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.

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.

This is an elimination reaction because refluxing conditions indicate high heat, an essential component for these reactions. We can eliminate answer choices that include substitution products, namely those containing methoxy groups, thus addressing answers IV and V.

Because the leaving group, bromide, is bonded to a tertiary carbon, this reaction will undergo an E2 mechanism. This means a carbocation will be formed at carbon one, and a subsequent deprotonation of an adjacent hydrogen will form the alkene. As answer choice III cannot be formed via deprotonation of an adjacent hydrogen, we can eliminate it.

By Zaitsev's rule_,_ we know that the most substituted alkene product of an elimination reaction will be the most stable, and thus most favorable, product. This allows us to see that the product formed in reaction I will be the most favorable.

This is an elimination reaction because refluxing conditions indicate high heat, an essential component for these reactions. We can eliminate answer choices that include substitution products, namely those containing methoxy groups, thus addressing answers IV and V.

Because the leaving group, bromide, is bonded to a tertiary carbon, this reaction will undergo an E2 mechanism. This means a carbocation will be formed at carbon one, and a subsequent deprotonation of an adjacent hydrogen will form the alkene. As answer choice III cannot be formed via deprotonation of an adjacent hydrogen, we can eliminate it.

By Zaitsev's rule_,_ we know that the most substituted alkene product of an elimination reaction will be the most stable, and thus most favorable, product. This allows us to see that the product formed in reaction I will be the most favorable.

All of the statements are correct.

SN1 substitution reactions take place in 2 steps, and SN2 substitution reactions take place on one step. Acetate is a better nucleophile than acetic acid because acetate is a negative ion, and therefore donates electrons as a nucleophile. The more hindered a strong base is, the more likely it is to produce an E2 reaction because the base will more easily remove a good leaving group to become more stable (done through elimination in one step via E2). In the absence of heat, strong bases, and good nucleophiles, tertiary alkyl halides will react via the SN1 mechanism because strong bases/good nucleophiles will always undergo SN1 mechanisms (substitution in two steps), with the exception of alkyl halides.

All of the statements are correct.

SN1 substitution reactions take place in 2 steps, and SN2 substitution reactions take place on one step. Acetate is a better nucleophile than acetic acid because acetate is a negative ion, and therefore donates electrons as a nucleophile. The more hindered a strong base is, the more likely it is to produce an E2 reaction because the base will more easily remove a good leaving group to become more stable (done through elimination in one step via E2). In the absence of heat, strong bases, and good nucleophiles, tertiary alkyl halides will react via the SN1 mechanism because strong bases/good nucleophiles will always undergo SN1 mechanisms (substitution in two steps), with the exception of alkyl halides.

Because both hydrogens neighboring the are anti-periplanar, both elimination products would be formed.

Because both hydrogens neighboring the are anti-periplanar, both elimination products would be formed.

Tertiary carbocations are stabilized by the induction of nearby alkyl groups. Since the carbocation is electron deficient, it is stabilized by multiple alkyl groups (which are electron-donating). Less substituted carbocations lack stability.

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.