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The bromonium ion is formed when alkenes react with bromine. When the π cloud of the alkene (acting as a nucleophile) approaches the bromine molecule (acting as an electrophile), the σ-bond electrons of Br2 are pushed away, resulting in the departure of the bromide anion.  In terms of the orbital picture, this displacement reaction corresponds to the interaction between the HOMO of alkene (nucleophilic electron pair) and the back-side lobe of the empty σ* orbital of Br-Br bond (back-side attack).
 
 

The species formed is a high-energy intermediate along the reaction path. It will react further with some nucleophile; in the simplest case with the bromide ion. In a way, it is a hidden version of the carbocation that normally forms when alkenes react with positively charged electrophiles such as H+. This participation of the lone electron pair of bromine in bonding can be represented by the resonance forms, showing that the positive charge is shared by the bromine atom (center) and the two carbon atoms (left and right).
 
 

Since the bromonium ion will react with the nucleophile in the next step, let's look at its LUMO.  The largest lobes of the LUMO are present at the "bottom" (back-sides) of carbon atoms and on the bromine atom. Not surprisingly this is exactly where the positive charges are in the resonance structures (see above). The best overlap with the HOMO of an approaching nucleophile, let's say Br−, can be obtained if that nucleophile attacks from the bottom face (attack on each carbon is equally likely).  Attack on the bromine from the top face would yield back the alkene and Br2.  The first option is preferred for thermodynamic reasons (C-Br bonds are stronger than the Br-Br and π-bond of the olefin).

 

How can one predict this peculiar shape for the LUMO of the bromonium ion? In fact, it is not that complicated. In its essence the bromonium ion is formed just by a combination of π electrons of the double bond and an empty p orbital of  Br+ as shown on the right.  The bonding combination is occupied by two electrons that were originally in the π system.  The antibonding combination is empty. It is the LUMO of the species. To better appreciate the 3D shape of this molecular orbital look at its spinning picture above.

 

If the alkene is substituted, let's say with the methyl group, the bromonium ion is no longer symmetric. The more substituted carbon (2o) can accommodate positive charge better than the less substituted carbon (1o), or in other words, resonance structure C contributes more to the description of the molecule than resonance structure A (see below). This situation is just the reflection of the increased stability of the secondary carbocations over the primary carbocations that is due to hyperconjugation and inductive effects.

 

As the consequence, the C(2o)-Br bond is longer (and weaker) than the C(1o)-Br bond.  If a substituent can stabilize the cationic center even more, as is in the case of the vinyl group replacing the methyl (when 1,3-butadiene is the starting olefin), the interaction between the bromine lone pair and the cationic center becomes so weak that a normal (delocalized) carbocation forms, rather than a bridged bromonium ion.

On the left,  you may see the geometry and the LUMO of the unsymmetrical bromonium ion. Look what just a simple methyl substituent will do!  The secondary carbon is almost planar and only weakly bonded to the bromine (the long bond). The LUMO is now mainly localized on that carbon (the bromine contribution notwithstanding), and it resembles very closely the LUMO of a secondary carbocation. You may recognize its familiar lobes of the p orbital on carbon (in the center of the structure) and small "wing" contributions from the hyperconjugating carbon-hydrogen bonds (on the right side).

The electronic structure of the bromonium ion has its consequences in the stereochemistry of the reaction where the bromonium ion is an intermediate.  Let's consider a simple example of addition of Br2 to propene in H2O. The unsymmetrical bromonium ion forms exactly as we have described above. In the reaction run in water, the water itself serves as a nucleophile, competing successfully with the bromide ion. Water molecules surround the bromonium ion from all sides, after all.

The attack by water takes place from the bottom face (back-side to the "departing" bromide) on the secondary carbon: that is where the LUMO lobe is the most accessible (see the picture above).  The back-side attack leads to the anti addition (after all is done, Br and OH groups end up on the opposite faces of the starting double bond), and the nucleophile (water, or after the follow-up loss of H+, the OH group) adds to the more substituted carbon (yes... that's Markovnikov's rule). The product stereochemistry (anti relation of Br and OH) and regiochemistry (OH on the more substituted carbon) are controlled by the LUMO of the bromonium ion.

For practice, you may write the remaining mechanistic details to answer the following questions: (a) How is HBr produced? (b) How is the mirror image molecule formed? (c) Is the product optically active?
 
Molecular Gallery Last updated 06/07/07 Copyright 1997-2008
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