There are several common projections used to visualize molecules in different perspectives.
Concept: Introduction to different projections.2m
So like I said, like Newman projections, there's actually a lot of different types of projections. As you can see, this one right here is called a Fischer projection. It's used mostly for sugars. So later on if we get into sugar chemistry and carbohydrates, we're using Fischer projections a lot. But they're also used in this chapter as well.
Two other common ways that are used are Haworth projection. The Haworth projection is just that 3D projection of a ring, that's the actual name, and it's usually used for rings. It's used to highlight what's at the top and what's at the bottom.
Then finally we have this one called a sawhorse projection. And this one's usually used for stereochemistry. It will basically say how these atoms are related to each other in terms of their orientation and their shape, their configuration.
Well, in all of these cases, whichever projection we're using, also remember that there's Newman as well, in all these cases we're going to have to convert them into bondline before analyzing them completely. What that means is that these projections are really good for analyzing certain types of things, but if we want to compare them against other normal molecules, we're going to have to convert them into bondline first because that's our metric system. That's our standardization.
We will need to know how to convert these into bondline so that we can analyze them.
Concept: How to convert Fischer projections into bondline structures6m
So basically you can see I have make a caterpillar, what the hell am I talking about. Well, really, a Fischer projection, even though it looks 2D, it's not really 2D. The way that it's really supposed to be interpreted is that every vertical bond is going into the page on a wedge. So these would be vertical and vertical. And then every horizontal bond coming off of it is a wedge. So those were dashes, I'm sorry. I might have said wedge. Those are dashes. And these are supposed to be wedges. So even though it looks 2D, it's really not 2D. That's just the way it's drawn to make it easier. But really the way it's supposed to be interpreted is with wedges and dashes.
So what I asked you guys to do is if we're going to convert this into bondline, we need to use this wedge and dash notation first. So what I say is if you're given a bondline, first, do what I just did and make it wedge and dash, like I just did. Then we're going to use an eyeball and we're going to pretend like we're looking at it from the side. What we're going to see is that it's going to make what I call a caterpillar.
Basically what that means is that this CHO here would be right here CHO. This CH2OH would be here. CH2OH. And what we would notice is that we have three different junctions: 1, 2, 3. And these are the places where bonds come off the top. If you think about it, maybe draw a line down here, this actually kind of looks like a caterpillar. Now this just got really goofy, but it kind of looks like a caterpillar. It's like on a leaf and it's like eating away and it's like munching away and it's got it's little hair sticking up. That's why I call it a caterpillar.
Basically, I don't want you guys to necessarily draw the whole caterpillar. You don't need to draw a face or anything. I'm not going to be strict about that. But what you should do is realize that this bond here, 1, relates to 1 right here. So what means is that then I look at the eyeball and I say according to that eyeball, what should be in the front and what should be in the back. Where does it seem closest to itself?
And what it's saying is there should be an H in the front because that's the closest one to the eyeball and there should be an OH in the back because that's the furthest one from the eyeball on the 1 carbon. So what I'm referring to here is that I have this H here and then I have this OH here. Is that making sense? So I'm looking at the 1 carbon and saying what's in the front, what's in the back.
In the same way I would work with the other ones. Then I'd say 2 should have an H in the front and an OH in the back. And then 3 should have an OH in the front and an H in the back. Is that making sense so far?
Once I have my caterpillar, then I have to do my last step and that's going to give me my bondline. So we're actually really close to the bondline. The thing is that bondline structures, are they ever like that where all the bonds are in a straight line? Usually not. Usually there's a zigzag pattern. So we need to restore this back to its zigzag pattern.
How do we do that? By rotating every other bond. Another thing I like to say is that by rotating every even bond. So what we're going to do here is I'm going to show you guys how to do this. Basically what we would do is we have 1 and 2 and what we would do is we'd rotate every other bond. So we would rotate 2 is my atom. I'm going to want 2 to face down. And if we rotate 2 to face down, that's going to restore my zigzag.
So now what I'm going to do is I'm going to draw this like this where I have 1 is here, 2 is here, 3 is here and then it goes down like this, so then the CHO is in the same exact place. The CH2OH is in the same exact place. In fact, 1 and 3 are in the same exact place. Notice that they were both pointing up before. 1 and 3 were both pointing up.
So that means that the groups that are on 1 and 3 should look exactly the same. That means that 1 should have an OH at the back and it means that 3 should have an OH at the front. Do I have to draw the H's? No, because this is bondline. Remember in bondline H's are omitted.
So far, so good. In fact, so far this is exactly the same molecule the only thing that's changed is that now I'm rotating 2 down. That means that whatever I had on 2, has to flip. That means that where should the OH go? Should it go on the back, the front, where should it go? Since it's rotating, the OH should now go on the front. Because of the fact that it rotated down, that means that that bond that was in the back is now going to rotate to the front.
So now what I've just done is I've just made my bondline. That is a bondline structure right there. All I did was I rotated every other bond, so meaning that I rotated this one. This one doesn't get rotated. See, this one is fine. That one doesn't get rotated. But then this one got rotated and then this one didn't. So notice that every other one I rotate. If this was a longer chain, then I would have also rotated atom 4 to go down. So I would have rotated atom 2, atom 4, atom 6, until my Fischer projection is done. And that would make the zigzag pattern.
So what I want you guys to do is just as free response convert the following Fischer projection into a bondline structure. Go ahead and try and solve it yourself and then I'll go ahead and step in and show you guys how to do it.
Concept: Convert the following Fischer projection into bondline structure.4m
Concept: R and S rule for Fischer Projections.3m
So this is basically the way it works. We want to determine the location of the lowest priority group. What I mean by location is we want to figure out is it vertical or is it horizontal. And the lowest priority group is always going to be 4. So basically I just want to know where is 4.
If 4 is vertical, if 4 is faced vertical then the chirality is exactly as it looks. I'm just going to go from 1 to 3. I'm going to draw that arrow and that's going to be it. Whereas if 4 is horizontal, then the chirality is just going to be flipped. So whatever you draw, you're just going to take the opposite sign.
So here I have two examples. This would be an example where H is vertical. So notice that my 4th priority group is vertical so that means that it's going to be as it looks. So I'm just going to say 1 to 2 to 3. It's going in an S direction so that's actually going to be the final answer. The final answer for this chiral center would just be S. Does that make sense so far?
Now let's look at this next one. Here's my chiral center. 4 as you can see is now horizontal so that means it's going to be flipped. So once again, I'm going to do 1 to 2, 2 to 3, 3 to 1. It looks like S, but it's actually going to be R because of the fact that it's horizontal.
Now keep in mind, for Fischer projections I'm never swapping groups. So I'm not swapping out 1 and 4 or anything like that. All I'm doing is I'm just taking the chirality and I'm flipping it if it happens to be horizontal or I'm keeping it if it happens to be vertical. In some ways, this is actually easier than what we learned for the other compounds. And when you have a very big Fischer projection that has a lot of chiral centers, you're going to be thankful that you have this method that you can just go really, really fast and figure it out.
So what I want you guys to do now is go ahead and determine the absolute configurations meaning the R and the S for this Fischer projection using this formula and using this format. Then when you're done, I'll go ahead and show you guys how to do it.
Determine location of lowest priority group:
Concept: Determine the absolute configurations for all chiral centers.6m
Draw structural formulas for the three compounds shown below. Make sure to include all pertinent stereochemical information.
Convert the following Fischer projection into a wedge/dash drawing.
Draw Fischer projection formulas for all stereoisomers of 2,4-dimethyl-3-hexanol, giving stereochemical details for each structure. As a bonus, include which compounds are enantiomers of each other and which are diastereomers.
Convert each of the following to a Newman projection along the C2-C3 bond.
Draw the Fisher Projection for the following compound.
Draw the Fisher Projections for the following compounds.
Convert the following to a Newman projection along the C2-C3 bond.
Draw the Fischer projections for the stereoisomers of
Clearly label pairs of enantiomers & diastereomers, if any. Drawings must be clear & correctly labeled. (You do NOT need to name the compounds)
Which statement about these Fischer projections is correct?
a) I and III are enantiomers.
b) II is a meso isomer.
c) II and IV are identical.
d) I and II are diastereomers.
Which of the following structures is a correct wedge/dash drawing of the following Fischer projection?
Convert the following bond line formula into a Fischer projection.
Draw the Fischer projection of the following compound.