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This is the answer to chapter 11, problem number 51 from the smith organic chemistry textbook.
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And this problem says, draw a stepwise mechanism for the following reaction and explain why a mixture of e and z isomers is formed.
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Okay.
00:22
And so we have two butine here, and we're treating it with hcl.
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And so we're getting the addition of a hydrogen and a chloride or a hydrogen and chlorine.
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This molecule, and we're getting a mixture of e and z isomers, and here is why.
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So the first step in this reaction is going to be the electrons of one of the pie bonds of the triple bond are going to grab this proton, and the hcl bond will break, and those electrons will go to the chlorine.
00:59
And so we will get this carbocation.
01:18
Okay.
01:18
And so as i said, we have a carbocation on this carbon now.
01:26
And basically the chlorine is then going to add to that carbocation.
01:32
It'll attack that carbocation.
01:35
And the reason that we get more than one product or a mixture of these enz isomers is, is because the chlorine can attack from either side.
01:45
There's nothing to create any sort of real preference here.
01:53
And so the chlorine can come from the top or the chlorine can come from the bottom...