Help with Substitution Reactions - Organic Chemistry

SN2 reactions undergo substitution via a concerted mechanism. Thus, no carbocation is formed, and an aprotic solvent is favored. Furthermore, tertiary substituted substrates have lowest reactivity for SN2 reaction mechanisms due to steric hindrance.

All of the given answers reflect SN1 reactions, except the claim that SN1 reactions are favored by weak nucleophiles.

SN1 reactions occur in two steps and involve a carbocation intermediate. The product demonstrates inverted stereochemistry (no racemic mixture). Tertiary substrates are preferred in this mechanism because they provide stabilization of the carbocation.

SN2 reaction mechanisms are favored by methyl/primary substrates because of reduced steric hindrance. No carbocation is formed via an SN2 mechanism since the mechanism is concerted; thus a strong nuclephile is used.

Substitution reactions—regardless of the mechanism—involve breaking one sigma bond, and forming another sigma bond (to another group).

is an extremely useful reagent for organic synthesis in instances where an alcohol needs to be converted to a good leaving group (bromine is an excellent leaving group). reacts selectively with alcohols, without altering any other common functional groups. This makes it ideal for situations in which a molecule contains acid-sensitive components that prevent the use of a strong acid to protonate a target alcohol.

While the mechanisms differ, reactions are similar to SN2 reactions in that they both invert the configuration at the site of attack. The configuration about the carbon adjacent to the alcohol in the given reactant is S. After substitution, the configuration of the major product is R, as is the case in molecule IV.

Based on the given reagents and the specification that the reaction takes place in a single step, it may be concluded that the reaction occurs by an SN2 or E2 mechanism. Since the compound lacks any moderately acidic hydrogen, an SN2 reaction is more likely. The absolute configuration at the reaction site in the initial compound is S, which is converted to R as a result of the "back-side attack" characteristic of all SN2 reactions. The major product is shown below:

This problem involves the synthesis of a Grignard reagent. Grignard reagents are easily created in the presence of halo-alkanes by adding magnesium in an inert solvent (in this case ). Once we have created our Gringard, it can readily attack a carbonyl. In this case, our Grignard attacks carbon dioxide to create our desired product.

An reaction is most efficiently carried out in a protic solvent. An inverted configuration site is characteristic of an reaction and the substituted nucleophile does not form a pi bond in an reaction.

In this question, we're given the reactant and product as well as the reagent being used in the reaction, and we're being asked to identify which reaction mechanism will correctly lead us from reactant to product.

To begin, it's important to notice that the reactant contains a tertiary bromine and the product contains a methoxy group in place of where the bromine was. Thus, we can conclude that a substitution reaction has taken place. If an elimination reaction had taken place, then there would have been a double bond in the product.

Now we need to identify which kind of substitution has occurred. Since the leaving group is attached to a tertiary carbon, we know that a stable carbocation will be generated upon dissociation. Therefore, we would expect this to be an reaction.

A solvolysis reaction is simply an reaction where the solvent acts as a nucleophile.

In this case, we start with a tertiary alkyl halide. Bromine, a stellar leaving group, leaves the substrate and leaves a carbocation intermediate. The methanol is then free to attack the carbon chain at the site of the carbocation to form an ether. The correct answer is 2-methoxy-2-methylpropane.

The correct answer is .

Alcohol is a horrible leaving group. is often employed to convert an alcohol group into a bromine group so that additional substitution and elimination reactions can ensue.

Cyanide is a weak base and a good nucleophile, and the solvent is aprotic; therefore, the product is favored. This involves 100% inversion of stereochemistry; therefore II is favored.

is a better leaving group because it is a larger atom and thus has its negative charge spread more evenly. It is more stable and a weaker base than . The molecule with the leaving group that would be most stable, or the weakest base, is the molecule that would react fastest in an SN2 reaction.

There would be more steric hindrance for a nucleophile in the 2-bromobutane because the leaving group, bromine, is located on a secondary carbon atom. A secondary carbon atom is attached to two other carbon atoms. The leaving group on the 1-propane is located on a primary carbon atom, which is attached to only one other carbon atom.

The double bond in 3-iodo-1-butene would stabilize the transition state through resonance. This would make the transition state lower in energy due to its increased stability. Thus, the reaction would be faster as reactions with lower activation energies proceed at faster rates.

This reaction would use an SN1 mechanism because the leaving group, bromine, is on a tertiary carbon, which is a carbon attached to three other carbon atoms. The bulk of these methyl groups would make SN2 impossible, but it would make the carbocation produced by an SN1 reaction very stable. The methyl group would lead to hyperconjugation, which is a type of resonance that stabilizes transition states.

This reaction would occur using an SN1 mechanism because the leaving group is attached to a tertiary carbon, a carbon atom with three of the carbon atoms attached to it. The rate laws for SN1 mechanisms do not depend on the nucleophile concentration. The slow-step occurs unimolecularly within the molecule with the leaving group.

The reaction shown is a nucleophilic substitution reaction. The molecule shown is a primary alkyl bromide, and the nucleophile that will be used is the iodide anion (). In this type of reaction the iodide will displace the bromide on the organic molecule, generating iodoethane ().

This reaction is an example of a nucleophilic substitution on a secondary alkyl halide. The nucleophile in this case is cyanide (), and the atom that attacks in the cyanide ion is the carbon. Because cyanide is a good nucleophile, the reaction will occur via mechanism. The answer is thus the molecule where the bromide is replaced with cyanide.

Note: when nucleophilic substitution is performed using a nucleophile that contains carbon (Grignard reagents, acetylide reagents, cyanide, etc.) it is often easy to incorrectly count the number of carbons in the final product.