is a better leaving group than  because it is a larger molecule and can distribute the negative charge over a larger area. SN2 works better with better leaving group and with less-substituted carbons (methyl > primary > secondary > tertiary)

In this question, we're presented with a scenario in which a chemistry student is trying to run a reaction. At first, the reaction doesn't work. But after the student adds a strong acid to the mixture, the reaction goes through. We're being asked to determine why this happens.

First and foremost, let's identify which kind of reaction is occurring. The student starts with  and ethanol, and ends up getting chloroethane. So, what has changed? The hydroxyl group on the ethanol has become replaced by a chlorine atom. As a result, we can identify this as a substitution reaction. Furthermore, because we know the hydroxyl group is attached to a primary carbon (a carbon that is only bound to one other carbon), we can categorize this as an SN2 reaction rather than an SN1 reaction. This is because the removal of the hydroxyl group would leave a primary carbocation, which is not likely to occur because this is very unstable.

Before addition of , no substitution reaction occurs. Why is that? What has to happen for the reaction to proceed? The answer is that the hydroxyl group needs to come off as a leaving group and be replaced with chloride. But, hydroxyl groups make very poor leaving groups. Compared to chloride ions, hydroxyl groups floating in solution are much more unstable. Thus, the hydroxyl group would rather stay attached to ethanol than to leave.

But, this all changes once  is added. The reduction in the pH of the solution causes the hydroxyl group to become protonated. Not only does this give the hydroxyl group a positive charge, but it also makes a much better leaving group. This is because when it leaves the hydrocarbon and enters solution, it will exist as , or water. Because water is much more stable than the chloride ion, chloride is able to attack the protonated ethanol and undergo a nucleophilic substitution reaction. Thus, it is the protonation of the leaving group that drives the reaction forward.

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.

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.

All are true for E2 reactions. E2 reactions are bimolecular, with the rate dependent upon the substrate and base. Both E1 and E2 reactions generally follow Zaitsev's rule and form the substituted double bond. The stereochemistry for E2 should be antiperiplanar (this is not necessary for E1).

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.

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.

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.

An E2 elimination reaction is a bimolecular reaction and its rate is dependent on the concentration of substrate and base. On the contrary, an E1 elimination reaction is a unimolecular reaction and its rate is dependent solely on the concentration of substrate.

An E2 reaction will form the less substituted product when the base is big and bulky. We are looking for an answer choice that is sterically hindered, and thus will remove a proton "more within it's reach," over removing the proton that forms the most stable (most substituted) product.

( t-butoxide) is a bulky base that is commonly used to favor the less substituted product. There are 3 methyl groups connected to the central carbon, and as such it is sterically hindered.

An E2 reaction requires a strong base, which eliminates water and  (acetic acid) as answer choices. and are both incorrect because they are not sterically hindered. The E2 reaction would result in the most substituted product.