CH310M/318M Organic I

Dr. Brian Pagenkopf

Alkene Bromination

A lot happens in the first step of this reaction. The bromine-bromine bond is very weak. Attack by the alkene nucleophile (1) causes the bromine-bromine bond to break (2) so that bromine doesn’t end up with more than 8 valence electrons. Experiments suggest that there is no discrete carbocation intermediate, so the bromine must donate a pair of electrons to the other end of the alkene as it is being attacked (3). Instead of an intermediate cation, there is an intermediate bromonium ion that is then attacked by Br¯. The significance of an intermediate bromonium ion and not a carbocation is that the reaction is stereospecific.

Why the bromonium ion? So why would bromine want to straddle two atoms and formally accept a positive charge? The Golden Rules remind us that charge delocalization is better than charges on a single atom, and a lot of partial positive charge remains on the carbon atoms bromine is attached to\. The other reason for formation of a bromonium ion is that bromine is huge compared to a hydrogen. If there’s a carbocation next to a bromine atom its lone pairs of electrons are already getting close enough to interact with the positive charge just because it is so big. Since the bromine atom is going to be interacting with an adjacent positive charge anyway, limiting the intermediate to a formal bromonium ion lowers the energy of the entire system because it allows more efficient orbital overlap and charge distribution.

How is this reaction stereospecific? If we look at the bromination of methylcyclohexene, of the four possible products A through D, only the enantiomers A and B are formed. Note the stereochemical relationship of the bromines, they are anti (or trans relative to each other). (Syn is analogous to cis in these cases).

The selectivity observed in this reaction can be explained by looking closer at the mechanism. Let’s use a simpler example, bromination of cyclopentene.

If we guessed that an intermediate carbocation was formed, then all four bromination products E through H would be expected. As we have already studied, nucleophiles can attack the carbocation from either side of the p orbital.

However, in every case attack only happens from the opposite face from the bromonium ion.

Why does the bromine always attack from the back side? Sterics alone isn’t enough to explain why the nucleophile will come from the back side in every case, a molecular orbital explanation can account for the perfect anti selectivity of this reaction. What orbital are the electrons from the Br¯ going into? Remember the electrons cant go into a filled orbital, so they must go into the C-Br anti-bonding orbital. The anti-bonding orbital points away from the C-Br bond, and therefore incoming nucleophiles always attack from the opposite or back side.

Mechanistically Similar Reactions:

Addition of ROBr (Br₂ and H₂O or Alcohols)

Oxymercuration and reduction.

Addition of HOBr. Let’s look at an example that’s similar to the addition of Br₂ to methylcyclohexene, except in this reaction the solvent will be methanol. The intermediate bromonium ion is intercepted by methanol to give the products J and K. We expect an anti-relationship between the MeO and the Br because of the bromonium ion intermediate, but note that none of the regioisomers L or M were formed. It also appears that the nucleophile MeOH went for the carbon with more steric hindrance!

When the two carbons of the halonium ion are not the same, more positive charge ends up on the carbon attached to the greatest number of carbons. Or, the bromine ends up on the carbon with the most hydrogens. That is, the addition of HOBr follows Markovnikov’s rule. The nucleophile (a base) attacks at the most Lewis acidic carbon, not the carbon that is easiest to get to.