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Example Questions
Example Question #1 : Substitution Mechanisms
Which SN2 reaction would proceed the fastest?
1-bromopentane and sodium iodide
tert-butyl bromide and sodium iodide
sec-butyl bromide and sodium iodide
1-chloropentane and sodium iodide
1-bromo-2-methylpentane and sodium iodide
1-bromopentane and sodium iodide
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.
Example Question #203 : Gre Subject Test: Chemistry
In reactions involving the alkylation of acetylide ions, it is preferred that the alkyl halide be primary. What is the reason for this?
The reactions generally occur in two steps
The mechanism for these reactions is SN1
An answer cannot be determined without more information about the reaction conditions
The mechanism for these reactions is SN2
The reaction involves a carbocation as intermediate
The mechanism for these reactions is SN2
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.
Example Question #1 : Help With Sn2 Reactions
Predict the major product of the given SN1 reaction.
I
IV
II
III
IV
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).
Example Question #1 : Help With Sn2 Reactions
What type of reaction is shown?
None of these
E1
E2
SN2
SN1
SN2
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.
Example Question #8 : Reaction Mechanisms, Energetics, And Kinematics
Which of the following reagents would complete this reaction with the proper stereochemistry?
and
and
and
The sterochemistry should be inverted for an SN2 reaction. The product is has S chirality , so the starting material should have R.
Example Question #501 : Organic Chemistry
Which of the following molecules would most readily undergo an SN2 mechanism?
III
V
II
I
IV
IV
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)
Example Question #2 : Help With Sn2 Reactions
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?
The addition of hydrochloric acid increased the energy of the transition state of the reaction, which resulted in enough energy to drive the reaction forward.
The addition of hydrochloric acid to the solution protonates the hydroxyl group on ethanol, allowing it be a better leaving group.
The addition of hydrochloric acid resulted in an increase in the concentration of chloride ion, which increased the reaction rate and drove the reaction forward.
The addition of hydrochloric acid protonated the solvent and, as a result, the reaction was able to move forward because DMF is a polar aprotic solvent.
The addition of hydrochloric acid to the solution protonates the hydroxyl group on ethanol, allowing it be a better leaving group.
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.