Chapter 8 is mostly about alkene reactions. That is, how one can transform alkenes into other functional groups. Most of these reactons are electrophilic additions, or the addition of electrophiles across the double bond. Several of these reactions amount to addition of water to the π-bond. The result is the transformation of alkenes into alcohols. We now look at the generalities of this type of reaction (electrophilic addition) and then we look at each reaction individually.
The C=C π-bond of alkenes is a source of electrons. It is considered a weak base or nucleophile. As such, it can react with strong electrophiles. We have already learned how to identify electrophiles, but the π-bond requires strong electrophiles to react with. These can be strong proton acids, or species containing atoms with incomplete octets (Lewis acids). Examples of strong proton acids are HBr and H2SO4. Examples of Lewis acids are BH3, transition metal salts such as HgSO4, and carbocations.
The addition of strong electrophiles to the C=C π-bond can be viewed as the opposite of the elimination reaction, as illustrated below.
Elimination results in net loss of HBr to form a new C=C bond.
Addition of HBr (a strong electrophile) across the π-bond forms a new functional group, in this case an alkyl halide.
According to Saytzeff’s rule, in elimination reactions where formation of several alkenes is possible, the most highly substituted alkene predominates as a product.
Conversely, Markovnikov’s rule says that in addition reactions of proton acids to alkenes, the proton of the strong acid preferentially bonds to the carbon in the π-bond that already holds the greater number of hydrogens on it. In the alkene shown below, C-2 has more protons attached to it (one) than C-1 (none). Accordingly, the hydrogen from HBr bonds to C-2 and the bromine bonds to C-1
Markovnikov's product, a tertiary bromide
This preferred orientation is referred to as Markovnikov orientation. This suggests the possibility that another product with opposite orientation, called the anti-Markovnikov product, might form. Given that this product does not normally form under ordinary conditions, the question then is, are there special conditions under which it could form? The answer is yes, but with a very limited scope. We’ll address that point later.
Anti-Markovnikov's product, a secondary bromide
Another way to state Markovnikov’s rule, in this case, is to say the the tertiary bromide will form preferentially over the secondary bromide. If the electrophilic part of the reactant (H+) adds to the least substituted carbon (C-2), then the nucleophilic part (Br-) adds to the most substituted carbon (C-1). The degree of substitution in this case refers to the number of alkyl groups originally attached to the π-bond. This is another way by which Markovnikov’s rule becomes the counterpart of Saytzeff’s rule. The two statements are complementary.
Modern understanding of ionic mechanisms provides the key to Markovnikov’s rule. In the first step of the reaction, which is protonation of the π-bond, the most stable carbocation forms. This leads directly to formation of the tertiary bromide, i.e. Markovnikov’s product.
Here is another example involving a cyclic alkene: