We've already talked about bromonium ion considering the alkene halogenation. It's a three terms cycle with a positive charge on the bromine atom. Remember that bromine (Br) belongs to the VIIth group of the periodic table, and therefore should have seven electrons in the outer shell:
Alkenes halogenation (treatment with Br2) takes place with the following mechanism:
When the bromonium ion is formed, the reaction isn't chemoselective and the result is the same whatever position is attacked by the bromide ion Br-.
What happens is different depending on the solvent because once the bromonium is formed the attack of different nucleophile gives different products. A competition is realized between bromide ion (halogenation reaction) and the solvent (ex: water, ethanol, etc.) itself. Considering that the amount of solvent is definitely greater compared with the bromide, only an irrelevant amout of dibromoalkane will forme compared with the alcohol or water addition product.
Solvent = alcohol
The reaction product of an alkene to elementary bromine is a compound called bromohydrin (halohydrin in general) while is an ether when the solvent is an alcohol.
We can notice that in both cases (but is actually true for alkene halogenation) the nucleophile attacks the more hindered end. Representing the reaction as an SN2 reaction can therefore seem a mistake (generally in an SN2 reaction the nucleophile attacks the less hindered end):
Why doesn't the reaction go through an SN1 mechanism and therefore through a tertiary carbocation?
Actually the reaction is not a pure SN2 reaction, but something in between an SN1 and SN2 mechanism. Was sperimentally determined that a real tertiary carbocation doesn't form but we can guess that the bromine atom starts leaving causing a partial positive charge on the more subsituted carbon. Consequently the weakening of the C-Br bond the nucleophile attack at the most hindered end is made easy, even if through a dynamic that is more similar to anSN2 reaction mechanism.
Here is the method currently considerated the most appropriate to represent this kind of reaction: is called "loose SN2 transition state"
Epoxides synthesis from bromohydrins (halohydrins)
Bromohydrines, got through addition of elementary bromien to an alkene in acqueous solvent, can be treated with a strong base as NaOH, in order to deprotonate the hydroxil group and obtain a particular kind of ether called epoxide.
- reaction mechanism
The reaction mechanism is quite simple. Usually treating an alcohol with an hydroxide we would get an alkoxide. Indeed the first step of the reaction is an acid-base, with deprotonation of the hydroxil group by the hydroxide (the halohydrin is even more acid than a normal alcohol because of the inductive effect) that gives an alkoxide. However the alkoxide has the possibility to give an intramolecular nucleophilic substitution. We have therefore the nucleophilic attack from O- on the electrophile carbon bonded to Br. Consequently, the bromide ion leaves.
Bromolactonization is a different reaction compared with the previous ones. Is one of the few intramolecular reactions whose dynamic is not controlled by the dimension of the new cycle formed but rather f by the stability of the carbocation.
First of all, the lactone forms in correspondence of the double bond. Subsequently or at the same time, the base (essential to conduct this reaction), deprotonates the carboxyl group. As we told previously, generally the reaction of an intramolecular reaction that brings to a cycle formation is under the number of atoms of the cycle controll. On the other side we have to consider that the carboxilate is extremely weak and would attack only a real carbocation therefore a very activated electrophilic center. Considering that the carbocation has to form before chronologically it will also control the development of the reaction.