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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2:30 minutes

Problem 23e,f

Textbook Question

Choose the member of each pair that will react faster by the SN1 mechanism.
e. 2-iodo-2-methylbutane or tert-butyl chloride
f. 2-bromo-2-methylbutane or ethyl iodide

Verified step by step guidance

Step 1: Understand the SN1 mechanism. The SN1 reaction proceeds via a two-step process: (1) formation of a carbocation intermediate after the leaving group departs, and (2) nucleophilic attack on the carbocation. The rate of the reaction depends on the stability of the carbocation formed and the nature of the leaving group.

Step 2: Compare the leaving groups in each pair. The leaving group ability is determined by how easily it can depart and stabilize itself. Iodide (I⁻) is a better leaving group than bromide (Br⁻) or chloride (Cl⁻) because it is larger and more polarizable.

Step 3: Evaluate the carbocation stability for each compound. Carbocation stability increases in the order: primary < secondary < tertiary. Tertiary carbocations are the most stable due to hyperconjugation and inductive effects from surrounding alkyl groups.

Step 4: Analyze pair (e): 2-iodo-2-methylbutane vs. tert-butyl chloride. Both compounds form tertiary carbocations upon losing their leaving groups, but iodide is a better leaving group than chloride, making 2-iodo-2-methylbutane react faster via the SN1 mechanism.

Step 5: Analyze pair (f): 2-bromo-2-methylbutane vs. ethyl iodide. 2-bromo-2-methylbutane forms a tertiary carbocation, while ethyl iodide forms a primary carbocation. Since tertiary carbocations are more stable, 2-bromo-2-methylbutane will react faster via the SN1 mechanism despite bromide being a slightly weaker leaving group than iodide.

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Key Concepts

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

SN1 Mechanism

The SN1 mechanism is a type of nucleophilic substitution reaction that involves two main steps: the formation of a carbocation intermediate and the subsequent attack by a nucleophile. The rate of the reaction depends primarily on the stability of the carbocation formed, which is influenced by factors such as the degree of substitution and the presence of electron-donating groups.

Carbocation Stability

Carbocation stability is a crucial factor in determining the rate of SN1 reactions. Tertiary carbocations are more stable than secondary or primary ones due to hyperconjugation and inductive effects from surrounding alkyl groups. Therefore, substrates that can form more stable carbocations will react faster via the SN1 mechanism.

Leaving Groups

The quality of the leaving group significantly affects the rate of SN1 reactions. Good leaving groups, such as iodide and bromide, can stabilize the transition state and facilitate the formation of the carbocation. In general, the better the leaving group, the faster the reaction will proceed, as it can depart more easily during the first step of the SN1 mechanism.

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Related Practice

Textbook Question

Give the substitution products expected from solvolysis of each compound by heating in ethanol.

(c)

(d)

Textbook Question

Propose a mechanism for the reaction of benzyl bromide with ethanol to give benzyl ethyl ether.

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.

(d)

Textbook Question

Predict the products of the following S_N2 reactions.

(c)

(d)

Textbook Question

When tert-butyl bromide is heated with an equal amount of ethanol in an inert solvent, one of the products is ethyl tert-butyl ether.

b. What happens to the rate if the concentration of tert-butyl bromide is tripled and the concentration of ethanol is doubled?

c. What happens to the rate if the temperature is raised?

Textbook Question

tert-Butyl chloride undergoes solvolysis in both acetic acid and formic acid. Solvolysis occurs 5000 times faster in one of these two solvents than in the other. In which solvent is solvolysis faster? Explain your answer. (Hint: Formic acid is more polar than acetic acid.)

Textbook Question

Rank the following alkyl halides from most reactive to least reactive in an S_N1 reaction:

2-bromo-2-methylpentane, 2-chloro-2-methylpentane, 3-chloropentane, and 2-iodo-2-methylpentane.

Textbook Question

Which reaction would be faster, the one with DMSO as the solvent or the one with ethanol (EtOH)?

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Choose the member of each pair that will react faster by the SN1 mechanism.e. 2-iodo-2-methylbutane or tert-butyl chloridef. 2-bromo-2-methylbutane or ethyl iodide

Key Concepts

SN1 Mechanism

Carbocation Stability

Leaving Groups

Watch next

Choose the member of each pair that will react faster by the SN1 mechanism.
e. 2-iodo-2-methylbutane or tert-butyl chloride
f. 2-bromo-2-methylbutane or ethyl iodide