Bromination of Alkenes
Last updated: November 18th, 2022 |
Bromination of Alkenes Gives anti Products
In a previous post we went through the key reactions of the carbocation pathway. It’s a family of reactions which proceed through 1) attack of an alkene upon an acid, forming a free carbocation, and 2) attack of a nucleophile upon the carbocation.
Although we saw that several key reactions of alkenes were consistent with this mechanism, it isn’t the case for all. Take the bromination of alkenes, for instance.
Treatment of an alkene with bromine (Br2) in a chlorinated solvent (CHCl3, and CH2Cl2 are popular choices; CCl4 is often cited in textbooks [Note 1]) leads to the formation of products containing two bromine atoms.
In this post we’ll describe the main observations that have been made about this reaction, and in the following post we’ll show the best theory we have for the mechanism.
Table of Contents
- Bromination Of Alkenes Observation #1: Only anti Products Are Observed
- Bromination of Alkenes Observation #2: The Reaction Is Stereospecific
- Observation #3: Rearrangements Are Never Observed
- Observation #4: Certain Solvents Can Affect The Reaction Products
- Summary: Bromination Of Alkenes
1. Bromination Of Alkenes Observation #1: Only anti Products Are Observed
Possibly the most interesting feature of this reaction is that the products follow a very predictable stereochemical pattern. For instance, in the reaction of cyclohexene with Br2, the two bromine atoms add to opposite faces of the alkene (“anti” stereochemistry). No “syn” products are observed. [Note 2]
2. Bromination of Alkenes Observation #2: The Reaction Is Stereospecific
What’s even more interesting is that the stereochemistry of the starting alkene is directly related to the stereochemistry of the product.
For instance if we treat cis-2-butene [aka (Z)-2-butene] with Br2, we get a mixture of enantiomers.
But if we treat trans-2-butene, we only get a single product (“meso” 2,3-dibromobutane), which is itself a diastereomer of (S,S)-2,3-dibromobutane and (R,R)-2,3-dibromobutane.
These starting materials, cis-But-2-ene and trans-But-2-ene, which differ only in the configuration of the double bond, lead to stereoisomeric products. This type of process is given the name, stereospecific.
What’s important about this? Two things.
First, given the product, it’s possible to work backwards to figure out which isomer of cis-But-2-ene was the starting material. (Expect to be tested on this).
Secondly, the fact that this happens means that the mechanism is inconsistent with a free carbocation! If there was a free carbocation, the stereochemistry of the starting alkene wouldn’t matter, since attack can come from either face. [Indeed, we know from labelling experiments that the reaction of H-Cl with cis or trans 2-butene is not stereospecific].
3. Observation #3: Rearrangements Are Never Observed
Another reason this reaction is consistent with the absence of a free carbocation is because rearrangements are never observed. For example, in the case below, we’d expect to see rearrangement (a 1,2-alkyl shift, to be precise) if a free carbocation was formed.
Instead, note that the methyl groups stay in the same place.
4. Observation #4: Certain Solvents Can Affect The Reaction Products
Here’s another experimental observation. The solvent matters.
When we use water as the solvent for this reaction, we get the product below.
[Note – this is called a “bromohydrin” since we have incorporated both bromine and water] .
Note that the stereochemistry is still “anti”, as before.
What this means is that somehow our solvent has intercepted a reactive intermediate in this reaction to produce the product above. (Note – this also occurs when we use alcohols as solvents; in these cases, ethers are obtained).
What’s even more interesting is that the reaction is regioselective. That is, when we have an unsymmetrical alkene, the major product is the one where water has added to the most substituted carbon of the alkene [most substituted = the sp2 carbon of the alkene directly attached to the fewest hydrogen atoms]. Such so-called “Markovnikov” selectivity was also observed in the reactions that proceed along the “carbocation pathway”.
5. Summary: Bromination Of Alkenes
So what’s going on? How can we explain these observations?
- Anti stereochemistry observed
- No carbocation intermediate (stereospecific, no rearrangements)
- Can be intercepted by nucleophilic solvent; attack occurs at most substituted carbon of the original alkene
What’s the mechanism for this process?
In the next post we’ll go through the best hypothesis we have for the mechanism of this reaction.
NEXT POST: Bromination of Alkenes – The Mechanism
Note 1. Off topic note: For some reason, textbooks continue to cite CCl4 as a common solvent for these reactions. Back in the day, CCl4 was a commodity chemical used for drycleaning (among other uses) and was a cheap, commonly available solvent. Since the discovery of its role in depletion of the ozone layer, the Montreal convention on CFCs has heavily restricted the availability of CCl4 to the point where legally obtaining CCl4 has become extremely difficult for labs in some countries. (Although it can often be substituted for other solvents, there are cases where nothing else will do. During my PhD in Canada several of us hoarded old, near-empty bottles of CCl4 the same way one might guard a precious bottle of 18-year old Scotch. )
Note 2. There are exceptions to the rule that only anti products are observed. If bromination occurrs on an alkene adjacent to an aromatic ring, some products that appear to have been produced from syn bromination are observed. This is particularly true in any case where a long-lived, stable carbocation can be formed (such as a benzylic carbocation).
13 thoughts on “Bromination of Alkenes”
Is the reaction electrophilic addition or something else?
Yes, bromine is an electrophile, this is electrophilic addition.
Because H2O attacks the bromonium ion in the second step as it is a better nucleophile than Br-.
H2O is not a better nucleophile than Br-, but if H2O is present in large excess (e.g. as the solvent or co-solvent) relative to Br- then its corresponding rate of reaction will be much higher, and this is why halohydrins are formed when bromination is performed with added water.
Hi, I have been trying to do a Bromination reaction. I start from D-Diethyl tartrate in CH2Cl2 and using PBr3 but I can’t make it work, I always end with yields really low (around 20%), any recommendation that I could get from you?, thank you :)
You didn’t give enough information. What product are you trying to obtain? When you activate the oxygen with PBr3 there is the possibility for anchimeric assistance from either the oxygen or the carbonyl in the ester.
Observation 2- I had this exact problem on an organic chemistry quiz. I put EXACTLY what you have written, yet it was marked as wrong. The teacher said it needed an OH group… please explain
Why alkene give alpha halo alcohol with bromine water while it gives bromo alkene with bromine and carbon tetra chloride??
Does it matter if the solvent is polar protic or aprotic for bromination to occur? Such as Br2, THF adding to the alkene?
If the “polar protic” solvent is water or an alcohol, then it will attack the bromonium ion and form , for instance, a “bromohydrin” in the case of water. Unless that’s what you’re intending, it’s best to use either chlorinated solvents (such as chloroform or CH2Cl2) or polar aprotic solvents (such as THF, ether, etc.)
For observation #1 and #4 I don’t see how the product that is shaded in grey is any difference the one that has been labeled “anti”. In both products for both examples, the Br and either the Br or the OH have added to opposite faces. One is wedges and one is dashed. Basically for #1 one bromine is up when the other is down in both products and same thing for the Br and the OH in #4
Or does the grey just mean it is the same product but flipped/rotated?
The grey product is not the same molecule…there is no way to rotate it in three-dimensional space so it matches up with the product shown in black. They are stereoisomers, and furthermore, they are enantiomers or mirror images of each other.