Substitution and Elimination Mechanisms - MCAT Biological and Biochemical Foundations of Living Systems

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.

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.

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.

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 SN2 reaction; only SN1 reactions use a carbocation intermediate.

Reaction 2 represents an E1 reaction. 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 2 is an E1 reaction, in which the rate limiting step is the formation of the carbocation intermediate. The carbocation intermediate has three single bonds and a positive charge on the central carbon; thus, it has sp2 hybridization, a planar structure, and free rotation about its single bonds. Bond rigidity is only observed with the presence of pi bonds.

Reaction 1 represents an SN2 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 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 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.

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.

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).

Reaction 1 is an SN2 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.

This question tests your understanding on 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.

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.

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.

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.

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.