Mechanism (show a mechanism using curved arrows..)

This is an S_N1 reaction, but the carbocation is special because it's allylic.

If the Br^- nucleophile attacks the 3º resonance form, the 1,2 product is formed. (This is the product that would have formed if resonance was not involved.)

If the Br^- nucleophile attacks the 2º resonance form, the 1,4 product is formed.

Arrows in organic chemistry always go from regions of high electron density to regions of low electron density. Most of the time this means arrows start from negative charges and go towards positive charges.

Because bromine is electronegative, the carbon directly bonded to it (also known as the alpha carbon) has a partial positive charge, and can be attacked by a nucleophile such as azide (N₃^-). 

Because this is an S_N2 reaction, no carbocation is formed; as the nucleophile attacks the alpha carbon, the leaving group (Br^-) leaves.

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).

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.

The first step in this reaciton, like many reactions, is an acid-base reaction- when you see an H⁺ (acid), the first step is usually something getting protonated. This starting material is a thiol (sulfur), but the same thing would happen with an alcohol or ether (oxygen).

When you have a protonated thiol (or alcohol, ether, etc.), you know the reaction isn't finished yet- products are rarely charged. There are two legal moves to get that positive charge off of the protonated thiol. Either the thiol deprotonates, or it takes off as a neutral leaving group. It can't deprontate because that would be going backwards, so your only "legal move" is for the HSCH₂CH₃ to act as a leaving group (See problem 518).

But then you're left with a carbocation, which is definitely not the final product. How can we get rid of it? (See problem 335 for general ways carbocations react). If the HSCH₂CH₃ attacked in an S_N1 type reaction that would be OK, but it would be going backwards! So the only way to get rid of the carbocation is to do a beta-elminiation (E1).

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.

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.