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 70

Chapter 7, Problem 70

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

Verified step by step guidance

Step 1: Analyze the structures of the alkenes provided. The first alkene has no substituents, the second has phenyl groups, and the third has ester and amine groups. Consider how these substituents might affect the rotational energy barrier.

Step 2: Consider the steric hindrance in each alkene. The first alkene has the highest rotational energy barrier due to the lack of substituents, which means there is no steric hindrance to stabilize the transition state. The second alkene has phenyl groups that can provide some stabilization through conjugation, lowering the energy barrier.

Step 3: Evaluate the electronic effects of the substituents. The third alkene has ester and amine groups, which can participate in resonance and hyperconjugation, significantly stabilizing the transition state and thus lowering the rotational energy barrier.

Step 4: Consider the transition states for each alkene. The transition state for the first alkene is less stable due to the absence of substituents. The second alkene's transition state is stabilized by the phenyl groups through π-conjugation. The third alkene's transition state is highly stabilized by resonance effects from the ester and amine groups.

Step 5: Conclude that the dramatic difference in rotational energy barriers is due to the varying degrees of steric and electronic stabilization provided by the substituents in each alkene. The more substituents that can stabilize the transition state, the lower the rotational energy barrier.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.

Video duration:

11m

Was this helpful?

Key Concepts

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

Rotational Energy Barriers

Rotational energy barriers refer to the energy required to rotate around a double bond in alkenes. This energy is influenced by steric hindrance and electronic effects, which can stabilize or destabilize the transition state during rotation. A higher barrier indicates greater resistance to rotation, often due to bulky substituents or unfavorable interactions in the transition state.

Transition States

Transition states are high-energy configurations that occur during the conversion of reactants to products in a chemical reaction. In the context of alkenes, the transition state during rotation involves the alignment of substituents around the double bond. The stability of this state is crucial in determining the rotational energy barrier, as more stable transition states correspond to lower energy barriers.

Steric Hindrance

Steric hindrance refers to the repulsion between bulky groups in a molecule that can impede certain reactions or conformational changes. In alkenes, larger substituents can create significant steric strain when attempting to rotate around the double bond, leading to higher rotational energy barriers. Understanding steric effects is essential for predicting the stability and reactivity of different alkene isomers.

Recommended video:

Guided course

02:53

Understanding steric effects.

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.

1666

views

Textbook Question

Write a mechanism that explains the formation of the following product. In your mechanism, explain the cause of the rearrangement, and explain the failure to form the Zaitsev product.

962

views

Textbook Question

One of the following dichloronorbornanes undergoes elimination much faster than the other. Determine which one reacts faster, and explain the large difference in rates.

1056

views

Textbook Question

The following reaction is called the pinacol rearrangement. The reaction begins with an acid-promoted ionization to give a carbocation. This carbocation undergoes a methyl shift to give a more stable, resonance-stabilized cation. Loss of a proton gives the observed product. Propose a mechanism for the pinacol rearrangement.

1787

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

972

views

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).

2590

views

comments