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. Give mechanisms to account for these products.

Structural representations of (2S,3R)-3-bromobutan-2-ol, its enantiomer, and trans-2-bromobut-2-ene.

Accepted Answer

All right. Hello everyone. So this question says that one positive or negative 23 droop pentane reacts with sodium hydroxide. The following forms. The following product proposed the mechanism that shows the formation of the products give it. So starting off with our initial compound here are starting material. Recall that the positive or negative symbol means that we have a reca mixture of RR and SS 23 debro mo painting. And our nuc file here sodium hydroxide is if you recall a strong Nile. So essentially one of the bromine atoms of the starting material gets replaced with a hydroxy group. And we can see this considering that two of our products here are in N two mes of three bromo pen 10 to all, we have two R, three S and 2 S3 R. Because of the fact that hydroxide or the hydroxide ion is a strong nucleo file. This means that the reaction in question, at least for that case is going to be essent two. Now SN two is a concerted mechanism that notably results in the inversion of stereo chemistry. So if the nucleic attack was on carbon two of the parent chain then the two R three S product came from the 2 S3 S starting material. And so the 2 S3 R product came from the two R three R starting material. On the other hand, our third product is trans talent to E, which is an all keen stemming from elimination. Now here, once again, because the hydroxide ion is a strong nucleo file, it also happens to be a strong base. So that's going to favor E two as the other mechanism here. So with this in mind, let's go ahead and draw out the shred materials. When it comes to the RS designation, recall that priority of all substituents on a given chiral carbon is decided based on atomic number. So if we focus on carbon number two of the parent chain, for example, oxygen having the highest atomic number is going to be first priority in both cases. And hydrogen is going to be the lowest priority. If however, you come across a tie between two atoms, for example, a tie between two carbon atoms attached to the same chiral center. Then the idea is to find a point of difference, you are to compare the next set of adjacent atoms until one of them takes priority over the other. For example, the carbon inside of the parrot chain is going to take higher priority than the carbon on the outside of the parent chain that's only attached to hydrogens aside from the chiral center. And so from there, you trace a path going in a circle from the first to the lowest priority. If that circle is going in a counterclockwise direction, while the lowest priority group is in a horizontal position, then that's our configuration. On the other hand, if the lowest priority group is in a horizontal position and that rotation is going clockwise, that's s configuration. And so with this in mind, let's go ahead and start by drawing out two R three, R 23 di bromo pente. As we can see in two R 3 S3 bromo pen 1020 carbon. Number two is in the R configuration when the hydroxy group is to the left side in a horizontal position. So here when drawing out the starting material, bromine should also be on the left to maintain that R configuration on carbon number three. However, Bromine should be to the right. So now for 2 S3 S 23 dibromm pentane, the bromine atom attached to carbon number two should be on the right and the one on carved number three should be on the left. And so from here, we can go ahead and start with the mechanism starting off with RSN two. Recall it in an SN two mechanism, the nuc file is going to attack at the same time, the leaving group is departing. And because of the fact that the leaving group is still attached when the Nile approaches the nucleo file is going to attack from the opposite side, which causes that inversion of stereo chemistry. So for both anant T mes, the hydroxy group is going to attack from the opposite side of the bromine atom. As a result of this bromine is going to go ahead and detach itself from carbon in the same step. This means that now the hydroxy group is going to be on the opposite side relative to where the bromy initially was. So if the bromine atom on carbon number two was initially on the left, the hydroxy group should be on the right and vice versa. And so now that we've discussed the substitution step, the S and two step, let's talk about elimination or E two. So for further discussion of the E two mechanism, we can convert the fisher projections that were given into a line angle formula. And we're going to do this by taking the carbon chain and laying it down from the left side. And so this means that substituents that were initially on the left side are depicted on a wedge. Whereas those on the right side are depicted on a dash. So at this point, now that we have the two line angle formulas of both starting materials or specifically both in and two mirrors, let's go ahead and rotate both structures so that they fall in line with the stereo chemical requirements of E two because E two is a concerted elimination, it's a single step elimination. But it's important to recognize that the leaving group, in this case, bromine and a hydrogen atom of an adjacent carbon must be antic coplanar to each other. So because of that, I'm going to rotate both molecules so that the living group, bromine and a hydrogen on an adjacent carbon are neither on a wedge nor a dash. And also to showcase their anti stereo chemistry, one of them should be pointing upwards and the other downwards. So from here in the elimination step hydroxide is going to behave as a base. And it's going to specifically detonate the hydrogen atom of a carbon adjacent to the one that contains a leaving group. These are described as beta carbons or beta hydrogens to be precise. So in both cases, beta hydrogen gets removed and that creates the double bond between the carbon atom containing the leaving group and the beta carbon. At the same time that this is happening, the leaving group must detach itself from carbon to make sure that carbon doesn't have too many electrons at once. So it's because of that coplanar requirement, that anti coplanar requirement for E two that we get a trans all keen and there you have it here is the completed mechanism or SN two on the left side of the screen and E two on the right side of the screen. So if you watch this video all the way through, thank you so very much for doing so. And I hope you found it 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. Give mechanisms to account for these products.

Key Concepts

Nucleophilic Substitution Reactions

Elimination Reactions

Stereochemistry and Enantiomers

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