# Determination of (R) and (S) configurations from a Haworth projection

Short version of question : How do I know which groups are equivalent to "out of and into the page (like in the Fisher projection)" for the Haworth projection (so that I can send the lowest priority group to the "back")?

Details:

I am comfortable enough finding R and S configuration of Fischer projections, but am unsure which groups are coming in or going out of the plane of the page in a Haworth display.

For example:

Figure 1

Taking carbon $3$ for example, the priorities are as shown below in blue:

Figure 2

By definition, horizontal groups are out of the page (i.e. towards us) and vertical groups are into the page, the lowest priority group (hydrogen) is out of the page. If hydrogen was into the page, carbon $3$ would have a clockwise R configuration, but as hydrogen is drawn out of the page, carbon 3 has the opposite, hence an anticlockwise S configuration.

If I now took carbon 3 in the Haworth for the structure above, I am assuming it is still an S configuration as the groups are still in the same position around the carbon 3 atom, relative to the other groups.

Update: After seeing the following image (annotated with numbered carbons from this source) I am not so sure about that anymore as carbon 4 has different configurations in the Fischer vs the Chair and Haworth.

Figure 3

But how would I determine the R and S configuration of carbon 3 without going back to the Fischer projection?

Mainly, how do I know which groups are equivalent to "out of and into the page (like in the Fisher projection)" for the Haworth projection (so that I can send the lowest priority group to the "back")?

I would think carbon 4 is "into the page" as it is further away from our point of view (which matches the Fischer projection, as it is drawn vertically there).

Based off this, carbon 2 would also be into the page (as its the same as Carbon 4 in the Fischer projection). Hence, the $\ce{OH}$ and $\ce{H}$ would be out of the page.

The priorities would be as follow:

Figure 4

If $\ce{H}$ was into the page, as draw it would be R, but as $\ce{H}$ is into the page, its the opposite and hence carbon 3 is an S

So I arrive at the same answer, but mainly due to comparing the Haworth off of the Fischer projection. How would I judge what was in and out of the page without the Fischer projection and only the Haworth?

• After reading this I'm not quite clear on what you don't understand. It seems you've done this correctly. It strikes me you may be asking how to imagine this Haworth diagram in three-dimensions. If this is the case it is time to get a model kit. Or at least use a 3d one online: molview.org/?cid=5793 – BiggChemT Nov 9 '16 at 3:31
• Thanks, that site is very useful. Well my issue was how to determine if the lowest priority groups in a Haworth were into or out of the page (as when writing R or S, we've always had to put the lowest priority group into the page). But I think I've figured it out and what I have to do for Haworth is treat groups pointing upwards as out of the page and groups pointing downward as into the page. – K-Feldspar Nov 9 '16 at 21:22

In a Haworth projection, the hexagonic shape of the ring already implies the direction of two bonds. The two bonds on the ring are always ‘bent ring-inwards’ when viewed from the outside. Thus, the two perpendicular bonds (typically vertical) must ‘point away’ from the ring: in a forward direction for the front carbons of a ring and in a backward direction for the rear side of the ring.

However, I notice that the question uses (R) and (S) a lot and attempts to argue a lot with it, so a note must be dropped on those stereodescriptors, too. It is important to realise that the (R) and (S) stereodescriptors can change even if the conformation of the carbon in question did not invert or otherwise change conformation. This is due to the way priorities are determined in the CIP system. Let me exemplify this.

First, remember that α-D-glucopyranose, β-D-glucopyranose and the aldehyde form of D-glucose are three different compounds with different physical properties. The α and β forms are diastereomers, more specifically epimers, more specifically anomers of each other and e.g. display different optical rotation. The chain-form of glucose is different altogether. I will construct all further discussion of this answer in a way that it does not matter whether we are dealing with α or β-D-glucopyranose, so you may always assume a mixture.

Now let’s consider the absolute configuration of carbon 3 in both the chain and pyranose forms.

• Pyranose

1. Obviously, the oxygen atom has highest priority.

2. Obviously again, hydrogen has lowest priority.

3. We have two carbon atoms of the (b) and (c) priorities; initially undistinguishable. Thus, we must go down one level.
Both carbons are attached to the atoms (in order of priority) $\ce{O, C, H}$. Again, this is a tie; we must go down one level.
Both oxygens are attached to a hydrogen; this is identical and the end of the sub-chain. We must go back to the next, lower priority of the former atom.
The two carbon atoms differ: one is the acetal group which expands to $\ce{O, O, H}$. The other is connected to $\ce{O, C, H}$ (carbon 5). Thus, we have resolved the issue.
Going towards lower carbon numbers is the higher priority.

Therefore, carbon $\ce{C{3}}$ has an (S) configuration.

• Chain form

1. Again, oxygen obviously has highest priority.

2. Equally again, hydrogen obviously has lowest priority.

3. The two carbons are initially equal.
Both are again attached to $\ce{O(H), C, H}$.
The two subsequent carbons differ again. This time, one is the aldehyde group which expands to $\ce{O, (O), H}$, the other is $\ce{C{5}}$ which expands to $\ce{O, C, H}$. Since a ghost oxygen atom has a higher priority, going up the Fisher projection has the higher priority.

We arrive at the same (S) conclusion as before.

Now, we will construct the same discussion for carbon $\ce{C{4}}$.

• Pyranose

1. Obviously, the oxygen atom has highest priority.

2. Obviously again, hydrogen has lowest priority.

3. We have two carbons, initially indistinguishable again. We go down a level.
Both carbons ($\ce{C{3}}$ and $\ce{C{5}}$) expand to $\ce{O, C, H}$. We must go down a level.
One of the two oxygens is connected to a carbon atom. Thus, $\ce{O(C), C, H}$ wins over $\ce{O(H), C, H}$.
Going towards $\ce{C{5}}$ has a higher priority.

Thus, carbon $\ce{C{4}}$ is (S) configured.

• Chain form

1. Bla, oxygen.

2. Bla, hydrogen.

3. Blabla, two carbons. Again, we descend a level to $\ce{C{3}}$ and $\ce{C{5}}$.
Both carbons expand to $\ce{O, C, H}$. We descend a level.
Both oxygens are attached to a hydrogen and nothing else. They are indistinguishable.
The two carbons we are now examining are $\ce{C{2}}$ and $\ce{C{6}}$. The former expands to $\ce{O, C, H}$, the latter to $\ce{O, H, H}$. Therefore, Going towards $\ce{C{3}}$ has a higher priority.

Thus, carbon $\ce{C{4}}$ is (R) configured.

Note that while the configuration of $\ce{C{3}}$ did not change, $\ce{C{4}}$’s absolute configuration changed from (R) to (S) upon formation of the pyranose ring. Thus, you cannot rely on absolute configurations to verify a correct Fischer to Haworth transformation for carbohydrates!

• Hi Jan, Could you please clarify this comment "In a Haworth projection, the hexagonic shape of the ring already implies the direction of two bonds. The two bonds on the ring are always ‘bent ring-inwards’ when viewed from the outside. Thus, the two perpendicular bonds (typically vertical) must ‘point away’ from the ring: in a forward direction for the front carbons of a ring and in a backward direction for the rear side of the ring.@ Which two perpendicular bonds are you referring to? Thank you. – K-Feldspar Aug 1 '17 at 20:57
• @K-Feldspar There is a six-membered ring which is drawn lying down in perspective as if going into and coming out of the plane. The bonds of this ring are assumed to be in the direction of the ring, so if you are looking at $\ce{C{2}}$ from outside the hexagon, the bonds to $\ce{C{1}}$ and $\ce{C{3}}$ are pointing behind you. Thus, the bonds that are drawn perpendicular to the plane of the hexagon (i.e. up and down) are pointing towards you. For a carbon atom on the back of the hexagon (behind the paper plane; e.g. $\ce{C{5}}$), the situation is reversed. – Jan Aug 7 '17 at 2:46
• Ah! Terrific. I assumed the vertical bonds where just straight vertical based on the way it was drawn. But this makes sense, it is just the opposite of how it is for a Fischer. I.e. horizontal = in to the page rather than out of the page for Fischer, and vertical = toward you (for a front carbon) rather than away from you for a Fischer. Thank you for explaining that Jan. – K-Feldspar Aug 7 '17 at 11:18
• @Jan "the bonds to C1 and C3 are pointing behind you" Did you mean pointing away from you instead? – Gaurang Tandon Mar 3 '18 at 14:34
• @GaurangTandon A long time ago™ and bad memory™, but I think yes, I used you instead of the paper plane for some reason … – Jan Mar 9 '18 at 1:06