Draw a Fischer projection for each compound. Remember that the cr...

Question

Draw a Fischer projection for each compound. Remember that the cross represents an asymmetric carbon atom, and the carbon chain should be along the vertical, with the IUPAC numbering from top to bottom.

(c) (S)-1,2-dibromobutane

(d) (R)-butan-2-ol

Accepted Answer

Hey everyone. Let's do this problem. It says draw Fischer projections for the following compounds. And tag any asymmetric carbon with an asterisk. Keep the carbon chain vertical with carbon one at the top. Alright, so let's review how to draw Fischer projections. And if we're going to draw the Fischer projections with S and R configurations, then we're going to also need to review the con and gold pre log nomenclature system so that we make sure our substitution groups are in the right positions. So for Fischer projections, we draw the longest carbon chain vertically. As the problem mentions, we wanted to be vertical and carbon one at the top is going to be the most oxidized carbon. Most oxidized carbon at the top. Alright then we have the horizontal lines and where those cross represent a carbon and we'll do that for carbons too. Through The 2nd to last carbon. So the first and last carbon, we write out the groups and the rest of the carbons in between. They are simply represented by those crossing lines. Alright then we'll add the hydrogen and substitution groups to the respective carbons. And we have to do that according to the R or S. Configuration. So that might take some trial and error. So I said we need to use con Ingle pre log. So let's quickly review that as well. So how do we assign our or S. Well, first we assign the priorities of the four groups. Right? An asymmetric carbon will have four different groups and we do that by atomic mess. So the highest priority is the highest atomic mass, lowest priority, lowest atomic mass. If we have a tie then we're simply going to look at the adjacent carbons. Or sorry, the adjacent Adams. And if we have a double bond or a triple bond we just count those atoms multiple times. So a double bond will count that adam twice. A triple bond will count that atom three times. And then when we're going to assign rRS we have to look at the orientation of our fourth priority group. If it is wedge or horizontal in a Fischer projection, right? Because the horizontal groups are wedged in official projection and we connect groups 12 and three. If their pathway is counterclockwise, that will be our configuration. If their pathway is clockwise that will be s configuration. And the opposite is true when group number four is dash or vertical, the counterclockwise pathway will be S and the clockwise pathway will be our But for Fischer projections I find that group number four is usually wedge or horizontal. Alright, so let's draw these compounds. So, compound A. We have S. Butane, 13 dial, so butane is four carbons. So let's do four carbons starting at the top vertically. One, 234. So those other two carbons are going to be crosses, right, we leave the first carbon and the last carbon written out. And on groups number sorry carbons number one and three, we're going to have an alcohol group. So carbon one will have C. H. 20. H. And carbon three will have an O. H. Group. Let's put it on this side And then the rest will be hydrogen. So hydrogen on this carbon hydrogen here and CH three group on the bottom. Okay, so now let's make sure that this orientation on this asymmetric carbon is correct. Right. This carbon has four different groups. So we have to check if the alcohol and hydrogen are in the correct positions. So let's assign their priorities. We have priority one is going to be the alcohol priority four is always hydrogen. If it's present then we have a methyl group and a carbon group that has another carbon. So this carbon group is going to be the next highest priority because it's attached to another carbon where this carbon just has three hydrogen. Okay, So in this case our group # four is wedge or dash. It's wedge, right? Because it's horizontal. So when we connect groups 12 and three, that is going in a clockwise position, which is going to make this s orientation, which is what the name says. So that means this is the correct structure. four. S. Butane 1, 3 dial. Okay, let's move on to be we have our 12 di bromo propane. So propane is three carbons. So we'll have one, two and then one cross 3 and we have one to dye bromo. So a bromine on carbon too. I'm sorry carbon one And a broom in here on carbon two. Let's put it on this side and we'll see what happens. So then the hydrogen is go on the rest of the positions. So let's assign our priorities. We have our asymmetric carbon here roaming is priority one hydrogen priority four. And then again we have two carbon groups but this carbon is attached to a bro me. Whereas this carbon just has three hydrogen. So this is going to be the next highest priority. And then the method group will be the third priority again, our fourth group is wedge. So when we connect groups 12 and three, this is going clockwise which is an S. Orientation but our name says it should be our orientation. So I know I need to switch these positions of grooming and hydrogen. Okay, so that's how we take care of that. So now we fixed it. So when we connect groups 12 and three, they are going counterclockwise. Which in this case is our which makes choice a the correct answer for this problem. That's it for this one. I hope this video is helpful and thank you for watching

Draw a Fischer projection for each compound. Remember that the cross represents an asymmetric carbon atom, and the carbon chain should be along the vertical, with the IUPAC numbering from top to bottom.a. (S)-propane-1,2-diolb. (R)-2-bromobutan-1-ol

Key Concepts

Fischer Projections

Chirality and Stereochemistry

IUPAC Nomenclature

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