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The following reaction is given:

enter image description here

I want to be able to tell which of the following stereochemical outcomes are possible. enter image description here

I am aware that there are other possible products than the ones shown here such as if the methyl shift did not occur or if the carbocation was placed on the lower carbon on the C=C. But for when the carbocation is on the top carbon in the C=C and then rearranged, I was thinking all 4 products given here are possible.

Here is my reason:

After the proton transfer, there can be a methyl shift. When the methyl shift occurs, either methyl group (wedged or dashed) on that top-most carbon could shift. Thus, if the wedged methyl shifted, the products A and B represent that process. If the dashed methyl shifted, and C and D represent that process. This is what I mean:

enter image description here

Next, after the methyl shift, the carbocation is formed. Cl can attack from either side. This explains the difference between 1 and 2 (where the wedged methyl shifted) in terms of the position of Cl, as well as the difference between 3 an 4 (where the dashed methyl shifted) in terms of the position of Cl.

enter image description here

Thus, all four products A, B, C, and D are formed. Is this correct or not? Thanks.

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    $\begingroup$ Yes to your question. By "dilute HCl" I hope you don't mean hydroxyl solvent (H2O, MeOH, etc.) because tertiary chlorides will undergo solvolysis. $\endgroup$
    – user55119
    Nov 3, 2021 at 22:19

1 Answer 1

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This is correct. Since the methyl rearrangement can occur at either side of the planar carbocation, and Cl- can attack either side of the resulting planar tertiary carbocation, all four possible diastereomers are formed with equal preference. Here, you don't need to consider 1,3-diaxial interactions of cyclohexane either since the reaction occurs at carbons 1 and 2.

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