Reaction 1 represents an SN2 reaction. The rate limiting step involves both reactants coming together to form a transition state. The rate of this reaction depends on the concentration of both the organic molecule and the nucleophile.

In contrast, reaction 2 is an E1 reaction, in which the rate limiting step is the removal of the leaving group to form a carbocation. In E1 and SN1 reactions, adjusting the concentration of the halide only is enough to affect the rate.

Tertiary alcohols react most rapidly via SN1 mechanisms because they form stable tertiary carbocations. Primary and secondary alcohols typically react most rapidly via SN2 mechanisms.

Of the available options, is the only one that contains a tertiary alcohol.

Reaction 1 represents an SN2 reaction. The rate limiting step involves both reactants coming together to form a transition state. The rate of this reaction depends on the concentration of both the organic molecule and the nucleophile.

In contrast, reaction 2 is an E1 reaction, in which the rate limiting step is the removal of the leaving group to form a carbocation. In E1 and SN1 reactions, adjusting the concentration of the halide only is enough to affect the rate.

Tertiary alcohols react most rapidly via SN1 mechanisms because they form stable tertiary carbocations. Primary and secondary alcohols typically react most rapidly via SN2 mechanisms.

Of the available options, is the only one that contains a tertiary alcohol.

The way the question is phrased, three answer choices must favor an SN2 reaction, while the "correct" answer is a factor that does not favor, or disfavors an SN2 reaction.

SN2 reactions are bimolecular, and thus their rate of reaction depends on both the substrate and the nucleophile, forming a high energy transition state in which the nucleophile will displace the substate's leaving group at an angle of 180o. The more sterically hindered the compound is, the higher in energy the transition state will be, and the slower the rate of reaction will be. Consequently, SN2 reactions are favored when the leaving group (a halogen in this case) is on a primary carbon center. Additionally, because the reaction is bimolecular, step two of the reaction will NOT occur without a good nucleophile to displace the leaving group. Finally, all SN2 reactions are favored by polar aprotic solvents.

Because SN2 reactions proceed via a transition state, no carbocation intermediate is formed (that happens in SN1 reactions) and therefore the formation of any carbocation favors an SN1 reaction, not an SN2 reaction.

Tert-butoxide is a large, sterically hindered, strong nucleophile that is often used in E2 reactions. Strong nucleophiles usually undergo the SN2 or E2 pathway, but tert-butoxide is much too large to undergo a substitution reaction.

For an E2 reaction to occur, there must be a hydrogen on the carbon adjacent to the carbon with the leaving group. Benzyl bromide contains no hydrogens on the carbon next to the carbon with the bromide, and would therefore undergo only a substitution reaction.

The way the question is phrased, three answer choices must favor an SN2 reaction, while the "correct" answer is a factor that does not favor, or disfavors an SN2 reaction.

SN2 reactions are bimolecular, and thus their rate of reaction depends on both the substrate and the nucleophile, forming a high energy transition state in which the nucleophile will displace the substate's leaving group at an angle of 180o. The more sterically hindered the compound is, the higher in energy the transition state will be, and the slower the rate of reaction will be. Consequently, SN2 reactions are favored when the leaving group (a halogen in this case) is on a primary carbon center. Additionally, because the reaction is bimolecular, step two of the reaction will NOT occur without a good nucleophile to displace the leaving group. Finally, all SN2 reactions are favored by polar aprotic solvents.

Because SN2 reactions proceed via a transition state, no carbocation intermediate is formed (that happens in SN1 reactions) and therefore the formation of any carbocation favors an SN1 reaction, not an SN2 reaction.

Tert-butoxide is a large, sterically hindered, strong nucleophile that is often used in E2 reactions. Strong nucleophiles usually undergo the SN2 or E2 pathway, but tert-butoxide is much too large to undergo a substitution reaction.

For an E2 reaction to occur, there must be a hydrogen on the carbon adjacent to the carbon with the leaving group. Benzyl bromide contains no hydrogens on the carbon next to the carbon with the bromide, and would therefore undergo only a substitution reaction.