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

7. Substitution Reactions

SN1 Reaction

Organic Chemistry

7. Substitution Reactions

SN1 Reaction

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3:04 minutes

Problem 26c

Textbook Question

Propose a mechanism involving a hydride shift or an alkyl shift for each solvolysis reaction. Explain how each rearrangement forms a more stable intermediate.
Hint: Most rearrangements convert 2° (or incipient 1°) carbocations to 3° or resonance-stabilized carbocations.
(c)

Verified step by step guidance

Step 1: Begin by identifying the leaving group in the substrate. In this case, the iodine atom (I) attached to the cyclohexene ring is the leaving group. Under solvolysis conditions, the bond between the carbon and iodine breaks, forming a carbocation intermediate.

Step 2: Analyze the stability of the initially formed carbocation. The carbocation formed directly after the iodine leaves is a secondary carbocation. Secondary carbocations are less stable than tertiary carbocations or resonance-stabilized carbocations.

Step 3: Propose a rearrangement mechanism to stabilize the carbocation. A hydride shift or an alkyl shift can occur to convert the secondary carbocation into a more stable tertiary carbocation. In this case, a hydride shift from an adjacent carbon atom can occur, moving a hydrogen atom along with its bonding electrons to the carbocation center.

Step 4: After the hydride shift, the carbocation is now tertiary, which is more stable due to increased hyperconjugation and inductive effects. This rearranged carbocation is the key intermediate for the subsequent reaction.

Step 5: The rearranged carbocation reacts with acetic acid (CH3COOH) to form the final product. The nucleophilic oxygen atom in acetic acid attacks the carbocation, leading to the formation of the ester products shown in the reaction.

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Key Concepts

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

Carbocation Stability

Carbocations are positively charged carbon species that can undergo rearrangements to achieve greater stability. The stability of carbocations increases in the order of primary < secondary < tertiary due to hyperconjugation and inductive effects from adjacent alkyl groups. In solvolysis reactions, rearrangements often convert less stable carbocations into more stable ones, such as transforming a secondary carbocation into a tertiary one.

Hydride and Alkyl Shifts

Hydride shifts involve the migration of a hydrogen atom with its bonding electrons from one carbon to an adjacent positively charged carbon, while alkyl shifts involve the movement of an alkyl group. These shifts are key mechanisms in carbocation rearrangements, allowing the molecule to form a more stable carbocation intermediate. Such shifts are often driven by the need to minimize the energy of the transition state and stabilize the resulting carbocation.

SN1 Mechanism

The SN1 mechanism is a type of nucleophilic substitution reaction characterized by a two-step process: the formation of a carbocation intermediate followed by nucleophilic attack. The first step is the rate-determining step, where the leaving group departs, forming a carbocation. The stability of this intermediate is crucial, as more stable carbocations lead to faster reaction rates and a higher likelihood of rearrangements, such as hydride or alkyl shifts, to achieve stability.

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Related Practice

Textbook Question

Predict the product(s) that would result when molecules (a)–(p) are allowed to react under the following conditions: (i) SOCl₂ ; (ii) PBr₃ ; (iii) SOCl₂ , NEt₃ (iv) 1. TsCl, Et₃N 2. NaCN; (v) 1. TsCl, Et₃N 2. NaOt-Bu (vi) H₂SO₄ (vii) HCl; (viii) HBr; (ix) PCC; (x) H₂CrO₄ , H₂O (xi) HOCl, H₂O (xii) HIO₄ If no reaction occurs, write 'no reaction.'

(f)

Textbook Question

Which compound is more reactive in an S_N1 reaction? In each case, you can assume that both alkyl halides have the same stability.

Textbook Question

a. Assuming that the two compounds shown below have the same stability, which one would you expect to be more reactive in an S_N1 reaction?

b. Draw the products that each would form when the solvent is ethanol.

Textbook Question

A reluctant first-order substrate can be forced to ionize by adding some silver nitrate (one of the few soluble silver salts) to the reaction. Silver ion reacts with the halogen to form a silver halide (a highly exothermic reaction), generating the cation of the alkyl group.

Give mechanisms for the following silver-promoted rearrangements.

(b)

Textbook Question

Choose the member of each pair that will react faster by the SN1 mechanism.

c. n-propyl bromide or allyl bromide

d. 1-bromo-2,2-dimethylpropane or 2-bromopropane

Textbook Question

Predict the compound in each pair that will undergo solvolysis (in aqueous ethanol) more rapidly.

(c)

(d)

Textbook Question

Allylic halides have the structure

a. Show how the first-order ionization of an allylic halide leads to a resonance-stabilized cation.

Textbook Question

Often, compounds can be synthesized by more than one method. Show how this 3° alcohol can be made from the following:

(d) a 3° alkyl bromide

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Propose a mechanism involving a hydride shift or an alkyl shift for each solvolysis reaction. Explain how each rearrangement forms a more stable intermediate.Hint: Most rearrangements convert 2° (or incipient 1°) carbocations to 3° or resonance-stabilized carbocations.(c)

Key Concepts

Carbocation Stability

Hydride and Alkyl Shifts

SN1 Mechanism

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Propose a mechanism involving a hydride shift or an alkyl shift for each solvolysis reaction. Explain how each rearrangement forms a more stable intermediate.
Hint: Most rearrangements convert 2° (or incipient 1°) carbocations to 3° or resonance-stabilized carbocations.
(c)