Question 1

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,
Luke

Thanks a lot,
Luke

Accepted Answer

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.

Question 2

At 11:21, 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 about 7:03, 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 at 11:21?

Accepted Answer

Short answer: You are looking down at the chirality center only at 11:21, and looking down at the bond between carbon 2 and 3 at 7:03 (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.

Question 3

At 11:49 shouldn'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?

Accepted Answer

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).

Question 4

Can you explain at 12:45 why it's R not S? I think I'm missing something.

Accepted Answer

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)

Question 5

At 12:30, why OH on chilary center have won? Why not HCOO on top? :O

Accepted Answer

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.

Question 6

At 05:36 , I don't know how to transform lactic acit to the saw horse? :O

Accepted Answer

The molecule shown at 5:36 is 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.

MCAT

Course: MCAT > Unit 9

Fischer projections

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