Draw structures for the following:
a. (S)-1-br...

Question

Draw structures for the following:

a. (S)-1-bromo-1-chlorobutane

b. (2R,3R)-2,3-dichloropentane

c. an achiral stereoisomer of 1,2-dimethylcyclohexane

Accepted Answer

Hey, everyone, let's do this problem and says, draw structures consistent with the names given below. And those names R two R three, R 23 die bro mo four methyl plantain are one bromo, one iota propane. And see says an A carol stereo is summer of 12 dimethyl cyclo plantain. So I notice that all of these names mentioned Cairo centers. And we're going to have to use our con mingled pre log rules which tells us how to assign RRS configuration in order to properly draw these structures. So let's quickly review those name, those rules. So first, when we have a Cairo Center, we assign the priorities of those four different groups attached to that Cairo center and we do this by atomic mass. So the higher the atomic mass, the higher the priority. What if there is a tie between two of the atoms, the immediate atoms touching that Carol center? Well, they must simply apply the rules to the adjacent atoms and go from there. What if we have a double bond or a triple bond? Well, then we count the double bond as if that, let's say we have a carbon double bonded to an oxygen. That counts as if we have a carbon, single bonded to two oxygen atoms. So we just count that double bond as two oxygen atoms in this case, and a triple bond is the same deal. So triple bond will count as three atoms. Now, when we're assigning R or S configuration, it depends on the orientation of the lowest priority group. If the lowest priority group is wedge, that is different than if the lowest priority group is dash. So when we connect groups 12 and three, if those go counterclockwise, that is going to be different than if they go clockwise. Okay. So we have a few different moving parts here. So if group number four is wedge and we connect groups 12 and three and those go counterclockwise, that will be our configuration going clockwise will be S configuration and vice versa for the dashed lowest priority group. So counterclockwise would be s clockwise would be our, I think most professors usually teach the lowest priority group going back or being dashed. So I'll highlight that as the rubric that we will likely use. That's the one I like to try and use and make my regular and then make the other one my exception. So then I know it's just the opposite. Does that make sense? So let's start with compound A. We have two R 3 R 2, 3 dye bromo four missile Pantene. So let's break this down from back to front, from the root to substitute parents. So, plantain is the root name here, which means we have five carbons. So let's draw a five carbon chain 12345 on carbon. Number four, we have a methyl group on carbons. Two and three. We have a bromine atom on each one. Dye bromo. And those borrowing groups looks like they're on Carville carbons because we're given two R, three R configuration on those Cairo centers. So let's add our substitution groups. We have a borrow mean on carbon to well make it which and I'm grooming on carbon three. Let's make that which and a method group on carbon four since carbon four is not Cairo, right? We have two methyl groups. We don't need to make those wedge or dash. But like we said, carbons two and three are Cairo. So we know there's a hydrogen there. So let's make that hydrogen wedge. So now our hydrogen or our lowest priority group is going to be dashed. Did I say wedge? Let's make those hydrogen dashed. That's what I meant. So again, we're going to follow that yellow rubric, follow the yellow rubric, follow the yellow rubric. So our priority, lowest priority group is dashed. So now let's assign the priorities for each carbon. So our second carbon, we have Romania's highest priority hydrogen as lowest priority. Now, the carbons, we have a methyl group. So that's three hydrogen on one carbon and the other carbon has a bro mean hydrogen and another carbon. So that carbon with the bromine is definitely going to be next highest priority. The methyl group will be third priority connecting groups 12 and three, that's going clockwise. And since we are in our dash configuration, that will be our orientation. So that is what we are looking for. So we'll keep it like that for our drawing. And for carbon number three, same priority groups pretty much, bro mean highest priority hydrogen, lowest priority. Now we have a carbon with the other browning that's going to be highest priority. And our carbon with the two methyl groups will be third priority, connecting groups 12 and three that goes clockwise as well, which is what we want. So that is our configuration. So let me redraw this without all the swirly arrows, 12345 carbons wedge, bro mean on carbon to wedge, bro, mean on carbon three And Metal Group on Carbon four. So that is our structure. Let me add the hydrogen as well. Dash hydrogen on both carbons two and three. So that is our structure for compound A. Alright. Moving on to compound B, we have our one bromo one I odo propane. So our route is propane, which tells us we have a three carbon chain and we have a bro mean and an iodine Adam both on carbon number one and there's only one Cairo center. So the our configuration is not specified. So let's see which carbon is Cairo. So we have three carbons, 123 and on carbon number one, we have bro, me and, and iodine. So let's make the iodine wedge and there's going to be a hydrogen there. So let's make that dash our lowest priority group in the back. So we have bro mean into the page iodine wedge hydrogen dash. There are four different groups on that first carbon. So that's our Cairo carbon. Let's see if this configuration is correct. So our highest priority group is going to be iodine, it has the highest atomic mass bromine will be next. Hydrogen is always last. So that means our carbon, the rest of our carbon chain is third priority, connecting groups 12 and three are hydrogen is dashed and connecting groups 12 and three goes counterclockwise. So that would be s configuration, right? But that's not what we want. We want our configuration. So what do we have to do? We have to switch the browning and the iodine. So let's redraw this with them switched and double check our configuration now. So we have iodine one roaming to carbon chain, three hydrogen four connecting 12 and three that goes clockwise, which is in fact our configuration. So this is the correct configuration. So lets redraw that with no writing on it. So that is the correct drawing for part B and let's move on to lastly part C which says draw an E Carroll stereo I summer of one to die metal cyclo painting. Okay. Well, I don't CRS configuration in this name. So maybe we're not using that for this one. Let's draw our structure and see what it means also before we do that. What does a car will mean? Well, Cairo is absence of symmetry. So that means an a Cairo compound will be symmetrical. Okay. So we want a symmetrical summer of our given name. So drawing 12 dimethyl cyclops contain cyclops contain is going to be a five carbon ring With two methyl groups On Carbons, one and 2. So we have a five member ring With two metal groups. So we can have two ice members of this, right. We can have a trans am er where the methyl groups, Our trans meaning one is wedge and one is dash or we can have a cyst where both of the methyl groups are either wedge or dash. I'll draw them both which, So which one is symmetrical? The cyst compound? Right? It has a plane of symmetry right down in between those two methyl groups. So that symmetry, that plane of symmetry is going to make the cis summer the A Carol stereo. So this sis structure is the correct structure for part C. So with all of that being said, that's going to make d the correct answer for this problem. That's it for this one. I hope This video is helpful and thanks for watching.

Draw structures for the following:
a. (S)-1-bromo-1-chlorobutane
b. (2R,3R)-2,3-dichloropentane
c. an achiral stereoisomer of 1,2-dimethylcyclohexane

Key Concepts

Stereochemistry

Chirality

Cycloalkanes and Isomerism

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