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Since it's in ether it is in an aprotic solvent which will only be sn2 or e2, so no carbocation or rearrangements. And it is sn2 because hcl is a strong nucleophile. Hope this helps
Yep you are correct!
Also, on that reaction, the Cl is less selective than br so you will get a mixture of products opposed to the Br that will have the most substituted carbon being the location the halogen (Br) is added.
True, I guess I just assumed they were the same since the carbocation and radicals get stabilized with the same effects. I bet you are right, the radicals must be a special case. Good call.