Question

Use the following \(\mathrm{p} K_{\mathrm{a}}\) data to decide whether sodium amide would be effective at producing vinyl anions \(\left(\mathrm{R}_{2} \mathrm{C}=\overline{\mathrm{C}} \mathrm{H}\right) \cdot \mathrm{p} K_{\mathrm{a}}\) : ammonia, 38 ; vinyl hydrogen, \(\sim 46\). Write the acid-base reaction for this process.

Short answer

Yes, sodium amide would be effective at producing vinyl anions.

Step-by-step solution

Identify the Relevant \(\text{p}K_{\text{a}}\) Values

Given: \(\text{p}K_{\text{a}}\) of ammonia is 38, and \(\text{p}K_{\text{a}}\) of vinyl hydrogen is approximately 46. These values will help determine if the reaction will favor the formation of vinyl anion.

Write the Acid-Base Reaction

The acid-base reaction involves sodium amide (\text{NaNH}_2) and vinyl hydrogen (\text{R}_2C=\text{CH}_2). The reaction can be written as: $$\text{NaNH}_2 + \text{R}_2\text{C}=\text{CH}_2 \rightarrow \text{Na}^+ + \text{R}_2\text{C}=\text{CH}^- + \text{NH}_3$$ Sodium amide (strong base) reacts with vinyl hydrogen (weak acid) to form vinyl anion and ammonia.

Compare the \(\text{p}K_{\text{a}}\) Values

\(\text{p}K_{\text{a}}\) of ammonia is lower (38) compared to vinyl hydrogen (46). Since ammonia is the conjugate acid of the amide ion, it indicates that the reaction will proceed towards forming vinyl anion. A higher \(\text{p}K_{\text{a}}\) value means a weaker acid.

Conclusion on the Effectiveness

Since vinyl hydrogen has a higher \(\text{p}K_{\text{a}}\) (implying it is a weaker acid) and ammonia has a lower \(\text{p}K_{\text{a}}\) (implying ammonia is a stronger acid), sodium amide is effective in producing vinyl anions.

Key concepts

pKa values

Understanding the concept of \( \text{p}K_a \) is key to solving problems involving acid-base reactions. \( \text{p}K_a \) measures the strength of an acid. The lower the \( \text{p}K_a \), the stronger the acid, and vice versa. For example, in the given problem, ammonia has a \( \text{p}K_a \) of 38 and vinyl hydrogen has a \( \text{p}K_a \) of around 46. This tells us that ammonia is a stronger acid than vinyl hydrogen because 38 is lower than 46. The difference in \( \text{p}K_a \) values helps us predict the direction of an acid-base reaction. Generally, a reaction will favor the formation of the weaker acid (higher \( \text{p}K_a \) value).

• Stronger Acid: Lower \( \text{p}K_a \)
• Weaker Acid: Higher \( \text{p}K_a \)
• Reaction favours formation of weaker acid.

Understanding these principles can help us determine whether a reaction will occur and which products will form.

Vinyl Anion Formation

Vinyl anion formation involves an acid-base reaction where a vinyl hydrogen atom is deprotonated. In our problem, this can be shown with the equation: \(\text{NaNH}_2 + \text{R}_2\text{C}=\text{CH}_2 \rightarrow \text{Na}^+ + \text{R}_2\text{C}=\text{CH}^- + \text{NH}_3\). This reaction uses sodium amide (a strong base) to remove a proton from the vinyl hydrogen. The formation of the vinyl anion \( \text{R}_2\text{C}=\text{CH}^- \) is driven by the relative \( \text{p}K_a \) values. Since vinyl hydrogen has a higher \( \text{p}K_a \) (meaning it is a weaker acid) compared to ammonia, the reaction proceeds to form the vinyl anion and ammonia. In simpler terms, sodium amide effectively pulls off the hydrogen, creating a negatively charged vinyl carbon (vinyl anion).

• Deprotonation of vinyl hydrogen
• Reaction with strong base: NaNH₂
• Product: Vinyl anion \(\text{R}_2\text{C}=\text{CH}^- \)

This knowledge is crucial for understanding why sodium amide is used in certain organic chemistry reactions to produce specific anions.

Sodium Amide

Sodium amide, \(\text{NaNH}_2\), is a strong base commonly used in organic chemistry for deprotonation reactions. Its strength as a base is due to the presence of the highly reactive amide ion (NH₂⁻). In the given problem, sodium amide is used to deprotonate vinyl hydrogen, forming a vinyl anion. This reaction can be represented by the equation: \(\text{NaNH}_2 + \text{R}_2\text{C}=\text{CH}_2 \rightarrow \text{Na}^+ + \text{R}_2\text{C}=\text{CH}^- + \text{NH}_3\). The strong basicity of \( \text{NaNH}_2 \) makes it effective at removing protons from weaker acids (higher \( \text{p}K_a \) values). In practical terms, sodium amide is an essential reagent for creating carbon-based anions required in various organic reactions. It's powerful, so it should be handled with care.

• Strong Base
• Reacts with weaker acids
• Creates essential carbon anions

Understanding the role and properties of sodium amide enhances our ability to predict and carry out organic reactions involving anion formation.