Given the following IUPAC names, draw the corresponding struct...

Question

Given the following IUPAC names, draw the corresponding structures.

(c) (S)-3-fluoropent-1-ene

Accepted Answer

Welcome back every once to another video, consider the compound with the IAC name as we loo 23 dimes Antoine in draw the structure. First of all, we want to draw the parent chain. And as the name suggests, we have pen, we know that the prefix pan means five. So we want to add five carbon atoms to our parent chain. Now, from here, we can enumerate that we can add 12345 from right to left or left to right. It doesn't matter also according to our name and one in it tells that the double bond starts at carbon number one. So we extend a pi bond from carbon number one to carbon number two. And this gives us our double bond between carbons one and two. OK. In addition to that, we notice that the name tells us that 23 dimethyl pant one in right. It basically means that there are two methyl substituents dimethyl bonded to positions two and three. So what we want to do in blue color is just draw two methyl substituents starting at carbons two and three, well done. And now also we know that there are three chloral. So not only we have a methyl substitution bonded to carbon number three, but we also have a chloral substituent. So we can draw an additional substituent, which is cl. Now, we could say that. All right, we are done here. But in addition to that, we have an s version of it, meaning there is a stereo center involved. And we noticed that carbon number three is a Cairo position because it has four unique substituents. So we want to assign an configuration. And for that purpose, we want to add the priorities to our carbon atoms. The first priority goes to chlorine. The reason is very simple. It has the greatest atomic number. We're comparing three carbon atoms bonded to the carl position against chlorine and chlorine has the greatest atomic number. Now, when we think about the remaining substitutions, the metal group will get priority number four. Once again, the reason is very simple, it is the shortest chain. If we have three carbon atoms bonded to the chiral position, then the shortest one will correspond to the lowest priority in this case. Now, considering the remaining two chains, they have equal lengths and we want to think about them. So if we think about the left chain, we clearly see that it's carbon bonded to two implicit hydrogen atoms. So we're going to draw two Hagen atoms and then it is bonded to the metal group. So we can just draw carbon on the right, we have a situation where there is a double bond. So there's a double bond in addition to that, we have a methyl group, right. So we are going to draw that specific carbon number two, which is bonded to another carbon from the methyl group and then it is bonded to a siege two using that double bond. So that means it is bonded to another two carbon atoms. We can treat a double bond to another carbon atom as two single bonds in our analysis of the Khan angle pre log rules. And now what we notice is that if we eliminate equivalent carbon atoms, we are comparing CH versus CC right two times. So that means the carbon atom with that double bond gets a higher priority because carbon is higher in priority than hydrogen based on the atomic number. And as a result, we can say that priority number two goes to the carbon atom with the double bond and priority number three goes to the carbon atom with a single bond. Now that we have our priorities, we can simply assume that one of our groups is pointing up and the other one is pointing down using wedges and dashes specifically, we want to focus on priorities one and four according to the king pre log rules, we want to make sure that priority number four is pointing down, right. So we can just assume that the muscle group is pointing down, meaning the chloral group is pointing up. So what we're going to do is just to draw a metal group on a dashed bond and the chloral group on a watch. OK. And now we want to check if our assumption is right, because we only have two possibilities. If we are not right, we are just going to swap them. So now let's think about the rotation. If we go and trace our path from 1 to 2 to 3, we notice that we would get a clockwise rotation, which is are, but we have instead. Well, what does that mean? It essentially means that we want to swap priorities of one and four. We want to make sure that priority number four is on a wedge instead. So now we know the answer, we can just redraw the structure so that it looks clean and go through the process. So we have our five member chain to begin with. As we said, we are adding our pi bond between carbons one and two. We're adding our substituents. Those are two metal groups bonded to carbons two and three. And we have our chloral group bonded to carbon number three. So what do we want to fix here? Those are the configurations more specifically the configuration of carbon number three. We said before that the metal group is pointing up, right, we had to reverse it. So we're going to draw our metal group pointing up on a wedge. The reason why we're doing this is because originally we added a dash and we said, if it's a dashed bond, then it's clockwise rotation. And we get R, we want to get s instead. So if we want to get that S we are simply going to draw that metal group on a wedge instead. So we get our methyl group and now we are going to add a dashed bond and the chloral group, right? Because we had to swap that wedge into a dash. And now, of course, we can just check if we are right. If we remember our priorities, that was 123, right? The metal group was number four. So now if we go along our path, 123 gives us clockwise rotation if priority number four is winding down. But if priority four is pointing up, we have to reverse it and the actual rotation would be counterclockwise, which corresponds to us and this is exactly what we wanted. So we're going to label our final answer because we have managed to obtain our structure and that would be it for today. Thank you for watching.

Given the following IUPAC names, draw the corresponding structures.
(c) (S)-3-fluoropent-1-ene

Key Concepts

IUPAC Nomenclature

Stereochemistry

Alkenes

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