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Which of the following substrates would have the fastest reaction rate for an SN1 mechanism?
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.
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Which of the following determines the general rate of an reaction?
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.
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What is the final product of the pictured reaction?
1.
2.
3.
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.
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Which of the following is not true for an SN1 reaction?
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.
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A student carried out a substitution reaction in the lab using ethanol as a solvent. The student began with an optically pure reactant (100% (R)-configuration) and finished with a racemic mixture of products (50% (R)-configuration, 50% (S)-configuration).
The reaction went through which of the following mechanisms?
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.
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Which SN2 reaction would proceed the fastest?
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.
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In reactions involving the alkylation of acetylide ions, it is preferred that the alkyl halide be primary. What is the reason for this?
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.
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Predict the major product of the given SN1 reaction.
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).
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What type of reaction is shown?
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.
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Which of the following molecules would most readily undergo an SN2 mechanism?
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)
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Which of the following reagents would complete this reaction with the proper stereochemistry?
The sterochemistry should be inverted for an SN2 reaction. The product is has S chirality , so the starting material should have R.
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Suppose that a chemistry student is trying to run a reaction in the lab. In his solution, he adds , ethanol, and dimethylformamide (DMF) as a solvent. However, no reaction takes place. To solve this problem, the student adds hydrochloric acid to the solution and, in doing so, a reaction takes place that produces the desired product, chloroethane.
What is the most likely reason for why the addition of hydrochloric acid to the solution allowed the reaction to proceed?
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.
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