The following reaction is given:
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:
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.
Thus, all four products A, B, C, and D are formed. Is this correct or not? Thanks.