Reaction Mechanisms, Energetics, and Kinematics - Organic Chemistry

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.

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.

Grignard reagents attack ketones at the site of the ketone's carbon to produce tertiary alcohols.

SN2 reactions require a strong nucleophile. If the the nucleophile is weak, SN1 is favored over SN2.

For SN2 reactions less substitution in the electrophile is favored (methyl > primary > secondary).

For SN2 reactions polar aprotic solvents are preferred (DMSO, acetone, DMF and ethers are common polar aprotic solvents). Polar protic solvents (ethanol, methanol) favor SN1 reactions.

The SN1 mechanism involves the formation of a carbocation intermediate in the rate-determining step. The most stable carbocation will produce the fastest reaction. We can immediately eliminate any answer choices that will produce primary or secondary carbocations, since a tertiary carbocation will be much more stable. When comparing tertiary carbocations, larger and more electronegative substituents will allow for more charge stabilization.

Since the tertiary carbocation formed by the dissociation of iodide from will the be most stable, this substrate will react the fastest.

The rate of an reaction is determined only by the concentration of the substrate. Unlike an reaction, where the addition occurs in one step and requires the activity of the substrate and the nucleophile, an reaction occurs in two steps and is only limited by the activity (i.e. leaving ability) of the substrate. Once the leaving group leaves the substrate, the nucleophile does not hesitate to attack the exposed carbocation.

Keep in mind that after the aldehyde is reduced into an alcohol, the molecule can undergo an intramolecular reaction, as alcohol is a good nucleophile and the halogen is a stellar leaving group.

A strong nucleophile is not required for SN1. A weak nucleophile may be used. Remember that the SN1 mechanism goes through a carbocation intermediate.

Rearrangements are possible for SN1 reactions (not SN2). A rearrangement will occur to create a more stable intermediate in the mechanism. For example, if the carbocation is secondary, a methyl shift may occur to make the carbocation intermediate tertiary.

A racemic mixture of products occurs when with the nucleophile may attack the carbocation from either the top face or bottom face. When a reaction goes through a carbocation intermediate, as in SN1, there may be a racemic mix of products.

SN1 is unimolecular, and the rate of the reaction is determined by the substrate and reaction constant.

SN1 reactions result in racemization when the nucleophile has a 50% chance of attacking the carbocation intermediate from the top face, and a 50% chance of attacking from the bottom face. SN1 reactions are favored in polar protic solvents, such as ethanol.

E2 and E1 are incorrect as they are elimination reaction mechanisms, and we are looking for a substitution mechanism. SN2 reactions result in inversion, not racemization. Additionally we know that SN2 is incorrect because SN2 is favored in polar aprotic solvents.

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.

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.

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.

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.