Question 13

[asfino12217]

i get how according to Q11 item II, that the answer would be 3,4-dibromo-2-butene, but according to markovnikov's rule, the Br should attach to the most substituted carbon. So wouldn't the first Br attach to either the second or third carbon since they are more substituted?

[jellywing_2058]

Markovnikov’s rule is when the alkene is attacked by an acid (HBr, HCl) but in this case, the alkene is attacked by Br2 and there is no formation of a carbocation. In this case, you get a “cyclic bromonium ion” instead and then the other Br (negatively charged) attacks one of the carbons forming the ring in order to break it and to place a Br on 2 different carbon atoms.

[oh_nivek1167]

I understand the concept of the bridged bromium ion. however, the explanation to Q11 states:

"Back to Reaction I: we now know how to create the intermediate. The double bond induces a dipole in Br2 resulting in one Br atom adding to the carbon with the most H's (Mark's rule), leaving a carbocation as an intermediate. Draw the reaction on your scratch paper.

For this question, B is the only possible answer since it is the only answer with carbocations as intermediates! You should have them in your reaction on your note paper."

so I'm confused...the reaction of an alkene with a Br2 proceeds via a carbocation???

[mcat_premed3832]

Yes, there is a carbocation intermediate.

This is an overview:

www.chem.ucalgary.ca/courses/350/Carey5th/Ch06/ch6-7.html

To imagine the carbocation, you must imagine when the reactant would have looked like a fraction of a second before the production of the cationic cyclic bromonium ion intermediate.