Step-by-Step Guidance for Alkynes (Organic Chemistry, Chapter 9)

Q1. How are alkynes prepared from dihalides?

Background

Topic: Preparation of Alkynes

This question tests your understanding of how alkynes can be synthesized from dihalides using strong bases, a fundamental transformation in organic chemistry.

Key Terms and Formulas

• Dihalide: A molecule with two halogen atoms attached to adjacent carbons.

• Alkyne: A hydrocarbon with a carbon-carbon triple bond ($\ce{C\equiv C}$).

• Strong Base: Commonly sodium amide ($\ce{NaNH2}$) is used for elimination reactions.

• E2 Mechanism: A bimolecular elimination process.

Step-by-Step Guidance

1. Identify the starting dihalide. It must be either geminal (both halogens on the same carbon) or vicinal (halogens on adjacent carbons).

2. Recognize that a strong base is required to promote elimination. Typically, $\ce{NaNH2}$ is used in excess.

3. Understand that two successive E2 eliminations occur: the first elimination forms an alkene, and the second elimination forms the alkyne.

4. Write the general reaction: $\ce{R-CHBr-CHBr-R + 2 NaNH2 \rightarrow R-C\equiv C-R + 2 NaBr + 2 NH3}$

Try solving on your own before revealing the answer!

Final Answer: Alkynes are prepared from dihalides by treatment with excess sodium amide, resulting in two E2 eliminations to form the triple bond.

The reaction proceeds via two consecutive eliminations, converting the dihalide to an alkyne.

Q2. What is the mechanism for dissolving metal reduction of alkynes?

Background

Topic: Reduction of Alkynes (Birch Reduction)

This question tests your ability to describe the stepwise mechanism by which an alkyne is reduced to a trans alkene using sodium metal and ammonia.

Key Terms and Formulas

• Birch Reduction: A method for reducing alkynes to trans alkenes.

• Radical Anion: An intermediate formed by electron transfer.

• Anti Addition: Addition of atoms to opposite sides of the molecule.

Step-by-Step Guidance

1. Start with the alkyne and sodium metal in liquid ammonia.

2. First, an electron is transferred from sodium to the alkyne, generating a radical anion intermediate.

3. Ammonia donates a proton to the radical anion, forming a radical intermediate.

4. A second electron is transferred from sodium to the radical intermediate, generating an anion.

5. Ammonia donates a proton to the anion, yielding the trans alkene.

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Final Answer: The dissolving metal reduction mechanism involves alternating electron transfers and protonations, resulting in anti addition and formation of a trans alkene.

The stepwise process ensures stereoselectivity for the trans product.

Q3. What are the products of ozonolysis of a symmetrical internal alkyne?

Background

Topic: Ozonolysis of Alkynes

This question tests your ability to predict the products formed when a symmetrical internal alkyne undergoes ozonolysis.

Key Terms and Formulas

• Ozonolysis: Oxidative cleavage of a triple bond using ozone ($\ce{O3}$) followed by hydrolysis.

• Symmetrical Internal Alkyne: An alkyne with identical groups on both sides of the triple bond.

• Carboxylic Acid: The typical product of alkyne ozonolysis.

Step-by-Step Guidance

1. Identify the symmetrical internal alkyne structure.

2. Apply ozonolysis conditions: $\ce{O3}$ followed by $\ce{H2O}$.

3. Recognize that the triple bond is cleaved, and each fragment is oxidized to a carboxylic acid.

4. For symmetrical alkynes, only one carboxylic acid product is formed, but two equivalents are produced.

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Final Answer: Ozonolysis of a symmetrical internal alkyne yields two equivalents of the same carboxylic acid.

The reaction cleaves the triple bond and oxidizes both sides to carboxylic acids.