Ch. 12 - Substitution and Elimination: Reactions of Haloalkanes

Mullins - Organic Chemistry: A Learner Centered Approach 1st Edition

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Ch. 12 - Substitution and Elimination: Reactions of Haloalkanes

Problem 63

Chapter 11, Problem 63

The introduction of elimination reactions provides a second way to synthesize alkynes in a two-step process starting with an alkene. Suggest a mechanism for both steps of this process.

Verified step by step guidance

Step 1: Analyze the starting material, which is an alkene. The first step involves the addition of bromine (Br₂) to the alkene. This reaction proceeds via an electrophilic addition mechanism, where the π-electrons of the alkene attack the bromine molecule, forming a bromonium ion intermediate.

Step 2: The bromonium ion is then attacked by a bromide ion (Br⁻), leading to the formation of a vicinal dibromide (a molecule with two bromine atoms attached to adjacent carbons). This completes the first step of the process.

Step 3: In the second step, the vicinal dibromide undergoes two consecutive elimination reactions in the presence of a strong base, sodium amide (NaNH₂). The base abstracts a proton from one of the carbons adjacent to a bromine atom, leading to the formation of a double bond and the elimination of a bromide ion.

Step 4: The second equivalent of NaNH₂ then abstracts another proton from the carbon adjacent to the remaining bromine atom, resulting in the formation of a triple bond (alkyne) and the elimination of the second bromide ion.

Step 5: The final product is an alkyne, specifically a terminal alkyne in this case, as the elimination reactions lead to the formation of a triple bond at the end of the carbon chain.

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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 or triple bond. In the context of synthesizing alkynes, elimination typically occurs in two steps: first, a halogen is added to an alkene, followed by a dehydrohalogenation step where a base removes a hydrogen halide, leading to the formation of an alkyne.

Mechanism of Bromination

The bromination of alkenes involves the addition of bromine (Br2) across the double bond, resulting in a vicinal dibromide. This reaction proceeds through a cyclic bromonium ion intermediate, which is then attacked by a bromide ion, leading to the formation of the dibrominated product. Understanding this mechanism is crucial for predicting the subsequent elimination step.

Base-Induced Dehydrohalogenation

In the second step of the synthesis, sodium amide (NaNH2) acts as a strong base to remove a hydrogen atom and a bromine atom from the vicinal dibromide, resulting in the formation of a triple bond. This process, known as dehydrohalogenation, is essential for converting the dibromide intermediate into the desired alkyne product, showcasing the importance of strong bases in elimination reactions.

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03:02

The dehydrohalogenation mechanism.

Related Practice

Textbook Question

Which of the following substitution reactions would you expect to occur more quickly? Explain your answer.

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Textbook Question

In Chapter 10, you learned how to make an alkyne by acetylide alkylation with a 1° haloalkane. Suggest a mechanism by which this reaction occurs.

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Textbook Question

When the reaction scheme in Assessment 12.63 is done on a monosubstituted alkene, at least three equivalents of base are needed. Reacting the product of step 2 with D–Cl (D is an isotope of H) incorporates deuterium at the terminal carbon. Explain these two observations.

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Textbook Question

Acetylide alkylation, from Assessment 12.61, fails to give the desired product with 2° haloalkanes. Why? What is the actual product of this reaction?

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Textbook Question

Give a mechanism for the following substitution and elimination reactions.

(a)

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Textbook Question

Suggest a bromoalkane and the conditions necessary to produce the alkenes shown.

(b)

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