Tert-butoxide is a large, sterically hindered, strong nucleophile that is often used in E2 reactions. Strong nucleophiles usually undergo the S_N2 or E2 pathway, but tert-butoxide is much too large to undergo a substitution reaction.

The way the question is phrased, three answer choices must favor an S_N2 reaction, while the "correct" answer is a factor that does not favor, or disfavors an S_N2 reaction.

S_N2 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 180^o. 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, S_N2 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 S_N2 reactions are favored by polar aprotic solvents.

Because S_N2反应过程eed via a transition state, no carbocation intermediate is formed (that happens in S_N1 reactions) and therefore the formation of any carbocation favors an S_N1 reaction, not an S_N2 reaction.

Reaction 1 is an S_N2 reaction. This type of substitution reaction prefers a polar, aprotic solvent. The polarity helps to solvate the nucleophile. Aprotic solvents help mediate the transition state and increase reaction rate.

Nucleophilicity increases to the left on the periodic table. Nucleophilicity will also generally increase with charge. The only equally charged ion in the answers that is present to the right of oxygen on the periodic table is the fluoride ion.

The addition of bromine gas () to the reaction vessel would likely result in the addition of one half of the diatomic bromine to each carbon, eliminating the double bond and resulting in an alkyl halide chain.

Halogenation reactions refer to reactions between a halogen and an alkane, while addition reactions occur between a halogen and an alkene (such as the product in reaction 2).

这个问题测试你的理解E2 elimination reactions. Because E2 reactions have no carbocation intermediates, the reaction proceeds through a transition state. The bromine atom (Br) must be displaced by the bond between an adjacent carbon and hydrogen, and that hydrogen must be anti-periplanar to the bromine. Compound 2, unlike compound 1 has one hydrogen that is anti-periplanar to the bromine, and thus undergoes an E2 elimination.

If compound 2 reacts, it is unlikely the reason that compound 1 would not react is because of steric hindrance. Additionally, aside from stereochemistry, there is nothing different enough about the two compounds that would suggest any reagent would be more favorable for one over the other.

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.

Reaction 2 represents an E₁reaction. The rate limiting step of reaction 2 involves the formation of a carbocation, whose stability is favored by the presence of substituents on the carbon involved. Carbocations are considered intermediates due to their relative stability compared to transition states.

Reaction 1 represents an S_N2 reaction. Such reactions depend, in part on the presence of strong nucleophiles, such as the hydroxide ion. Water can be a nucleophile as well, but it is weaker. Using water in place of hydroxide would cause reaction 1 to proceed far more slowly.

The carbocation forms spontaneously with the loss of the chlorine atom. This is the rate determining step, thus, the concentration of the halide is the most important determinant of reaction rate. Carbocations form spontaneously in these reactions, and do not use the strong base to remove the halogen.