Markovnikov's Rule and Acid Catalyzed Hydration of Alkenes

Sascha Vongehr                          15-Oct-03

1.) Markovnikov's Rule:

When reagent X-Y acts on an alkene, the least e-negative reagent atom (H in case of H-Y) predominantly binds the sp2 carbon with the most H already (= least substituted one).

This is called regioselectivity: Y can be many a group of atoms, yet of two possible constitutional isomers, one always occurs predominately. The rule applies to ClBr, ICl or hypohalous addition, acid catalyzed hydration of alkenes, and hydrohalogenation.  E.g., in acid catalyzed hydration of alkenes the reagent in the balanced equation is simply H₂O, i.e. for isopropanol versus propanol:

         Me-CH=CH₂  +  H₂O →     Me-CH(OH)-Me           (not  Et-CH₂-OH)                                (1.1)

The H ion of H-Y attaches using the double bond's Pi e^-. The sp2-C become e^- deficient. The more the charge on a C can disperse (spread out), the more stable the intermediate and the more likely it is formed. A larger structure eases the dispersion over the close by surface area of the molecule. The three C directly attached to a tertiary C can disperse charge through the sigma bonding (referred to as "hyper-conjugation") more easily than the only two C of a seconday C. Hence, tertiary so called carbocations are more stable than secondary ones and so on. Benzylic and allylic carbocations moreover delocalize the positive charge due to the conjugation of charge with a conjugated Pi e^- system. So the balanced equation from above may be put as follows:

         Left      →       {Me-CH⁺-Me/Et-Me⁺}  +   Y^- →     Me-CH⁺-Me     +          Y^- (1.2)

                                             Me-CH⁺-Me     +          Y^- →       Right                                        (1.3)

As an aside, neither in (1.2) nor in (1.3) is the newly added H at the site the positive charge's plus sign is at, due to the e^--distribution. Writing "[Et-Me]⁺ " to conserve the bad "H=+"-shortcut would at some point result in a careless application of "H=+", (mis)leading to Me-CH-Me⁺ and Et⁺-Me, both very wrong.

2.) Acid Catalyzed Hydration of Alkenes

Markovnikov's Rule predicts correctly that the acid catalyzed hydration of alkenes is regioselective in that the more highly substituted alcohol is very predominant. If the less substituted one is initially produced to some extend, rearrangements, where an H or Me along with both e^- in the bond move to an adjacent C, often occur - but not in all cases, which indicates that the reaction indeed involves a carbocation intermediate. Tertiary carbocations never undergo molecular rearrangement because of the highly stable nature of the tertiary carbocation.

The hydration of alkenes needs strongly acidic conditions. To explain the acidic catalyse balanced equations are insufficient. The reagent X-Y in the balance of educts and products is H₂O yet actually the first step in the reaction mechanism has to use H₃O⁺, i.e. Y seems to be neutral H₂O, not OH^-:

         Me-CH=CH₂    +          H₃O⁺ →       {Me-CH⁺-Me/Et-Me⁺}  +          H₂O

→    Me-CH(H₂O⁺)-Me                                            (2.1)

Via a lone pair the carbocation binds to a water molecule's O, charging it positively.  (2.1) is still very much the structure as presented above. That familiar scheme though will be destroyed now such that the reagent X-Y seems to be different in the end from what it appeared to be when starting out.

         Me-CH(H₂O⁺)-Me        +          H₂O →       Me-CH(OH)-Me           +          H₃O⁺ (2.2)

In the final step, a molecule H₂O pulls one H ion off the O. The resulting alcohols may undergo molecular rearrangement, so it is not possible to use this reaction to get the other isomer.

Quite unsatisfactory in this treatment is that a practically simple example illustrating the Markovnikov's Rule cannot answer the question of what the reagent that the rule is all about, actually is. Neither can it point to the catalyst of this supposedly catalytic reaction. Lastly though maybe gravest: The general statement and the statement of a straightforward realisation of it are not the same. A satisfactory presentation should be mnemotic, the example accesable via substitution of the more general terms.

3.) Recovering Reagent and Formal Structure by Identification of the Catalyst

The formal structure as presented earlier in 1.) is best conserved via assigning one H⁺, say the one that was added when the acid dissociated and one,  the role of the catalyst. The catalyst modifies the reaction partners. It binds them, maybe cracks them, unites them,  but in all this it never changes and it survives the whole unmodyfied.  We write it in large font and bold as H⁺, so it will appear as static surface, just like many catalysts are. An example is a Pt-plate adsorbing H₂ and cracking it into two H radicals such that O₂ will react with it readily. Electrochemically, the Pt-plate is an electrode that may have positive applied voltage. With this in mind, H⁺ binds a water molecule at its oxygen and cracks it into H⁺ and OH^-. The alkene is also captured, whenever or say after the cracking via its reaction with the liberated proton.

Me-CH=CH₂  +  H₂O--H⁺ →     {Me-CH⁺-Me/Et-Me⁺}--H⁺--HO^-

→       Me-[CH(OH)--H⁺]-Me                                      (3.1)

The reagent is now the one expected from the balancing equations, namely H₂O.  Equation (2.2) reveals itself here as the catalyst desorbing the reaction product and grabbing onto the next reagent molecule as it is entirely expected from a catalyst. While (2.2) based on "the final step, a molecule H₂O pulls one H ion off the O" is unmotivated and has to be remembered, it can now be written down easily:

         [Me-CH(OH)-Me]--H⁺  +          H₂O →       Me-CH(OH)-Me           +          H₂O--H⁺ (3.2)

Of course, this resolution breaks along with the iso-symmetry (isospin , isotope). "The catalyst" D⁺( = ²H⁺) from deuterated acid is lost to undeuterated water and will be partially recovered in the product.