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uysha [10]
3 years ago
5

Draw the product formed by the reaction of potassium t-butoxide with (1s,2s)-1-bromo-2-methyl-1-phenylbutane (shown below). (dra

w the correct stereoisomer of the product.)

Chemistry
2 answers:
Scilla [17]3 years ago
4 0

Answer:

(Z)-(2-methylbut-1-en-1-yl)benzene

Explanation:

In the E2 elimination reaction, we will have a <u>single-step mechanism</u>. In this single step, we need an <u>anti-peri-planar geometry</u>. In other words, we need a specific configuration for the <u>leaving group</u> and the <u>H that would be removed</u>. These 2 atoms must have an opposite configuration when the base attacks the hydrogen.

With this in mind, we have to <u>rotate</u> the molecule in order to obtain the desired geometry (see figure). When we do this, the <u>methyl group</u> would be placed on <u>top</u> and the <u>ethyl group</u> at the <u>bottom</u>.  

Finally, when the elimination takes place (<u>blue arrows</u>) the configuration of the final alkene would be "Z" due to the previous rotation that we did in order to obtain the correct configuration for the elimination.

VikaD [51]3 years ago
3 0
 <span>You have to use a Newman projection to make sure that the H on C#2 is anti-coplanar with the Br on C#1. (Those are the two things that are going to be eliminated to make the alkene.) 

My Newman projection looks like this when it's in the right configuration: 
Front carbon (C#2) has ethyl group straight up, H down/left, and CH3 down/right 
Back carbon (C#1) has H straight down, Ph up/left, and Br up/right. 

Then when you eliminate the H from C#2 and the Br from C#1, you will have Ph and the ethyl group on the same side of the molecule, and you'll have the remaining H and CH3 on the same side of the molecule. 

This is going to give you (Z)-2-methyl-1-phenyl-1-butene.</span>
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