Markovnikov’s Rule Has Nothing to Do With H-Atoms and Everything to Do With Carbocation Stability

Markovnikov’s Rule Has Nothing to Do With H-Atoms and Everything to Do With Carbocation Stability

As an organic chemistry student you will learn many ‘tricks’ for understanding what happens in responses. One of these is how Markovnikov’s rule helps you clarify the products of alkene addition responses.

Understanding Markovnikov’s rule is meaningful to quickly identifying products in alkene and alkyne responses.

I’ve heard many mnemonics over the years that help you memorize without really understanding They do help so I won’t mock them. In fact, here are the two that I think are most useful

  1. The (H) high get (H) richer
    Translation: The carbon atom containing more H-atoms at the start of the reaction gets the new H-atom
  2. H goes to the less substituted carbon
    Translation: Once again, the carbon atom with less bonds to carbon (therefor more bonds to hydrogen) gets the hydrogen atom

These tricks are OK if you’re in a crunch and need answers fast. However, without understanding why the reaction proceeds as it does, you risk forgetting the ‘trick’ and consequently missing a question on your next organic chemistry quiz or exam

If you don’t understand why this happens, you will find yourself confused when faced with a reaction that doesn’t follow Markovnikov’s rule. Or perhaps a reaction that has a tricky intermediate step.

So here’s my bold statement:

Markovnikov’s rule has nothing to do with H-atoms.

Instead, it has everything to do with carbocation stability.

The speed of an organic reaction is determined by the stability of its intermediates. The more stable an intermediate, the more likely it will form, therefor the faster the reaction.

Let’s see how this plays out using a simple reaction of H-Br adding to an alkene

Nucleophilic pi electrons reach out for the slightly positive electrophilic H-atom on the H-Br molecule

The electrons which used to form the pi bond now form a sigma bond between one of the former sp2 carbon atoms and hydrogen

The question is WHICH ONE?

Don’t answer this just however, instead think of what happens to the other carbon atom. It used to have 4 bonds and consequently a complete octet. Having lost the pi bond it now has just 3 bonds and 6 electrons in its octet. This deficiency makes the carbon unstable, unhappy, and results in a net charge of +1

Carbon doesn’t like a +1 charge and will try to prevent the reaction from forming. That’s why the carbocation formation step is referred to as the rate-calculating step (very important for SN1/E1 responses)

If the 2 carbon atoms are equivalent there is no concern for Markovnikov’s rule. However if 1 carbon is more substituted than the other, the more substituted carbon will be much happier holding the pi bond

Back to Markovnikov’s rule, What he’s really saying is as follows:

A molecule prefers to undergo a reaction with a more stable carbocation intermediate. This refers to the more substituted carbocation intermediate

When breaking the pi bond, choose the more stable carbon to be the carbocation.

H winds up on the other carbon atom BY DEFAULT.

Do you see the difference? It’s not about the product.

Instead it’s about the stability of the intermediate.

Since we started with the H-Br example let’s complete the mechanism.

Opposites attract, and so the negative nucleophilic Br- will attack the carbocation since it’s positive. It’s not that Bromine wants to add to the more substituted carbon. Instead, it’s about Br as a negative ion wanting to attack a positive carbon.

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