Let's talk about how to properly draw monosaccharides using a representation that was designed specifically for sugars and that representation is called the Fischer projection. So, guys in 1891 Emil Fischer who was the most prolific organic chemists of the 19th century, he's essentially the guy has discovered carbohydrates and discovered carbohydrate chemistry, he devised a representation called Fischer projection specifically for the purpose of depicting carbohydrates. So, even though you've learned about Fischer projections previous to this topic, this is really the reason he design them, it was so that he could better understand and better illustrate the chirality of sugars, okay? Now, it's going to be really important, because of this it's getting very important that we understand how to go from bond lines of Fischer using, with sugars, and also how to go from Fischer to bond line, okay? Now, one thing that you need to know right away is that the way that you're supposed to align your monosaccharides, your Fischer projection, is with the most oxidized carbon, the most oxidized atom on the top of your Fischer projection. So, here what you see is that I have this monosaccharide that's represented on the left as just a bond line and it is much more difficult to deal with the monosaccharide in that form, it's much more difficult to illustrate the chirality to compare it to another one. So, Emil Fischer said, why don't I draw them like this, that's going to make them so much easier compared to each other because you just see all the OH's, you can compare OH's much easier this way, and what you see is that this top carbon would be considered more oxidized because it has two bonds to O and this bottom one would be considered less oxidized because it only one bond to O, okay? So, you always put the most oxidized part on the top, okay? So, this is D-ribose, we talked about ribose already before but now you know that that's actually what the bond line looks like, okay? So, bond lines of Fisher, this is the first essential tool you need, okay? So, similar to our lessons in Orgo1. Remember, that whenever we wanted to turn a bond line into a Fisher, we use this Johnny patented method called the caterpillar method, right? And what the caterpillar method taught us is that what we want to do is we want to rotate all the bonds so you have essentially at the back of a caterpillar. So, where does the caterpillar come from, I think of this as being a caterpillar and this is like its face, it's just like happy chewing on a leaf or something, and then those things at the top are like the hairs on the back of the caterpillar. So, you can go back and learn about the caterpillar method there but I'm just going to remind you guys that what the caterpillar method says is that every substituent that's already facing up stays the same. So, we notice is that carbon 2 stays exactly the same, the OH is faced towards the back here and it's still faced towards the back but in order to make the other substituents face up, in order to rotate them to the up position you also need to rotate the wedge and dash information so that means that in order to rotate this OH to face up I need to face it towards the back on carbon one and on three I need to face it towards the back as well because it rotated, okay? Then once you have it lined in that way you can easily draw your Fischer projection because then all you need to do is look at it, think, okay, in the front, what do I see? I see H, H, H. So, that'd be this one, this one, this one and that would be on this side and then on the back what do I see OH, OH, OH, and that would be back here, okay? So, that is the most surefire way to get it right, okay?
Now, alternatively once you get good at this method, what you'll notice is that all you really have to do as a shortcut, is swap the stereochemistry of all downward facing alcohols, okay? So, what that means is that if I wanted to kind of take a shortcut and bypass the Caterpillar and go straight to the Fischer how could we do that? let's see, can we actually do this? Well, what you'd say is here's my eyeball, I'm looking at it this way, okay? And, what I see is that on the 2 carbon, I have an OH faced away from me, okay? This one right here, I'm going to make it yellow. So, I would already put it face away from me on this side as if my eyeball was looking this way, okay? So, let's say, well it's faced away, those two red eyeballs are in the same position just with different representations. So, it's faced away. Now, on the red eyeball on the left it looks like those OH's are faced towards me, so it looks like I should put them on this side of the Fischer projection but no, they're faced downwards. So, if it's down you have to flip it. So, those two OH's on one and three they look like they're going to face this way but I actually because they're faced downwards I have to swap them to this side, meaning that this one here and this one here even though they're both towards me, towards the eyeball, I have to draw them away from me because they swapped, okay? So, it's a kind of just a little shortcut, the downward facing positions you would swap, cool. Now, go from Fischer to bond line. So, from Fischer to bond line the most, again surefire way, the most reliable way to convert it would be to use a reverse caterpillar. So, I'd say, hey, we've got this molecule, this is what it looks like from my eyeball here, what was the caterpillar look like? Well, the caterpillar would be a straight back with positions 1, 2 and 3 and everything that's closest to the eyeball, you can't really see the whole eyeball, sorry, let me draw it again, it's better, okay? So, everything that's closest to the eyeball should be on a wedge so that would be this group here, this group here, should be here and here, right? And everything that's far from the Iball should be on a dash, especially this one here, okay? So, that is the reverse caterpillar, I just made a caterpillar. Now, how do I turn this into a bond line? Well, turns out that when you do that you're going to get two possible answers, they're both the same molecule but they're just rotated differently because it just depends if I'm starting off with a zigzag that starts going up or a zigzag that starts going down you're going to it two different possibilities, okay? So, let's start off with the one that is going up, okay? So, the one that's going up, what I would do is I would just draw this aldehyde exactly the same and I would say at this position 1, 2, 3 and then this is your CH2OH, are you guys fine with that? this CH2OH is this guy right here, okay? Similarly over here, my aldehyde would now face up. So, this would be positioned 1, 2, 3 and then this is my CH2OH where it's just face down now, okay? Now, how would we actually drawn the OH's? Well, notice you would only need to flip the ones that are faced down now. So, in the first one, which positions can stay exactly the same? which positions are still facing up? one and three. So, what I could is I could draw one as a wedged OH and I could draw three as a dashed OH. why? Because those are the same exact positions that they were in on the caterpillar, by the way, I'm not going to draw H's because they can be implied, right? Now on two, two is facing downwards. So, now I need to flip it, meaning that two should actually have a dashed OH and now I'm done with that molecule, it's done, cool? Awesome. Now, let's see how it would differ with the second one. So, in the second one, which position stayed the same in terms of they're still facing up? Now it's two, so, two I can face the same direction. it should be an OH that's wedged. Now, one and three because the way it drew my zigzag. Now, one and three have to be changed. So, now one would be a dashed OH and three would be a wedged OH, okay? So guys, even though these two molecules look very different they're the same exact molecule and both of these answers would be correct answers on a test, your professor would give you full credit for both of them. The reason I draw both of them is because if I drew the first and you do the second one, I'm going to get so many questions of, Johnny is mine the same as yours. So, I want to make sure to draw both ways just that you know, you're always going to get two different looking answers but both of them are the same if you just use the correct rules, alright? So, now we know how to draw Fischer projections and how to also kind of undraw them and go back to the bond line, let's go ahead and do some practice related to Fischer projections.
Problem: Convert the following monosaccharide into its Fischer representation. Is it a D or L-isomer?5m
Problem: Convert the following monosaccharide into a bondline representation.5m
Draw the Fischer projection of the following sugar.
Identify the anomeric carbon of the following sugar.
The configuration of R-(+)-glyceraldehyde is as follows: What is the absolute configuration of (-)-lactic acid?
A) D and L configuration
B) L configuration
C) R and S configuration
D) R configuration
E) S configuration