From the IUPAC name, draw the corresponding structure.
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Question

From the IUPAC name, draw the corresponding structure.

(b) (1R,2S)-1-chloro-2-methylcyclobutane

Accepted Answer

Hello everyone. Let's do this problem. It says draw the skeletal structure that corresponds to the name one R two S one bromo two methyl Cyclohexane. Let's review the Iupac naming rules and we will apply them to going from the name to the structure rather than the structure to the name. So it is the same order first, we usually identify and number the parent chain. And in this case, going from the name to the structure, we look for the parent chain at the end of the name. All right, and we want to look for any modifiers. So say if we had a double bond, the parent chain would end in, in, if we had a triple bond, it would end in I and there are other functional group modifications as well that we would see all in the end of the parent chain name. OK. I don't see that in this name. So I won't discuss that much further here. Then we will look at the, the substituent groups which are listed out front of the parent chain name and their Logan or the number of the carbon where those substituent are located is in front of the prefix for that substituent, right. And lastly, we need to address configuration. So do we have E or Z configuration for any double bonds? Do we have RS configuration for any chiral carbons? And in this specific case, I see that we are dealing with R and S configuration. So we have to use the con engle pre log nomenclature rules or configuration rules rather. So first, we have to assign priorities of the four different groups attached to that carbo carbon and we do this by atomic mass. So the higher the atomic mass, the higher the priority if we have, we don't have any double bonds in this case. So we don't need to worry about that. If we have a tie, then we simply move on to the adjacent atoms and compare those. All right, then once we are done assigning priorities, we are almost ready to assign RRS configuration. But it depends on the orientation of the lowest priority group. So its group number four wedge or is it dashed? And this matters because we will draw a circle connecting groups 12 and three. And if that circle is counterclockwise, we will have one configuration. If it is clockwise, we will have a different but it changes depending on group number four. So if group number four is dash and we have clockwise, that is R configuration, counterclockwise, S configuration. If group number four is wedge clockwise is S counterclockwise is R all right. So let's apply these rules to draw our structure. So let's start at the end of our name to find the parent chain. We have cyclo hex. A cyclo tells us we have a ring. So we have a six membered ring hex, right? And there are no double bonds since we just have that A&E ending. So let's number our ring and then let's look at our substituent. We have one bromo and two methyl. So we have bromium carbon one methyl on carbon two and we have our configuration. So one R and two s so that tells me these are two chiro carbons. One is supposed to be R and two is supposed to be s so let's draw our substituent groups, either wedge or dash. I'll just draw them both wedge for now. And if the, if we find out the configuration is wrong, we will just switch it around later. So let's start with carbon number one, there's also a hydrogen attached there, right? Which is the fourth group, it's actually also the lowest number, the lowest priority group. So it is actually group number four, the bromine is the highest priority group. And then now we have two carbon groups, but carbon number two has another carbon attached to it. So that is going to be the second highest priority group. And carbon number six will be third priority. So in this case, group, number four is dashed. If we connect groups 12 and three, they go clockwise, dash and clockwise is in fact our configuration. So that is correct having bromine wedge on carbon number one. Now, what about carbon? Number two, we have the methyl grouped wedge as well. So let's draw our dash. Hydrogen. Hydrogen is lowest priority again. And let's see which is next or which is highest priority. So we have carbon, our methyl group has three hydrogens, that's going to be the next lowest priority because there is mostly hydrogens attached to that carbon. Whereas carbon number three has two hydrogens and another carbon attached. So that will be second priority and highest priority will be the carbon with that bromine atom, right? So connecting groups 12 and three, we are going counterclockwise group number four is dashed. So that is s configuration and the name says two s so that is correct as well. So this is the correct structure with bromine and methyl both wedge. So this is the correct skeletal structure for one R two S one bromo, two methyl cyclo heine. That's it for this one. I hope this was helpful and thanks for watching.

From the IUPAC name, draw the corresponding structure.
(b) (1R,2S)-1-chloro-2-methylcyclobutane

Key Concepts

IUPAC Nomenclature

Stereochemistry

Cycloalkanes

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