Bromination of Alkenes: The Mechanism
Last updated: November 28th, 2022 |
Bromination Of Alkenes: The Mechanism
How does bromination of alkenes work?
In the previous post (See Bromination of Alkenes ) we showed how the mechanism for bromination of alkenes can’t possibly go through a carbocation intermediate.
Why not? We have at least 3 good reasons:
- It’s stereoselective. The two atoms that form new bonds to carbon add to opposite faces of the alkene (“anti” stereoselectivity)
- The reaction is stereospecific – (Z)-2-butene gives different product(s) than does (E)-2-butene (in fact, the products are stereoisomers of each other).
- No rearrangements are observed, as they are for, say, the reactions of H-Cl with certain alkenes
None of those results are consistent with the reaction going through a planar carbocation.
There is one other detail to explain: when using water as a solvent, the reaction proceeds with “Markovnikov” regioselectivity. That is, the C-O bond is formed on the most substituted carbon of the alkene. (See post: Markovnikov’s Rule)
Table of Contents
- Bromination Proceeds Through A 3-Membered “Brominium Ion” Intermediate
- When Water Or Alcohols Are The Solvent, “Marknovnikov” Regioselectivity Is Observed
- Reaction Mechanisms Can Never Be “Proven”, But…
- (Advanced) References and Further Reading
1. The Bromination Mechanism Proceeds Through A 3-Membered “Brominium Ion” Intermediate
Back in 1937, Roberts and Kimball [Ref] used similar observations to point out the deficiency of a free-carbocation mechanism proposed earlier by Robinson [Note 1]. They argued that a free carbocation would allow for rotation about the carbon-carbon bond, and that this was inconsistent with the products observed in this reaction.
Instead, they invoked the intermediacy of a 3-membered ring structure with a positive charge on the halogen, which is “isoelectronic” (i.e. has the same electronic configuration) as a member of the oxygen family [epoxides, for instance]. Importantly, they said:
“Since the two carbons in [the structure] are joined by a single bond and by a halogen bridge, free rotation is not to be expected”
Furthermore, this proposal also explains the stereochemistry of the reaction:
This second step, which may be the addition of either a halogen ion X- or some other atom or molecule, is probably a simple “three-atom” reaction… In this case the new atom will approach one of the carbon atoms from the side opposite to the X atom already present. A bond to this carbon will be formed while the bond from the original X to the carbon is broken, with simultaneous neutralization of the charge of the ion. This process will always lead to trans addition
The final issue – what we now call “regioselectivity” – was not specifically addressed in this paper, but makes sense when you consider that it is the carbons which are electrophiles in this structure, not bromine [despite the positive formal charge on bromine, recall that bromine is more electronegative than carbon!] and the reaction will proceed at the carbon best able to stabilize positive charge.
Here’s what it looks like graphically:
So in this reaction, the alkene acts as a nucleophile, attacking the electrophilic bromine, giving rise to a 3-membered ring intermediate. This is then attacked from the back side [similar to the backside attack in the SN2 Mechanism] at the carbon best able to stabilize positive charge, to give the trans product.
[Note that, like a flat coin that can land on “heads” or “tails” with equal likelihood, the bromine can “land” on either face of the alkene – this gives rise to enantiomers in this case].
2. When Water Or Alcohols Are The Solvent, “Marknovnikov” Regioselectivity Is Observed In The Formation of Bromohydrins
If we use a solvent which can potentially act as a nucleophile such as water or an alcohol, the intermediate halonium ion can be “trapped” by solvent, giving rise to “halohydrin” products in the case where the solvent is water. Again, note that the water molecule attacks the most substituted carbon:
Why the most substituted carbon? Because that’s the carbon best able to stabilize positive charge, and since it has the most partial positive character, it will also be the most electrophilic of the two carbons.
[Worth noting: bromination of alkenes is technically an oxidation reaction, because each carbon goes from being bound to another carbon (0) to bromine (–1). The oxidation state of each carbon in ethene is +2; the oxidation state of each carbon in dibromoethane is +1. ]
3. Reaction Mechanisms Can Never Be “Proven”, But… The Bromination Mechanism Is On Very Solid Ground
Alright – so this “halonium ion” proposal seems to explain these results well enough. It’s a nice idea. But is there any hard evidence that has come to light since 1937 that helps us to know if they really exist?
As a matter of fact, yes: halonium ions have actually been observed in the wild! The halonium ion from the reaction below is particularly stable. [Think about it for a second – why might it be unusally stable? Note 2]
Furthermore, R.S. Brown and co-workers succeeded in using X-ray diffraction to obtain a crystal structure of this molecule. Here’s the iodonium ion portion of the structure:
In chemistry, obtaining an X-ray crystal structure of a molecule is considered to be near-definite “proof” of its existence. Although mechanisms can never be “proven”, this crystal structure represents about as close to a conclusive piece of evidence as you can get.
In the next post we’ll expand this idea of intermediate bridging ions to explain the results of a whole host of other reactions.
NEXT POST: Alkene Addition Pattern #2 – The “3 Membered Ring” Pathway
Note 1. 1952 Nobel Prize winner Robinson, (of Robinson Annulation fame), was a brilliant man, but he was wrong about the mechanism of this reaction according to a reference that I cannot obtain at the moment but is apparently referenced here [Ingold, Chem. Rev., 15, 225, 1934] This is instructive: looking backward, it seems obvious that the halonium proposal is correct, but in science we move forward gazing through a foggy windshield.
See: Roberts and Kimball http://pubs.acs.org/doi/abs/10.1021/ja01284a507
Note 2. The extremely bulky “adamantyl” groups make the approach of any nucleophile on the backside of the halonium ion very difficult, which gives this halonium ion an unusually long life.
(Advanced) References and Further Reading
- THE RELATIVE RATES OF BROMINATION OF THE OLEFINS
Harold S. Davis
Journal of the American Chemical Society 1928, 50 (10), 2769-2780
An early paper studying the kinetics of alkene bromination under a variety of conditions.
- The Halogenation of Ethylenes
Irving Roberts and George E. Kimball
Journal of the American Chemical Society 1937, 59 (5), 947-948
One of the earliest descriptions in the literature of a three-membered bromonium ion, accounting for the anti stereochemistry of this reaction.
- Stable carbonium ions. LXII. Halonium ion formation via neighboring halogen participation: ethylenehalonium, propylenehalonium, and 1,2-dimethylethylenehalonium ions
George A. Olah, J. Martin Bollinger, and Jean Brinich
Journal of the American Chemical Society 1968, 90 (10), 2587-2594
This is an early paper on the characterization by NMR of halonium (3-membered chloronium, iodonium, and bromonium) ions by ionization of 1,2-dihaloethanes in superacid medium (SbF5/SO2).
- The question of reversible formation of bromonium ions during the course of electrophilic bromination of olefins. 2. The crystal and molecular structure of the bromonium ion of adamantylideneadamantane
H. Slebocka-Tilk, R. G. Ball, and R. Stan Brown
Journal of the American Chemical Society 1985, 107 (15), 4504-4508
This paper describes the X-ray crystal structure of an isolated, stable bromonium ion. This is significant because it proves the intermediacy of these three-membered cyclic bromonium ions in the electrophilic addition of bromine to alkenes.
- Stable Bromonium and Iodonium Ions of the Hindered Olefins Adamantylideneadamantane and Bicyclo[3.3.1]nonylidenebicyclo[3.3.1]nonane. X-Ray Structure, Transfer of Positive Halogens to Acceptor Olefins, and ab Initio Studies
R. S. Brown, R. W. Nagorski, A. J. Bennet, R. E. D. McClung, G. H. M. Aarts, M. Klobukowski, R. McDonald, and B. D. Santarsiero
Journal of the American Chemical Society 1994, 116 (6), 2448-2456
Cyclic iodonium ions, analogous to the bromonium ions, can also be isolated and characterized. The parent alkene in these studies, adamantylideneadamantane, is prepared using a McMurry reaction with 2-adamantanone.
- Investigation of the Early Steps in Electrophilic Bromination through the Study of the Reaction with Sterically Encumbered Olefins
R. S. Brown
Accounts of Chemical Research 1997, 30 (3), 131-137
An account by R. S. Brown describing the research he carried out in interrogating bromonium ion intermediates by a variety of methods.
- ALKYNE via SOLID-LIQUID PHASE-TRANSFER CATALYZED DEHYDROHALOGENATION: ACETYLENE DICARBOXALDEHYDE TETRAMETHYL ACETAL AND ACETYLENE DICARBOXALDEHYDE DIMETHYL ACETAL
Rufine Akué-Gédu and Benoît Rigo
Org. Synth. 2005, 82, 179
- Principles of an Electronic Theory of Organic Reactions.
Christopher K. Ingold
Chemical Reviews 1934, 15 (2), 225-274
- —The relative directive powers of groups of the forms RO and RR′N in aromatic substitution. Part IV. A discussion of the observations recorded in parts I, II, and III
James Allan, Albert Edward Oxford, Robert Robinson, and John Charles Smith
J. Chem. Soc. (Res.) 1926, 129, 401-411
13 thoughts on “Bromination of Alkenes: The Mechanism”
I’m studying for a quiz and the problem I came across was an alkene addition with Br-Br and NaCl. I know that the Cl should add first but I can’t seem to figure out the mechanism for it.
No, what happens is that you form the bromonium ion with Br2 and then Cl(-) is the active nucleophile which attacks the resulting bromonium ion at the most substituted position.
Kind of a strange question though, because they probably don’t mention solvent. NaCl is not going to dissolve in an organic solvent. Better choice would have been Bu4N+ Cl- .
Good day Sir. I understand that halogenation reactions of alkenes involve the bridge ion as intermediate, and that no rearrangement occurs. But I don’t get how a 1,4 addition product is produced when conjugated dienes are involved as starting material. Since the intermediate is a bridge ion, how can resonance be formed? Thank you in advance for your reply.
See this post: https://www.masterorganicchemistry.com/2017/04/11/more-on-12-and-14-additions-to-dienes/
How can i synthesis only cis dibromo alkane from an alkene ?
Something I don’t understand (my professor couldn’t explain it well either): logically I would assume that the second attack be on the less substituted carbon as there would be less steric hindrance. Also, being more substituted (more neighboring carbons) means that these carbons would donate more of their electron density to that same carbon, which would make it more negative (hence less electrophilic and less reactive).
I noticed you didn’t go into more detail than “[…] the reaction will proceed at the carbon best able to stabilize positive charge.” Is there a deeper explanation to this?
Hi Idan. The reason the second attack occurs on the more substituted carbon, instead of the less substituted, is explained by the following. In the transition state leading to the product, one C-Br bond in the bromonium ion is already partially broken. Therefore, one of the two C atoms will bear a partial positive charge–it will be cation-like, in other words. Now, think about the relative stability of carbocations. Since more substituted carbocations are more stable, the more substituted carbon will undergo partial C-Br bond cleavage, and therefore nucleophilic attack will preferentially occur at the tertiary carbon in the above example.
While you’re right that the more substituted carbon is more sterically hindered, this effect is less important than the electronic effect of carbocation stability.
This is complicated by the fact that the major product isn’t 1,2-dibromoethane. The water also gets involved in the reaction, and most of the product is 2-bromoethanol.
Is there any reason why in case of water, H2O reacts before Br- does? H2O should be a weaker nucleophile than Br-.
Water is the solvent, so it is present in tremendously high concentration (pure water has a concentration of 55 M !). This explains why water attacks preferentially over Br-, which otherwise is a much better nucleophile.
Hey Prof., I get everything and you explain everything in a simpler way, but I don’t understand the part about “the reaction will proceed at the carbon best able to stabilize positive charge?” Could you perhaps show me that it is that C with greater partial charge in the corresponding example by using partial charge symbols?
It would be the most substituted carbon – think about carbocation stability, which follows the trend tertiary (most stable) > secondary > primary (least stable)
I was wondering about this, too. Thanks for taking time to explain it.