Lesson 6 video 4

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Lesson 6 video 4

Postby ArielM » Wed Nov 15, 2017 7:45 pm

This is a simple question to something that I have never truly understood. I guess I have just gone along by and by in my organic chemistry classes, blindly believing what my teacher says without actually truly understanding what he was talking about, specifically with regards to Diagnosing Isomers.

In Lesson 6 Video 4, Carbons 1, 2, 3, and 4 are labeled R, S, R, and R, respectively.
1. I understand that we assign priority to the ligand with the highest atomic number.
2. I understand that priorities are assigned as 1, 2, and 3 by starting with (1) the ligand with the greatest atomic number and ending with (3) the ligand with the smallest atomic number.
3. Based on this "1,2, and 3" numbering system, I understand that the directions ("clockwise" and "counter-clockwise") are given in accordance to the respective increase in the numbers (i.e. 1,2,3, going to the right = "clockwise", like a clock; 1,2,3, going to the left = counterclockwise).
4. After the numbers have been assigned and the direction of the molecule's chirality has been elucidated- I understand that the lowest priority substituent is placed in the back of the molecule (usually a hydrogen atom).

I understand the numbering and the designation of directions for the carbons numbered 1 and 2, and designated R and S, respectively.

However, I do not understand the numbering and directions given for carbon atoms that are numbered 3 and 4.

I understand that, for the carbon labeled "3", the -OH substituent is given the number 1 priority due to the oxygen atom's atomic number of 8, which is the highest atomic number of any atom found on this respective chiral carbon. However, as I travel down the vertical lines of the molecule, carbon by carbon, I see that if I go out by one carbon in each direction, each carbon atom that I encounter carries an equal atomic number, 6.

Therefore, I take another step, in both directions, to the next two atoms on the D-glucose molecule. I see that, again, I encounter two carbon atoms with identical atomic numbers. This is inconclusive; therefore, I travel a third time to the next two atoms on the molecule. In the northbound direction on the D-glucose molecule, I encounter another carbon atom, C2, (or for this particular diagnosing question, this is the carbon numbered "1" and given the "R" designation). In contrast to the C2 carbon, the base carbon, (I suppose this would be the "C6", or the 6th carbon in the entire D-glucose molecule), is attached to an -OH group. As we know, Oxygen has a higher atomic number (8) than that of Carbon (6).

Because of this, I am convinced that carbon 3 should be designated as S. Why should I NOT say, "the "O" in -OH found on the base carbon has a greater atomic number than that of the C2 carbon (assigned as carbon number "1" and given the "R" designation); thus, I will assign the base carbon attached to -OH with second priority and the C2 carbon, that seems just like all the other carbons that I encountered with this question, with third priority", rearrange the molecule by putting the Hydrogen in the back of the molecule, then designate the molecule as "S", instead of "R"?

The same phenomenon perplexes me when I try and diagnose the fourth carbon. Why am I wrong?

I have attached the slide for this video to this message for reference.
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Re: Lesson 6 video 4

Postby NS_Tutor_Clara » Fri Nov 17, 2017 12:23 pm

Hi Ariel,

Great question! It sounds like you have a very good understanding of the fundamentals here – but this is a difficult example, and students often get tripped up by questions like these.

You're entirely correct through this statement that you made:

In the northbound direction on the D-glucose molecule, I encounter another carbon atom, C2, (or for this particular diagnosing question, this is the carbon numbered "1" and given the "R" designation). In contrast to the C2 carbon, the base carbon, (I suppose this would be the "C6", or the 6th carbon in the entire D-glucose molecule), is attached to an -OH group. As we know, Oxygen has a higher atomic number (8) than that of Carbon (6).

Let's stick with the typical numbering conventions we'd use for a glucose molecule, just to keep things simple. So, we're comparing C2 with C6. You are absolutely correct that C6 is attached to an -OH group. However, C2 is also attached to an -OH group, shown immediately to its right. Since both carbons are attached to an -OH, but C2 is also attached to the carbon above it (C1), the substituent leading up to C2 is given higher priority than the substituent leading down to C6. This results in the prioritization, and overall R configuration, described in the lesson video. I've attached a rough drawing of this situation, with positions that are directly being compared described in the same color.

The main takeaway is to look at every substituent attached to a given position - rather than getting too used to looking either "up" or "down" the Fischer projection, and missing substituents depicted horizontally. I believe this was probably the issue with that last carbon as well.

Please let me know absolutely anytime if I can help with anything else!

Lesson 6 part 4 - isomers.png
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Re: Lesson 6 video 4

Postby ArielM » Mon Nov 20, 2017 9:25 pm

Hi Clara,

Thank you for taking the time to explain this question. What I understand is that the C2 and the C6 carbons both share -OH groups.
C6 is attached to an -OH group. However, C2 is also attached to an -OH group.

So at this point we cannot determine which carbon should get the greater priority because, at this point, they both share the same substituents. However, the difference between the two carbons is that C2 is attached to a Carbon that is with an extra aldehyde group, is this correct?

The main takeaway is to look at every substituent attached to a given position.

Okay, just a quick question, could we say that the entire C2 substituent could be considered an ethyl dioxide group and that the C6 substituent could be considered a methane hydroxide group? Therefore, ethyl dioxide is a greater substituent than methane hydroxide and thus, the greater priority goes to the carbon attached to the greater substituent? I've included an attachment.

-Ariel Morrow
Lesson 6 part 4 - isomers.pdf
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Re: Lesson 6 video 4

Postby ArielM » Fri Jan 05, 2018 6:54 pm


I just have a few questions about isomers again. For some reason, I just do not feel comfortable with this topic.

I was watching Lesson 6 Video 10, a later video (I have included an image of the slide). On the molecule featured furthest left, the C1 carbon is rotating counterclockwise in the S-configuration. The C2 carbon (immediately to the right of the C1 carbon), is initially rotating in the clockwise direction (the R-configuration); however, when we invert the molecule (sending the H+ atom into the plane), the rotation switches from the R-configuration to the S-configuration (or you could also say, from clockwise to counter-clockwise). So after inverting the C2 carbon, C1 and C2 are now both rotating in the S-configuration.

My questions:
1) Is this always the case when you invert a molecule? If a molecule is rotating in the S-configuration (counterclockwise), does inverting the molecule result in the opposite rotation of the molecule? (I.e. A molecule first rotating counterclockwise in the S-configuration will, after inversion, rotate in the opposite direction, or the R-configuration (clockwise)?) I hope my question makes sense. I just want to know if this is a trend or a pattern? Is safe to assume that if you invert a molecule, it will rotate in the opposite direction? Please correct me if I am wrong.

2) Also, it seems to me that in the right-most molecule, C1 and C2 are both rotating in the S-configurations. I do not want to challenge Dr. Anthony in any way whatsoever, but I just do not understand why it is R and not S, could you please help me?

3) And if possible, if you wouldn't mind, could you please provide a brief explanation about the difference between the S & R configurations and the L & R configurations?
Thank you for your patience.

-Ariel Morrow
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Re: Lesson 6 video 4

Postby NS_Tutor_Andrew » Tue Jan 09, 2018 10:07 pm

Hi Ariel,

Before getting into a more detailed response, I think it might be useful to step back and make sure that the terminology is clear. I wasn't quite sure what you meant by "rotating" in the S configuration. Although we might mentally do some rotation to help us visualize the molecule more clearly and to correctly apply the priority rules needed to determine R vs. S, the molecule itself isn't actually doing anything -- R vs. S is just a way of formally specifying how a given molecule is oriented in space. I would highly, highly, highly recommend buying a used molecular model kit or just DIY-ing your own simple molecule of a tetrahedral molecule to help visualize all of this concretely.

The distinction between R vs. S is known as absolute configuration, because if we follow the rules of determining R vs. S correctly, we have a single, universally true description for how the substituents are arranged around a chiral carbon. This is an important misconception to address. The only way to change from R to S would be to literally break and reform the bonds. Rotation won't do the trick.

So to recap the rules for assigning R vs. S:

(1) Assign priorities to all 4 substituents using the Cahn-Ingold-Prelog rules.
(2) Rotate the molecule mentally so that the lowest-priority substituent faces into the page. This is usually H, but doesn't have to be.
(3) Draw a circle connecting the top-priority substituent to the second-priority substituent and the second-priority substituent to the third-priority substituent. If you go clockwise to do so, the orientation is R, while if you go counterclockwise, it's S.

The whole concept of "inverting" the molecule has to do with a shortcut that you can use in some situations. Take a look at the image that I've attached:
R and S sketch.png
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. The key thing here is to realize is that if you look at the molecule from the opposite side (i.e., going from having the weakest substituent face into the page to having it face out of the page), you will switch the direction in which you connect the substituents. The molecule is the same, but you're looking at it from the opposite direction.

Therefore, imagine that you just have a tough time visualizing the molecule with the lowest-priority substituent facing into the page, but it's easy to imagine it facing out of the page. You can go ahead and connect the substituents, but you have to correct for the fact that you were viewing it from the "wrong" side for the purposes of R/S determination. So, for instance, if we have the lowest-priority substituent facing out of the page and we connect the substituents in a clockwise direction, we have to realize that we need to flip that to determine R/S -- if we were looking at it from the opposite side, with the lowest-priority substituent facing into the page, the substituents would be connected in a counterclockwise direction, so the configuration must be S.

I hope the sketch I've attached will help, but if not, draw this out yourself or use a 3-dimensional tool to practice this until you "see" it.

Now, to turn to your questions:

1) I think the above discussion answers 1), but keep in mind that all that's happening is that you're looking at the same molecule from the opposite side. We're not switching from R to S or vice versa. The absolute configuration is absolute; the only thing that changes is how it relates to a clockwise vs. counterclockwise connection of the substituents given how they are oriented.

2) The right-hand molecule is shown in a Fisher projection, where by convention the horizontally placed bonds are understood to be pointing out of the page. Therefore, we have to apply that "inversion" trick. In this context, counterclockwise connectivity = R configuration because we have to "flip" it to account for the fact that R and S have to be assigned from the perspective of the -H pointing into the page.

3) This is potentially a big question, but I'll try to keep it concise. R and S are descriptions of the absolute configuration of the molecule's structure. The terms levorotatory and dextrorotatory (l and d; technically with a small l and a small d, although care is often not taken) refer to the direction that a chiral center rotates plane-polarized light, and is not directly predictable from a molecule's structure. L and D (with small caps, technically) are used in carbohydrate chemistry, based on the orientation of the chiral center furthest from the carbonyl group (see more details here).

Hope this helps clarify things!
Andrew D.
Content Manager, Next Step Test Prep.

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