Elimination and Addition Flashcards
True or false: The use of any Lewis base makes it very likely that the substrate will undergo an E1 reaction
This statement is false. Lewis bases are defined as electron donors, whether they are particularly likely to abstract protons from a substrate or not. By contrast, Brønsted-Lowry bases are those that must abstract protons from their environment. Use of a Brønsted-Lowry base can provide the correct chemical environment for an E1 reaction to occur. We want to take the proton!
Students are trying to determine by which elimination reaction mechanism two different reactions are proceeding. The rate of reaction A seems to depend only on the concentration of the substrate, while reaction B seems to depend on both the concentration of base and concentration of substrate. Can a determination be made from this information, and if so, what is it?
The first reaction is likely E1. Like SN1, the E1 kinetics is dependent on the substrate concentration. Like SN2, the second reaction which is more likely and E2 reaction is dependent on the substrate and base concentrations. Blue Print Explanation: E1 reactions obey a unimolecular rate law (rate = k[A]) while E2 reactions obey a bimolecular rate law (rate = k[A][B]). Given that it is already known that elimination reactions are occurring, we can rather confidently assume that reaction A is proceeding through an E1 mechanism while reaction B is proceeding through an E2 mechanism
Predict the major product(s) of the reaction above, assuming it follows Zaitsev’s rule.
Zaitsev’s rule is particularly applicable to E1 elimination reactions, and worth being tangentially aware of for the MCAT. It states that in an elimination reaction, the more substituted product will be favored. IMPORTANT: notice there are no wedges or dashes on this molecule. recall that sp2 hybridized carbons have a trigonal planar molecular geometry. Two adjacent sp2 hybridized carbons connected by a double bond will be restricted in their rotation, and all their substituents should share a single plane.
What is the most likely outcome of this reaction?
While for reasons outside the scope of the MCAT, the actual product is likely to be entirely the trans isomer but the cis technically could form. Tertbutoxide is a strong base and will likely lead to an elimination.
The mechanism by which this would occur is likely E2. Reactions carried out in a strongly basic environment, such as that created by tert-butoxide, strongly favor the E2 mechanism.
How can you identify nucelophilic addition versus substitution?
If no substituents leave and a new bond is formed, we can infer that a nucleophilic addition reaction is almost certainly taking place. Addition just adds another group where substitution replaces a group.
Is this nucelophilic addition, substitution, or neither.
Addition.
What characteristics should a good E1 substrate have?
Like SN2, E1 rely on the formation of a carbocation intermediate. A more substituted substrate will be more stable when forming a carbocation intermediate so a tert-butyl chloride is a tertiary alkyl halide which would do better than a primary alkyl halide.
E1 or E2:
E1
The E1 mechanism is analogous to the SN1 mechanism in that it requires the formation of a carbocation intermediate and occurs in two steps.
E1 or E2:
E2
The E2 mechanism is analogous to the SN2 mechanism in that it occurs in a single, concerted step in the presence of a strong nucleophile.
True or False: Cyclization of an aldohexose (or ketohexose, through an intermediate tautomerization step) is an example of an intramolecular nucleophilic addition mechanism.
True
This would form a ketal
Hemiacetal
Is the following considered nucleophilic addition?
NO. The formation of an acetal from a hemiacetal - as is pictured here in this glycosidic bond formation mechanism between two molecules of glucose to form trehalose - is a nucleophilic substitution reaction, as one -OH substituent is replaced in the process.
Zaitsev’s Rule
the more substituted c are preferred for stability, trans is more stable than cis bonds as well. This rule m
