Reaction 1 will experience greater steric hindrance with the addition of a methyl group, in place of a hydrogen, on the central carbon of the reactant. The result of this is increased activation energy, and a reduced rate of reaction in unchanging temperature and with no addition of a catalyst.

Intermediates are relatively stable, while transition states are unstable and transient. The transition state (not the intermediate) of reaction 1 is a planar uncharged structure; however, only relatively stable species such as carbocations are considered intermediates. Reaction 1 does not have an intermediate, and is an example of an S_N2 reaction; only S_N1 reactions use a carbocation intermediate.

The Zaitsev product is the most stable alkene that can be formed. This is the major product formed in E1 elimination reactions, because the carbocation can undergo hydride shifts to stabilize the positive charge. The most stable alkene is the most substituted alkene, and thus the correct answer.

Reaction 1 represents an S_N2 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 S_N1 reactions, adjusting the concentration of the halide only is enough to affect the rate.

 reactions, also known as unimolecular nucleophilic substitution reactions, occur in two steps. Here, we are concerned with the first and second (rate-determining) steps, in which the leaving group breaks off of the molecule to form a carbocation. Alkanes that form the most stable carbocations are most likely to undergo  reactions. Tertiary carbocations are the most stable, followed by secondary. Primary and methyl carbocations are very unstable and unlikely to form at all. The tertiary alkane, , will form a very stable tertiary carbocation compared to the other answer choices.

An alcohol can be oxidized and form a ketone only if the alcohol is a secondary alcohol. Tertiary alcohols cannot be oxidized and primary alcohols are oxidized to form aldehydes and carboxylic acids.

3-pentanol is a secondary alcohol, while 2-methyl-2-butanol is a tertiary alcohol. Methanol and propyl alcohol are both primary alcohols.

Reduction of an ester to an alcohol requires a strong reducing agent, such as .

is not a strong enough reducing agent for this reaction. Hydrogenation and reaction with zinc would reduce the nitro group to an amino group instead of affecting the carboxylic acid. is a Grignard reagent, and would not function as a reducing agent.

The conversion of compound B to compound C results in the elimination of water, which, by definition, is a dehydration reaction. The hydroxyl group on compound B is protonated by the sulfuric acid, generating an leaving group and allowing the formation of the alkene product in compound C.

Hydroboration is the oxidation of an alkene with a borohydride (usually sodium borohydride) reagent, to produce an alkane. Hydration involves the use of a water reactant, usually producing an alcohol product. Hydrogenation is the oxidation of alkene double bonds with the use of a palladium interface to produce an alkane product. Decarboxylation results in product carbon dioxide from a carboxylic acid or carbonic anhydrase reactant.

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.

Noncompetitive inhibitors are a specific type of mixed inhibitor that binds to both the free enzyme and the enzyme-substrate complex with equal affinities, resulting in the same binding affinity (K_m value) but a decrease in the maximum rate of reaction (V_max value).