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Bromomethylcycloheptane has been prepared in 92% yield by the reaction shown. Write a stepwise mechanism and use curved arrows to show electron flow. The reaction was carried out in water, so use H,O* as the proton donor in your mechanism. Is the rate- determining step unimolecular (SN1) or bimolecular (SN2)? OH + HBr Br H₂O

Bromomethylcycloheptane has been prepared in 92% yield by the reaction shown. Write a stepwise mechanism and use curved arrows to show electron flow. The reaction was carried out in water, so use H,O* as the proton donor in your mechanism. Is the rate- determining step unimolecular (SN1) or bimolecular (SN2)? OH + HBr Br H₂O

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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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Question 5.42
**Question 5.42** Bromomethylcycloheptane has been prepared in 92% yield by the reaction shown. Write a stepwise mechanism and use curved arrows to show electron flow. The reaction was carried out in water, so use H₃O⁺ as the proton donor in your mechanism. Is the rate-determining step unimolecular (Sₙ1) or bimolecular (Sₙ2)? **Reaction:** Cycloheptanol (with an OH group attached) reacts with HBr to form bromomethylcycloheptane (with a Br group replacing the OH group) and water (H₂O). --- **Explanation of Diagram:** The diagram shows a chemical equation: 1. **Reactants:** - A cycloheptane ring with a hydroxyl group (OH) attached. - Hydrobromic acid (HBr). 2. **Products:** - A cycloheptane ring with a bromine atom (Br) attached. - Water (H₂O). The equation depicts the substitution of the OH group in cycloheptanol with a Br atom to form bromomethylcycloheptane, with water as a byproduct. The mechanism involves the use of H₃O⁺ to facilitate the protonation necessary for this substitution reaction.
Transcribed Image Text:**Question 5.42** Bromomethylcycloheptane has been prepared in 92% yield by the reaction shown. Write a stepwise mechanism and use curved arrows to show electron flow. The reaction was carried out in water, so use H₃O⁺ as the proton donor in your mechanism. Is the rate-determining step unimolecular (Sₙ1) or bimolecular (Sₙ2)? **Reaction:** Cycloheptanol (with an OH group attached) reacts with HBr to form bromomethylcycloheptane (with a Br group replacing the OH group) and water (H₂O). --- **Explanation of Diagram:** The diagram shows a chemical equation: 1. **Reactants:** - A cycloheptane ring with a hydroxyl group (OH) attached. - Hydrobromic acid (HBr). 2. **Products:** - A cycloheptane ring with a bromine atom (Br) attached. - Water (H₂O). The equation depicts the substitution of the OH group in cycloheptanol with a Br atom to form bromomethylcycloheptane, with water as a byproduct. The mechanism involves the use of H₃O⁺ to facilitate the protonation necessary for this substitution reaction.
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