Ch. 10 - Alkynes: Electrophilic Addition and Redox Reactions

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

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Mullins 1st Edition

Ch. 10 - Alkynes: Electrophilic Addition and Redox Reactions

Problem 30

Chapter 9, Problem 30

Long, polarized bonds are also reducible in the same way that the C―C π bond of an alkyne is. Show a mechanism by which a C―Br bond might be reduced by sodium metal.

Verified step by step guidance

Step 1: Recognize that the C-Br bond is polarized due to the electronegativity difference between carbon and bromine. Bromine is more electronegative, making the bond susceptible to reduction.

Step 2: Understand that sodium metal (Na⁰) acts as a reducing agent. Sodium donates electrons to facilitate the cleavage of the C-Br bond.

Step 3: In the first step of the mechanism, sodium donates an electron to the C-Br bond, resulting in the formation of a carbon radical and a bromide ion (Br⁻). This is known as a single-electron transfer (SET).

Step 4: In the second step, another sodium atom donates an electron to the carbon radical, converting it into a carbanion (C⁻). This carbanion is highly reactive and will quickly abstract a proton (H⁺) from a solvent or proton source, forming the final alkane product.

Step 5: The bromide ion (Br⁻) formed in the first step combines with sodium ions (Na⁺) to form sodium bromide (NaBr), which is a byproduct of the reaction.

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Key Concepts

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

Reduction Reactions

Reduction reactions involve the gain of electrons or the decrease in oxidation state of a molecule. In organic chemistry, this often pertains to the conversion of a compound with a polarized bond, such as C―Br, into a less oxidized form. Understanding the principles of reduction is essential for predicting the outcome of reactions involving halides and metals.

Mechanism of Metal Reduction

The mechanism of metal reduction typically involves the transfer of electrons from a metal, such as sodium, to the substrate. In the case of reducing a C―Br bond, sodium donates an electron, leading to the formation of a radical or an anion. This process is crucial for understanding how metals can facilitate the breaking of strong bonds in organic compounds.

Radical Intermediates

Radical intermediates are highly reactive species that contain an unpaired electron. In the reduction of a C―Br bond, the formation of a carbon radical can occur after the initial electron transfer from sodium. Recognizing the role of radicals in reaction mechanisms is vital for predicting the pathways and products of organic reactions.

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Related Practice

Textbook Question

The alkyl carbocation is estimated to be 15 kcal/mol more stable than the alkenyl carbocation. If this is also the difference in the energies of the transition state leading to each, what is the expected rate difference?

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Textbook Question

We have studied the following reactions in previous chapters. For each, (i) indicate which reaction sheets they should appear on, (ii) show the best structure to use to represent them, and (iii) write the notes you could put in the margin so that the mechanism is implied.

(b) Radical halogenation of an alkane

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Textbook Question

Predict the major products resulting from the addition of one equivalent of HX to the following alkynes.

(b)

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Textbook Question

When sodium generates electrons in the presence of ammonia, these electrons persist in solution, giving the blue color. However, electrons do not persist when sodium is added to water. Why?

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Textbook Question

The following deprotonation step occurs during the trans reduction of an alkyne. Calculate K_eq for this reaction.

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Textbook Question

Sodium amide, the base we use to deprotonate terminal alkynes, is synthesized by reducing ammonia via a mechanism similar to the reduction of alkynes in Figure 10.21. Suggest a mechanism for this reaction.

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