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

Ranking Acidity

Organic Chemistry

3. Acids and Bases

Ranking Acidity

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Problem 9

Textbook Question

Imidazole is a heteroaromatic base. Which nitrogen, a or b, is most basic?

Verified step by step guidance

Step 1: Understand the structure of imidazole. Imidazole is a five-membered heteroaromatic ring containing two nitrogen atoms. One nitrogen (a) is part of the aromatic system and has a lone pair that participates in the aromaticity. The other nitrogen (b) is bonded to a hydrogen atom and has a lone pair that does not participate in the aromatic system.

Step 2: Analyze the basicity of nitrogen 'a'. The lone pair on nitrogen 'a' is delocalized into the aromatic ring, contributing to the aromatic stability of imidazole. Because this lone pair is involved in resonance, it is less available to accept a proton (H⁺), making nitrogen 'a' less basic.

Step 3: Analyze the basicity of nitrogen 'b'. The lone pair on nitrogen 'b' is localized and not involved in the aromaticity of the ring. This makes the lone pair more available to accept a proton (H⁺), increasing the basicity of nitrogen 'b'.

Step 4: Compare the two nitrogens. Nitrogen 'b' is more basic than nitrogen 'a' because its lone pair is free and localized, while nitrogen 'a' has a lone pair that is delocalized and contributes to the aromatic stability of the ring.

Step 5: Conclude that nitrogen 'b' is the most basic nitrogen in imidazole due to the availability of its lone pair for protonation.

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Key Concepts

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

Heteroaromatic Compounds

Heteroaromatic compounds are cyclic structures that contain at least one atom other than carbon in the ring, such as nitrogen, oxygen, or sulfur. In imidazole, there are two nitrogen atoms in the five-membered ring, which contribute to its aromaticity and influence its chemical properties, including basicity.

Basicity and Protonation

Basicity refers to the ability of a compound to accept protons (H+ ions). In the context of imidazole, the basicity of the nitrogen atoms is determined by their electron density and the availability of lone pairs to bond with protons. The more basic nitrogen will be the one that can stabilize the positive charge after protonation.

Resonance and Electron Delocalization

Resonance involves the delocalization of electrons across multiple atoms in a molecule, which can stabilize charged species. In imidazole, the resonance structures can affect the electron density on the nitrogen atoms, influencing their basicity. The nitrogen that participates less in resonance will generally be more basic, as it retains more electron density.

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Related Practice

Textbook Question

Which of the following indicated atoms would you expect to be most basic?

(b)

Textbook Question

The nitrogen screened in purple in Figure 4.43 is not protonated at physiological pH. Why?

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Textbook Question

Which is the most stable base in each pair?

(b) HS^– and HO^–

Textbook Question

Identify the stronger base in each pair. Explain your choice. Citing pK_a values is not an acceptable answer.

(d)

Textbook Question

Without using pK_a values, pick out the least reactive (most stable) base in each pair. Explain your answer.

(b)

Textbook Question

Using qualitative reasoning for the acid–base reactions shown,

(i) which is stronger, the acid or the conjugate acid?

(ii) Which side of the reaction is favored?

(iii) Would you expect a K_eqgreater than, equal to, or less than 1?

(b)

Textbook Question

Identify the most stable conjugate base in each pair. Tell which structural features you analyzed and why you weighted them as you did in picking one answer.

(a)

Textbook Question

Identify the most stable conjugate base in each pair. Tell which structural features you analyzed and why you weighted them as you did in picking one answer.

(f)

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