Table of contents

1. A Review of General Chemistry5h 5m
2. Molecular Representations1h 14m
3. Acids and Bases2h 46m
4. Alkanes and Cycloalkanes4h 17m
5. Chirality3h 39m
6. Thermodynamics and Kinetics1h 22m
7. Substitution Reactions1h 48m
8. Elimination Reactions2h 30m
9. Alkenes and Alkynes2h 9m
10. Addition Reactions3h 19m
11. Radical Reactions1h 58m
12. Alcohols, Ethers, Epoxides and Thiols2h 42m
13. Alcohols and Carbonyl Compounds2h 17m
14. Synthetic Techniques1h 26m
15. Analytical Techniques:IR, NMR, Mass Spect7h 3m
16. Conjugated Systems6h 13m
17. Ultraviolet Spectroscopy51m
18. Aromaticity2h 34m
19. Reactions of Aromatics: EAS and Beyond5h 1m
20. Phenols55m
21. Aldehydes and Ketones: Nucleophilic Addition4h 56m
22. Carboxylic Acid Derivatives: NAS2h 51m
23. The Chemistry of Thioesters, Phophate Ester and Phosphate Anhydrides1h 10m
24. Enolate Chemistry: Reactions at the Alpha-Carbon1h 53m
25. Condensation Chemistry2h 9m
26. Amines1h 43m
27. Heterocycles2h 0m
28. Carbohydrates5h 53m
29. Amino Acids4h 20m
30. Peptides and Proteins2h 42m
31. Catalysis in Organic Reactions1h 30m
32. Lipids 2h 50m
33. The Organic Chemistry of Metabolic Pathways2h 52m
34. Nucleic Acids1h 32m
35. Transition Metals6h 14m
36. Synthetic Polymers1h 49m

3. Acids and Bases

Acid Base Equilibrium

Organic Chemistry

3. Acids and Bases

Acid Base Equilibrium

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6:34 minutes

Problem 17b,c

Textbook Question

Ethyne (HC≡CH) has a pK_a value of 25, water has a pK_a value of 15.7, and ammonia (NH₃) has a pK_a value of 36. Draw the equation, showing equilibrium arrows that indicate whether reactants or products are favored, for the acid–base reaction of ethyne with
b. ⁻NH₂.
c. Which would be a better base to use if you wanted to remove a proton from ethyne, HO⁻ or ^-NH₂?

Verified step by step guidance

Step 1: Understand the acid-base reaction. Ethyne (HC≡CH) acts as an acid, and the base in question is −NH2 (amide ion). The reaction involves the transfer of a proton (H⁺) from ethyne to −NH2, forming the conjugate base of ethyne (HC≡C⁻) and the conjugate acid of −NH2 (NH3).

Step 2: Write the chemical equation for the reaction. The equation is:

HC≡CH + NH_2^- 
ightleftharpoons HC≡C^- + NH_3

. Use equilibrium arrows to indicate the direction of the reaction.

Step 3: Compare the pKa values to determine the direction of equilibrium. The pKa of ethyne is 25, and the pKa of ammonia (NH3) is 36. Since the pKa of ammonia is higher, it is a weaker acid compared to ethyne. The equilibrium will favor the side with the weaker acid and weaker base, which in this case is the products (HC≡C⁻ and NH3).

Step 4: Address part c of the problem. To determine whether HO⁻ or −NH2 is a better base for deprotonating ethyne, compare their conjugate acids' pKa values. The conjugate acid of HO⁻ is water (pKa = 15.7), and the conjugate acid of −NH2 is ammonia (pKa = 36). Since a stronger base has a weaker conjugate acid (higher pKa), −NH2 is a stronger base than HO⁻.

Step 5: Conclude that −NH2 is the better base to use for removing a proton from ethyne because it is stronger than HO⁻, as indicated by the higher pKa of its conjugate acid (NH3).

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Key Concepts

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

pKa and Acid-Base Strength

The pKa value is a quantitative measure of the strength of an acid in solution; lower pKa values indicate stronger acids. In acid-base reactions, the equilibrium favors the formation of the weaker acid and base. Understanding pKa values helps predict the direction of the reaction and which species will act as the acid or base.

Equilibrium in Acid-Base Reactions

Acid-base reactions can be represented by equilibrium arrows, indicating the relative stability of reactants and products. The position of equilibrium is influenced by the strengths of the acids and bases involved. In this context, the reaction between ethyne and -NH2 will favor the formation of the weaker acid and base, which can be determined by comparing their pKa values.

Comparing Basicity of HO− and -NH2

To determine which base is better for deprotonating ethyne, one must compare the basicity of HO− (hydroxide ion) and -NH2 (amide ion). The stronger base will have a higher tendency to accept protons. Since -NH2 has a higher pKa than HO−, it is a stronger base and more effective at removing a proton from ethyne.

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Related Practice

Textbook Question

Complete the following proposed acid–base reactions, and predict whether the reactants or products are favored.

(a)

(b)

Textbook Question

Show how you would use extractions with a separatory funnel to separate a mixture of the following compounds: benzoic acid, phenol, benzyl alcohol, aniline

Textbook Question

Phenols are less acidic than carboxylic acids, with values of pK_a around 10. Phenols are deprotonated by (and therefore soluble in) solutions of sodium hydroxide but not by solutions of sodium bicarbonate. Explain how you would use extractions to isolate the three pure compounds from a mixture of p-cresol (p-methylphenol), cyclohexanone, and benzoic acid.

Textbook Question

For the following acid–base pairs, (iii) predict the favored side of equilibrium;

(e)

Textbook Question

For each of the acid–base reactions in [Section 2.3], compare the pK_a values of the acids on either side of the equilibrium arrows to prove that the equilibrium lies in the direction indicated.

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2. <IMAGE>

Textbook Question

For each of the acid–base reactions in [Section 2.3], compare the pK_avalues of the acids on either side of the equilibrium arrows to prove that the equilibrium lies in the direction indicated.

3. <IMAGE>

4. <IMAGE>

Textbook Question

Ethyne (HC≡CH) has a pK_a value of 25, water has a pK_a value of 15.7, and ammonia (NH3) has a pKa value of 36. Draw the equation, showing equilibrium arrows that indicate whether reactants or products are favored, for the acid–base reaction of ethyne with

a. HO^-.

Textbook Question

Given the data in Problem 47:

a. What pH would you make the water layer to cause the carboxylic acid to dissolve in the water layer and the amine to dissolve in the ether layer?

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