Acetylide: Videos & Practice Problems

Terminal alkynes exhibit unique acidity due to their sp hybridization, allowing them to act as strong nucleophiles when deprotonated by strong bases like sodium amide. This process generates acetylide ions, which can participate in nucleophilic substitution reactions with primary alkyl halides, leading to the formation of new carbon-carbon bonds. Additionally, dehydrohalogenation of vicinal dihalides can yield alkynes through elimination reactions, further expanding synthetic pathways in organic chemistry.

Understanding how to convert terminal alkynes to alkynides.

Triple bonds, specifically in hydrocarbons known as alkynes, can be transformed into strong nucleophiles through a unique property of terminal alkynes. Terminal alkynes possess a hydrogen atom at one end and exhibit unusual acidity due to their hybridization. This acidity can be understood through the concept of hybridization effects, which influence the acidity of compounds. In this context, terminal alkynes have sp hybridization, meaning they have 50% s character. This high s character contributes to their acidity, with a pKa value of 25, which is relatively low for hydrocarbons.

To utilize this acidity, a strong base is required to deprotonate the terminal alkyne, resulting in the formation of a strong nucleophile known as a sodium alkynide. Common strong bases for this reaction include sodium amide (NaNH₂) or sodium hydride (NaH). When a strong base is introduced, it can effectively remove the acidic hydrogen from the terminal alkyne, leading to the generation of a negatively charged species at the carbon atom involved in the triple bond.

The reaction can be illustrated as follows: when sodium hydride dissociates, it produces Na⁺ and H^-. The H^- ion acts as the base, reacting with the hydrogen of the terminal alkyne. This process results in the formation of a sodium alkynide, characterized by a negative charge on the carbon of the triple bond. This transformation allows the previously stable hydrocarbon to act as a potent nucleophile, opening up new pathways for chemical reactions.

The mechanism of akynide synthesis.

In organic chemistry, the concept of multistep synthesis is crucial, particularly when working with alkanides. One significant reaction involving sodium alkynides is their interaction with primary alkyl halides. This process begins with a terminal alkyne reacting with sodium amide (NaNH₂), which acts as a strong base. The reaction initiates an acid-base interaction where the negatively charged amide ion (NH₂^-) abstracts a proton from the terminal alkyne, resulting in the formation of a negatively charged alkynide ion.

The mechanism can be illustrated as follows: the negative charge from the alkynide pulls off the acidic hydrogen, transferring the electrons to the triple bond, thus generating a nucleophile known as an alkynide. This alkynide is often associated with sodium, hence the term sodium alkynide.

Once the alkynide is formed, it can further react with primary alkyl halides through a substitution reaction. In this context, the alkynide acts as a nucleophile, attacking the carbon atom bonded to the halogen (e.g., bromine). This results in the displacement of the halogen, which is characteristic of a substitution reaction, specifically an SN2 mechanism due to the nature of the primary alkyl halide.

To visualize the final product, one can depict a triple bond connected to a new carbon chain, illustrating the successful substitution of the halogen with the alkynide. This reaction not only highlights the utility of sodium alkynides in organic synthesis but also emphasizes the importance of understanding nucleophilic substitution mechanisms in the broader context of organic reactions.

Using double dehydrohalogenation to perform alkynide synthesis.

In the process of dehydrohalogenation of a vicinal dihalide, we start with a compound containing two bromine atoms on adjacent carbon atoms. The reaction utilizes sodium amide, a strong base, which plays a crucial role in the deprotonation steps. The first step involves the removal of a beta hydrogen from the less substituted carbon, following Zaitsev's rule. In this case, the left carbon has one hydrogen, while the right carbon has two. The amide ion preferentially deprotonates the carbon with two hydrogens, leading to the formation of a double bond and the elimination of one bromine atom, resulting in an alkene.

After the formation of the alkene, the remaining bromine atom allows for further reaction. The second mole of sodium amide again targets the hydrogen on the carbon that now has the leaving group (the bromine). This time, the amide ion removes the hydrogen from the carbon adjacent to the bromine, forming a triple bond and expelling the bromine atom. The resulting structure is an alkyne, which is more acidic due to the presence of the triple bond.

In the final step, the negatively charged terminal carbon of the alkyne reacts with methyl iodide through an SN2 mechanism. The nucleophilic carbon attacks the methyl carbon, displacing the iodine atom. This reaction results in the formation of an internal alkyne, where a methyl group is added to the previously formed triple bond.

Overall, this sequence of reactions effectively transforms a vicinal dihalide into an internal alkyne, showcasing the utility of strong bases and nucleophiles in organic synthesis.

Predict the final product of the following reaction sequence.

In this reaction, we are examining an SN2 mechanism, which is characterized by a backside attack by a nucleophile on a carbon atom that is bonded to a leaving group. The nucleophile, carrying a negative charge, approaches the carbon from the opposite side of the leaving group. This attack leads to the formation of a new bond between the nucleophile and the carbon, while simultaneously breaking the bond with the leaving group. This process is essential because a carbon atom can only form four bonds; thus, the formation of a fifth bond necessitates the displacement of the leaving group.

In this specific example, we start with a carbon backbone containing eight carbon atoms. After the nucleophilic attack by the alkanide (in this case, a methyl group from CH₃I), we add one more carbon to the structure, resulting in a total of nine carbon atoms. This transformation illustrates a fundamental concept in organic synthesis: the ability to construct larger, more complex molecules from smaller, readily available ones. Organic chemists often focus on this process, as it is crucial for the development of pharmaceuticals and other valuable compounds.

It is important to always keep track of the number of carbon atoms in your structures during these reactions. In this case, we started with eight carbons and added one from the methyl group, confirming that the final structure contains nine carbons. This simple arithmetic—8 + 1 = 9—can help prevent mistakes in drawing chemical structures, which is a common pitfall for many students. Therefore, careful counting and verification of carbon atoms is a vital practice in organic chemistry.

Here’s what students ask on this topic:

What makes terminal alkynes uniquely acidic?

Terminal alkynes are uniquely acidic due to their sp hybridization. In sp hybridization, the carbon atom has 50% s-character, which increases the acidity of the hydrogen attached to it. This is because the s-orbital is closer to the nucleus, making the bond more polar and easier to break. The pKa of a terminal alkyne is around 25, which is unusually low for hydrocarbons, indicating higher acidity. This property allows terminal alkynes to be deprotonated by strong bases, forming acetylide ions that can act as strong nucleophiles in various organic reactions.

How are acetylide ions formed from terminal alkynes?

Acetylide ions are formed from terminal alkynes through deprotonation by a strong base. The process involves using a strong base like sodium amide (NaNH₂) to remove the acidic hydrogen from the terminal alkyne. The reaction can be represented as:

${R-C\equiv C}_{-}+{\mathrm{Na}-\mathrm{NH}}_2\to {R-C\equiv C}_{-}+{\mathrm{Na}+}_{}+{\mathrm{NH}}_3$

This reaction generates a negatively charged acetylide ion, which is a strong nucleophile.

What is the role of sodium amide in the formation of acetylide ions?

Sodium amide (NaNH₂) acts as a strong base in the formation of acetylide ions. It deprotonates the terminal alkyne by removing the acidic hydrogen atom. The reaction can be represented as:

${R-C\equiv C}_{-}+{\mathrm{Na}-\mathrm{NH}}_2\to {R-C\equiv C}_{-}+{\mathrm{Na}+}_{}+{\mathrm{NH}}_3$

The sodium amide provides the amide ion (NH₂^-), which abstracts the proton from the terminal alkyne, resulting in the formation of the acetylide ion (R-C≡C^-).

How do acetylide ions participate in nucleophilic substitution reactions?

Acetylide ions, being strong nucleophiles, can participate in nucleophilic substitution reactions, particularly with primary alkyl halides. The negatively charged acetylide ion attacks the carbon atom bonded to the halide, displacing the halide ion in an SN2 mechanism. The reaction can be represented as:

${R-C\equiv C}_{-}+{R\text{'}-\mathrm{Br}}_{}\to {R-C\equiv C}_{-}+{\mathrm{Br}}_{-}$

This reaction forms a new carbon-carbon bond, extending the carbon chain and allowing for the synthesis of more complex molecules.

What is dehydrohalogenation of vicinal dihalides and how does it lead to alkyne formation?

Dehydrohalogenation of vicinal dihalides involves the elimination of hydrogen halides (HX) from adjacent carbon atoms, leading to the formation of alkenes or alkynes. Using a strong base like sodium amide (NaNH₂), the process can be driven to form alkynes. The reaction proceeds in two steps:

1. The base removes a β-hydrogen, forming a double bond and eliminating a halide ion.

2. A second equivalent of the base removes another β-hydrogen, forming a triple bond and eliminating the second halide ion.

The overall reaction can be represented as:

${R-\mathrm{CH}\left(\mathrm{Br}\right)-\mathrm{CH}\left(\mathrm{Br}\right)-R}_{+}$

This reaction is useful for synthesizing alkynes from dihalides.