What stereochemistry do you expect for the alkene obtained by E2 elimination of (1 R, 2 R) 1,2-d

step 1 So we have the following organic molecule. We have 1R, 2R, 1 2 diBromo, 1 2 diifanthal ethane. A

What stereochemistry do you expect for the alkene obtained...

Key Concepts

E2 Elimination Mechanism

The E2 elimination is a concerted reaction in which a base removes a proton while the leaving group simultaneously departs, forming a double bond. This mechanism requires a specific geometric arrangement for proper orbital overlap, and its stereospecific nature means that the relative configuration of the substituents in the starting material directly influences the configuration of the product.

Anti-Periplanar Geometry

In E2 eliminations, the most favorable conformation is one in which the hydrogen to be abstracted and the leaving group are positioned anti-periplanar, meaning they are 180° apart in the same plane. This orientation minimizes steric hindrance and allows effective orbital alignment, which is essential for the elimination to occur and typically leads to the formation of a trans (E) alkene.

Newman Projection

A Newman projection is a diagram used to visualize the spatial arrangement of substituents around a particular carbon-carbon bond. In the context of elimination reactions, these projections are particularly useful for illustrating the anti-periplanar relationship required for a successful E2 elimination, helping to rationalize the stereochemical outcome of the reaction.

Stereochemistry of Alkenes

The stereochemistry of alkenes is defined by the relative positions of the substituents attached to the double-bonded carbons. In E2 eliminations, the requirement for anti-periplanar geometry often leads to the formation of the more stable trans (E) alkene due to reduced steric interactions, a concept that is critical in understanding the product distribution of such reactions.

Transcript

00:01 So we have the following organic molecule.

00:03 We have 1r, 2r, 1 2 dibromo, 1 2 diifanthal ethane.

00:08 And we want to draw a pneumic projection.

00:10 And we also want to show what would happen if we performed an e2 elimination on this organic compound.

00:17 We just want to double check to make sure that we have the right configuration.

00:23 So let's go ahead and look at the left carbon here and assign our priorities.

00:28 So quickly going here, we know that our burmob is going to be our priority one.

00:31 We know that this carbon is going to be priority two because it has a connecting bromine to it.

00:36 The phenol is going to be priority three, and our hydrogen is going to be priority four.

00:50 All right, so that looks like it's going counterclockwise, which would appear s.

00:55 However, since we want the hydrogen going away from us, and right now we have it going towards us, we would say it's the opposite, so it's actually in our configuration.

01:05 If we do the same thing for the right carbon, we would assign our highest priority group here, which is the bromine.

01:13 Label that one.

01:15 This carbon would be 2 because of the connecting bromine.

01:18 The pheno would be 3 and the hydrogen would be 4.

01:22 Here it looks like it's also going counterclockwise but since again we have the hydrogen going towards us but we wanted to go away from us it's actually an r as well.

01:33 So now that we have established that it is indeed the right configuration, how would we draw a new and projection of this? so let's say we have an eye here and we're looking down the first.

01:51 Organic molecule in that direction, looking it down that bond.

01:55 So the resulting newman projection will look something like this.

02:01 So we'll go ahead and draw the front carbon donated by a period here.

02:06 And we know that the bonds are 120 degrees from each other.

02:10 So it looks something like this.

02:12 Looks like the phenyl group is pointing upwards.

02:15 The bromine group is pointing to the left and the hydrogen group is pointing to the right.

02:22 Now we want to draw the carbon in the back.

02:24 And its resulting groups as well, or it's connecting groups.

02:29 So you denote that with the circle and the following lines.

02:36 So it appears that the bromine is on our left.

02:40 The hydrogen's on the right.

02:42 And this kind of group is going down.

02:46 All right.

02:47 So we want to perform an e2 reaction or an e2 elimination reaction.

02:53 And an important thing to know is when we're doing e2 elimination reaction is that we want the reacting hydrogen and alkio -halide, which is in this case bromine, to be an anti -parriplanar geometry from each other.

03:06 So we want them to be 180 degrees from each other, in which case we see that this particular bromine circle, and this particular hydrogen circled in red, are 180 degrees from each other, and this particular bromine and this particular hydrogen, circled in red, are also 180 degrees from each other.

03:26 But since we know that there is, a line of symmetry here, we can essentially say that these are essentially the same groups...

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