Reaction Mechanisms, Energetics, and Kinematics - Organic Chemistry

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.

The elimination reaction (as opposed to the SN2 because of the big bulky base reagent) only proceeds with the anti-periplanar product.

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

For E2 reactions a tertiary electrophile > secondary electrophile > primary electrophile. A polar aprotic solvent favors E2 (remember that polar protic solvents favor E1). A strong base is needed to undergo E2.

SN2 reactions result in an inversion of products, thus R-configured reactants become S-configured products and vice versa. Polar aprotic solvents such as ether favor SN2 reactions.

E1 and E2 are elimination mechanisms and the question asks for a substitution mechanism. SN1 results in racemization of products, not inversion. Thus, if the reaction had gone through SN1 we would see a mixture of R and S-configuration in the products. Additionally, polar protic solvents (not aprotic like ether) favor SN1 reactions.

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.

SN2 reactions involve a backside nucleophilic attack on an electrophilic carbon. As a result, less steric congestion for this backside attack results in a faster reaction, meaning that SN2 reactions proceed fastest for primary carbons. In addition, beta-branching next to a primary carbon results in a slower reaction, as does a poorer leaving group (i.e. chloride instead of bromide).

1-bromopentane has a good leaving group (bromine) attached to a primary carbon with no beta-branching, meaning it will proceed the fastest.

The reason that the alkyl halide is preferred to be primary is because the mechanism for these reactions is SN2. SN2 indicates a substitution reaction that takes place in one step. A primary alcohol is preferred to prevent steric congestion caused by the simultaneous binding of the nucleophile and release of the leaving group. This reaction mechanism is faster because it omits the formation of a carbocation intermediate.

In contrast, SN1 reactions take place in two steps and involve the formation of a carbocation intermediate.

SN1 reactions are characterized by two distinct steps. The first step, which determines the rate of the reaction, is the dissociation of the leaving group. This step leaves behind a carbocation intermediate.

As opposed to SN2 reactions, in which nucleophilic substitution occurs in one step, the temporary formation of a carbocation in SN1 reactions allows for carbocation rearrangement, which serves to stabilize the positive charge.

The major product, molecule IV, results from the shift of a hydrogen atom from the adjacent carbon, moving the positive charge to a carbon with greater alkyl substitution. Electron density is inducted to the secondary carbocation (bound to two alkyl groups), stabilizing the positive charge. Carbocation rearrangement occurs extremely fast, usually before a nucleophile (in this case water) may bind.

As a result, molecule IV is the major product instead of molecule II (the SN2 product).

This is an SN2 reaction. When there is a methyl halide with a strong nucleophile, the nucleophile will force the halide group to leave. Strong nucleophiles dictate SN2 reaction mechanisms.

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)

The sterochemistry should be inverted for an SN2 reaction. The product is has S chirality , so the starting material should have R.

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.

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).

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.

Regarding chemical reactions, the activation energy is the minimum energy which must be available in a chemical system in order for reactants to participate in a chemical reaction. Catalysts speed up reactions by lowering this minimum energy. Raising it would have the opposite effect. Catalysts do not alter the of a reaction. Remember this--it is a common error to make!