The justification or rather intuition behind Markovnikov's rule is the stability of intermediate carbocation. The more substituted the carbocation the more stable the product. But based on this ideas how can we satisfactorily explain the following reaction?
$$\ce{CH2=CHCl + HCl -> CH3-CHCl2}$$


1 Answer 1


Markovnikov's rule is usually stated in terms of subsitution on the alkene, for example:

In the addition of HX to an alkene, the hydrogen atom always attaches at the less substituted carbon atom (the carbon atom which already has more hydrogen atoms).

This version of the rule works well most of the time. It works in your case. The hydrogen atom of $\ce{HCl}$ end up on the carbon atom that already has two hydrogen atoms. The chlorine atom ends up on the other carbon atom:

$$\ce{CH2=CHCl +} {\bf \ce{H-Cl}} \ce{-> {\bf H}-CH2-CHCl-{\bf Cl} }$$

A better version of the rule has to do with the stability of the carbocation intermediate (and can be generalized to electrophiles other than $\ce{H+}$.

An alkene reacts with electrophiles in such a way as the electrophile attaches to the carbon atom which will result in the more stable carbocation.

The order of stability for simple carbocations is $3^\circ >2^\circ >1^\circ >\text{methyl}$, however there can always be more complex carbocations (de)stabilized by heteroatoms and/or resonance. The original wording of Markovnikov's rule may fail in these more complex cases.

For example, the addition of $\ce{HX}$ to $\ce{CH3O-CH=C(CH3)2}$. In this case the major product (before it decomposes into the aldehyde) is $\ce{CH3O-CHCl-CH(CH3)2}$. Two possible carbocations form, and the one where the proton attaches at the more substituted position is the one that is more stable.

First carbocation, $2^\circ$, but stabilized by resonance (more stable):

$$\ce{CH3-O-CH+ -CH(CH3)2 <-> CH3-O+ =CH-CH(CH3)2}$$

Second carbocation, $3^\circ$, but not stabilized by resonance (less stable):


The first carbocation goes on to form the product. Can you extend this to what happens to your molecule?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.