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

6. Thermodynamics and Kinetics

Enthalpy

Organic Chemistry

6. Thermodynamics and Kinetics

Enthalpy

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Problem 7a

Textbook Question

In the following reactions, identify the bonds formed and the bonds broken.
(a)

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Step 1: Analyze the reactants and products. The reactant is benzene (C6H6), which contains a six-membered ring with alternating double bonds (aromatic structure). The product is cyclohexane (C6H12), which is a six-membered ring with only single bonds.

Step 2: Identify the bonds broken. In the reaction, the three π-bonds (double bonds) in the benzene ring are broken. These π-bonds are part of the aromatic system.

Step 3: Identify the bonds formed. For each π-bond broken, two new σ-bonds are formed between the carbon atoms and hydrogen atoms. Since 3 H2 molecules are added, a total of 6 C-H σ-bonds are formed.

Step 4: Recognize the type of reaction. This is a hydrogenation reaction, where hydrogen gas (H2) is added to an unsaturated compound (benzene) to convert it into a saturated compound (cyclohexane).

Step 5: Summarize the bond changes. Three π-bonds in benzene are broken, and six C-H σ-bonds are formed, resulting in the conversion of benzene to cyclohexane.

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Key Concepts

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

Hydrogenation

Hydrogenation is a chemical reaction that involves the addition of hydrogen (H2) to an unsaturated compound, such as alkenes or alkynes, converting them into saturated compounds. In the context of the provided reaction, benzene (an aromatic compound) undergoes hydrogenation to form cyclohexane, a saturated cyclic alkane. This process typically requires a catalyst, such as palladium or platinum, to facilitate the reaction.

Bonds Broken and Formed

In chemical reactions, bonds are both broken and formed. In the hydrogenation of benzene, the π bonds in the benzene ring are broken, allowing for the addition of hydrogen atoms. Simultaneously, new σ bonds are formed between the carbon atoms of the ring and the hydrogen atoms, resulting in the formation of cyclohexane. Understanding which bonds are broken and formed is crucial for predicting the products of a reaction.

Aromaticity

Aromaticity refers to the special stability and unique properties of certain cyclic compounds, such as benzene, due to their delocalized π electrons. Aromatic compounds follow Huckel's rule, which states that a compound is aromatic if it has (4n + 2) π electrons, where n is a non-negative integer. In the reaction shown, the loss of aromaticity occurs as benzene is converted to cyclohexane, which is non-aromatic and saturated.

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Related Practice

Textbook Question

Calculate ∆H° for the following reactions.

(a)

Textbook Question

Using bond-dissociation energies, identify the most stable radical. Justify the difference in stability based on the structure.

(a)

Textbook Question

Using bond-dissociation energies, identify the most stable radical. Justify the difference in stability based on the structure.

(c)

Textbook Question

Using bond-dissociation energies, identify the most stable radical. Justify the difference in stability based on the structure.

(d) I• vs •OH

Textbook Question

In the following reactions, identify the bonds formed and the bonds broken.

(c)

Textbook Question

Teflon-coated frying pans routinely endure temperatures that would cause polyethylene or polypropylene to oxidize and decompose. Decomposition of polyethylene is initiated by free-radical abstraction of a hydrogen atom by O₂. Bond-dissociation energies of C—H bonds are about 400 kJ/mol, and C—F bonds are about 460 kJ/mol. The BDE of the H—OO bond is about 192 kJ/mol, and the F—OO bond is about 63 kJ/mol. Show why Teflon (Figure 7-5) is much more resistant to oxidation than polyethylene is.

Textbook Question

Use the bond-dissociation enthalpies in Table 4-2 (page 167) to calculate the heats of reaction for the two possible first propagation steps in the chlorination of isobutane. Use this information to draw a reaction-energy diagram like Figure 4-8, comparing the activation energies for formation of the two radicals.

Textbook Question

When a small amount of iodine is added to a mixture of chlorine and methane, it prevents chlorination from occurring. Therefore, iodine is a free-radical inhibitor for this reaction. Calculate ΔH° values for the possible reactions of iodine with species present in the chlorination of methane, and use these values to explain why iodine inhibits the reaction. (The I―Cl bond-dissociation enthalpy is 211 kJ/mol or 50 kcal/mol.)

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In the following reactions, identify the bonds formed and the bonds broken.(a)

Key Concepts

Hydrogenation

Bonds Broken and Formed

Aromaticity

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In the following reactions, identify the bonds formed and the bonds broken.
(a)