What is the configuration of each of the asymmetric centers in...

Question

What is the configuration of each of the asymmetric centers in the Fischer projection of

e. D-sorbose?

Accepted Answer

Hey, everyone and welcome back, determine the absol configuration of each of the chiral centers in the following structure. And the structure that we're given is the man, it's actually the official projection that we're looking at. And we have a total of four chiral centers which are the intersection points of our bonds within the fi projection. And those four carbon atoms have four different substitutions. We want to assign the absolute configurations to each of our chiral centers. And those are whether R or S configurations. In other words, to determine these configurations, we have to analyze each of our carbon atoms separately And we're going to start with the top one to assign the absolute configuration. We, first of all, first of all want to give priorities, see the substitutions bonded to our chiral position. And these priorities depend on the atomic number. So I can state that atomic number will be considered. We have to look at all of the atoms bonded to our chiral carbon atom and we have carbon oxygen, another carbon and hydrogen. Because oxygen has the highest atomic number, it gets the highest priority number one and hydrogen has the lowest atomic number. So it's it gets priority four. Now, between the two remaining carbon atoms, we have to think about the bonding pattern. The top substation is an aldehyde group. So it has C O double bond and hydrogen. Now at the bottom of the trend has carbon hydrogen, then it has an O H substi and then it's also bonded to another carbon. Now to make it really easy for us to compare, we're going to consider the aldehyde group as though it had two C O single bonds. And we want to exclude equivalent parts. So we have C O. In each case, we don't need to worry about hydrogen. We can just exclude CEO from each case. And then we have a comparison between C and C each. We can exclude these. And now we noticed that the only remaining atom is oxygen and carbon. And because oxygen has a greater atomic number than carbon than the aldehyde substi get priority number two and the bottom substi guess priority number three, Now that we have our priorities, we want to remember that we also want to make sure that priority number four is pointing down. In that case, we can trace our fat directly. But in fiss projections, horizontal bonds are been on wedges pointing up. And because that's the case, we are going to reverse our path to an opposite one because we want to make sure that priority four point hang down. So if it's pointing up instead, that means we will just reverse our thought process. So if we trace our path from substitute 1-2-3, we notice that the rotation that we get is clockwise. But because we have to reverse that it would actually be counter clockwise and counterclockwise rotation corresponds to an S configuration. So that's how it's done. Well done. We have got an S configuration for the first center and we are going to erase all of our priorities so that we can clearly see what happens for the carbon. So let's assign s for the first one. Now, moving forward, once again, we have oxygen, that's priority one hyen which gets priority four. And now we once again have to distinguish between the two substitutions that have carbons. So if we look at the top carbon, it has hyen and hydroxy, the bottom one has hygiene and hydroxy. So these are equivalent parts. And then we have a comparison between Cho group and Cho H and we have already said that an aldehyde will get a higher priority than Cho H. So that said once again, we have priority number two for the aldehyde versus priority number three. For Cho H Once game priority # four is pointing up. So that means we will have to reverse our path Going from 1-2-3. We once again have our clockwise rotation. So we have to reverse it into counterclockwise rotation, which is S once again. OK. So we got the absolute configuration of the second center And now we are ready to move on to the 3rd 1. We have oxygen priority number one, we have hydrogen priority number four. And now we have to compare our carbons. So on the bottom, we have CHO H on top, we have CHO H as well. So then if we move down, we have C H2O H versus CHO H which also has another carbon bond set. So to make it really simple for us, we are just going to draw the C H2O H substituent. Specifically, we have our carbon, what two hydrogen atoms and then oxygen for the bottom part and for the top part, we have carbon bonded to oxygen bonded to hydrogen and then bond it to another carbon coming from the aldehyde. If we exclude equivalent parts. In other words, oxygens, hydrogens, then we have a comparison between carbon and hydrogen and carbon has a high atomic number. That means the top substituent has a high priority number two. And the bottom one gets priority # three. Hydrogen priority # four is winding up. We once again will have to reverse our assignment. So if we trace our path 123, that goes counterclockwise, and because we have to reverse that the actual rotation will be clockwise instead, which would be R. Remember that the clockwise rotation is R well done. So we only need to consider the final chiral position, oxygen gets priority one. Hydrogen gets priority four. And now we have to compare our remaining substitutions. We already know that we have C H T O H and the other one is cho H, so we can throw it out. That would be C H O H. So we can just add oxygen which is once again bonded to another carbon. So we have an equivalent scenario to the previous one. We will have a comparison between oxygen and hydrogen. So we can state that the top substituent will have a higher priority number two And the bottom one will have priority number three. Hydrogen is pointing up on a wedge. We will have to reverse our assignment. When we trace our path 123 that goes counterclockwise, reversing it. We obtain a clockwise rotation which is our well done. So we have obtained our absolute configurations for the four stereo centers. And from top to bottom, those would be S S R and R. We are going to label them in a green box is our final answer and conclude our video. Thank you for watching. And I hope to see you in the next one.

What is the configuration of each of the asymmetric centers in the Fischer projection of
e. D-sorbose?

Key Concepts

Fischer Projections

Chirality and Asymmetric Centers

D and L Notation

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