Textbook: Loudon 5th Ed. (2009)

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.

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

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.

Note that this is the reverse of problem 519. Instead of going from thiol (alcohol) to carbocation to alkene, we're going from alkene to carbocation to thiol. In each step ask yourself "what arrows can I draw?" and choose the step that doesn't go backwards. 

The first step is the alkene picks up a proton to form the more stable carbocation. (See problem 334 if you don't understand why only the 3º carbocation is formed). To get rid of the carbocation, we can either do a beta-elimination (E1) to form an alkene, or an addition reaction (S_N1) to form a protonated thiol. Since forming the alkene would be going backwards, the only choice is addition.

Then you're left with a protonated thiol. How do we get rid of the positive charge on the sulfur? There are two legal moves- either the RSCH₂CH₃ acts as a leaving group, or the sulfur deprotonates. Since the RSCH₂CH₃ leaving would form a carbocation and take us backwards, the only option is for the sulfur to deprotonate.

Because the sulfur added to the more substituted carbon, this was a Markovnikov addition.

a) There are five different carbons in 2-methylbutane, so hypothetically, you should get five different products, 20% A, 20% B, 20% C, 20% D, and 20% E.

But notice that A and B are the same product! So only four products are possible.

b) Based on statistics alone, we would expect product A (or B) to be the major product, encompassing 40% of all the products (20% A + 20% B).

But this is not what happens in real life because A is primary, and for free-radical halogenation reactions, more substituted products are favored. So the major product will be C, which is tertiary.

This happens because the intermediate is a radical, and radical stability goes 3° > 2°  > 1°.

c) Chlorine reacts faster than bromine, and so is less selective than bromine.

For example, we know that the 3° halide will be the major product. But with chlorine, the proportion might turn out to be ~70% 3° , 20% 2°, and 10% 1°.

But since bromine reacts more slowly, you will get better selectivity. The proportion of products will might be something like ~95% 3° , 4% 2°, and 1% 1°.

While on this topic of reactivity, it's interesting to note that iodine reacts too slowly to be useful, and fluorine reacts so violently it's dangerous!

This is a hydrogenation reaction. The product is an alkane. But this product has two stereocenters.

Normally, the product would be a racemic mixuture of enantiomers (equal amounts of each stereoisomer). But in this case, the starting material is optically active (chiral but not racemic), so the products won't be equal and opposite as usual.

Since there is a substitutent blocking the top face of the molecule (isopropyl is a wedge), the hydrogens will preferentially add to the opposite face of the molecule (dashes).