5
$\begingroup$

Recently, in my chemistry lesson, we were discussing the mechanism of the bromination of alkenes. The teacher brought up the involvement of the cyclic bromonium intermediate. She also mentioned that the nucleophile attacks the more substituted carbon atom in the bromonium ion. However, this raised many questions among my classmates. Firstly, based on steric reasons, the nucleophile is more likely to attack the less-substituted carbon since there is less hindrance. Hearing this, the teacher then said that the reason for the preference is based on electronic factors. However, this makes little sense as well, considering that the presence of more alkyl groups will decrease the partial positive charge on the more substituted carbon, making it less attractive to the nucleophile (be it $\ce {Br^-}$ or $\ce {H2O}$).

I went to seek clarification by looking at more authoritative texts, such as Advanced Organic Chemistry: Part A: Structure and Mechanisms by Carey and Sundberg, as well as Guidebook to Mechanism in Organic Chemistry by Peter Sykes. However, they were citing very weird reasons that were not very intuitive. Here are some extracts from relevant chapters of the two texts:

From the former,

Unsymmetrical alkenes nevertheless follow the Markovnikov rule because the partial positive charge that develops is located predominantly at the carbon that is better able to accommodate an electron deficiency...

From the latter,

With an unsymmetrical alkene, e.g. 2-methylpropene (32), the more heavily alkyl-substituted carbon will therefore be preferentially attacked by the residual nucleophile, $\ce {Cl^-}$.

Why do the alkyl groups result in greater cationic character when they are in fact, electron releasing?

Improve this question
$\endgroup$
3
  • $\begingroup$ Clayden has a section on acid-catalysed epoxide opening, I think. The same principles apply here. $\endgroup$
    – orthocresol
    Aug 20, 2018 at 12:45
  • $\begingroup$ @orthocresol Is it possible to rationalise this using MO theory? Has it got something to do with the hyperconjugative interactions of the alkyl groups C-H and C-C sigma MOs with the C-Br sigma * MO? $\endgroup$
    – Tan Yong Boon
    Aug 20, 2018 at 14:36
  • $\begingroup$ Hmm... that's not the way it's conventionally taught, but it is possible. You can see some involvement of the CH bonds in the LUMO that Martin provided, consistent with the mixing that you described. I think it's important here to distinguish between discussing the ground state (i.e. the bromonium ion itself) and the transition state (for ring opening). Martin's answer talks about the GS, which is actually very convincing (to me), but the textbooks usually go for an explanation based on the TS. Although perhaps they are just two sides of the same coin... $\endgroup$
    – orthocresol
    Aug 20, 2018 at 18:06

1 Answer 1

Reset to default
11
$\begingroup$

With bromination, the first step is the formation of an adduct complex, which can be explained that the π-bond polarises the dibromide. The next step is the formation of the bromonium ion, which carries an overall positive charge. A symmetry-changing induction like substituents will always result in asymmetric carbon-bromine bond lengths in the bromonium ion. Since it is a cation, similar rules for the stability/instability as for carbocations apply. Therefore the higher substituted carbon will have the longer $\ce{C-Br}$ bond length, and the higher charge (more positive/less negative). This makes it a better target for the residual nucleophile $\ce{Br-}$ to attack there.

Here is the bromonium ion for the reaction $$ \ce{\underset{alkene}{H3C-CH=CH2} + Br2 -> [\underset{bromonium ion}{H3C-CH(Br)CH2}]+ + Br- -> \underset{$trans-$alkane}{H3C-CHBr-CH2Br}} $$ calculated at the DF-b97d3/def2-SVP level of theory, charges with NBO6, to illustrate the concept:

bromonium ion charges

Additionally the HOMO has the higher coefficient at the lesser substituted carbon, while the LUMO has the higher coefficient at the higher substituted carbon.

bromonium ion HOMO bromonium ion LUMO

From this, you can see that both quotes from the textbooks are correct.

Improve this answer
$\endgroup$
6
  • $\begingroup$ What could have caused the C-Br bond to the more substituted carbon to be longer? Orbital interactions? Electronic effects? $\endgroup$
    – Tan Yong Boon
    Aug 20, 2018 at 23:08
  • $\begingroup$ Why does the calculation show that the carbons are bearing negative charges? Shouldn't they bear a partial positive charge? $\endgroup$
    – Tan Yong Boon
    Aug 20, 2018 at 23:13
  • $\begingroup$ @Tan The molecular structure is always by a delicate balance of electronic effects and orbital interactions. The molecule will adopt the structure, which is lowest in energy (or all structures accessible at a certain temperature). I am not a big fan of calling things +/-I or +/-M effects, as that are zeroth order approximations. We know that there will always be hyperconjugation from a methyl group, which will certainly the case here, too. And that much should be clear, that a positive charge will be stabilised by it. It follows basically the same trend as carbocations. $\endgroup$
    – Martin - マーチン ♦
    Aug 21, 2018 at 8:56
  • $\begingroup$ @Tan No they should not. You might be confusing formal charges with partial charges. There is a significant difference in electronegativities for carbon and hydrogen, most of the electron density of the hydrogen is polarised to the carbon. Even such an over-simplified model like oxidation numbers would predict this trend. The values are certainly within expectation. $\endgroup$
    – Martin - マーチン ♦
    Aug 21, 2018 at 9:00
  • $\begingroup$ Are you certain that there will be hyperconjugation? The geometry of bromonium is slightly different from that of the conventional carbocation. Maybe the orbitals (i.e. C-Br sigma * and C-C sigma) may not have the right symmetry/orientation? Could you illustrate more quantitatively the hyperconjugation that is taking place? I am very curious... $\endgroup$
    – Tan Yong Boon
    Aug 21, 2018 at 9:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.