Stereochemistry - R S Configuration & Fischer Projections

horizontal of the Fischer projection is in front, so you can redraw the structure this way if you want, they come off the page and are therefore on the solid wedge. These two groups are on the dash or hatch wedge. So when you go to the page, you can see them as if they are on the back, so make sure you understand that now the hydrogen is usually on the side, which means that as group number four, you will have to reverse it when determining the

configuration

. Keep this in mind whenever group number four is in front, you need to reverse it, so let's start by determining the

configuration

of this chiral center.

The hydroxyl group will be group number one. Oxygen surpasses these two carbon atoms. Now we compare those two carbon atoms. they are the same but here we have a bromohere there is hydrogen so this group has a higher priority than the methyl group so this will be group number two the methyl will be three and then the hydrogen will always be four if it is there, so going from one to two and three we go in a clockwise, so it's r, but h remember that h when it's on a horizontal part is like being on a solid wedge. h is in front, so we have to invert it, so we will get s, so we have the s configuration for that chiral. center now for the next one, so we're going to focus on this chiral center.

Bromine will be the number one group and has the highest priority. Next, we'll compare the methyl group to this group, so we have carbon to carbon and then hydrogen. to oxygen oxygen wins so this whole group will be number two methyl is three h is four so when we go from one to two to three it seems that we are going in the s direction but if we reverse it it will be r so We have the r configuration here , so now that we have assigned the absolute configuration to each cryogenic center, we can go ahead and name this particular molecule, so the hydroxyl group will take priority over the bromine group, so let's count it.

In a way that we give the hydroxyl group the lowest number, we want to give it a 2 instead of a 3. So we're going to count it in this direction, so on carbon 2 we have the s configuration, so this will be 2s and on carbon 3 we have the r configuration so 3r and then we have a bromine on carbon 3 so it will be three bromine and the alcohol will be part of the main name and it is on carbon 2 so it will be two butanol because we have a four carbon chain, so 2s3r 3-bromo 2-butanol that's what we can name this particular Fischer projection.

Now let's try another example, let's go ahead and name this Fischer projection and determine the configuration at each chiral center just like we did. I did it before, so let's start with this one, bromine will be group number one, this group with the chlorine atom, we know it will be two methods, three h is four, so going from one to two and three ignoring four, this goes counterclockwise. direction, so that's s, but h is in front, so we need to reverse it, so we're going to get r for the first, now for the second chiral center.

Chlorine will be group number one, this whole group is number two if you compare carbon to carbon it's a tie and then carbon to bromine, bromine wins so this is number two, ethyl is number three, h is four, so going from one to two and three skipping four, it appears to be r, but h is in front, so we reverse and we get the configuration s, so now that we have the configuration at both chiral centers we can name it, so no We want to count in this direction because this would be three and that would be four, but we want to count in this direction, so this will be two and this will be three, so we have a five carbon chain, so we are dealing with pentane, but first let's focus in the configuration uh so the configuration is r on carbon two so we're going to have 2r and then it's s on carbon 3 so 3s and then we have a br on carbon 2 and we have to put it in alphabetical order b comes before c so it's going to be two bromine and then we have a cl on carbon three then three chlorine and then for the five carbon chain that's going to be pentane so that's the nomenclature or the iupac nomenclature for this projection Fischer's in particular, is 2r3s 2-bromo 3-chloropentane.

Now let's work on a more challenging problem, so let's say this chiral center has an ethyl group, a methyl group in front and let's put a chlorine group in the back and a hydrogen, let's go ahead and assign the configuration to the chirocarbon, is it rs? Now this problem is different because hydrogen group number four is neither in front nor there. is in the back, so how is the configuration assigned in this situation? Well let's start, we know that chlorine has the highest atomic number, so it will be group number one. Ethyl has higher priority than methyl, so ethos two methyl is three h is four in In a situation like this, there is a technique you can use to assign the configuration, whatever group is behind, put it in a circle, so I'm going to put one in the circle and you could put a little subscript b to indicate that in the circle is in the back sometimes it could be in the front, but for now whatever is circled in this problem is in the back, now the other numbers 2, 4 and 3 you want to arrange them in a triangle shape now if you notice that 2 is on the top, so let's put that on the top. 4 is at the bottom left with respect to 2, so let's put here 3 is at the bottom.

Now what we're going to do is rotate this. molecule such that the number 4 will be at the top where the 2 is, so let's rotate it 120 degrees clockwise so that one is still in a circle four has replaced two two is now where was three and three is where four was now the next thing we're going to do is flip the molecule as we flip it two and three will move up, group number four will push out group number one, so now four is in a circle, one is below and then three and two are above now four group number four is behind because what is circled represents what is behind and then we can count it from one to two to three and this gives us the s isomer so the configuration in this chiral center is s, so that's the technique you can use as long as group number four is not in front or behind.