3. For this E2 reaction leading to the Zaitsev product,...

3. For this E2 reaction leading to the Zaitsev product, redraw the substrate structure in a conformation that is required for reaction, using wedge/hash bonds to show three dimensions and appropriate bond angles. Draw in the base on the side that it should be approaching the substrate from. Draw arrows showing bond forming and breaking. Draw the product with correct stereochemistry. (2.5 pts) 4. This series of reactions includes an SN2 reaction. (2.5 pts) (1) In the space below, draw product "X" showing three dimensions with using wedge/hash bonds. (2) Draw in the nucleophile on the side that it should be approaching the substrate from. Draw arrows showing bond forming and breaking. Draw the product "Y" with correct stereochemistry.

Transcript

00:01 Okay, we've gotten the e2 reaction to consider, and we're starting with the sawhorse diagram, bromine down, wedge to hydrogen, a dash to a ch3, and then we'm just going to write et here for ethel, so we don't have to drop the whole chain, a wedge to a ch3.

00:27 In fact, i'm just going to write these in as me for methyl.

00:31 That means the same thing as ch3, and a dash to a a hydrogen.

00:38 And the question is we need to redraw the substate structure and a confirmation that's required for the reaction.

00:43 All right.

00:44 So we need to basically rotate about that bond to give whatever structure would be necessary for an e2 reaction to happen.

00:59 Let's see.

01:00 Draw in the base from the side that should be approaching the substrate from, draw arrow showing the bond breaking, forming and breaking.

01:05 Okay.

01:06 And then draw the product of this correct stereo chemistry.

01:08 So recall, an e2 reaction to take place, we need the following.

01:19 We need to have our leaving group, which in this case is bromine, to be anti -periplaner to hydrogen, because a base is going to come in and abstract that proton, which will create the double bond at the exact moment that bromine leaves.

01:38 And because this necessitates the overlapping or the movement of these electrons into the anti -bonding orbital of the bromine, right, which is going to stick out in that direction.

01:53 Well, it wouldn't be, it would have a node.

01:55 Sorry, it's going to be like something like that.

02:01 Yeah, and for that reason, we need these electrons to be in an orbital that are aligned with that orbital.

02:08 Right.

02:08 So it's just necessary to have this anti -periplanner configuration.

02:12 So we need to rotate about this bond, such that this hydrogen is pointing straight up.

02:20 So let's do that.

02:24 I erase this.

02:27 And then we'll just kind of have to think about what would happen to the other groups if we've made that rotation.

02:34 So i'm going to redraw the right hand side exactly the same.

02:44 And i'm going to redraw this, but with the hydrogen anti -barriplanner to the bromine.

02:49 And let's think about it.

02:50 So i think the easiest way for me to conceptualize hydrogen going straight up is to rotate this way, right, 120 degrees.

03:00 That is, so the ethyl is going to be moved down into the position that the methyl previously was.

03:09 So this will be the ethel.

03:11 And the methyl will be moved back into the position that the hydrogen previously was.

03:18 Right...

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