When (E)-3-methylhex-3-ene undergoes hydroboration–ox...

Question

When (E)-3-methylhex-3-ene undergoes hydroboration–oxidation, two isomeric products are formed. Give their structures, and label each asymmetric carbon atom as (R) or (S). What is the relationship between these isomers? What is the relationship between the products formed from (Z)-3-methylhex-3-ene and those formed from (E)-3-methylhex-3-ene?

Accepted Answer

Hey everyone and welcome back in today's video, we are going to tackle. The following problem too is um eric products are produced in the hydro operation oxidation of E three muscle pants in. Number one give the structures of the products and determine the absolute configuration of each stereo center. In each of the structures you proposed number to identify the relationship between the two is americ products and number three determine the relationship between the products formed in the hydro operation oxidation of Z three metha pen to in and those formed in the hydro operation oxidation of E three metal pen to in. Now there are many things to do in this problem. So let's just break it down. Right, let's start with part one for part one. As the problem suggests, we are starting with an L. Keen. We have E three metal pen to in So because configuration really matters here, we are going to start with our double bond at position number two, we are going to add one more carbon atom here at position number one. And because the parent is panting, we need two more carbon atoms. So we may simply draw an ethyl group on the opposite side of the double bond. Because we wish to have the highest priority groups to be on opposite sides of the double bond. Right? We have an ei zimmer meaning because position number three has a metal group. So, we have a method group at this position. Over here, there must be a hydrogen atom at this position and this is indeed an E I Zimmer because the highest priority groups, the metal group and the ethyl group are on opposite sides of the double bond. The problem states that in this particular case, we are performing hydro operation oxidation. Right? So number one, we are adding borough hydrate and tetra hydro Furin. And number two, we are treating the mixture with hydrogen peroxide and hydroxide. Now, let's recall that hydro operation oxidation follows the NC more cognitive edition. That's why peroxide here. Really important. And what we're going to do, we're going to locate our double bond and across the double bond, the hydrogen atom will be added to a more substituted position because we're performing the N. C. More cosmic of reaction. So retail chemistry here is really important. We want to make sure that hydrogen is added at this position and the alcohol group is added at a less substituted carbon. Right? Meaning we are ready to draw one of our products. Recall that This is also syn addition, meaning that hydrogen and the alcohol group will be in sin arrangement or on the same side with respect to the play. Now, if we draw the original carbon carbon bond and if you want to create a certain arrangement, we have two ways to do so. But first of all, let's just fill out our substitutions on top, there's hydrogen and there's an ethyl group and in particular what we can do here, if we're adding h and a wage from the front side, they will both they will be both pointing out, right? So we may say that there will be two wedges because that's in addition and hydrogen goes to a more substituted position. So we have hygienic disposition and a wage at a less substituted position and the original two substitutions. They were two method groups. So they will both be pointing down a methyl group there and here. However, if we perform that syn addition from the opposite side or basically from the back of the molecule, we will essentially produce another alcohol. So we will fill out our substitutions hydrogen. And now if we add these to you from the back side, they will be both pointing down hydrogen and a wage as you notice on letters in addition and the two metal groups will be pointing up. So these are our two structures, our two products essentially based on the problem. We provided the right structures and now we want to determine the absolute configuration of each stereo center. Now, let's see that using the con Ingle pre log rules, recall that we need to apply the con in gold where you log rules if we analyze each stereo center separately. Starting with this stereo center, hydrogen gets priority number four because it has the lowest atomic number. And now we are thinking about three L kill substitute prints. We have a methyl group, a longer ethel substitution. And then we also have this group which is C E O. Right? Because carbon is directly bonded to oxygen here for our substitution. This gets priority number one. Then we have a competition between a longer ethyl group and a shorter methyl group. So we may say that ethel will get priority number two and muscle will get priority number three. Now, if we trace our path from 1-2, I'm sorry, 1 to 2 to three, we will notice that the rotation that we get is counterclockwise. Right? So we may assume that okay, this should be us. But we call that hydrogen priority number four must be pointing down. So we are reversing our assignment. Instead of having counterclockwise rotation, we actually have clockwise rotation, which would imply that the first stereo center has an R configuration. Now, what about the second one? For the second one? We are going to assign priorities once again in a different color. And let's suppose that the first priority goes to O. H. Because carbon is bonded to oxygen and auction has a greater atomic number than carbon or hydrogen here or another carbon. Right, So that makes sense. As we know, hydrogen gets the lowest priority because it has the lowest atomic number. And now, between that metal group and a longer substitution which has two carbon atoms bonded to it as well as hydrogen. This longer L kill chain will get a higher priority. So we can say that be priority number two and the methyl group would be priority number three. Now we assigned the rotation, we want to make sure that the lowest priority group is pointing away from the viewer, right? So we may simply think about it this way. What if we draw the front view? In other words, what if we simply look along that carbon hydrogen bond? Right? And essentially what we will see is this we are seeing a wage pointing up then the metal group on the right and on the left, we have our L kill chain. And in particular we already know that oh, age is priority number one. The method group here Is priority number three in this C we can just say see or basically al killed group are okay, we already know that this is priority number two. If we look at this rotation 123, we trace our path right? And that's counterclockwise. So you may say that the second stereo center has an S. Configuration. And we are done. We have our stereo centers now thinking about this structure because hydrogen is not pointing up but it's pointing down. Instead we're basically flipping our configuration and we're going to say that this stereo center is us. And similarly here notice that how we flipped our Oh, age group priority number one. That was pointing up now is pointing down meaning we're flipping us and that'll be our at this position. So we got our configurations right for our products and we may now identify the relationship between the two eyes of eric products notice that because as we previously said, we flipped our configurations right from our to us and from us to our we basically understand that we have two has three are. And to our three S recall that whenever you're changing your configuration to the opposite one from S to R. And R two S. As long as you perform that step to each of your stereo center, then you're flipping all the configurations, meaning you get a pair of finance commerce. So you may say that These two structures that we got in part one are in an trimmers. We may label these sour products first of all. And for number two, we know that these two are in and humors. Now let's think about the third part of the problem. And for the third part we want to determine the relationship between the products formed in the hydro operation oxidation of Z three metal penso in and those formed in the hydro operation oxidation of E three metals, benzene, all that we're going to do. Now we're just going to change our alky into Z is more And to do that, we're simply going to rotate that bond. Actually, we cannot rotate it right, we are just going to change the position of an ethyl group to that pointing down basically right because we want to make sure that the highest priority groups are on the same side of the double bond. So we will just flip it and we will say that metal will be pointing up and the ethnic group will be pointing down. So let's draw our C. I. Zimmer, That would be part # three. Howard ci Zimmer would have a double bond hydrogen, that methyl group. And now because we are changing the position or basically the spatial arrangement of the ethnic group. Now it's pointing down with respect to the double bond and above the double bond we have a method group. So that's our Ci Zimmer. And according to the problem, the reaction is exactly the same. Right? We're thinking about hydro operation oxidation, meaning we have two steps, number one borough hydride in tetra hydro Furin and number two hydrogen peroxide hydroxide. Right, as we already know, we're going to add hydrogen to a more substituted position. So, basically over here and the older age group goes at a less substituted position because that's the ante. Markov. Markov addition. Right, okay, so basically also let's keep in mind that that this reaction will have a sin arrangement of the two groups. H and O. H. Will be on the same side. So, we may now just use the previously indicated steps, we may throw the central carbon carbon bond then on top here we have hydrogen. The other carbon contains a metal group. And because this is sin arrangement, we have two possibilities either add hydrogen and a wage from the front side or the back side. If we do it from the front side, we will notice that here we will have T wedges where hydrogen is added at a more substituted position and alcohol is added to be less substituted position. And the ethyl group will now be pointing down. Now regarding the right part, we still have a metal group that will now be pointing down. And similarly if we add them in sin arrangement from the back side, that would be hydrogen muscle group. And now what we're seeing here is basically two dashes. one of them will have an a wage group, the other one hydrogen. So against an arrangement right there on the same side and what we have is the remaining group of fl actually let's change the direction. So that's ethel and the other group would be the method group. So that's our structures. We got our products. We simply want to assign the absolute configurations to them just as we did before. So let's recall the con angle pre log rules. And let's notice that these two are stereo centers because they have four different substitutions, hydrogen gas priority number four because it has the lowest atomic number. This part gets priority # two just as we said before. The longer al kill chain, which is ethyl gets priority number three. I'm sorry, actually this part gets priority number one. Right, because we start with priority number four, not one. So this part gets priority number one, priority number two goes to ethel because it's longer than muscle. And finally Metal gets priority # three. Right? So now if we perform our rotation that we 123, that's our clockwise. But again we call that hydrogen is pointing up so we want to reverse our assignment. Since priority number four must be pointing down, meaning instead of getting our the actual configuration is as And similarly, if we assign priorities to our next carbon, we already did. So that's priority number one. Priority number four than priority number two on the left and three for the muscle group. Now, if we think about the bond rotation just with it before. Oh, age would be pointing up, right. If we imagine looking along that carbon hydrogen bond, we will see a wage pointing up. There will be a methyl group on the right and there will be the alcohol substation on the left. So, if we think about the bond rotation, that's counterclockwise, Right? So it's another S and here we are just flipping our configurations in particular, hydrogen was pointing up. Now it's pointing down so that we are. And here, oh, H was pointing up now it's pointing down so that be ours while we are changing it from S to r. Now we got our absolute configurations, right? We now wish to determine they they are in And first of all, right, Because we're going from us as to our our. So they are a pair of finance humors. But the question was asking us determine the relationship between these products and the products obtained in the previous reaction notice how we got to us. We are and now we have to us three S. For the first one and then for the second one we got to our three R. If you compare these notations, you will notice that you are keeping one of your centers fixed. So it's us at position number two. But you are inverting your your configuration of position number three from our to us. If you are reversing not all of your configurations, it means that you're getting Dia serial murders right previously. Was that if we are flipping all of them were getting a pair of finance tumors. But if we are not flipping all of them were keeping one or more of them fixed or at the same configuration. Just as we have for two S. While we're changing the configuration of the third position. That means these two our diocese to emerge just as these two because we have two R. Three R. And previously we had two R. Three S. So again, we're flipping the configuration of position number three. So we can say that the relationship between them is that they form a pair of die Astoria marks. So amy label that. And that would be it. Now we may conclude that the correct answer choice to this multiple choice question is D. D. However, if you have any questions, don't hesitate to leave your comment down below in the comment section and thank you for watching.

Key Concepts

Hydroboration-Oxidation

Stereochemistry and Chiral Centers

Isomerism and Relationship Between Isomers

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