Skip to main content
Ch. 7 - Structure and Synthesis of Alkenes; Elimination
Wade - Organic Chemistry 9th Edition
Wade9th EditionOrganic ChemistryISBN: 9780135213728Not the one you use?Change textbook
All textbooksWade 9th EditionCh. 7 - Structure and Synthesis of Alkenes; EliminationProblem 66
Chapter 7, Problem 66

A chemist allows some pure (2S,3R)-3-bromo-2,3-diphenylpentane to react with a solution of sodium ethoxide (NaOCH2CH3) in ethanol. The products are two alkenes: A (cis-trans mixture) and B, a single pure isomer. Under the same conditions, the reaction of (2S,3S)-3-bromo-2,3-diphenylpentane gives two alkenes, A (cis-trans mixture) and C. Upon catalytic hydrogenation, all three of these alkenes (A, B, and C) give 2,3-diphenylpentane. Determine the structures of A, B, and C; give equations for their formation; and explain the stereospecificity of these reactions.

Verified step by step guidance
1
Step 1: Analyze the reaction conditions. The reaction involves sodium ethoxide (NaOCH2CH3) in ethanol, which is a strong base. This suggests that the reaction proceeds via an E2 elimination mechanism, where the base abstracts a proton from a β-carbon, leading to the formation of an alkene.
Step 2: Consider the stereochemistry of the starting materials. The compound (2S,3R)-3-bromo-2,3-diphenylpentane has specific stereochemistry at the 2nd and 3rd carbons. In an E2 elimination, the anti-periplanar geometry is required, meaning the β-hydrogen and the leaving group (Br) must be in opposite planes. This stereochemical requirement will influence the products formed.
Step 3: Predict the alkenes formed from (2S,3R)-3-bromo-2,3-diphenylpentane. The anti-periplanar elimination leads to two possible alkenes: A (a cis-trans mixture due to rotation around the double bond) and B (a single pure isomer, likely due to stereospecificity). The stereochemistry of the starting material dictates the geometry of the double bond in B.
Step 4: Analyze the reaction of (2S,3S)-3-bromo-2,3-diphenylpentane. This compound has different stereochemistry at the 3rd carbon compared to the first starting material. The anti-periplanar requirement will again dictate the elimination pathway, leading to alkenes A (cis-trans mixture) and C (a single pure isomer with distinct stereochemistry).
Step 5: Explain the stereospecificity and hydrogenation results. Upon catalytic hydrogenation, all alkenes (A, B, and C) are reduced to 2,3-diphenylpentane, indicating that the double bonds are in the same carbon framework. The stereospecificity arises from the anti-periplanar geometry required in the E2 elimination, which determines the configuration of the double bonds in B and C.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
19m
Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Stereochemistry

Stereochemistry is the study of the spatial arrangement of atoms in molecules and how this affects their chemical behavior. In this question, the stereochemistry of the starting materials (2S,3R and 2S,3S) is crucial for understanding the formation of the alkenes A, B, and C. The terms 'cis' and 'trans' refer to specific geometric isomers that arise from the restricted rotation around double bonds, which is essential for predicting the products of the reactions.
Recommended video:
1:38
Polymer Stereochemistry Concept 1

Elimination Reactions

Elimination reactions involve the removal of atoms or groups from a molecule, resulting in the formation of a double bond. In this case, sodium ethoxide acts as a base, facilitating the elimination of bromide from the bromoalkane to form alkenes. Understanding the mechanism of elimination, including the E2 pathway, is key to predicting the structures of the resulting alkenes and their stereochemical outcomes.
Recommended video:
Guided course
00:40
Recognizing Elimination Reactions.

Catalytic Hydrogenation

Catalytic hydrogenation is a reaction that involves the addition of hydrogen (H2) across a double bond, typically using a metal catalyst. This process converts alkenes into alkanes, and in this question, it is used to illustrate that all three alkenes (A, B, and C) yield the same product, 2,3-diphenylpentane. Recognizing the role of hydrogenation helps in understanding the relationship between the alkenes and their saturated counterpart.
Recommended video:
Guided course
05:21
General properties of catalytic hydrogenation.
Related Practice
Textbook Question

Protonation converts the hydroxy group of an alcohol to a good leaving group. Suggest a mechanism for each reaction.

(b)

3
views
Textbook Question

a. Design an alkyl halide that will give only 2,4-diphenylpent-2-ene upon treatment with potassium tert-butoxide (a bulky base that promotes E2 elimination).

b. What stereochemistry is required in your alkyl halide so that only the following stereoisomer of the product is formed?

Textbook Question

Pure (S)-2-bromo-2-fluorobutane reacts with methoxide ion in methanol to give a mixture of (S)-2-fluoro-2-methoxybutane and three fluoroalkenes.

a. Use mechanisms to show which three fluoroalkenes are formed.

3
views
Textbook Question

When (±)−2,3−dibromobutane reacts with potassium hydroxide, some of the products are (2S,3R)-3-bromobutan-2-ol and its enantiomer and trans-2-bromobut-2-ene. Give mechanisms to account for these products.

7
views
Textbook Question

Protonation converts the hydroxy group of an alcohol to a good leaving group. Suggest a mechanism for each reaction.

(a)

8
views
Textbook Question

Pure (S)-2-bromo-2-fluorobutane reacts with methoxide ion in methanol to give a mixture of (S)-2-fluoro-2-methoxybutane and three fluoroalkenes.

b. Propose a mechanism to show how (S)-2-bromo-2-fluorobutane reacts to give (S)-2-fluoro-2-methoxybutane. Has this reaction gone with retention or inversion of configuration?

5
views