Orgo elimination question

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blackberryman

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Cyclic = no sn2 can occur here. (1) the bromine leaves due to a strong base as seen in traditional E2 reactions (2) the hydrogen connected to the carbocation moves into the ring to form a double bond (3) the double bond creates a carbon (the carbon with the second bromine) to be connected to 5 other atoms (4) the second bromine leaves so that the carbon can return to a formal charge of 0

Hope this helps
 
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Aromatic = no sn2 can occur here. (1) the bromine leaves due to a strong base as seen in traditional E2 reactions (2) the hydrogen connected to the carbocation moves into the ring to form a double bond (3) the double bond creates a carbon (the carbon with the second bromine) to be connected to 5 other atoms (4) the second bromine leaves so that the carbon can return to a formal charge of 0

Hope this helps

I am confused, as this reactant isn't aromatic?
 
1. The electron pair bond between the C-Br moves to the Br, forming Br− and setting it free from cyclohexane
2. The nucleophile (-OH) gives an electron pair to the adjacent axial H, setting H free and bonding to it to create H2O
3. The electron pair bond between the adjacent axial H moves to the bond between the two C-C making it C=C

Majority of the time this is an E2 rxn, while SN2 can occur. I was confusing EAS and E2 cyclic rxns
 
I believe that there is a mistake in the answer. First of all KI is a good nucleophile and therefore prefers SN2. Secondly, in order ti do Elimination you need a strong base. You do not have a strong base here. How can 2 Bromines leave using only KI? This must be a mistake. You were right with your answers.

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