Markovnikov’s Rule explained

[BACK]

In the addition of an acid (say, HCl) to an ‘ene’ group, the proton (electrophile) first attacks the p electrons of the double bond, forming a new s bond with one of the carbons, leaving the other carbon with one too few electrons (carbocation). Then, in the second step the nucleophile Cl^– attacks the carbocation, filling it’s empty orbital and forming a new bond.

Reaction 1: Example of an addition of an acid to a symmetrically substituted alkene. Markovnikov’s rule doesn’t apply here.

The transition state in both these first steps have to be stable enough to stick around long enough for the Nucleophile to attack. If it were to decompose too quickly, the reaction would never reach completion. Since both sides of the ‘ene’ group are identical in this example, both products will be equally likely. Here, they are indistinguishable anyway.

In the second step, the reactive part of the transition state is the carbocation (it has the charge). Anything we do to stabilize the carbocation (reduce the charge) will result in that particular reaction path being favored because it will help the transition state ‘stick around’ long enough for the second step to happen.

If we put on one of the carbons a substituent that pushes electron density onto that carbon, then that carbon will be stabilized if it becomes the carbocation. If, we put an electron-withdrawing (highly electronegative) substituent on a carbon then it will be destabilized as a carbocation. It turns out that R groups (alkyl groups) are electron donating (they have lower electronegativity than a Carbon with double bonds) while a halogen would be electron withdrawing (it has high electronegativity).

Thus, for a hydrocarbon-only compound with an asymmetrically substituted ‘ene’ group, the carbon with more R groups will be stabilized relative to the carbon with more hydrogens (since the R group will push electron density onto the carbocation, distributing the positive charge and stabilizing it.)

Reaction 2: The traditional formulation of Markovnikov’s rule: for acid addition to asymmetrically substituted alkenes, the carbon of the ‘ene’ group that has the most hydrogens will get the hydrogen (since the other one will form the more stable carbocation and so get the nucleophile) seems to work here for the hydrocarbon-only case. Below, we’ll see a situation where the rule appears to be broken.

In the case of non-alkyl groups on the ‘ene’ carbon, we must first see if it is electron donating (like an alkyl) or electron withdrawing. If we have Cl as a substituent, the opposite reaction path becomes favored. The highly electronegative Cl tries to pull electron density away from the already positively charged carbon. The carbocation will be thusly destabilized, i.e., the transition state will quickly dissociate back to the starting materials, making the second step of that reaction path less likely to occur.

Reaction 3: In this case, we see that the product is configured opposite to what the traditional version of Markovnikov’s rule states. It appears to be an anti-Markovnikov product. However, recently, Markovnikov’s rule has been restated in a way that takes into account the differences in the substituents.

Markovnikov’s Rule NEW FORM: In the addition of an acid to an alkene, the proton will go to the side that leaves behind the more stabilized carbocation.

Now, using this new definition, both reactions 2 and 3 can be seen as obeying Markovnikov’s rule. There is no anti-Markovnikov reaction.

Last edited: 31 July 2003.