These conditions are perfect for an  reaction. We see a secondary halide so first instinct might be to say that an  reaction would ensue. However, anytime we have a secondary halide reacting with a strong, bulky base, the laws of sterics dictates that an  reaction will ensue.

The given conditions are perfect for an  reaction. Here we have a primary haloalkane reacting with a big, bulky base. Big, bulky bases such as , and alkene products are classic indicators that the reaction was an  reaction.

Treatment with strong base indicates E2 mechanism. The halide has two beta-carbons (II and III), and potassium tert-butoxide is a bulky base. The favored product is the Hofmann product (less substituted alkene). Therefore III is the hydrogen that's removed. 

E2 reaction mechanism requires antiperiplanar orientation between the leaving group and the hydrogen. Given the stereochemistry of the methyl group, forming an alkene bond (to give the more substituted product) is impossible.

This is an acid-catalyzed dehydration of a tertiary alcohol. The first step is protonation of the alcohol oxygen which will create a good leaving group (), and subsequent loss of water from the molecule. The molecule will then contain a carbocation (which will not move due to the positive being on the most substituted carbon atom in its immediate vicinity), and the conjugate base (water in this case) will remove a hydrogen from a carbon next to the carbocation. The base will preferentially remove a hydrogen from the one that will produce the more substituted alkene, as this is generally the more stable product. This is known as Zaitsev's rule for the formation of alkenes.

An elimination reaction occurs when there is a release of atoms in a given compound to produce two or more products. In the reaction given a hydrogen and chloride atom are eliminated from the original compound to form hydrochloric acid and ethylene.

Heterolytic bond breaking occurs in polar compounds to form to products of opposite charges. In these types of reaction two electrons from the original bond stays with one fragment upon cleavage. In the reaction given, the bond between the hydrogen and chlorine atom is broken with the two electrons from the original bond staying with the chlorine atom. The resulting products are hydrogen ion and chloride ion. 

Here we see an  reaction with rearrangement. The bromine, an excellent leaving group, leaves the carbon chain and a carbo-cation (positively charged carbon) is formed on that carbon. A positive charge is more stable on a more substituted carbon, and so the positive charge rearranges itself onto the branched carbon. Essentially, the positive charge and a hydrogen on the branched carbon switched positions. The methanol was then free to attack the branched carbon to form the major product shown.

This reaction adds  and  (eliminate II). The reaction is Markovnikov (Eliminate I). A hydride shift occurs putting the carbocation on the more substituted carbon before addition of (eliminate III and V).