**Question:**See Figure 11-5. The reaction shown would not happen as written because:1. ○ It would require the formation of a primary carbocation which is unstable.2. ○ The proton would go on the primary carbon to make the secondary carbocation.3. ○ The secondary carbocation that forms in the first mechanistic step would likely rearrange to form the more stable tertiary carbocation.4. ○ All of the above. ### Figure 11-5The image illustrates the chemical reaction of a cyclopentene compound with HBr to produce a bromocyclopentane compound.#### Reaction PathwaysA) **Stepwise Reaction:**1. \( \text{Hg(OAc)}_2, \, \text{H}_2\text{O}, \, \text{H}^+ \)2. \( \text{NaBH}_4 \)3. \( \text{HBr} \)B) **Hydroboration-Oxidation and Tosylation:**1. \( \text{BH}_3 \)2. \( \text{H}_2\text{O}_2, \, \text{OH}^-, \, \text{H}_2\text{O} \)3. \( \text{TsCl} \)4. \( \text{Br}^- \)C) **Anti-Markovnikov Addition and Tosylation:**1. \( \text{HBr} \)2. \( \text{OH}^-, \, \text{DMF} \)3. \( \text{TsCl} \)4. \( \text{HBr} \)D) **Oxymercuration-Demercuration and Halogenation:**1. \( \text{Hg(OAc)}_2, \, \text{H}_2\text{O}, \, \text{H}^+ \)2. \( \text{Br}_2, \, \text{H}_2\text{SO}_4 \)The figure illustrates different reaction pathways to introduce bromine into a cyclopentane ring through various chemical transformations, including hydroboration, oxymercuration, and halogenation. Each step involves specific reagents that lead to the intermediate or final product formation.

In the given reaction, an alkene is converted to an alkyl bromide.

See Figure 11-5. The reaction shown would not happen as written because It would require the formation of a primary carbocation which is unstable. The proton would go on the primary carbon to make the secondary carbocation. The secondary carbocation that forms in the first mechanistic step would likely rearrange to form the more stable tertiary carbocation All of the above.

See Figure 11-5. The reaction shown would not happen as written because It would require the formation of a primary carbocation which is unstable. The proton would go on the primary carbon to make the secondary carbocation. The secondary carbocation that forms in the first mechanistic step would likely rearrange to form the more stable tertiary carbocation All of the above.

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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