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 and says draw a Fischer projection for each of the following compounds, antagonize asymmetric carbon with an asterisk. Keep the carbon chain vertical with carbon one at the top. So like the problem mentions we're going to keep the carbon chain vertical. So that's the first step withdrawing a Fischer projection. Is that vertically? You want to have the longest carbon chain and most oxidized carbon will go at the top. Okay then horizontally, you're going to start at carbon too and go to the one before the last with crossing the lines horizontally. And those will be substitute prints and hydrogen. And then lastly, we have to decide where exactly we're putting those groups if we have our or s configuration, which we do here. So we do that First by assigning the priorities of the four groups which goes by atomic mass. So the higher atomic mass, the higher the priority, which implies that hydrogen will always be lowest priority. Right? If there is a hydrogen, if we have a tie, go to the next carbon or the next adam might not be a carbon, go to the next atom. If we have a double bond or a triple bond, those count as two times or three times respectively. And then once we have our priorities we will assign R or S. So if the lowest priority group is dash or in this case vertical, right? Because in a Fischer projection the vertical groups are dashed and the horizontal groups are wedge, then we will have s counterclockwise and our clockwise when we connect Groups 1, 2 and three, right, excluding group four. And it's the lowest priority group is which or horizontal, which I think is more common in Fischer projections, especially if it's a hydrogen. Right? That's not going to be in the vertical position. Then we will have the opposite 12 and three going counterclockwise will be our and 12 and three going clockwise will be S Alright, so let's apply these steps to the compounds we have here. So first we have our beauty mean one to dial And let me just write this out. So I don't have to scroll be we have S. two bromo beauty. All right, so let's break down these names. So we have butane which is a four carbon chain. And we said we're going to keep one carbon at the top and bottom vertical position. So that means we'll have to crossing carbons. Right? And we have so four carbons one to dial. So the problem said we want carbon one at the top. So we're going to have those alcohol groups, one at the first carbon and one at the second. And we have in our configuration at our asymmetric carbon. Okay, So let's draw this out. one, 2, 3, 4 Carbons. The bottom carbon will just be ch three carbon number one has an O. H. Group. So that means it will have 10. H group and two hydrogen is right because we know carbon likes to have four bonds. So 1234. And now we only have another alcohol group on carbon number two. So at carbon number three will get to hydrogen and here will be a hydrogen and an alcohol group. But which side do we put them on? Right because this is our asymmetric carbon, it has four different groups, hydrogen and alcohol. This carbon with another alcohol group and this carbon with the rest of the carbon chain. So that is going to be where our our our configuration is. So let's see, we can test it by putting hydrogen here and alcohol here and then if it doesn't work we know we can switch them. So let's give our priorities. We know hydrogen will be number four and the alcohol group will be number one because of that oxygen. Now the next carbon has an alcohol group. And oh and this carbon is only attached to another carbon for a potential high priority. So alcohol trumps the carbon. So this group will be second highest priority and this carbon will be the third highest priority. Okay, so when we do that is our low priority group wedge or dash. It's horizontal. So it is wedge. Right? So that means if we connect 12 and three going this way, that is counter clockwise which will be our again, since our low priority group is wedge and that fits the name. So this is actually the correct orientation. So we will leave it just like that. Okay, let's do the next one. So again we have butane, So four carbon chain and we have a bro mean on carbon number two and this time we have S configuration, so we'll have one, 234, nope, no substitution groups on our end carbons, right. Only a bromine on carbon number two. So these will both be ch three groups. Nothing on carbon number three. So those are hydrogen and carbon number two has a bro mean, which means we'll have a bro mean and a hydrogen on this carbon and this carbon number two again is going to be our asymmetric carbon but we don't know if these positions are correct for hydrogen and bromine, right? We have to make sure that this s configuration is correct. So let's test it out and if it's wrong we'll switch it. So hydrogen again is our lowest priority bromine is our highest priority. And now we have again, two carbon groups. Well, this carbon group has three hydrogen. This carbon group has another carbon. So this carbon is going to be our next highest priority and this will be our third. So again, our low priority group is wedged. So see how I mentioned that it's usually going to be wedge or horizontal in a Fischer projection. So when we connect 12 and three this time it's going clockwise. But since group four is wedge, this is going to be S So this is the correct orientation For S. two Bruma Butane. This makes be the correct answer choice. That's it for this problem. I hope this video was helpful, and thanks 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.c. (S)-1,2-dibromobutaned. (R)-butan-2-ol

Key Concepts

Fischer Projections

Chirality and Stereochemistry

IUPAC Nomenclature

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