The reaction shown is a nucleophilic substitution reaction. The molecule shown is a primary alkyl bromide, and the nucleophile that will be used is the iodide anion (). In this type of reaction the iodide will displace the bromide on the organic molecule, generating iodoethane ().

This reaction is an example of a nucleophilic substitution on a secondary alkyl halide. The nucleophile in this case is cyanide (), and the atom that attacks in the cyanide ion is the carbon. Because cyanide is a good nucleophile, the reaction will occur via  mechanism. The answer is thus the molecule where the bromide is replaced with cyanide.

Note: when nucleophilic substitution is performed using a nucleophile that contains carbon (Grignard reagents, acetylide reagents, cyanide, etc.) it is often easy to incorrectly count the number of carbons in the final product.

 is a better leaving group because it is a larger atom and thus has its negative charge spread more evenly. It is more stable and a weaker base than . The molecule with the leaving group that would be most stable, or the weakest base, is the molecule that would react fastest in an S_N2 reaction. 

There would be more steric hindrance for a nucleophile in the 2-bromobutane because the leaving group, bromine, is located on a secondary carbon atom. A secondary carbon atom is attached to two other carbon atoms. The leaving group on the 1-propane is located on a primary carbon atom, which is attached to only one other carbon atom. 

This reaction would occur using an S_N1 mechanism because the leaving group is attached to a tertiary carbon, a carbon atom with three of the carbon atoms attached to it. The rate laws for S_N1 mechanisms do not depend on the nucleophile concentration. The slow-step occurs unimolecularly within the molecule with the leaving group. 

E1 reactions occur in two steps. First, the leaving group is removed, yielding a carbocation. Second, a weak base removes a proton from the carbon adjacent to the carbocation carbon. Thus, to favor E1, a protic solvent is desired in order to stabilize the carbocation. Weak bases favor an E1 mechanism.

E2 reaction occur in only one step. The strong base removes the beta-hydrogen, while the leaving group simultaneously leaves. The double bond forms simultaneously. 

Elimination reactions involve the use of bases, which remove hydrogen atoms. The leaving group, which is bound via a sigma bond is removed, along with a hydrogen (thus two sigma bonds are broken). The result is a double bond, which consists of one sigma bond and one pi bond. 

E1 reactions occur in two steps, forming a carbocation intermediate, which is most stable if there is a protic solvent present. Furthermore, the use of a weak base favors E1.

E2 reactions occur in one step; thus no carbocation intermediate is formed, and an aprotic solvent is favored. E2 reactions are favored by strong bases and higher temperatures.