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.

(a)

Accepted Answer

Hi, everybody. Welcome back. Here's our next problem. It says draw the line angle structure of the following fisher projection and assign R and S at each chiral center. And my molecule has a coo H at the top. Next carbon down has H on the left oh on the right, next carbon has oh on the left H on the right. And then the bottom group is Cho I'm presented with four multiple choice options. A through D showing those line angle structures with the stereo chemistry indicated. I'll go into those in more detail as we work out what the structure will be. So to convert this to official projection, I like to use the method in the Pearson channels video that Johnny shows called that he calls the reverse caterpillar method. So the way he usually draws it, he puts that top group. So the Coh on the left sort of the head of the caterpillar and the bottom group on the right. So then I'll label the um two carbons in the middle one and two going down from the top. So add one and two on the back of the caterpillar and then I draw a little eyeball looking from the left side of the molecule. And from that perspective, carbon one has the oh group facing away. So a dashed wedge and on carbon two, the oh group facing towards the viewer saw a solid wedge. However, the issue is when I'm going to compare this to my multiple choice options, they all have the cho group on the left and the coo H on the right, that's going to make it difficult to compare because I'm going to have to do these mental rotations of my molecule through space. So for the pur just for the purposes of comparing to my multiple choice options, I'm going to rotate my uh fissure projection 90 degrees to the right instead of 90 degrees to the left. So I will draw that on the right side here in red. So draw my little caterpillar here with my cho group on the left to correspond to my answer choices coo H on the right. So then the carbon on the left hand side will be carbon two. The carbon on the right hand side will be carbon one on the back of the caterpillar. And since I've rotated the other direction that my eyeball needs to be looking from the right. And from that perspective, carbon one, the oh is towards the viewer. So carbon one on the right of my caterpillar would be a solid wedge and then carbon two the oh is facing away from the viewer. So in carbon two, I'll have a dashed wedge. Now, I'm going to transfer that to my zigzag structure. And again, I could draw my zigzag either way, starting up or starting down because that R and S notation is going to serve as that orientation. The reason it's so useful is no matter which way I draw my zigzag, how I rotate my molecule. The R and S is always going to tell me how the substituents are arranged and allow me to compare different types of structures. So again, I'm going to try and correspond to my multiple choice options which show that cho group starting on the downward zigzag. So I'll draw OHC and then my zigzag going up and that will be carbon two. Then zigzag down to carbon one zig zag back up to the coo h now my upward zigzags I maintain the steri chemistry. I drew my caterpillar. So the carbon two has the oh on a dash draw that and carbon one, it's now a downward zigzag. So I need to rotate those groups 100 80 degrees. So now the oh is a dash wedge rather than solid on carbon one. So with this far along before I get to my R and si can actually see whether there are any structures I can eliminate. So I have the two ohs going on dashes down away from the screen. So choice A shows the second oh on a solid wedge, that's incorrect. So I can eliminate choice A on that basis. Choice B shows both on dashes. That's OK. Choice C both on dashes but choice D has one on a solid wedge and one on a dash. So eliminate twisty. Now let's just figure out our RNS chemistry. So what we need to look at is where's our lowest priority functional group on each carbon that each has a hyd one hydrogen. So it will be the hydrogen. So on carbon two, so I'll write over here two, the hydrogen is in the front since the oh is in the back. So I put h in front. So now let's think about the priority of our other groups. My molecules draw in red. So I'll do my priorities in blue. Top priority will obviously be the o then my carbon on the left, the co group, remember to count double bonds as two bonds. So that carbon has bonds to there's a double bond in oxygen. So it will be to oo and then h now carbon one is on the right, so that carbon has bonds to a carbon. Oh, let's start with the hire party has bonds, oxygen, carbon and hydrogen och. So that means that on the left hand side, that's priority two. Since there are two bonds to oxygen, the right hand side, that's priority number three. So to go from 1 to 2 to 3, I'm going counterclockwise my hydrogen, my lowest party group is in the front. That means if I am going counter clockwise, I wrote CCW, then the stereo chemistry here is R and then for carbon one there, my oh, also is in the back. So my hydrogen's in the front. Let's look at the priority groups here. I have my O group is party one. Now, I've already labeled that but looking from the other direction now, on the left hand side, that carbon is bonded to an oxygen to another carbon and the hydrogen. So I've got OC and then on the right hand side, I've got this carboxylic acid group so that carbon is an ooh. So my priority group two will be the carpet on the right side. And priority grade group three will be on the left. There's already a three there. So I'll leave it and to go from 1 to 23, I'm moving counterclockwise again. So counterclockwise CCW I have the hydrogen in front. So I have oops switch that to red. So it matches counterclockwise. So my stereo chemistries also are for carbon one. So back to my answer choices, I see on choice B I have the two R's at both chiral centers. So that will be my answer. Choice. C has R and S the chiral centers. So that's why that is incorrect. That second, that s should be a second R. So C is wrong. So we worked our way through this, we drew a line angle. Structure. And our answer here is Choice B with my two alcohol groups on dash wedges and the stereo chemistry at both Carol centers being R. See you in the next video.

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

Key Concepts

Fischer Projections

Line-Angle Drawings

Cahn-Ingold-Prelog (CIP) Priority Rules

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