Tuesday, December 10, 2013

Ch 9: Question 56 (p. 409)


Given the conditions, these reactions will be both SN1 and E1 reactions, meaning they will happen in more than one step, and form carbocations. The first step is the same in each of the products, making a carbocation where the Br was and a Br ion. For the first set of products, we then add CH3OH to the carbocation, making the SN1 and E1 product as shown. In the second set of products, there is a 1,2 alkyl shift making the cyclohexane a cyclopentane and moving a carbon up. This shifts the carbocation. From there the CH3OH is added and makes the SN1 and E1 products shown. In the third set of products, there is a 1,2 methyl shift, where the cyclohexane remains a cyclohexane, but one of the methanes is shifted. This too moves the carbocation, making the SN1 and E1 products shown when CH3OH is added.

1 comment:

  1. This is a very good explanation of the problem, and I can see the thought process behind it. This is a good differentiation between E1 and SN1

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