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Cyclohexene is reacted with bromine in carbon tetrachloride in the dark. The product of the reaction is:

enter image description here

I know that reaction with bromine in $\ce{CCl4}$ will result in vicinal dibromocyclohexane. So the answer is either A or B. But how to decide between A and B? Since its an SN2 reaction it would lead to inversion of the reacting compound. Does this have to do anything with the answer?

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    $\begingroup$ Philipp's answer is good, but structures A and B are both drawn in misleading ways. Both appear trans (one bromine is up and one is down), but neither correctly shows the conformation (should be either diequatorial or diaxial). $\endgroup$ – Ben Norris Aug 24 '13 at 16:09
  • $\begingroup$ I updated my answer. $\endgroup$ – Philipp Aug 24 '13 at 16:27
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    $\begingroup$ I just feel compelled to add: the drawings of the chair conformers in that question are frankly appalling. $\endgroup$ – Greg E. Aug 24 '13 at 20:18
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To answer this question you have to look at the reaction mechanism. Since the reaction is conducted in the dark, you don't have any radicals to worry about. Under these condition bromine reacts with an electron-rich alkene in a electrophilic addition reaction in the course of which a cyclic bromonium ion is formed. This bromonium ion is highly reactive and in the absence of other nucleophiles reacts with the leftover $\ce{Br-}$ ion in an $\text{S}_{\text{N}}2$ reaction which leads to the opening of the 3-ring and formation of the trans product.

enter image description here

Update:

As @BenNorris pointed out, the ring structure in the question are drawn wrongly. So, I will add the correct version: The dibrominated product will initially be formed in a diaxial conformation (this is always the case due to the ring geometry and the mechanism of the $\text{S}_{\text{N}}2$ reaction). This is usually a very unfavourable comformation so the ring will subsequently flip (if possible) to the diequatorial conformation.

enter image description here

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