Bromination of alkenes

In regard to the mechanism of halogen addition, consider the situation when we add a halogen (say, bromine) to an unsymmetrical alkene about the double bond (say, propene) in the presence of $\ce{CCl4}$.

I knew the mechanism would be something of this sort:

A cyclic bromonium ion would be formed with the two carbons of the double bond. The second bromide ion (acting as a nucleophile) attacks on the more substitued carbon from the side opposite to the bromonium ion to cause ring opening and formation of a vic-dibromide.

My teacher however tells me that other than ethene, no other alkene forms bridges like these, because of unstability of bromonium complex due to repulsion between substituents on carbon and the bridged element.

I have two questions in this regard:

1. Which of the two is the correct mechanism?
2. If the former is indeed the correct mechanism, then would the nucleophile attack the least substituted carbon? (seeing as how the partial positive charge on the more substituted carbon is nuetralised to a greater extent due to $+I$ effect of attached methyl/ethyl groups).

1 Answer

Yah, there are two mechanisms for this, one is as you wrote and the other one is th stereospecific reaction where stereoisomeric starting substrate produce products which are stereoisomers of each other.

1. First $\ce{Br2}$ will dissociate into ions:-

$\ce{Br2-> Br+ (electrophile) + Br-(nucleophile)}$

2. $\ce{H2C=CH2 + Br+ -> H2(Br)C-C+H2(carbocation)}$ slow
3. $\ce{+CH2-CH2Br + Br- -> CH2Br-CH2Br}$

Symmetrical alkenes and Symmetrical reagents, addition reaction gives:-

$\ce{cis ~alkene + syn ~addition -> meso}$

$\ce{trans + syn~addition -> racemic}$

$\ce{cis + anti~ addition -> racemic}$

$\ce{trans +anti ~addition -> meso}$

You can watch this video (http://m.youtube.com/watch?v=5GQelnluHzE)

My teacher however tells me that other than ethene, no other alkene forms bridges like these, because of unstability of bromonium complex due to repulsion between substituents on carbon and the bridged element.

For this I have not heard any thing as such and also couldn't find it.

For the most common mechanism which i was taught was the first one with bromonium ion complex formation.

As for your doubt about nucleophilic attack, negative bromide anion is attracted to the slight positive charge on the carbon atoms. It is blocked from nucleophilic attack on one side of the carbon chain by the first bromine atom and can only attack from the other side.

Personally how I solve it, is to check for the stability of the carbocation formed just before the nucleophilic attack practically because its $\ce{SN2}$ mechanism carbocation is not formed but I just use it like this best way is to check for high electron density. Though when it comes to aryl group, it is seen that it attack to the place with most substituted C-atom.

Here the stability was checked by hyperconjugation. Inductive effect is not the most effective way. You should know that the priority order of these effects to check the stability,

$$\mathbf{aromatisation~\Rightarrow~resonance~\Rightarrow~hyperconjugation~\Rightarrow~inductive effect}$$

• The attack of the second bromine is an Sn2 mechanism,right? In that case, how does a carbocation form? – Daipayan Mukherjee Oct 11 '15 at 13:32
• I have corrected my answer. Actually what i meant was that just before, for the fraction of second, when the bond breaks, to know where the nucleophile would attack, the stability of carbocation is checked though it might be miss leading as carbocation is originally not formed, substrate reactivity will increase with increase of electron density. Its anti electrophilic addition. That occurs here. And that, again its my own way so please don't think much about that, best and most logical according to $\ce{SN2}$ reaction is, that it will attack to more substituted group. – shaistha Oct 11 '15 at 16:15
• So what im actually doing is, that if cation were to be formed just before the nucleophile attaches to ANY one side, then where would it get attached. For example for propene, if you use $\ce{Br2 and NaCl}$, then here $Cl$ will be the nucleophile. So, what would it form (major) 2bromo-1-chloro propane or 1-bromo-2-chloro propane? The answer is the second one. Hone – shaistha Oct 11 '15 at 16:20
• (-1) Since randomly bolding stuff seems to be prevalent in this answer and the comments, bromine does not dissociate into ions, especially not in a non-polar solvent... – orthocresol Jan 10 '16 at 8:03