Use resonance forms of the conjugate bases to explain why methanesulfonic acid (CH 3 SO 3 H, p K a = –2.6) is a much stronger acid than acetic acid (CH 3 COOH, p K a = 4.8).

Welcome back everyone to another video, identify the areas of high and low electron density in the compounds below using resonance structures. And we're given two examples. So let's begin. First of all, we're going to look at the first example. And what we notice is that the first example contains two carbon fragments, right? We have two carbonel functional groups. And this tells us that if we have pi bonds present, we can possibly deal with resonance structures. What we notice is that the car bono group has one sigma and one pi bond, we can always break our pi bond and delocalized those electrons in the form of a long pair on an electron atom such as oxygen. So this gives us our next resonance structure. Let's go ahead and draw it out if we place those electrons onto oxygen, oxygen is going to become electron ridge, right? And it is going to get a formal negative charge leaving carbon electron deficient. So carbon gets a formal positive charge and we can see the same with the other carbel right. We can essentially break the pi bond and shift the long pair of electrons on to oxygen, which essentially gives us our third resonance structure. Even though these are not stable, we are just interested in the formal positive and formal negative charges. So we're going to leave it as they are. We're not interested in major resonance contributors, right? And now let's go ahead and look at the second example before we answer the main question, what we notice is that we are given oxygen and en energy and pi bond electrons shift from high electron density to low electron density. So what we're going to do here essentially take the long pair of oxygen and form a pi bond out of it by breaking the adjacent pipe bond and forming a long pair on carbon. Let's go ahead and to the second resonance form based on the arrows. What we're noticing is that we are getting oxygen with a total of three bonds. And that means oxygen gets a formal positive charge and carbon with the resultant long pair of electrons gets a formal negative charge. And these formal charges essentially tell us which areas are high and low electron density areas. When we go back to the first example, we notice that both oxygen atoms have formal negative charges, right. So we can essentially tell that those are high electron density areas because there's a formal negative charge within our resonance form. So let's introduce the labels. We're going to show two arrows and state that the t oxygen atoms are high electron density areas. And what we notice is that the two carbon atoms from the car bono groups are low electron density areas because they have normal positive charges. Coming to the second example, we noticed that oxygen has a formal positive charge. So it is it is electron density area and the formal negative charge belongs to the double bonded carbon, meaning it is a high electron density area. So we have our answers. Let's label them and conclude the video. Thank you for watching.

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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

Ranking Acidity

Organic Chemistry

3. Acids and Bases

Ranking Acidity

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Problem 32

Textbook Question

Use resonance forms of the conjugate bases to explain why methanesulfonic acid (CH₃SO₃H, pK_a = –2.6) is a much stronger acid than acetic acid (CH₃COOH, pK_a = 4.8).

Verified step by step guidance

Identify the conjugate bases of methanesulfonic acid (CH3SO3H) and acetic acid (CH3COOH). The conjugate base of methanesulfonic acid is the methanesulfonate ion (CH3SO3⁻), and the conjugate base of acetic acid is the acetate ion (CH3COO⁻).

Recall that the stability of the conjugate base is a key factor in determining the strength of an acid. A more stable conjugate base corresponds to a stronger acid because the equilibrium of the acid dissociation reaction shifts more toward the ionized form.

Analyze the resonance structures of the methanesulfonate ion (CH3SO3⁻). The negative charge on the oxygen atom is delocalized over three oxygen atoms through resonance, which significantly stabilizes the conjugate base. Write the resonance forms of CH3SO3⁻ using MathML:

{CH 3}_{({SO 3}_{) -}}

Compare this to the resonance structures of the acetate ion (CH3COO⁻). The negative charge on the oxygen atom is delocalized over only two oxygen atoms, which provides less stabilization compared to the methanesulfonate ion. Write the resonance forms of CH3COO⁻ using MathML:

{CH 3}_{({CO 2}_{) -}}

Conclude that the greater resonance stabilization of the methanesulfonate ion (CH3SO3⁻) compared to the acetate ion (CH3COO⁻) makes methanesulfonic acid (CH3SO3H) a much stronger acid than acetic acid (CH3COOH). This is reflected in their respective pKa values, where a lower pKa indicates a stronger acid.

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Key Concepts

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

Resonance Structures

Resonance structures are different ways of drawing the same molecule that illustrate the delocalization of electrons. In the context of acids, resonance can stabilize the conjugate base by distributing negative charge over multiple atoms, making the acid stronger. For methanesulfonic acid, its conjugate base can be represented by several resonance forms, enhancing stability compared to the conjugate base of acetic acid.

Acid Strength and pKa

The strength of an acid is often measured by its pKa value, which indicates the tendency of the acid to donate a proton (H+). A lower pKa value signifies a stronger acid, as it more readily donates protons. Methanesulfonic acid has a pKa of -2.6, indicating it is a much stronger acid than acetic acid, which has a pKa of 4.8, reflecting its weaker tendency to lose a proton.

Conjugate Bases

The conjugate base of an acid is what remains after the acid donates a proton. The stability of the conjugate base is crucial in determining the strength of the corresponding acid. In this case, the conjugate base of methanesulfonic acid is more stable due to resonance stabilization, while the conjugate base of acetic acid is less stable, contributing to the significant difference in their acid strengths.

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