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
Alright guys so let's go ahead and go step by step, the first step would be to redraw this with the wedge and the dash so I would put this on a dash and I would put this on a dash, I would put these guys on wedges, B-R, B-R, H H is that cool so far? Now what I want to you is I'm going to draw my angle because that's going to help me remember what things look like, OK? Notice that I have carbon 1 here carbon 2 here, these are the back of the caterpillar so now when I convert this into a caterpillar what it's going to look like it's like this? Where I have atom 1, atom 2 that's the back of the caterpillar what it's going to have here is COOH, what it's going to have one over here is CH2 and H2 now I just to figure out what's on the top? What's on the hairs, OK? So it should actually be really simple I should just have for my wedges I should have H and H, OK? Because those are the ones that are closest to the eyeball. For the back what I should have is BR BR, is that making sense so far? Cool, now I think a question that some of you guys might have is that I notice that before up here I was drawing the Hs on the right side and the front the wedge on the right and the dash on the left and then here I was drawing the wedge on the left and the dash on the right, it does not matter you can draw them however you want as long as the thing that's in the front is still in the front, the thing that's in the back is still in the back, alright? So now we have our caterpillar so now how do we convert this into bond line? All I do is I rotate every other bond or what I want to make sure every other atom is faced down, every even atom is faced down so what that means is that it's going to be atom 2, OK? So atom 2 is going to be the one that has to face down, OK? So let's go ahead and convert this what that means is that now this is going to turn into this, this and that where this is now atom 1, this is atom 2, OK? So now I'm going to actually subtract myself from here take myself out so that I don't get in the way so now what should be coming off of 1? What should be coming off of 1 is the same thing as before, I should have the H in the front but I don't have to draw Hs so I'm just going to draw the BR in the back, is that cool? So what should be coming off of the one that's on the left of 1, the one right here what should be coming off of there? Well what I should have there is just the COOH just like before, OK? Cool, what should be coming off of the carbon over here? This should be the CH2 and 2, OK? So then finally what should be coming over right where my head should be? What should be at the 2? Well because of the fact that 2 had to switch down that means that the BR has to switch its location so that means that may BR should actually face towards the front and there you go that is bond line structure, OK? So basically this is how you make a bond line structure from a fisher projection, notice that you can't just do it in one step you have to do multiple steps to get here but if you're consistent I think you guys will be able to do this just fine.
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
Alright guys so this wasn't an easy problem this is actually a pretty complicated problem if you got it right power to you, OK? But this is not easy so first of all here's my first Chiral center and I need to recognize what these substituents are, I need to figure out priorities so my first priority obviously has to be my oxygen because oxygen is the higher atomic weight but then what I have here is carbon and also that's my blue carbon and my red carpet and my blue carbon, OK? Which one is going to get higher priority? well in order to understand this I need to know what the CHO functional group is, do you guys remember what that is? I hope you guys remember that is an aldehyde so what that actually looks like I'm just going to scratch this out is like this, double bond O and H, OK? That's what CHO means I told you guys when I taught you all those functional groups that you have to memorize that, OK? So there you go that's your that's one functional group the other one is a carbon with a chlorine and oxygen so if you can't immediately tell which one is going to win you have to do the playoff system and if you do the playoff system what you would find is that the C that's in blue would be attached to an O and O and an H whereas the C that's in the red would be attached to a CL, an O and a C, OK? So which one is going to win just the one that has the heaviest atom on it so red has to win because red has a CL on it so this one is going to be 2 then this is going to be 3 and then finally this would be my 4, OK? So that was a little bit tricky, OK? Now I'm going to say OK is my 4 vertical or horizontal? It is horizontal is that means whatever I draw has to get flipped so I'm going to go ahead and go 1 to 2, 2 to 3, 3 to 1, this looks counterclockwise but really this is going to be clockwise, OK? So that's my first one it has a lot of work let's move on, OK?
So let's go on to this next Chiral center so for this next one I have that this is going to be my one, because chlorine is heavier than oxygen this is my 2 then I have 3 and 4 which is this blue carbon and this green carbon, OK? So which one is going to win there? Once again I mean this one you're just going to have to do playoffs I'm going to do blue that C what is that attached to? It it's attached to an O, it's attached to a C and it's attached to an H. Let's go down to this green one, this green one is attached to an O, it's attached to a C and it's attached to an H so which one is going to win, I actually don't have a winner yet they're both the same so what happens if I get to a playoff and they're both the same I have to keep going down the longest Carbon chain, alright? So that means that now I'm actually going to go to each carbon that it was attached to so notice that this blue was attached to another carbon that's this carbon, now I'm going to say what are the 3 things that that carbon is attached to? well we already did this one it's attached to O, O, O and H, the reason I'm counting O twice is because that is an aldehyde so it has a double bond O, I'm going to do the same thing for this C this has to do with this C so this C is attached to what? It's actually attached to just an O and an H and an H, why is that? Because it's a CH2O, OK? So now do we have a winner? Yes we do, now we have a winner this one has to be 3 and this one has to be 4 so it's going to be 3 and it's going to be 4, OK? So that one like I said this one is just a really tricky question.
So now is my fourth priority vertical or horizontal? It is vertical, OK? So what that means is that I'm just going to draw is exactly how it looks 1 to 2, 2 to 3, 3 to 1 it looks like an S and it is an S, OK?
So now let's go on to this last one, this last one would have this is my 1, this is my 4 and we're already did this before in terms of we know the CL is going to be better so this is my 2 and this is my 3, is my last priority horizontal or vertical? It's horizontal so I'm going to flip whatever I get 1 to 2, 2 to 3, 3 to 1 it looks like an S but actually it's an R, OK? And those are my 3 Chiral centers, now I know this seems like it took forever but it wasn't because of actually figuring out the rotation, the part that took a long time was that this was just a hard problem in terms of figure priorities so no matter what strategy you use this is going to take forever, OK? It's going to take a while but I think that way you're going to notice is that using the horizontal and vertical thing is going to save you guys a lot of time and just you know you can't there actually is no alternative if you don't want to use this horizontal vertical thing the only other alternative to get it right every time is to convert it completely to bond line and then do it from there and that's just like I said that's a bitch that's a massive bitch that would take forever so this is going to be a lot easier hopefully save us a lot of time, alright? So let's go ahead and move on to the next topic.
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.
Draw the Fisher projection of (2R,3S,5R)-5-bromohexa-2,3-diol.
Name and draw two diastereomers of the above compound.
Name and draw the enantiomer of the compound.
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.
Fischer projections are a 2D representation of the atomic connectivity and stereochemistry of 3D molecules. Originally developed in 1891 by Emil Fischer, they help visualize carbohydrates, amino acids, and other compounds in Organic and Biochemistry.
Here we’ve got Fischer projections of L- and D-glucose. They’re enantiomers of each other, and there are plenty of other stereoisomers of the monosaccharide.
The minimum variation required to be considered a stereoisomer is to swap the horizontal groups in one chiral center, so that means that glucose has tons of diastereomers.
Galactose, shown above, differs from glucose in the configuration of just one chiral center. It's got the same chemical formula and atomic connectivity as glucose but different stereochemistry, so it's a diastereomer.
bondline structures to Fischer projections and vice-versa using the caterpillar method as seen above. Double checking your Fischer projection’s R and S to see if it’s been drawn properly is always good practice!It’s fairly straightforward to convert
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