When (±)−2,3−dibromobutane reacts with potassium hydroxide, so...

Question

When (±)−2,3−dibromobutane reacts with potassium hydroxide, some of the products are (2S,3R)-3-bromobutan-2-ol and its enantiomer and trans-2-bromobut-2-ene. Why is no cis-2-bromobut-2-ene formed?

Accepted Answer

All right. Hello everyone. So this question says that when positive and negative 23 dibromm pentane reacts with sodium hydroxide. The following forms the following product. So our first two products are en numer of three bromo pent 201 is two R three S and the other is 2 S3 R. The third product is trans two Brummel P two. In explain why there is no formation of sis two bromo I informed. All right. So first, let's start by drawing the structures of the starting materials. Recall that the symbol positive and negative here in front of 23 dibromm pentane means that we have a reca mixture where one in an tome is RR and the other is SS. Now, when considering the structures of the first two products, we can see that the bromine atom attached to carbon number two has been replaced with a hydroxy group, which does imply that these first two products are because of substitution. Now, speaking of substitution, our Nile in that case would be the hydroxide ion which if you recall is a strong Nile as well as a strong base. So here, the reaction in question is specifically going to be scent two, which means that the configuration of carbon number two has been inverted. There was an inversion of stereo chemistry on carb number two, which is characteristic of SN two reactions. And so what that means is that two R 3 S3 bromo pen 1020 was created from the 2 S3 S and an tumor. By contrast, 2 S3 R three bromo pen 20 was created from the two R three R in anir. Now our third product, on the other hand is an all keen and because hydroxide can behave as a base trans two bromo pent two E is the elimination or E two product. So now that we have some additional context in mind, let's briefly talk about the rules behind R and S configuration. Before we go ahead and draw out the starting materials, recall that priority is given to the four groups attached to the chiral carbon according to their atomic number. The higher the atomic number, the higher priority that substituent has. This means that if we consider carbon number two of both anant two me here or both anant two of three, Bromma Penan 20, the oxygen of the hydroxy group is going to take first priority because oxygen has the highest atomic number. This means that hydrogen is going to be the lowest priority. But now there's a tie between two carbon atoms that are attached to the same chiral center. So from here, whenever there is a tie, the idea is to compare the next set of adjacent atoms until you find a point of difference. For example, let's compare the two carbon atoms branching off of the Cairo center, which is carbon number two of the parrot chape. One of them is a methyl carbon, which means that it's connecting to three other hydrogens. Whereas the other is an internal carbon that's connecting to a B proline as well as an additional carbon. Therefore, the carbon atom inside of the parent chain is going to take higher priority than the methyl carbon. So I would label them accordingly. And so from here, the ideas should now trace a path from the first to the lowest priority. And when I trace the path in the first anatomy here, from first to the lowest priority, it's worth mentioning that in both cases, the lowest priority group is in a horizontal position. So when the lowest priority is in a horizontal position, and that path that we're tracing is going in a counterclockwise direction, that's our configuration. By contrast, if you're going in a clockwise direction, that's s configuration. If the lowest priority group is in a vertical position instead, then those rules are flipped. So now we can go ahead and draw our starting materials starting with two R, three R 23 di bromo painting to make sure that both chiral centers are going to be in the R configuration. The bromine atom on carbon two is going to be to the right, excuse me, to the left. And on carbon three, it's going to be to the right. So now to create its anant tumor, that's 2 S3 S 23 dibromm pentane, I would simply swap the location of both bromine atoms on both chiral centers. So now on the left, we have two R three R and on the right, we have 2 S3 S. But because our focus is specifically on the elimination reaction and the elimination product, let's convert these fissure projections into line angle structures. So first, I'm going to start by drawing out the carbon chain and then adding my substituents here as a wedge or a dash. Right now, I drew the two R three RN a tumor. So the bromine atom on carbon number two is on a wedge and the one on carbon three is on a dash which means it forts. And to me, I'm going to go ahead and swap them and here is the 2 S3 S. And so from this, we can go ahead and discuss one more detail about the E two reaction. Recall that in order for the E two reaction to take place, the carbon atom that's attached to the living group must have a carbon adjacent to it that has at least one proton. The proton of the adjacent carbon that is to be removed is known as the beta proton or beta hydrogen now, e two elimination is concerted, which means that it only requires one single step, but the leaving group and the beta hydrogen must be anti coal cleaner to each other. So now we're going to take the line angle formulas that we just drew and rotate them to showcase that anti coplanar requirement. So this time around for both A and two years, one bromine atom and its beta hydrogen are going to be neither on a wedge nor a dash and they're going to be pointing in opposite directions. So now the original stereo chemistry is still maintained, but now it's going to be easier to visualize that anti coplanar requirement. So now for e two elimination, the hydroxide ion is going to detonate its going to remove the beta hydrogen and the pair of electrons in between the beta carbon and the beta hydrogen goes in between the beta carbon and the one containing the leaving group. This is going to force the leaving group namely Broen to detach itself from carpet and both and an mes are ultimately going to produce a trans product. Now, the reason for this is because the two bromine or the two bromine atoms in both and tumors are trans relative to each other. This means that only a trans product can be produced after the reaction. If we want to produce a cis product, then the two bromine atoms must instead be cis to each other. Now, one approach is to instead proceed via the E one mechanism which will produce a mixture of cis and trans all KS. But e one mechanisms always introduce the possibility of rotation after the cruel cion is formed. And so with this in mind, the final answer reads as follows, this initial compound had bromine atoms at positions that favored a trans all keen and there you have it. So with that being said, thank you so very much for watching. And I hope you found this helpful.

When (±)−2,3−dibromobutane reacts with potassium hydroxide, some of the products are (2S,3R)-3-bromobutan-2-ol and its enantiomer and trans-2-bromobut-2-ene. Why is no cis-2-bromobut-2-ene formed?

Key Concepts

Elimination Reactions

Stereochemistry and Alkene Formation

Regioselectivity and Zaitsev's Rule

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