I know that NBS is a selective brominating agent especially on the allylic position. And its reaction mechanism is of free radical.

But I can't figure out the reason for the change in position of the double bond in the products.

It would be helpful if someone could give the mechanism along with its reason.

reaction of 4,4-dimethyl cyclohexene with NBS at its respective allylic positions (A)and(B)

This is what I'm thinking, that might be happening there but still I don't understand how and why something like this would happen.

for A for B

My mistake, number 3 and the product which I have marked as not formed are same.

2)Which one among these three products is the major product?

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    $\begingroup$ You're going through an allylic radical intermediate, right? How is the allylic radical stabilised? $\endgroup$ – orthocresol Oct 27 '15 at 11:10
  • $\begingroup$ The stabilization process for the free radical is the same as the carbocation, right.Because both are electron deficient. But I really can't understand the displacement of double bond in this case. $\endgroup$ – shaistha Oct 27 '15 at 11:34
  • 1
    $\begingroup$ Can you draw a diagram to indicate the stabilisation? $\endgroup$ – orthocresol Oct 27 '15 at 11:46
  • $\begingroup$ I have added it to my answer,and I don't understand why that is not formed when acting upon allylic site (B) $\endgroup$ – shaistha Oct 27 '15 at 12:55
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    $\begingroup$ 1) The triradical structures look horrifying, rather you would draw appropriate fish-hooks to move the double bond in one step. 2) Yes it is formed. Look closely. Look harder. $\endgroup$ – Jan Oct 27 '15 at 13:00

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