 4 on a meso sugar must have opposite steroechemistry. Then, TBR uses this logic to show that if carbon 4 is R, then carbon 2 must also be R. This seems contradictory to me.

It is because the simple "OH on right side = R" rule does not work when you have diols?

 

[techfan]

Well since the sides of a Fischer projection are "wedges" and top and bottom are "dashes" I would've switched the H and CH2OH on C2 and labeling -OH as 1, C3 as 2 and the CH2OH (now on the left side of C2) as 3 then seeing that it is clockwise (Right thumb points into the page) label it R, but switching to S since I switched two groups.

As for C3 I would've switched CH2OH and H -OH is 1 C2 is 2 and CH2OH as 3, this time it goes counterclockwise (right thumb points out) so it is S, but switches to R because I switched two groups.

So C2=S C3=R which is opposite.

I think you are getting D and L mixed with R and S. For D and L labeling when you have the most oxidized carbon on the top, if the last (or bottom) chiral center has -OH on the right side it is D and if -OH is on the left it is L.

 

[justadream]

@techfan

Oh, so is there no easy shortcut to determine R/S in Fisher projections as there is in L/D?

I just have to figure it out individually (like you did)?

 

[techfan]

I don't know of a shortcut, but I'm not very good with stereochemistry.

 

[justadream]

@techfan

Also, your R/S assignments make sense to me but TBR says that this is: 2-S, 3-R

How do they get that?

This is TBR OChem II page 130 #47

 

[techfan]

I have the 2012 edition and don't have that question. If that's what they said, it doesn't make any sense to me. Assuming that C2 is the C in the cross at the top, I would've said 2-R 3-R.

Now if there was an aldehyde at C1 then I would've said 2-S, 3-R (switch HC=O and H and OH=1 HC=O=2 and C3=3 then get thumb into the page for R, but switch to S)

 

[justadream]

@techfan

That's weird! I have 2012 edition too.
This is Ochem Book Two Passage VII "Unknown D-aldohexose elucidation"

Both questions (#45 an #47) are from it.

The answer explanation for #45 also says "All sugars drawn on the right in a Fischer projection have R stereochemistry".

This makes no sense because in a meso compound, you need symmetry. So let's say you have a sugar with all chiral carbons having OHs facing rightwards in the Fischer projection. If all of them face right, they are all supposedly "R". But in meso compounds, you need OPPOSITE R/S (TBR asserts this in the answer explanation).

 

[techfan]

I thought my book was 2012 (it has a copyright for that date), but I got it off amazon so who knows and the version I have removed that passage. I pulled up a PDF of an earlier version, and it was in there. If the attached picture is the same then I think I know what's going on. Since this passage is an elucidation of an unknown compound you have to use the reactions given. In compound X you only know that it is D, so that fixes the position of the -OH group on the 2nd to last C to the right. The parentheses signify that you don't know which way the -OH groups are facing.

When you add HNO3 to (nitric oxide sugar oxidation) you are pretty much just turning both the top and bottom Cs into COOH so that the compound will show whether it is meso or not (plane of symmetry within the molecule). In this case it says that Z is optically active which means at least one -OH is not on the same side as the rest which doesn't really help us much.

When you do the Ruff degradation (Br2/H2O followed by Ca(OH)2/H2O followed by H2O2/Fe3+) you are essentially removing the top C from the cross (C2). Then they add HNO3 (for reasons stated above), and you get an optically inactive compound which means that the old C3 has to be the same as the old C5 (to get a mirror image). Since the -OH on C5 is fixed as right (due to D), the C3 has its -OH on the right also.

It says that the stereochemistry is opposite for Compound W not X, X is not meso (due to it having optical activity when you add HNO3). In W the old C3 and the old C5 have opposite stereochemistry (with C3 you curl your fingers up {the C=O is at the top} and with C5 you curl down {C=O is at bottom} then switch so you get R and then S). For compound X C3 has -OH on right and you swap H and the C2 curl fingers up, thumb points out for S then switch to R. For C5 switch CH2OH and H curl fingers up and then switch to get R.

The last sentence "All sugars drawn on the right have R stereochemistry" should have been written, for D (right) sugars the last chiral carbon has R stereochemistry. I think it is making the assumption that you knew that only the last C determines D and L and that D is R and L is S.

As for 47. When you do an reverse aldose reaction, with compound X, you are cutting the alpha and beta carbons (C1 and C2) off and C3 becomes the aldehyde carbon Then you reduce the compound so you can see if it is meso (make C3=C6). You don't know which way the -OH on C4 is pointing, and since it is it optically active you find that the -OH on C4 is opposite to the -OH on C5. C5 is right so C4 is left. Switch H on right with C3 and then -OH on left is 1 C3 on right is 2 and C5 is 3, your fingers curl up and thumb points into the page for R, but since you switched 2 groups switch to S.

So 4 has S. and 3 has R (because it is tied to 5 due to compound W being meso).

In case the solution you have is different than the PDF, I attached it.

 

[justadream]

@techfan

Thanks for writing that up.

I followed it but my problem remains:

#45
Old C3 and old C5 are both R. Here, both the OH groups are facing rightwards. But in meso compounds, shouldn't you have OPPOSITE R/S?

#47
The answer is that 4 is S and 5 is R. I follow how they got that. But since you have something that is optically ACTIVE, shouldn't you have the SAME R/S?


Here is what I mean by typical meso compounds having OPPOSITE R/S:

 

[techfan]

Compound X is NOT meso, if it was then when you treated it with HNO3 (i.e. compound Z) it would NOT be optically active.

Compound W IS meso, and as such C2 (C3 on compound X) and C4 (C5 on compound X) have different stereochemistry.

47. What do you mean if it's optically active it has the same R and S? Optically active just means that it isn't meso (as long as you have chiral centers).

I think you're using the 2 stereocenter example too much. C4 and C5 don't depend on each other for optical activity. Think about folding X over in half and you'll see that the -OH on C5 and C3 have to match and the OH C4 and C2 have to match in order for it to be optically inactive. Thus C3 and C5 have to have opposite stereochemistry, and C2 and C4 have to have opposite stereochemisty.

 

[justadream]

@techfan

Sorry, I meant that Compound W is meso. However, you said that "For compound X C3 has -OH on right and you swap H and the C2 curl fingers up, thumb points out for S then switch to R. For C5 switch CH2OH and H curl fingers up and then switch to get R"

That means that in Compound X: C2 is R. C4 is also R.

How are these opposite stereochemistry?

I'm trying to say that in meso compounds, you have OPPOSITE R and S across the plane of symmetry.

For example, if you have a molecule with 2 chiral centers you will get something like: R | S

If you have something with 4 chiral centers, you will get something like: R R | S S

The point is that the stereochemistry (whether it is R or S) is OPPOSITE when you "reflect" half the molecule to the other side.

 

[techfan]

Unless you are taking into consideration #47 (the reverse aldol) you don't know what the stereochemistry for C2, or C4. When you do the Ruff degradation (and then adding the HNO3) the plane of symmetry goes right through C4 so you can't make any judgments about C4 and C2 is cut out so you can't make any judgments on that either. Those 2 carbons have variable stereocenters.

I agree that meso compounds have opposite stereocenters across the plane of symmetry, but again you can't make any judgments about C2 and C4 (notice how in the solution they left the -OH group of those 2 in parentheses).

 

[justadream]

@techfan

I thought you DO know the stereochemistry for C2 in Compound W. Isn't that what #45 is asking (#45 asks for stereochemistry of carbon 3 in compound X which is equivalent to carbon 4 in compound W).


But anyways, let's just look at the structure of Compound X after undergoing the reverse aldol reaction (this structure is in the answer key).
In that structure, the two middle carbons have OPPOSITE stereochemistry (something we've established is typically associated with being MESO) but yet are actually optically active here. Can you help me reconcile that?

 

[techfan]

You know the stereochemistry for C2 in W (which is C3 in X, sorry I should've been more clear), but I was referring to C2 from X which is turned into the aldehyde in Ruff (C1 is cleaved off).

C3 from X is equivalent to C2 from W, not C4. Ruff cleaves off the C=O carbon (C1) and then oxidizes C2 to form a C=O. so C2 from X becomes C1 in W, C3 from X becomes C2.

So since you know that W is meso and that C4 (X C5) is D and thus the OH group is fixed as facing right, the plane of symmetry goes through C3 (XC4), thus C2 (XC3) must have the -OH group facing right.

So for WC2 -OH on the right, curl fingers up and thumb goes into the page, but H is in wedge, so switch for S. WC4 curl fingers down (because after adding HNO3 you get C=O at the bottom), curl fingers down for S, but H is in wedge so switch to R.


For something to be optically inactive ALL the stereocenters have to opposite across the plane of symmetry. Having it be optically active (assuming we know the above part) only tells us that C2 and C4 have the same stereochemistry (either both S or both R), it doesn't tell us which. We don't have the info to get the stereochemistry of C4 OR C2! It's like having 4 unknowns and 2 equations.

 

[justadream]

You know the stereochemistry for C2 in W (which is C3 in X, sorry I should've been more clear), but I was referring to C2 from X which is turned into the aldehyde in Ruff (C1 is cleaved off).

C3 from X is equivalent to C2 from W, not C4. Ruff cleaves off the C=O carbon (C1) and then oxidizes C2 to form a C=O. so C2 from X becomes C1 in W, C3 from X becomes C2.

So since you know that W is meso and that C4 (X C5) is D and thus the OH group is fixed as facing right, the plane of symmetry goes through C3 (XC4), thus C2 (XC3) must have the -OH group facing right.

So for WC2 -OH on the right, curl fingers up and thumb goes into the page, but H is in wedge, so switch for S. WC4 curl fingers down (because after adding HNO3 you get C=O at the bottom), curl fingers down for S, but H is in wedge so switch to R.


For something to be optically inactive ALL the stereocenters have to opposite across the plane of symmetry. Having it be optically active (assuming we know the above part) only tells us that C2 and C4 have the same stereochemistry (either both S or both R), it doesn't tell us which. We don't have the info to get the stereochemistry of C4 OR C2! It's like having 4 unknowns and 2 equations.

@techfan

First of all, thank you for helping me through all of this!

In your last paragraph, you refer to something as being optically active? What molecule are you referring to? I assume compound X (because that is clearly not meso)?

And I think I follow what you are saying about Compound W. It is meso and it does indeed have OPPOSITE chiral centers across the line of symmetry.

========

Ok but now look at this molecule:

Would you say the stereochemistry at the second and third carbons are R, and R, respectively? This is a realization I just made.

If so, that would make sense since this is an optically ACTIVE molecule. Also, if so, then I think I finally understand everything.

 

[techfan]

Yeah, no problem!

I was referring to Z, but I think X would also be optically active. Again, you need to have the same groups on the top and bottom (like you showed in your pic)
=====
I agree that C2 and C3 are both R. So yes, it should be optically active.

Great, but I still hope that something like this doesn't show up on the MCAT, this problem would suck.

 

[justadream]

@techfan

So I noticed for sugars (as long as you have the C=O group at the top and the alcohol group at the bottom), if the OH is on the right, it is always R. If the OH is on the left, it is always L.

However, if you don't have C=O on top and OH on the bottom (like the structure I drew in my earlier post), then you have to determine R or S manually.

 

[kraskadva]

Here is a diol created from a meso sugar:

It has an internal plane of symmetry.

What are the R/S configurations of C2 and C3?

Aren't they both "R" since the OH's are both on the right side.

However, I thought meso compounds needed to have chiral centers with OPPOSITE R/S configuration?

Forget this R on the right nonsense. Totally not true.

Here's a no fail stereochem trick...
(though you need to understand ranking, so look up the rules to that if the numbering below doesn't make sense)

With your molecule above, the OH groups are each ranked as 1, the bond across the plane of symmetry is ranked as 2, the end groups (-CH2OH) are ranked as 3, and the H is 4.
The OHs and Hs are coming out of the paper, the end groups are going into the paper.
Imagine a circle drawn from 1 to 2 to 3 with the H sticking out from the middle.
Stick out your hands- thumbs up, fingers curled.
Imagine the H at the tip of your thumbs. 1 should be at the base of your palm, 2 at the base of your fingers, 3 at the tips of your fingers.
Turn your hands around as necessary to fit the circle, only one hand will match the direction you've drawn it.
Which hand does the circle match? Does it curl in the same direction as the fingers on your right hand or your left hand?
If it's your right hand, it's R.
If it's your left hand, it's S.

C2 is S, C3 is R.
which makes it meso.

Model kits can also help you get the hang of visualizing this. If you don't have one, get one. They're cheap on Amazon and ebay.