Organic Chemistry

When HBr reacts with in an alkene in the presence of peroxides (which are free radical initiators), the reaction goes through a free radical mechanism and adds in an anti-Markovnikov manner. The product is an alkyl bromide.

The reaction goes anti-Markovnikov because the more substituted radical intermediate is only formed if the bromine adds to the less substituted carbon.

This is a hydroboration reaction (also called hydroboration-oxidation in some textbooks). The product is an alcohol.

Boron bonds to the less substituted carbon, and is replaced with an oxygen after the borane intermediate is treated with peroxide, resulting in an anti-Markovnikov addition. 

Note that one equivalent of BH₃ reacts with three equivalents of alkene; the borane intermediate is not BH₂(alkyl) but rather B(alkyl)₃.

Addition of a chlorine, bromine, or iodine to an alkene goes through a halonium ion intermediate, which results in an anti-addition (trans product).

This is a acid-catalyzed hydration of an alkene, which is a Markovnikov addition. The product is an alcohol.

You know this is an elimination reaction because no nucleophile is present; H₃PO₄ and H₂SO₄ are non-nucleophilic acids. IF the reagent were HCl or HBr on the other hand, this would be a substitution reaction.

Since acid is present, a carbocation will probably form, so we know that this is an E1 mechanism.

The 2º carbocation that is initially formed will undergo a 1,2-hydride shift to become a 3° carbocation, but that doesn't affect the final product of the reaction; with either the 2° or 3° carbocation, the most stable alkene product is 2-methyl-2-butene.

The 2° carbocation formed immediately undergoes a 1,2-methyl shift (a rearrangement) to form the more stable 3° carbocation, so the product is the 3° alkyl chloride instead of the 2º alkyl chloride, which would have formed in the absence of rearrangement.

Due to Markovnikov's rule, only the more substituted alkyl halide is formed.

For a), the product is neutral and so you are done after the nucleophile (Cl-) attacks the carbocation.

For b), the the intermediate is a protonated alcohol, and so you must do a proton exchange step (also called a hydrogen exchange or deprotonation) to get the final alcohol product, which is neutral.

For c), there are two different types of beta hydrogens, and so two different alkenes are possible.

The more stable alkene is the one that will form, and this will always be the most highly substituted alkene. This is Zaitsev's rule. The rationale for this is hyperconjugation: neighboring carbon atoms stabilize an alkene.

3° carbocations are more stable than 2º carbocations, so only the 3º carbocation is formed.