For each of the following compounds, draw the product that for...

Question

For each of the following compounds, draw the product that forms in an E2 reaction and indicate its configuration:

a. (1S,2S)-1-bromo-1,2-diphenylpropane

Accepted Answer

Hello and welcome. Let's take a look at the next problem. It says draw the product for the E two reaction of one S two S that's in parentheses, one bromo, 12 diphenyl pentane and C three on a indicate the configuration, we have four answer choices labeled A through D with different possible products. Note they're all alkenes as we would expect for an elimination reaction. Yeah. And it's indicated whether they are in the E or Z configuration. So let's think about how nature reaction works and what we would expect. So we recall in general that E two reactions are a concerted elimination of a beta hydrogen as the leaving group leaves and that beta hydrogen must be anti coplanar to the living group. So before we can see how this is going to happen, we need to draw our original product based on its name. So our base chain is a pen tame. So we'll draw a five carbon chain. So we'll draw it the other way. 12345. So those are five carbons and it's one bromo. So we'll add a bromine and 12 dienel. So we've got on carbon one, we've got a bromine and a fennel group. So it's a benzene ring, all that on there. And then also on carbon two, we have a py group. So that's our original molecule with its substituents. But we also need to indicate its stereo stereo chemistry so that we have everything arranged in the right place to see what kind of product we'll end up with whether we have an E or a Z configuration. So do we have, are, are we drawn these things in the right way? Well, we'll need to add on our solid wedges and dash wedges to indicate which things are coming out of the plane of the screen towards us and which ones are going away from us behind the plane of the screen. So we know that one carbon one and two are both have the S configuration. So we need to draw that. Let's look at carbon one. So for carbon one, just write that up on the right there. Uh Priority number one will obviously be our bromine. So number one, bromine, priority number two, what we have um the priority number four will always be hydrogen and then we have two carbons as well as the bromine. So our first carbon is part of the benzene ring. So it has a C double bond to do ac and then a second carbon. And then priority number three will be carbon two, which is a carbon and it has bonds too two other carbon ce c and then a hydrogen ach. So you can see why it's priority three. It has uh a hydrogen rather than a third carbon there. And then priority number four, of course, is the hydrogen. So let's look at whether these are in the clockwise or counterclockwise arrangement, our first three party groups. So from bromine to our phenyl carbon, we go in a counterclockwise direction. So since we have an S configuration, it's going in the counter parque direction, that will mean our priority four group will be in the back. So I'm going to erase the bond to the bromine and that will become a solid wedge and then we'll show a bond to a hydrogen and with a dashed wedge going back. So let's look at carbon two, we have another s configuration but C two, our fourth priority group again is a hydrogen and then we have three bonds to carbon. Well, our top priority group will, we might be tempted to say it's the C group C in the fenny ring, but it's actually carbon one because that's a carbon bonded to a bromine and a carbon and a hydrogen roaming being the highest molecular weight. So that will actually be carbon one will actually be priority one there. And then our second priority will be the phenyl carbon. So that would be carbon with CC double bond and another carbon. And then our third priority will be carbon three which has bonds to another carbon and then two hydrogens. So let's look at what direction that goes in. If we start, we're looking at carbon two. So we start with carbon one as the top priority. Second priority is a phenyl carbon that's going counterclockwise. So that means again, the hydrogen must be in the back. So we're going to erase the bond to the fennel ring. So let's draw that solid wedge to our fennel ring and then another dash switch to the hydrogen. So we've got our stereo chemistry indicated. We just need to think about how our concerted E two reaction takes place to let us know what the stereo chemistry of the product will be like. So let's simplify this a little bit to kind of have a zoom in on our arrangements around the bond that's going to become a double bond. So I'll draw that CC bond and then I can draw, I have the bromine coming forward, it's a solid wedge, the hydrogen going back and then I have a fennel group. So I'm just going to put ph there. So I don't have this big bulky ring sitting there. Now, it's important sometimes I need to remember that's a big bulky ring. If uh the ster hindrance, there might prevent something from happening. But I don't have a very hysterically hindered base. I just have this meth oxide. So then on the other side of this single bond, I have going down with solid wedge, the fennel group. So just ph going down the dashed wedge is the hydrogen and then going up on just a solid line is the rest of the chain. So I can actually just write that as CH two, ch two ch three. Now when we look at our leaving group, and I'm going to go ahead and draw the electrons from that the bromine bond going on to the bromine just so that we can focus on, we know where a leaving group is. So we need to look at the beta carbon and which hydrogen will be taken. Sometimes we need to think about what the product will be decided between hydrogens. But in this case, when we look at the beta carbon, it only has one hydrogen. So no debate about which one will be removed. And we see that it is indeed an anti, an anti cop cleaner uh orientation with regards to the bromine. So we just need to show that these electrons from the ch bond will then come on to make the new double bond. So then if we draw a resultant alkene from that elimination, so we'll draw that double bond, what we have left the bromine is gone, we have a hydrogen going up and then a fenny group going down on the right side, we've removed the hydrogen, we have the rest of the molecule going up. So ch two ch two, ch three, two and then going down on the right, we have the fenny group there still. So when we look at this molecule, we see that we have the Z configuration because our two identical py groups are on the same side of the double bond. So let's look at our answer choices and find one that matches this in choice. A. We do have the Z configuration, we have two phenyl groups on the same side. However, there is no double bond indicated there and the rest of the chain isn't there. On the other side, from the phenyl groups is just uh two hydrogen molecules. It's a choice. A incorrect. Let's look at choice B. Choice B has it final groups in the E configuration. So that's incorrect. Now, let's check. So C we have our final groups correctly in a Z configuration. The double bond is present the new double bond and we have the remainder of the molecule at CH two ch two ch three on the opposite side of the double bond from the two py groups. So choice C is our correct molecule and we've already identified choice C as the correct answer. But to be thorough, let's look at choice D which shows the correct molecule. But saying that we'll get a mixture of the E and Z configurations. However, we won't see that because again, we'll remove the beta hydrogen that's in an antic coplanar orientation and it's a concerted reaction. So there's no moving around it all happens at once. And so we'd see just the Z configuration. So choice D is not correct. So once again, the product for the E two reaction of our given molecule with sodium meth oxide and indicating its configuration, we have a new double bond form between carbon one and carbon two with the elimination of the bromine and a beta hydrogen. And we have our two phenyl groups and the Z configuration around this new double bond. See you in the next video.

For each of the following compounds, draw the product that forms in an E2 reaction and indicate its configuration:
a. (1S,2S)-1-bromo-1,2-diphenylpropane

Key Concepts

E2 Reaction Mechanism

Stereochemistry and Configuration

Substituent Effects on Elimination Reactions

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