Ch. 7 - Structure and Synthesis of Alkenes; Elimination

Wade9th EditionOrganic ChemistryISBN: 9780135213728Not the one you use?Change textbook

Wade 9th Edition

Ch. 7 - Structure and Synthesis of Alkenes; Elimination

Problem 68b

Chapter 7, Problem 68b

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. Why is no cis-2-bromobut-2-ene formed?

Verified step by step guidance

Step 1: Analyze the reaction mechanism. The reaction involves the elimination of HBr from (±)-2,3-dibromobutane using potassium hydroxide (KOH). This is an E2 elimination reaction, which requires the β-hydrogen and the leaving group (Br) to be anti-periplanar (i.e., opposite sides of the same plane).

Step 2: Consider the stereochemistry of (±)-2,3-dibromobutane. The molecule exists as a pair of enantiomers, and the anti-periplanar geometry is crucial for the elimination process. The anti-periplanar arrangement determines which β-hydrogen can be eliminated along with the bromine atom.

Step 3: Formation of trans-2-bromobut-2-ene. In the anti-periplanar geometry, the elimination of HBr leads to the formation of the trans isomer because the substituents on the double bond are positioned opposite to each other, minimizing steric hindrance.

Step 4: Explain why cis-2-bromobut-2-ene is not formed. The cis isomer would require the substituents on the double bond to be on the same side, which is not possible due to the anti-periplanar requirement of the E2 mechanism. The geometry of the starting material does not allow for the formation of the cis isomer.

Step 5: Summarize the stereochemical outcome. The reaction produces trans-2-bromobut-2-ene as the sole alkene product due to the stereochemical constraints of the E2 elimination mechanism. Additionally, (2S,3R)-3-bromobutan-2-ol and its enantiomer are formed via substitution reactions.

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Key Concepts

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

Elimination Reactions

Elimination reactions involve the removal of atoms or groups from a molecule, resulting in the formation of a double bond. In the case of (±)−2,3−dibromobutane reacting with potassium hydroxide, the reaction proceeds via an E2 mechanism, where the base abstracts a proton while a leaving group departs, leading to the formation of alkenes. The stereochemistry of the starting material influences the product distribution.

Stereochemistry and Alkene Formation

Stereochemistry refers to the spatial arrangement of atoms in molecules and is crucial in determining the types of products formed during reactions. In the formation of alkenes, the orientation of substituents around the double bond can lead to different isomers, such as cis and trans. The trans isomer is favored in this reaction due to steric factors and the anti-coplanar arrangement required for the E2 elimination mechanism.

Regioselectivity and Zaitsev's Rule

Regioselectivity describes the preference of a chemical reaction to yield one structural isomer over others. Zaitsev's Rule states that in elimination reactions, the more substituted alkene is typically the major product. In this case, the formation of trans-2-bromobut-2-ene over cis-2-bromobut-2-ene can be attributed to the stability of the trans isomer and the steric hindrance that would arise from forming the cis isomer.

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Defining Zaitsev’s Rule

Related Practice

Textbook Question

When 2-bromo-3-phenylbutane is treated with sodium methoxide, two alkenes result (by E2 elimination). The Zaitsev product predominates.

a. Draw the reaction, showing the major and minor products.

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Textbook Question

Explain the dramatic difference in rotational energy barriers of the following three alkenes. (Hint: Consider what the transition states must look like.)

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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.

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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.

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Textbook Question

When 2-bromo-3-phenylbutane is treated with sodium methoxide, two alkenes result (by E2 elimination). The Zaitsev product predominates.

b. When one pure stereoisomer of 2-bromo-3-phenylbutane reacts, one pure stereoisomer of the major product results. For example, when (2R,3R)-2-bromo-3-phenylbutane reacts, the product is the stereoisomer with the methyl groups cis. Use your models to draw a Newman projection of the transition state to show why this stereospecificity is observed.

c. Use a Newman projection of the transition state to predict the major product of elimination of (2S,3R)-2-bromo-3-phenylbutane.

d. Predict the major product from elimination of (2S,3S)-2-bromo-3-phenylbutane. This prediction can be made without drawing any structures, by considering the results in part (b).

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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?

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