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Example Question #1 : Help With E2 Reactions
Which of the following compounds could NEVER undergo an E2 reaction when treated with potassium tert-butoxide?
Benzylbromide
Cyclopentylbromide
3-methyl-3-iodopentane
Bromoethane
Cis-2-bromo-1-methylcyclohexane
Benzylbromide
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.
Example Question #1 : Help With E2 Reactions
Which of the following statements concerning substitution and elimination reactions is true?
Acetate is a better nucleophile than acetic acid
In the absence of heat, strong bases, and good nucleophiles, tertiary alkyl halides will react via the SN1 mechanism
The more hindered a strong base is, the more likely it is to produce an E2 reaction
SN1 reactions follow a 2-step mechanism; SN2 reactions follow a 1-step mechanism
All of these are true statements
All of these are true statements
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.
Example Question #1 : Help With E2 Reactions
If the following disubstituted cyclohexane is refluxed (reacted in boiling solvent) in THF and sodium methoxide, which of the following will be the major product?
IV
V
I
II
III
I
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.
Example Question #1 : Help With E2 Reactions
Which of the following represents the general rate law for E2 elimination reactions?
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.
Example Question #1 : Help With E2 Reactions
A student is carrying out an E2 reaction in the lab on a tertiary substrate. Which of the following bases should the student use to obtain the least substituted product?
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.
Example Question #1 : Help With E2 Reactions
Which of the following products will result from this reaction?
The elimination reaction (as opposed to the SN2 because of the big bulky base reagent) only proceeds with the anti-periplanar product.
Example Question #6 : Help With E2 Reactions
What is the product of the following reaction?
No reaction.
More than one of these.
Because both hydrogens neighboring the are anti-periplanar, both elimination products would be formed.
Example Question #1 : Help With E2 Reactions
Which of the following sets of conditions most strongly favors an E2 elimination reaction?
Strong base, polar aprotic solvent, tertiary electrophile
Strong base, polar aprotic solvent, primary electrophile
Weak base, polar aprotic solvent, tertiary electrophile
Strong base, polar protic solvent, secondary electrophile
Strong base, polar aprotic solvent, primary electrophile
Weak base, polar aprotic solvent, tertiary electrophile
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.
Example Question #1 : Help With E2 Reactions
A student carried out a substitution reaction in the lab using ether as a solvent. The student began with an optically pure reactant (100% (R)-configuration) and finished with an optically pure product (100% (S)-configuration).
The reaction went through which of the following mechanisms?
E1
E2
Either SN1 or SN2
SN1
SN2
SN2
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.
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