Reaction Mechanisms, Energetics, and Kinematics - Organic Chemistry

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.

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.

Here we have a classic electrophilic addition reaction.

We must remember that a double bond is an area in a molecule that has a high electron density. As a result, double bonds can act as nucleophiles and react with good electrophiles.

In this problem, the alkene attacks the dibromide to form a bromonium ion, which is rather unstable. The bromonium ion pulls carbon's electrons toward itself, causing the carbons in the bromonium complex to have a slightly negative charge. As a result, the negatively charged bromine can attack either of the carbon atoms and, thus, we have added two bromines to our carbon chain.

The nitro substituent is strongly deactivating, ketones are moderately deactivating substituent, halides are weakly deactivating, alkyl groups are weakly activating, and amine groups are strongly activating.

The five resonance structures of the given molecule are shown below:

Most organic reactions are carried out in multiple steps. The rate equation can be derived from the process that occurs in the slowest step of the mechanism. An E1 reaction's slow step is when a leaving group separates from the hydrocarbon and a carbocation is formed. The only reactant in this step is one molar equivalent of the hydrocarbon. Thus, the rate equation only depends on that substance. The molar equivalent determines the exponent of each reactant in the equation.

By and large, organic compounds contain mainly covalent bonds. Covalent bonds are typically harder to break, which is why organic reactions happen at a relatively slower rate than inorganic 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.