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

16. Conjugated Systems

Orbital Diagram:5-atoms- Allylic Ions

Organic Chemistry

16. Conjugated Systems

Orbital Diagram:5-atoms- Allylic Ions

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4:58 minutes

Problem 37a

Textbook Question

The pentadienyl radical, H₂C=CH–CH=CH–CH₂•, has its unpaired electron delocalized over three carbon atoms.
a. Use resonance forms to show which three carbon atoms bear the unpaired electron.

Verified step by step guidance

Step 1: Begin by identifying the structure of the pentadienyl radical, H₂C=CH-CH=CH-CH₂·. The radical is located on the terminal CH₂ group, and the molecule contains conjugated double bonds, which allow for resonance delocalization.

Step 2: Draw the first resonance structure. Move the unpaired electron from the CH₂ group to the adjacent carbon (C4), and simultaneously shift the π-electrons of the C3=C4 double bond to form a new π-bond between C2 and C3. This results in a new resonance structure where the unpaired electron is now on C4.

Step 3: Draw the second resonance structure. Move the unpaired electron from C4 to C3, and shift the π-electrons of the C2=C3 double bond to form a new π-bond between C1 and C2. This results in a resonance structure where the unpaired electron is now on C3.

Step 4: Summarize the resonance delocalization. The unpaired electron is delocalized over C3, C4, and C5 through the conjugated π-system, as shown by the resonance structures. This delocalization stabilizes the radical.

Step 5: Highlight the importance of resonance in stabilizing radicals. Explain that resonance allows the unpaired electron to be spread over multiple atoms, reducing the energy of the system and increasing stability.

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Key Concepts

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

Radicals

Radicals are species that contain an unpaired electron, making them highly reactive. In the case of the pentadienyl radical, the unpaired electron is crucial for understanding its reactivity and stability. The presence of this unpaired electron allows for resonance stabilization, where the electron can be delocalized over multiple atoms, reducing the overall energy of the radical.

Resonance Structures

Resonance structures are different ways of drawing the same molecule that illustrate the delocalization of electrons. For the pentadienyl radical, resonance forms help visualize how the unpaired electron can be distributed across the three carbon atoms. This concept is essential for predicting the stability and reactivity of radicals, as it shows that the radical can exist in multiple forms, contributing to its overall character.

Delocalization of Electrons

Delocalization of electrons refers to the spreading of electron density across multiple atoms rather than being localized between two. In the pentadienyl radical, the unpaired electron is delocalized over three carbon atoms, which stabilizes the radical. This delocalization is a key factor in understanding the behavior of radicals in chemical reactions, as it influences their reactivity and the types of products formed.

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Related Practice

Textbook Question

The pentadienyl radical, H₂C=CH–CH=CH–CH₂•, has its unpaired electron delocalized over three carbon atoms.

g. Remove the highest-energy electron from the pentadienyl radical to give the pentadienyl cation. Which carbon atoms share the positive charge? Does this picture agree with the resonance picture?

Textbook Question

The pentadienyl radical, H₂C=CH–CH=CH–CH₂•, has its unpaired electron delocalized over three carbon atoms.

h. Add an electron to the pentadienyl radical to give the pentadienyl anion. Which carbon atoms share the negative charge? Does this picture agree with the resonance picture?

Textbook Question

Answer the following questions for the MOs of 1,3-butadiene:

a. Which are $π$ bonding MOs, and which are $π$ * antibonding MOs?

b. Which MOs are symmetric, and which are antisymmetric?

c. Which MO is the HOMO and which is the LUMO in the ground state?

Open Question

Predict the molecular orbitals and identify the HOMO and LUMO orbitals of the following cation.

Textbook Question

The pentadienyl radical, H₂C=CH–CH=CH–CH₂•, has its unpaired electron delocalized over three carbon atoms.

b. How many MOs are there in the molecular orbital picture of the pentadienyl radical?

c. How many nodes are there in the lowest-energy MO of the pentadienyl system? How many in the highest-energy MO?

d. Draw the MOs of the pentadienyl system in order of increasing energy

Textbook Question

The pentadienyl radical, H₂C=CH–CH=CH–CH₂•, has its unpaired electron delocalized over three carbon atoms.

f. Show how your molecular orbital picture agrees with the resonance picture showing delocalization of the unpaired electron onto three carbon atoms.

Textbook Question

The pentadienyl radical, H₂C=CH–CH=CH–CH₂•, has its unpaired electron delocalized over three carbon atoms.

g. Remove the highest-energy electron from the pentadienyl radical to give the pentadienyl cation. Which carbon atoms share the positive charge? Does this picture agree with the resonance picture?

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The pentadienyl radical, H2C=CH–CH=CH–CH2•, has its unpaired electron delocalized over three carbon atoms.a. Use resonance forms to show which three carbon atoms bear the unpaired electron.

Key Concepts

Radicals

Resonance Structures

Delocalization of Electrons

Watch next

The pentadienyl radical, H₂C=CH–CH=CH–CH₂•, has its unpaired electron delocalized over three carbon atoms.
a. Use resonance forms to show which three carbon atoms bear the unpaired electron.