Ch. 10 - Alkynes: Electrophilic Addition and Redox Reactions

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

Mullins1st EditionOrganic Chemistry: A Learner Centered ApproachISBN: 9780137566471Not the one you use?Change textbook

Mullins 1st Edition

Ch. 10 - Alkynes: Electrophilic Addition and Redox Reactions

Problem 29

Chapter 9, Problem 29

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.

Verified step by step guidance

Step 1: Begin by recognizing that sodium amide (NaNH₂) is formed by the reaction of ammonia (NH₃) with metallic sodium (Na⁰). The reaction also produces hydrogen gas (H₂) as a byproduct.

Step 2: Understand that metallic sodium (Na⁰) acts as a reducing agent. Sodium donates an electron to ammonia, initiating the formation of a radical intermediate. This is similar to the reduction mechanism seen in alkynes.

Step 3: Propose the first step of the mechanism: Sodium donates an electron to ammonia (NH₃), forming an ammonia radical anion (NH₃⁻•). This radical anion is highly reactive.

Step 4: The ammonia radical anion (NH₃⁻•) undergoes proton transfer, where it loses a proton to form the amide ion (NH₂⁻) and hydrogen gas (H₂). This step is facilitated by the presence of another ammonia molecule acting as a proton donor.

Step 5: Finally, the amide ion (NH₂⁻) combines with sodium cation (Na⁺) to form sodium amide (NaNH₂). This process repeats for the second ammonia molecule, resulting in the overall reaction: 2 NH₃ + 2 Na⁰ → 2 NaNH₂ + H₂.

Verified video answer for a similar problem:

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

Video duration:

Was this helpful?

Key Concepts

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

Deprotonation of Terminal Alkynes

Deprotonation of terminal alkynes involves the removal of a proton from the terminal carbon atom, resulting in the formation of an alkyne anion. This process is typically facilitated by strong bases like sodium amide (NaNH₂), which can effectively abstract the acidic hydrogen due to the high acidity of terminal alkynes compared to other hydrocarbons.

Reduction Mechanisms

Reduction mechanisms in organic chemistry refer to the processes that involve the gain of electrons or hydrogen, or the loss of oxygen. In the context of synthesizing sodium amide from ammonia, the reduction of ammonia involves the transfer of electrons from sodium metal to ammonia, resulting in the formation of sodium amide and hydrogen gas.

Sodium Amide Synthesis

Sodium amide (NaNH₂) is synthesized through the reduction of ammonia (NH₃) using sodium metal (Na). This reaction is significant in organic synthesis as sodium amide serves as a strong base for deprotonating terminal alkynes, enabling further reactions such as nucleophilic substitutions or eliminations in organic synthesis.

Recommended video:

Guided course

01:33

Amide Nomenclature

Related Practice

Textbook Question

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.

views

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?

<IMAGE>

views

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

views

Textbook Question

Suggest alkynes that might be used to make the following trans-alkenes.

(c)

views

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?

views

Textbook Question

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

views