Carbocation Fundamentals (Stability, Rearrangement, SN1)

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.

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.

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).

The more substituted the carbocation, the most stable it is. This is because of the inductive effect- adjacent carbon atoms donate some of their electron density to neighboring carbocations (which are electron deficient), making them closer to neutral and more stable.

So the 3° carbocation is the most stable.