- Nucleic acids, lipids, and carbohydrates questions
- Nucleic acid structure 1
- Antiparallel structure of DNA strands
- Saponification - Base promoted ester hydrolysis
- Lipids - Structure in cell membranes
- Lipids as cofactors and signaling molecules
- Carbohydrates - Naming and classification
- Fischer projections
- Carbohydrates - Epimers, common names
- Carbohydrates - Cyclic structures and anomers
- Carbohydrate - Glycoside formation hydrolysis
- Keto-enol tautomerization (by Sal)
- Disaccharides and polysaccharides
Fischer projections are a way to represent three-dimensional molecules in two dimensions. By following specific rules for drawing these projections, one can depict complex carbohydrates such as glucose and fructose in a way that conveys their structural information. Created by Jay.
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- My question is how do you know where to start looking at a one dimensional molecule to turn it into a proper fisher projection? Particularly with the double chiral center molecules. Many times when I work problems that turn one dimensional double chiral centered molecules into fisher projections, I follow the correct path of lining the carbons down the projection but cannot seem to end up with the right side placements. It seems like each one needs a different technique to align to the correct R or S config. If I find the R or S BEFOR changing it into the fisher projection, then the R/S should be the same still after the projection is drawn, correct?
Thanks a lot,
- Excellent question, Luke. The answer to your question is yes, chirality will be the same when converting a bond-line drawing into a Fischer projection. You are actually getting the correct answer without knowing why! Don't worry, you're not the only one confused by this. In Fischer projections, the convention is that the lines going up and down on a page are going away from you in space (into the desk below the page), and the lines going left and right are coming out toward you (as if to hug you). Remember each chirality center is determined (R) or (S) individually. So too that in a Fischer projection, each chirality center is drawn individually. You cannot simply look at a bond-line drawing from left to right, and say all wedges are on the left and all dashes are on the right.
Here is a trick: The sp3 hybridization of a carbon atom bonded to 4 other atoms makes the most accurate representations of bond-line drawings look like a jagged line or a series if peaks and valleys (/\/\/\/\/\/\/\/\/\/\). It may seem strange, but notice that the bonds are going away from you as you look down (from the top of the page) at one of the peaks (or chirality centers), and they are also going away from you when you look up at one of the valleys (from the bottom of the page). If you keep your orientation (for example, top of your head pointing toward the left), you can progress from left to right--switching between looking down at each peak (wedges would be left, dashes right, and looking up at each valley (wedges now on the right, dashes left)--drawing the Fischer projection of the molecule. Determining (R) and (S) is an excellent check for this system. As Jay mentioned, practice and visualization are keys to understanding Fischer projections.(9 votes)
- At11:21, Jay said the CHO bond is going away from him, which I understand completely. It's the other carbon, the other one attached to a C, H, and CH2OH group, that I get confused on. How is that going away from the observer exactly? Earlier in the video, at about7:03, the two main carbons are on a plane and not going towards or away from the observer. Why is it different in the second scenario at11:21?(4 votes)
- Short answer: You are looking down at the chirality center only at11:21, and looking down at the bond between carbon 2 and 3 at7:03(Jay wasn't drawing a Fischer projection at the time).
Long answer: When drawing Fischer projections, the convention is that the bonds coming toward us are depicted as left and right, and the bonds going away are up and down. When assigning (R) or (S) configuration, knowing where the atoms are in space is critical, so he draws carbon 3 going away (remember our chirality center at carbon 2 is sp3 hybridized and tetrahedral). If it helps, imagine looking from an angle slightly more toward the CHO (around the top of the molecule looking back and down). Models help.(7 votes)
- How can I know 3-D structure of the molecules just by looking at bond line structure? (Meaning - whether OH is coming out, H is pointing further away etc)(4 votes)
- Good question, the short answer is, the Fischer projection alters the conformation of the molecule such that all the H and OH are pointing out and all the carbons on the carbon chain are pointing in, no matter which carbon you view it from, so the carbon chain is effectively changed from a naturally stable zig-zag into an actually very unstable curve, or circle, depending on how long the chain is. The long answer is, well, really long and I put under the tips section.(1 vote)
- Can you explain at12:45why it's R not S? I think I'm missing something.(2 votes)
- I think you are confused on the "trick" that he mentioned. When determining R or S, the lowest priority substituent needs to be facing towards the back, which can be done "turning" the molecule at the chirality center that the hydrogen is attached too. Some people can't visualize, or understand, how this happens so the trick is that whenever the lowest priority (this time hydrogen) is not facing the back, the answer R or S will be opposite. So if lowest priority is facing the back (dashed line), figure priorities of surrounding substituents, and determine R (clockwise/right) or S (counterclockwise/left/sinister)(5 votes)
- At11:49shouldn't the 4th carbon be CH2OH, rather than just C? Does this effect the R/S chirality at carbon 3 since both OH groups (attached at C2 and C4) will be 1 carbon away?(3 votes)
- You must remember that Jay is only talking about the absolute configuration of the chirality center at carbon 2. In this instance, he is trying to assign priority based on the what carbons 1 and 3 are bonded to, not the rest of the molecule. The carbon above the chirality center (carbon 1) does not have 2 oxygens bonded to it but you count it as 2 because of the pi bond when assigning absolute configuration. Thus carbon 1 (O,O,H) takes priority over carbon 3 (O,C,H) . The rest of the molecule is irrelevant because the priorities can be assessed by finding the first point of difference at that level. Watch the video on Cahn-Ingold-Prelog System for those rules. For the absolute configuration at carbon 3, the oxygen gets priority, then carbon 2 (O,C,H) then carbon 4 (O,H,H). Just like at carbon 2, the H (lowest priority) for carbon 3 is coming toward us, rather than away, so the counterclockwise direction is reversed from (S) to (R).(3 votes)
- At12:30, why OH on chilary center have won? Why not HCOO on top? :O(2 votes)
- The atoms attached directly to C-2 are O, C, C, and H. These are the only atoms we consider. O has top priority because it has a higher atomic number than either C or H.(3 votes)
- At05:36, I don't know how to transform lactic acit to the saw horse? :O(0 votes)
- The molecule shown at5:36is not lactic acid. Lactic acid contains only three C atoms. This is an aldehyde with four C atoms. Its common name is erythrose.
A sawhorse projection represents a carbon-carbon bond that you are looking at neither exactly from the end nor from the side. You are looking at it at an angle such that all bonds and atoms are visible. Note that the lower carbon in a sawhorse projection is more or less toward your eye, and the other is away from your eye (The C atoms are usually not shown in a sawhorse projection.) A Fischer projection is always in the eclipsed conformation.
To draw the sawhorse projection of an eclipsed conformation, start out by first drawing a line representing the carbon-carbon bond sloping upward and to the right (Pause the video as you trace these steps). Then draw one bond in front and one in back exactly straight down. Next, add the two remaining bonds at each carbon, so that all three added bonds are at angles of 120°. The picture should look like the letter “Y” at each end of the sloping line. Finally, add the symbols and formulas for the attached groups.
In your Fischer projection, the first sloping line you drew represents the bond between two carbon atoms. The other carbon atoms are attached to the bottoms of the vertical lines (the bottoms of the two Ys).
Thus, the functional group of highest priority (the aldehyde group) goes on the bottom of the Y furthest away from you. The other carbon (the CH2OH group) goes on the bottom of the Y closer to you. The H and OH groups go on the outstretched arms of each Y.
In the video, they just happened to pick a stereoisomer that has the OH groups on the right and the H atoms on the left. Your picture should now look something like the one in the video.
Hope this helps.(7 votes)
- I have a question: how do you do Fischer Projections for compounds with a carbon-carbon double bond? Is it even possible to do so?(3 votes)
- My textbook says "High priority group lies at the top of the vertical line. (Numbering starts from the top)" I don't know what to make of this. Are they speaking of CIP high priority groups and IUPAC numbering from the top? Also, since the lowest priority group according to CIP rules has to be away from the observer, the compound may need to be rotated. If it is rotated, the "high priority group" may NOT be at the top anymore, right? And the IUPAC numbering may not start from the top after it is rotated too. So what are they saying?(2 votes)
- [Voiceover] Fischer projections are another way of visualizing molecules in three dimensions, and let's use the example of lactic acid. It's called lactic acid, sometimes a carboxylic acid functional group over here on the right. And this is the only chirality center in lactic acid, it's an sp three hybridized carbon with four different subsituents attached to it, so with only one chirality center, we would expect to have two stereoisomers for this molecule. And those stereoisomers would be enantiomers of each other. Over here, I picked one of those enantiomers, and I've just drawn it in this fashion. Let's see which enantiomer we have over here. Well, this is my chirality center, the one attached to my OH, and if I were to assign absolute configuration to that chirality center, I look at the first atom connected to that chirality center. Well, that's oxygen versus carbon, versus a carbon over here in my carbonyl, so obviously oxygen's going to win, so we can assign oxygen a number one priority since it has the highest atomic number. And when I compare these two carbons to each other, I know the carbon on the right is double bonded to an oxygen, so that's gonna give it higher priority than the carbon over here on the left, since that's bonded to hydrogens. And then my other hydrogen attached to my chirality center is going away from me in space, so when I'm assigning absolutel configuration, I look at the fact that it's going one, two, three, it's going around this way, it's going around clockwise, therefore this is the R enantiomer of lactic acid. So that's all from a previous video. Now, if I wanted to draw a Fischer projection of R lactic acid, what I would do is I would put my eye right here, and I would stare down at my chirality center. So I would stare down at my chirality center, and I would draw exactly what I see. Well, if I'm staring down this way, I could draw a line right here to represent my flat sheet of paper, and I can see that both my hydrogen and my OH are above my sheet of paper, whereas my carboxylic acid and my CH3 are below my sheet of paper. So this carbon is my chirality center carbon, and I have my OH coming out at me, and this is actually going to be on the right side, so if you take out your molecular model set, you will see this OH will be coming out at you, and it will be on the right side of you. And this hydrogen will be coming out at you, it'll be on the left side of you, so that hydrogen would go out over here like that. This carboxylic acid functional group, this is the top of my head right here, then that would make this go at the top of what I'm looking at, and so, that is going away from me in space, so we would use a dash to represent that, and so we could go ahead and draw our C double bond to an O and then an OH going away from me, and then if I look at this CH3 group over here, it's also going away from me, it's going down in space, so I can represent it going down in space like that. And, this is the viewpoint of a Fischer projection, so if I'm going to convert this into a Fischer projection, a Fischer projection is just drawing across like that, and then at the top, you have your C double bonded to an O, and then an OH is just a way of abbreviating this carboxylic acid functional group, and then I have a hydrogen over here, and then I have an OH group over here, and then I have a CH3 here, so this is a Fischer projection, this is the Fischer projection for R lactic acid, so this is R lactic acid. And Fischer projections were invented by Emil Fischer, who won the Nobel Prize in chemistry for one of the things was for his research in carbohydrates, and he drew Fischer projections to help him draw carbohydrates, and so that's where you'll see Fischer projections used most often, even though some chemists don't really like them very much. So this is the Fischer projection for R lactic acid, and if I wanted to draw the Fischer projection for S lactic acid, I would just reflect this molecule in a mirror. So let's see if I can fit my mirror in over here. And I would have my OH reflected in my mirror, and then I'd go ahead and draw my Fischer projection, and then my methyl group would be over here, my hydrogen would be over here, and my carboxylic functional group would be right there. So, this would be S lactic acid on the right, and R lactic acid on the left. S lactic acid is the type of lactic acid that you find in the build up of muscles after extreme exercise, and the type of lactic acid that some people heard of from milk is actually a racemic mixture, so the bacteria in sour milk will break down the lactose into a 50 percent mixture of R, and a 50 percent mixture of S lactic acid. Let's take a look at a carbohydrate since Fischer used Fischer projections for carbohydrates specifically, so here I have a carbohydrate, and if I were to number this carbohydrate this carbonyl would get a number one and then this will get a number two over here, a number three and a number four this is a four carbon carbohydrate. How many stereoisomers does this carbohydrate have? Well, this carbon number two is a chirality center, and carbon number three is a chirality center, so two chirality centers, so I use the formula of two to the n, where n is the number of chirality centers so I would expect two squared or four possible stereoisomers for this molecule. So you could draw four different stereoisomers for this molecule, we'll draw them in a few minutes. For right now, I've gone ahead and drawn one of them, as a saw horse projection. So here I have a saw horse projection of one of the possible stereoisomers. And what we're going to do, is we're going to put our eye right up here, and we're going to stare straight down at this bond right here, and we're going to see if we can draw the Fischer projection for this molecule, so, what do we see? Well, let's start with this carbon right up here, so we'll make that carbon this one, and you can see that the OH attached to that carbon is going to the right, and it's going up at us, so that OH is going to the right, and it's going up at us, and then if I look at this hydrogen over here it's on the left, and it's going up at us. So my hydrogen is on the left, and it's going up at us. And this aldehyde functional group, this CHO, you can see, is going down. So this aldehyde functional group is going away from us, so we can go ahead and represent that aldehyde as going away from us in space like that. Well, this chirality center carbon is connected to this chirality center carbon, so we'll go ahead and draw a straight line, since we're looking straight down at it, and once again, we will find that our OH group is on the right, coming out at us. Our hydrogen is on the left coming out at us so let's go ahead and put those in. OH group is on the right coming out at us, hydrogen is on the left, coming out at us, and then of course, we have this CH2 OH down here, is going away from us in space, so we go ahead and draw that CH2 OH going away from us in space like that. So that would be the Fischer projection translated. Let's go ahead and make it to an actual Fischer projection where we just go ahead and draw straight lines, and the intersection of those straight lines are where our chirality centers are. So this would be an H, this would be an OH, this would be an H, this would be an OH, this would be our CH2 OH, and then at the top we have our aldehyde, CHO. So this is one of the four possible stereoisomers. And Fischer projections just make it much easier when you're working with carbohydrates. So this is one of the four. Let's go ahead and redraw the one we just drew and let's get the other three to get our total of four on here. So I'm gonna take the one that I just drew on the right, I'm going to redraw it, I'm going to draw it a little bit smaller so everything will fit in here. So this is one possible stereoisomer. I have my OHs on the right, I have my hydrogens, I have my CHO, I have my CH2OH. Okay if I wanted to draw the enantiomer to this molecule, I would just have to reflect it in a mirror. So I could just do this. I could reflect the molecule in the mirror, and I would have the enantiomer, so this would be the enantiomer to the stereoisomer that I just drew. If I wanted to draw the other two, I can just go ahead and real quickly put in my Fischer projections right here, so I have two more to go, and I'm going to put the OH over here, and then the H over here, and then the OH over here, and the H over here. So this is yet another possible stereoisomer, and I'll draw the mirror image over here on the right, so I have to have a hydrogen right here, and then my OH must be on this side, and then I have an OH, it must have an OH right here, and then a hydrogen on the other side, and then a CHO for my aldehyde, and a CH2OH. So here I have my four possible stereoisomers for this carbohydrate. And I'm gonna go ahead and label them, I'm gonna label this first one here, stereoisomer A, stereoisomer B, stereoisomer C, and stereoisomer D. Well, C and D are mirror images of each other, so they are enantiomers of each other so these are enantiomers. A and B are mirror images, so they are enantiomers to each other, and then we talked about in the diastereomer video, if I took one of the ones from A and B, so let me just go ahead and circle that. If I just took A, if I took one of the ones from A and B, and one of the ones from C and D, I'll just take C, then A and C are diastereomers of each other. They are non-superimposable, non-mirror images of each other. So those are enantiomers and diastereomers to review what we covered in an earlier video. Let's do one more thing with Fischer projections. Let's assign absolute configurations to one of these stereoisomers, so let's just choose the first one, A, so we've been talking about A, and let's go ahead and redraw it really fast. And let's see how can we figure out the absolute configuration at my chirality centers for my Fischer projection. So it just makes it a little bit tricker than usual, so here I have my Fischer projection and your aldehyde is gonna get a one and then two, three, four, in terms of numbering your carbon chain. I want you to figure out the absolute configuration at carbon two here. So at carbon two, what do I have? I know a Fischer projection tells me that if it's a horizontal line, everything's coming out at me, so my OH is coming out at me, and my hydrogen is coming out at me. Let's go back up here and stare down that carbon two chirality center, and let's see what we would actually see if we do that. So here is carbon two right here. I'm gonna stare down, right here, this time. So I have my OH coming out at me, my hydrogen coming out at me. That makes this bond and this bond actually go away from me in space. So the aldehyde is going to go away from me in space, like that. So I'm gonna go ahead and draw my aldehyde. Now, I'm actually gonna go ahead and show the carbon bond to one hydrogen. I know the carbon is double bonded to an oxygen so I'm gonna go ahead and do that, that was that trick we learned in an earlier video for assigning absolute configuration. And then the rest of the molecule is actually going down in space, right, so this would be a carbon here, bonded to a hydrogen. And this carbon is bonded to an oxygen and a carbon, so what is the absolute configuration of this carbon here. Well, if I think about, this is my chirality center, what are the atoms directly attached to that carbon? Well I have a hydrogen, a carbon, an oxygen and a carbon. Well, immediately, I know that my oxygen is going to win, so I can go ahead and assign a number one to my oxygen right here. And then I think about what's next priority. Well it would be carbon versus carbon, so the top, I have oxygen, oxygen, hydrogen. The bottom carbon, I have oxygen, carbon, hydrogen. So we saw in an earlier video, you go for first point of difference. So oxygen versus oxygen, no one wins, then I go oxygen versus carbon, and oxygen wins. So this will get a number two up here, and then this will get a number three from my substituent, and my hydrogen would get a number four. So I'm going around this way. I am going around this way if I ignore my hydrogen. So I'm going counter-clockwise, so it looks like it's S, but remember, the hydrogen is actually coming out at me, so in the little trick I showed you in the earlier video, if the hydrogen is coming out at me, all you have to do is reverse it. So it looks like it's S, but since the hydrogen is coming out at me, I can go ahead and say with certainty, that it is R at that chirality center. So at carbon two, at this carbon it is R. So you can do the same thing with the chirality center at the third position. So you could do the same thing with this one. And if you do that, you will find that it is also R. So you can go ahead and say, for this carbohydrate, it is R at carbon two, and it is R at carbon three, so it is two R, three R, and it's a two R, three R stereoisomer. And you can do that for all four of these stereoisomers that we drew for this carbohydrate, and you can then compare enantiomers and diastereomers that way as well. So that's a quick summary of Fischer projections. Practice and use your molecular model set to help you with the visualization aspect.