Ace Your Next Organic Chemistry
Exam.

With these Downloadable PDF Study Guides

Our Study Guides

Alkene Reactions

By James Ashenhurst

Bromination of Alkenes: The Mechanism

Last updated: July 5th, 2019 |

Bromination Of Alkenes: The Mechanism

How does bromination of alkenes work?

In the previous post 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

0-brom

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.

Table of Contents

  1. Bromination Proceeds Through A 3-Membered “Brominium Ion” Intermediate
  2. When Water Or Alcohols Are The Solvent, “Marknovnikov” Regioselectivity Is Observed
  3. Reaction Mechanisms Can Never Be “Proven”, But…
  4. Notes

1. The Bromination Mechanism Proceeds Through A 3-Membered “Brominium Ion” Intermediate

Back in 1937, Roberts and Kimball used similar observations to point out the deficiency of a free-carbocation mechanism proposed earlier by Robinson [see note]. 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 carbonwhich 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:

1-brom

 

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] 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:

2-brom

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 this be?]

4-adamantyl

 

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:

Screen shot 2013-03-04 at 2.03.45 PM
Figure adapted from R.S. Brown et. al., Journal of the American Chemical Society, 1994, 116, 2448-2456. Used with permission.

 

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


Notes

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

Related Posts:

Comments

Comment section

13 thoughts on “Bromination of Alkenes: The Mechanism

  1. 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?

    Thanks,
    Soji

    1. It would be the most substituted carbon – think about carbocation stability, which follows the trend tertiary (most stable) > secondary > primary (least stable)

  2. Is there any reason why in case of water, H2O reacts before Br- does? H2O should be a weaker nucleophile than Br-.

    1. 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.

  3. 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.

  4. 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?

    1. 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.

  5. 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.

  6. Hello,

    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.

    1. 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- .

Leave a Reply

Your email address will not be published. Required fields are marked *

This site uses Akismet to reduce spam. Learn how your comment data is processed.