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Example Questions
Example Question #1 : Help With Elimination Reactions
Which of the following reaction conditions favors an E1 reaction mechanism?
Weak base
Weak nucleophile
Aprotic solvent
Strong nucleophile
Strong base
Weak base
E1 reactions occur in two steps. First, the leaving group is removed, yielding a carbocation. Second, a weak base removes a proton from the carbon adjacent to the carbocation carbon. Thus, to favor E1, a protic solvent is desired in order to stabilize the carbocation. Weak bases favor an E1 mechanism.
Example Question #1 : Help With Elimination Reactions
Which of the following conditions favors an E2 mechanism?
Weak base
Strong base
Strong nucleophile
Weak nucleophile
Strong electrophile
Strong base
E2 reaction occur in only one step. The strong base removes the beta-hydrogen, while the leaving group simultaneously leaves. The double bond forms simultaneously.
Example Question #2 : Help With Elimination Reactions
In an elimination reaction __________.
two pi bonds are broken and one sigma bond is formed
one sigma bond is broken and one sigma bond is formed
one sigma bond and one pi bond are broken, and two sigma bonds are formed
two sigma bonds are broken, and one sigma and one pi bond are formed
one pi bond is broken and one pi bond is formed
two sigma bonds are broken, and one sigma and one pi bond are formed
Elimination reactions involve the use of bases, which remove hydrogen atoms. The leaving group, which is bound via a sigma bond is removed, along with a hydrogen (thus two sigma bonds are broken). The result is a double bond, which consists of one sigma bond and one pi bond.
Example Question #3 : Help With Elimination Reactions
Which of the following characteristics does not reflect an E1 reaction mechanism?
Formation of a carbocation intermediate
Tertiary substrate
Unimolecular reaction kinetics
Aprotic solvent
Weak base
Aprotic solvent
E1 reactions occur in two steps, forming a carbocation intermediate, which is most stable if there is a protic solvent present. Furthermore, the use of a weak base favors E1.
Example Question #3 : Help With Elimination Reactions
Which of the following characteristics does not reflect an E2 reaction mechanism?
Strong base
Two-step mechanism
Bimolecular reaction kinetics
Tertiary substrate
Aprotic solvent
Two-step mechanism
E2 reactions occur in one step; thus no carbocation intermediate is formed, and an aprotic solvent is favored. E2 reactions are favored by strong bases and higher temperatures.
Example Question #4 : Help With Elimination Reactions
The molecule shown reacts with . Which type of reaction will ensue?
These conditions are perfect for an reaction. We see a secondary halide so first instinct might be to say that an
reaction would ensue. However, anytime we have a secondary halide reacting with a strong, bulky base, the laws of sterics dictates that an
reaction will ensue.
Example Question #5 : Help With Elimination Reactions
Chloropropane reacts with to form propene.
Through which mechanism did this reaction occur?
The given conditions are perfect for an reaction. Here we have a primary haloalkane reacting with a big, bulky base. Big, bulky bases such as
, and alkene products are classic indicators that the reaction was an
reaction.
Example Question #3 : Help With Elimination Reactions
Which of the following hydrogens are removed upon treatment of the pictured molecule with potassium tert-butoxide?
IV
III
I
II
III
Treatment with strong base indicates E2 mechanism. The halide has two beta-carbons (II and III), and potassium tert-butoxide is a bulky base. The favored product is the Hofmann product (less substituted alkene). Therefore III is the hydrogen that's removed.
Example Question #5 : Help With Elimination Reactions
What is the product of the reaction shown?
I, II, and III
I and II
II only
I only
III only
II only
E2 reaction mechanism requires antiperiplanar orientation between the leaving group and the hydrogen. Given the stereochemistry of the methyl group, forming an alkene bond (to give the more substituted product) is impossible.
Example Question #61 : Organic Concepts
What is the major product in the following reaction?
This is an acid-catalyzed dehydration of a tertiary alcohol. The first step is protonation of the alcohol oxygen which will create a good leaving group (), and subsequent loss of water from the molecule. The molecule will then contain a carbocation (which will not move due to the positive being on the most substituted carbon atom in its immediate vicinity), and the conjugate base (water in this case) will remove a hydrogen from a carbon next to the carbocation. The base will preferentially remove a hydrogen from the one that will produce the more substituted alkene, as this is generally the more stable product. This is known as Zaitsev's rule for the formation of alkenes.
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