A chemist allows some pure (2S,3R)-3-bromo-2...

Question

A chemist allows some pure (2S,3R)-3-bromo-2,3-diphenylpentane to react with a solution of sodium ethoxide (NaOCH₂CH₃) in ethanol. The products are two alkenes: A (cis-trans mixture) and B, a single pure isomer. Under the same conditions, the reaction of (2S,3S)-3-bromo-2,3-diphenylpentane gives two alkenes, A (cis-trans mixture) and C. Upon catalytic hydrogenation, all three of these alkenes (A, B, and C) give 2,3-diphenylpentane. Determine the structures of A, B, and C; give equations for their formation; and explain the stereospecificity of these reactions.

Accepted Answer

everyone. And welcome back in today's video we are going to tackle the following problem. The reaction of two R. Three R. Three bromo 23 diphenyl hexane with sodium a mite in liquid ammonia yields three products to cis trans eyes MERS A and a pure eyes Moby Another substance to S. Three S. Three bromo 23 di final hacks in under the same conditions gives a differences. Trans make sure C. And a pure eyes Mardi determine the structures of all these products. And explain why the two reactions produce two different pure eyes. Merz B and D. Now there's a lot to do in this problem and let's actually concentrate on getting our substrates right Because their configuration involved at two different Cairo centers two and three in particular, we may start by drawing kexin because that's how our parents. So we are drawing six member chain 12345 and six. We may state that this is position number two. This is position number three. Position number two has a final substitution, position number three has a bromo substation and a final substitution. So there's one bromo group and one final group. We may also expand This implicit hydrogen. A position number two. Because we will be assigning priorities. Recall that according to the conning gold pre log rules, priorities are assigned based on the atomic number. Starting with position number two, hydrogen has the lowest priority priority number four because it has the lowest atomic number. Now we have the compared some be between a methyl group and a final group. But also we have this longer al kill chain. So if we compare them for the final group, it only contains C. C bonds versus CBR bond here because bro ming is greater an atomic number. It gets the first priority followed by CC double bonds. We have a longer L kill chain in particular, we have a longer double bond chain, carbon carbon double bonded over here in the final group which is longer and less saturated than a metal group. So if we think about the metal group, ch would be lower in priority than CC. So that's sufficient to understand to give final priority number two, Which leaves the metal group with priority number three. And the center must be to our we will assume that priority number four is pointing away from the viewer and we will try to assign a rotation directly. And if we do so if we trace our path 123 goes counterclockwise that would be to us if hydrogen was pointing away. But if we make that hydrogen point up we would just reverse our process and we will get to our So we are good now meaning ha jin is pointing up and the final group would be pointing down. That's how we are done with the first center. Now, regarding the second center, if we change the color, we have four different substitue ints and one of them is roaming roaming in particular again gets priority number one because it has the greatest atomic number. Then we have a comparison between C. C. H. And C. So this carbon in particular has hydrogen, another carbon from the final group and another carbon from the methyl group. While here we have a fennel which only has carbon atoms bonded to it. So even though it has double bonds, we may just draw double bonded carbons as two single bonded carbons. So it only has three carbon atoms. If we eliminate equivalent parts, we notice that we have a competition between CH and CC. So fennel is definitely greater priority than this substitution over here which has hydrogen bond. It's it meaning priority number two goes to fennel and now we are comparing the left substitute which has C. With one C. And two H. S. So you may say there are two hydrogen atoms bond it's it and then it's bonded Senator carbon. And here we have C H C C C C. So C H C C N C C. Now, if we eliminate equivalent parts, we noticed that again, we have a competition between C. C N. C. H. So definitely this substation on the riot is greater in priority. So it gets priority number three. And priority number four goes here. Now this is a bit complicated because priority number four is actually on a solid line. We're not going to you know use it as a wedge or a dash. We want to leave it as a solid line to represent our parent. So we are going to essentially draw a partial newman projection. You don't don't necessarily need to think about it as a Newman projection, but you just want to make sure that you're essentially looking along that bond in particular list those that you're looking down. That bond. Now, priority number four in that case would be pointing away and at the front you want to see an ast rotation. In other words, two substitutions will be pointing up and one substitution will be pointing down that substitution that is pointing down already has priority number three. And if you want to make sure that this is an ast rotation, you must be going counterclockwise. So that must be too let's see If we go counterclockwise, that would be 123. Right, so that would be 12. This means priority number one must be pointing away from the viewer and simply speaking, it essentially implies that bro mean must be on a dash, right? Because it's pointing away while fennel group must be at position number two. So it must be pointing towards the viewer or pointing up. So this is our final structure. Now let's redraw it really clean. So we clearly see what's happening in the reaction and let's start thinking about products A and B. So we have hexane position, number two has hygiene pointing up and pointing down Position. Number three has fennel pointing up bro mean pointing down we want to think about the reaction in which this structure is treated with sodium a mine in liquid ammonia. Now recall because we have a good leaving group and sodium a mite essentially it contains a mite which is a strong base. We will be eliminating beta hydrogen. Is this will be an E two reaction. So we may classify this as an E two reaction because we are using a strong base on our substrate and we will be producing al kings. Also that stated in the problem, we will have Cis trans eyes more. So that implies that it's consistent with our assumption that this must be an E two reaction. Having identified our alpha carbon which has the leaving ravana to it, we noticed that their beta hydrogen in particular. There are two beta high jen's in this position and also there is this beta hydrogen because there are two beta high jen's here, we would have two different products. And if we look back at the context of the problem, we are getting to cis Trans Sizemore's a meaning to get to Cis Trans Sizemore's will be eliminating these hydrogen. Let's see how that looks like we want to make sure that hygiene is anti co planer to the leaving group. So if we remove this hygiene and form a double bond, get rid of the leaving group. We will get the following products. So we may redraw it as follows. There will be a six member chain. Now we are forming a double bond at this position, Removing one of the Hydrogen Atoms. Now we still have a fennel group because bromide has been removed so there will be a fennel group. And in addition to that we have another final group. Nothing changes regarding configuration here because this center did not participate in the reaction. So we may just say that hygiene is pointing up, fennel is pointing down. However, we have another hydrogen here, it's not antico planer so we have to perform bond rotation and if we do so we will be able to eliminate that hydrant as well. So let's rotate our bond and redraw our structure. Now, if we start with this carbon we would have final roaming hi to jen and now if we rotate that ethyl group in particular it would now be pointing down and this allows us to identify one of the beta hydrogen in antico planer arrangement. And now if we remove it we would form a double bond at this position, we would get rid of the leaving group and we would form another is um er of the previous structure. So let's throw it out again. We're using the same set of regions sodium Amytal in liquid ammonia. Now what we're getting is pretty much the same but this was trans, the two alcohol substitutes are on opposite sides. Now we will get sis, there will be a final group over here, we still have hygiene pointing up fennel pointing down. And now all that we have to do is basically make sure that it says an arrangement. So we are going to draw That's ethnic group. That's it. So that's our next product. We understand that these are cis trans size matters right? Because the cl kill groups or in particular it will be more accurate to call them E. N. Z. Right? Because the highest priority groups are on opposite sides. And here they are on the on the same side of the double bond. So we got our structures A. And B. Actually that would be just A. Because as the problem suggests this trans eyes Mazar A. So these two are our A. Now if we think about the removal of the other hydrogen atom, we would form our next product. It's already antico planer in its arrangements. So if we remove that hydrogen in particular forming a double bond and removing broom in would give us product B. And we can draw that product be let's use the original structure. There is no need to rotate that bond. So if you visualize we're removing this hygiene forming a double bond here and getting rid of roaming the already anti an arrangement. So we may draw our six. Remember chain Now because we're forming a double bond at this position. Final group would be in plain. This final group would also be in plain. There will be double bond and that's it. That's our structure B. We have shown that there's only one pure eyes M R B. Because there's only one beta hygiene here. That is an anti co planar arrangement. There's something else we can see or removed. So we're good with that. But now the problem suggests that there is another structure which is two S three S. Version of the original structure. So let's draw that two S 3 s compound of the same structure. If we do that, We are going to draw six carbon atoms in our chain and essentially we are just going to flip the configuration of the second center. If hygiene was pointing up now it will be pointing down because now it says to us instead of to our So hye jin is pointing down and fennel is pointing up, we have reversed the configuration at that center and now the third center is exactly the same. It says tweet us so we would still have final pointing up and romaine pointing down now, similarly to the previous problem, we are going to think about the beta high genes that we could remove. Now there are two beta hae jin's just as discussed before. One of them is pointing up. The other one is pointing down and for this one we actually have to perform bond rotations to eliminate it. But let's start with these two. If we eliminate these two we will have two different products. Right? One of them will be insist arrangement and another one will be in trans arrangement after the bond rotation. So using the previously indicated logic, let's draw the trans one. Or the one. We may draw A six member chain at position number three. We would have a double bond now than a final group and here nothing changes regarding configuration. Hygiene is pointing down, fennel is pointing up. But now if we perform bond rotations to eliminate the other hygiene we will just have a Z. I. Zimmer. So we may now draw it. Starting with a double bond. We will simply draw that ethyl group pointing down. And now here nothing changes regarding configuration. There'll be hygiene and a final group. So we get our E. N. C. I's mars this will be our products see and we are ready to look at the final product. To eliminate that beta hydrogen, we need to perform another bond rotation. If we perform bond rotation to make sure that hydrogen is anti complainer to the leaving group, we will get the following structure essentially We made troll Fennell was pointing up. Now it will be in plain hydrogen will be pointing up and a methyl group will be pointing down and similarly here we have exactly the same configuration as basically grooming and fennel. So we have performed the bond rotation to make sure that hydrogen now is an anti co planer arrangement. And if we remove that hydrogen during elimination, we will form a double bond at this position and get rid of the leaving group and we will get our purisima de The group is pointing down there say double bond formed and the two final groups are on the same side of the double bond. That's our product D. Well done. So we got all of our products. Let's just label them. This is our product D. These are our E. N. Z. Heismans for see this is our product B. And in addition to that, we also got two products for A. We may label these as our products for A. So now that we got our products, we may answer the main question. The reason why we're giving while we're getting two different eyes. Merz B and D. Is simply because we have to perform bond rotation on that other structure due to the requirement of an anti co planer arrangement between hydrogen and the leaving group. We had to perform bond rotation to get a differently looking eyes Mardi than be for the B. One, we did not need to rotate that bond because the anti co planer arrangement had already been satisfied. But in the final case we needed to perform bond rotation, which in turn gave us a different. We may conclude that the correct answer to this multiple choice question is C. And if you have any questions, don't have states, leave your comment down below in the comment section and thank you for watching.

Key Concepts

Stereochemistry

Elimination Reactions

Catalytic Hydrogenation

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