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

18. Aromaticity

Aromatic Heterocycles

Organic Chemistry

18. Aromaticity

Aromatic Heterocycles

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7: minutes

Problem 59

Textbook Question

Which of the following compounds could be protonated without destroying its aromaticity?

Verified step by step guidance

Step 1: Recall the criteria for aromaticity. A compound is aromatic if it satisfies Huckel's rule, which states that it must have a planar, cyclic structure with a conjugated π-electron system containing (4n + 2) π-electrons, where n is an integer.

Step 2: Analyze pyrrole. Pyrrole is a five-membered ring with a nitrogen atom. The nitrogen contributes two electrons from its lone pair to the π-system, making the total π-electron count 6 (4n + 2, where n = 1). Protonating the nitrogen would remove its lone pair from the π-system, disrupting aromaticity.

Step 3: Analyze pyridine. Pyridine is a six-membered ring with a nitrogen atom. The nitrogen's lone pair is not part of the π-system; it resides in an sp2 orbital perpendicular to the ring. Protonating the nitrogen does not affect the π-electron count, which remains 6 (4n + 2, where n = 1), preserving aromaticity.

Step 4: Compare the two compounds. Protonation of pyrrole destroys its aromaticity because the lone pair on nitrogen is essential for the π-system. Protonation of pyridine does not affect its aromaticity because the lone pair on nitrogen is not part of the π-system.

Step 5: Conclude that pyridine can be protonated without destroying its aromaticity, while pyrrole cannot.

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Key Concepts

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

Aromaticity

Aromaticity is a property of cyclic compounds that exhibit enhanced stability due to the delocalization of π electrons across the ring structure. For a compound to be aromatic, it must follow Hückel's rule, which states that it should have (4n + 2) π electrons, where n is a non-negative integer. This delocalization allows for resonance, contributing to the compound's stability and unique reactivity.

Protonation

Protonation is the addition of a proton (H⁺) to a molecule, which can alter its chemical properties and reactivity. In the context of aromatic compounds, protonation can disrupt aromaticity if it occurs at a position that affects the delocalized π system. Understanding where protonation can occur without disrupting aromaticity is crucial for predicting the behavior of these compounds in chemical reactions.

Comparison of Pyrrole and Pyridine

Pyrrole and pyridine are both nitrogen-containing heterocycles, but they differ significantly in their aromaticity and reactivity. Pyrrole, with a nitrogen atom contributing to the π system, is less stable and can be easily protonated, which disrupts its aromaticity. In contrast, pyridine, where nitrogen is part of the ring but does not contribute to the π system, can be protonated at the nitrogen without losing its aromatic character, making it a more stable base.

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Related Practice

Textbook Question

Consider the following compound, which has been synthesized and characterized:

a. Assuming this molecule is entirely conjugated, do you expect it to be aromatic, antiaromatic, or nonaromatic?

b. Why was this molecule synthesized with three tert-butyl substituents? Why not make the unsubstituted compound and study it instead?

Textbook Question

The following molecules and ions are grouped by similar structures. Classify each as aromatic, antiaromatic, or nonaromatic. For the aromatic and antiaromatic species, give the number of pi electrons in the ring.

(d)

Textbook Question

(g)

Textbook Question

Hexahelicene seems a poor candidate for optical activity because all its carbon atoms are sp² hybrids and presumably flat. Nevertheless, hexahelicene has been synthesized and separated into enantiomers. Its optical rotation is enormous: [α]_D = 3700°. Explain why hexahelicene is optically active, and speculate as to why the rotation is so large.

Textbook Question

Explain why each compound is aromatic, antiaromatic, or nonaromatic.

(d)

(e)

(f)

Textbook Question

Draw resonance forms to show the charge distribution on the pyrrole structure.

Textbook Question

How would you convert the following compounds to aromatic compounds?

(d)

(e)

(f)