Textbook: Brown 6th Ed. (2011)

Protonation is a common first step in mechanisms. When you have a protonated alcohol, it's likely that either:

1. The alcohol will deprotonate to form a neutral alcohol (right), or
2. The alcohol will leave as a neutral molecule to form a carbocation.

Either one is possible. If you're not sure which one happens, remember that when writing out mechanisms, never go backwards- don't draw structures you have drawn before.

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.

Be on the lookout for a 1,2-shift when you have a carbocation adjacent to a carbon atom that is more substituted (ex: a 2° carbocation next to a 3° or 4° carbon).

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

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.

Because the chlorine is more electronegative than iodine, the iodine will have a partial positive charge and will be attacked by the alkene. It forms the more stable carbocation as normal (as in problem 335). The nucleophile (HSCH₂CH₃) will eventually attack the carbocation, but the iodine does something special first- it forms a cyclic intermediate.

The formation of the cyclic halonium ion is probably the most important concept you will learn in first semester organic chemistry. With iodine, it's called an iodonium ion. (with chlorine or bromine, it's called a chloronium or bromonium ion). 

The HSCH₂CH₃ will attack the carbon that was the carbocation, but it's forced to attack from the side opposite to that of the iodine. So the product will always have anti stereochemistry (you get the trans product).

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).