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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.

enter image description here

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 (disregarding stereochemistry) is the major product?

enter image description here

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    $\begingroup$ You're going through an allylic radical intermediate, right? How is the allylic radical stabilised? $\endgroup$
    – orthocresol
    Commented Oct 27, 2015 at 11:10
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    $\begingroup$ Can you draw a diagram to indicate the stabilisation? $\endgroup$
    – orthocresol
    Commented Oct 27, 2015 at 11:46
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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
    Commented Oct 27, 2015 at 13:00
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    $\begingroup$ Please give some more hint or explanation, I'm trying but I don't seem to get anything $\endgroup$
    – shaistha
    Commented Oct 27, 2015 at 13:10
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    $\begingroup$ SciFinder doesn’t find any reaction of 4,4-dimethylcyclohexene with NBS, for your reference(s). So the entire exercise is academic. $\endgroup$
    – Jan
    Commented Jun 1, 2016 at 19:46

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

This problem is a worked example in McMurry.

The first step is abstraction of a hydrogen atom (radical), leading to an allylic radical. These show resonance:

enter image description here

As a consequence, a bromine atom (radical) can form a bond in two positions for the top, and in two position for the bottom resonance structure (for the bottom, there is symmetry, so you end up with the same molecule). All of the products are chiral, so products come in pairs of enantiomers, in a 50:50 ratio because everything else is achiral.

[OP] Which one among these three products (disregarding stereochemistry) is the major product?

For other product mixtures, where the degree of substitution at the double bond differs between possible products, the one with the highest degree of substitution is favored (see e.g. this video). Here, there is no such difference. There might be a slight preference for abstracting the hydrogen further from the dimethyl group. There might also be a preference for not having the bromine end up right next to the dimethyl group (as in 3-bromo-4,4-dimethylcyclohexane).

In the absence of any of these preferences, the product formed from the symmetric allylic radical (3-bromo-5,5-dimethylcyclohexane) would be favored 2:1:1, just because of the statistics.

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