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 Hydrocarbons

Organic Chemistry

18. Aromaticity

Aromatic Hydrocarbons

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2:46 minutes

Problem 69d

Textbook Question

Which ion in each of the following pairs is more stable?
d.

Verified step by step guidance

Step 1: Analyze the chemical structures provided in the image. The two ions appear to be cyclopropyl carbocations, with one structure having conjugation with a double bond and the other lacking conjugation.

Step 2: Recall the concept of resonance stabilization. Conjugation between a carbocation and a double bond allows for delocalization of electrons, which stabilizes the ion.

Step 3: Evaluate the stability of the carbocations. The ion with the double bond adjacent to the carbocation can participate in resonance, spreading the positive charge over multiple atoms. This makes it more stable.

Step 4: Consider the lack of resonance in the second structure. The carbocation in the second structure is isolated and cannot delocalize its positive charge, making it less stable.

Step 5: Conclude that the ion with conjugation (the structure on the left) is more stable due to resonance stabilization, while the ion on the right is less stable due to the absence of resonance.

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Key Concepts

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

Aromatic Stability

Aromatic compounds are particularly stable due to their cyclic structure and the delocalization of π electrons across the ring. This stability arises from fulfilling Huckel's rule, which states that a compound must have 4n + 2 π electrons (where n is a non-negative integer) to be considered aromatic. The resonance energy associated with this delocalization contributes significantly to the overall stability of the molecule.

Resonance Structures

Resonance structures are different ways of drawing a molecule that illustrate the delocalization of electrons. In the context of aromatic compounds, these structures help to visualize how electrons are shared across multiple bonds, leading to increased stability. The actual structure of the molecule is a hybrid of these resonance forms, which stabilizes the compound by lowering its energy.

Inductive and Mesomeric Effects

Inductive effects refer to the electron-withdrawing or donating effects of substituents through sigma bonds, while mesomeric effects involve the delocalization of electrons through pi bonds. Both effects influence the stability of ions and molecules. In the comparison of the two structures in the question, the presence of electron-donating or withdrawing groups can significantly affect the stability of the resulting ions, making it essential to consider these effects when determining which ion is more stable.

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Related Practice

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.

(c)

Textbook Question

One of the following compounds is much more stable than the other two. Classify each as aromatic, antiaromatic, or nonaromatic.

Textbook Question

Which ion in each of the following pairs is more stable?

Textbook Question

Which ion in each of the following pairs is more stable?

Textbook Question

Biphenyl has the following structure.

a. Is biphenyl a (fused) polynuclear aromatic hydrocarbon?

b. How many pi electrons are there in the two aromatic rings of biphenyl? How does this number compare with that for naphthalene?

Textbook Question

Biphenyl has the following structure.

c. The heat of hydrogenation for biphenyl is about 418 kJ/mol (100 kcal/mol). Calculate the resonance energy of biphenyl.

d. Compare the resonance energy of biphenyl with that of naphthalene and with that of two benzene rings. Explain the difference in the resonance energies of naphthalene and biphenyl.

Textbook Question

How would you convert the following compounds to aromatic compounds?

(a)

(b)

(c)

Textbook Question

Some of the following compounds show aromatic properties, and others do not.

1. Predict which ones are likely to be aromatic, and explain why they are aromatic.

(n)

(o)

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