Convert the Fischer projections into line-angle drawings and a...

Question

Convert the Fischer projections into line-angle drawings and assign (R) and (S) at each chiral center.

(c) Fischer projection of a chiral molecule with hydroxyl and hydrogen groups, illustrating chiral centers for R and S designation.

Accepted Answer

Hi, everyone. Welcome back. Our next question says, the given fissure projection on the left is equivalent to the line angle structure with labels S and R at each chiral center on the right, true or false. So as my question indicates, I'm given a fissure projection and a li angle structure and I need to see if they're equivalent. So let's draw my own li angle structure and see if it matches. So I'm going to use the method in the channel's video that Johnny gives called that he calls the reverse caterpillar method, but he usually rotates the fissure projection 90 degrees to the left. I want the one I'm drawing to correspond to the one I'm given. Obviously that makes it easier to tell if they match. And in the one I'm given note that on the fissure projection, the top group is the carboxylic acid and the bottom is a methyl. And in the lining structure, the carboxylic acid is on the right and the methyl group on the left. So it's rotated 90 degrees to the right. So I just have to remember that I'm rotating and doing this in the opposite direction and that means when I'm deciding where my groups are going to be placed on the solid um wedge or on the dash that I want to have the eye looking at my molecule, my fissure projection from the right rather than the left. In Johnny's video, he shows it on the left since I've rotated it the other direction. So I draw my caterpillar drawing the methyl group and then a line going up, then the back of the caterpillar and then the end of it. So line going down to the carboxylic acid group, I have four carbons in the middle. And I'm going to go back to my F projection and number them 1234 going down. Now that's going to make them go left to right on my caterpillar. So I have to keep that in mind. So I'm going to put that on my structure before I put four dots and I'll label them 4321 again, just keeping straight going from structure to structure. So if I start with carbon one, so on my caterpillar, that's on the right there. If my eye is looking at this structure from the right, the fisher structure, the O group, I have an oh group that's on the right of the fisher projection. So that means I draw it as a wedge coming out towards me and then I have an H in the back. I'm not going to draw that because that's implied. Basically going to carbon two again, I have an H on the left oh on the right. So I will draw the O on the solid wedge. Carbon three hoh on the right H on the left. So the oh also on that solid wedge and carbon four has the H on the right and the oh on the left. So there my oh will be on the dashed wedge rather than a solid. So I have all these orientations on the caterpillar. But now I need to convert it into the zig zag. So basically, every group that's on the up portion of the zigzag will remain in the same orientation that I've drawn here. Every group that's facing down, I rotate around. And again, if I were just generating my own line inco structure, I could start in the down position or it could start in the up position either way. Because again, the convenience of the RS labeling is that I have two equivalent structures as long as they're labeled R and S my stereo chemistry will be correct. But I'm trying to compare it to a given structure. And so I want to keep it the same in the given structure. The ch three group is in the down position. So I will start that way myself. So I've got my methyl group and then let's draw my first line, my first bond going upward. So in that upward position, that would be ca carbon four from my caterpillar that had the oh group on the dashed wedge since this is the opposition, I leave it the same the oh group in. Now, next, I have carbon three which is in a downward position. So I need to rotate around 180 degrees. So my oh was on the solid wedge in the down position. It's on the dash tw now I have carbon two in the up position. So I maintain the solid wedge for the oh group that I had drawn and then carbon one is in the down position. So my solid wedge for my oh flips around and becomes a dashed wedge. And then last of all in the up position, I have my Coh. So next, I'll have to sign R and S but before I go through all that, let's double check that this part matches. And indeed the line angle structure I'm given has the from left to right. I have the methyl group and then I have an oh on a dashed wedge oh H on a dashed wedge oh H on a solid wedge oh on a dashed witch. And then the carboxylic acid group. So, so far I am matching now, I just need to figure out R and S for each of the chiral centers. So all of my internal carbons. So again, I'll go from left to right. So for carbon four, I'm just gonna write this down to keep track of it all since my oh group is facing back, my lowest priority, my H is my lowest priority. Each of these carbons has a hydrogen implied in the structure. That will be the lowest priority group. So my H is in the front. So I'll note that down. Now, I need to think about whether I need to go clockwise or counterclockwise to go from 1 to 3. So I know that my top party group will be my o age group. Now, I need to distinguish between the two carbons. Well, this one's pretty straightforward, I have a methyl group on one side and then a carbon bonded to an alcohol on the other. So carbon three is going to be my second priority group and the methyl group, carbon is my third priority. So you can see that I need to rotate or I need to go clockwise to go from 1 to 2 to 3. So I'll right clockwise and I like to write this down because again, I have to think about where is my lowest priority group. So my lowest priority group, my H is in the front, I have to go clockwise. That means it will be BS as my label. Again, if I, my lowest part is in the back, then clockwise means R but if my lowest part is in the back and the front, excuse me clockwise means s now let's move on to carbon three. In this case, the oh group is also in a dashed wedge. So once again, my hydrogen is in the front to label the priorities. I know that my o age group again is number one. Now, let's look at the two carbons on either side, this is a little trickier because in some respects, they're ID, they're identical right at the location of the carbon, they each have an oh group, a hydrogen and another carbon. So we have to go back even one step further. And on the left, in the case of carbon four, the next bond after that is to a methyl group, in the case of carbon two, the next bond after that is to another carbon with an oxygen. So there you see we get to an oxygen which is higher priority than the methyl group there. So that makes hydrogen number two, excuse me. Carbon number two getting turned around here as priority two. I'm actually going to erase the numbers from my previous carbon. We saw how to do it and it's going to get really, really confusing and cluttered in here if those numbers stay. So we'll just focus on one at a time. So we've labeled again that carbon two as priority two and then carbon four will be priority three. So now we see that we need to go counterclockwise. So back to my note notes here, I have hydrogen in front counterclockwise. I'll write CCW since my hydrogens in front that gives me R as my label. So now we'll again erase my numbers here to make room for the next carbon. So now we're looking at carbon two. So carbon two, the oh is on a solid wedge. So in this case, my H is in the back oh group is priority one. Now which of the carbons is next, we have both with an oh group, but carbon one has a carboxylic acid group while carbon three has that carbon with an alcohol group. So carbon four is going to be priority two and carbon two will be priority three. So I need to go clockwise. So going back to my notes, the hydrogen in back clockwise, that will be R OK. Now we've moved on to our very last Haron carbon one. And we see in this case, oh is on the dash wedge and therefore hydrogen is in the front priority one. Again, the o group priority two will be the carboxylic acid carbon and priority three carbon two. So we have to go counterclockwise. So go back to my notes hydrogen in front, counter clockwise. So that means I have R here. So going from left to right, my chiral centers are srrr and when I look at my structure, I see that yes, that is the arrangement there srrr from left to right. So are these two things is the Fisher projection equivalent to the line angle structure? And the answer is yes, it is. So this is true. As my answer to this problem, see you in the next video.

Convert the Fischer projections into line-angle drawings and assign (R) and (S) at each chiral center.
(c)

Key Concepts

Fischer Projections

Line-Angle Drawings

Cahn-Ingold-Prelog Priority Rules

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