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Multiple Answer 2 points Reaction of (R)-2-chloro-4-methylhexane with excess Nal in acetone gives racemic a) an \( \mathrm{S}_{\mathrm{N}} 2 \) reaction has occurred with inversion of configuration b) racemization followed by an \( \mathrm{S}_{\mathrm{N}} 2 \) attack c) an \( \mathrm{S}_{\mathrm{N}} 1 \) reaction has taken over resulting in inversion of configuration d) an \( \mathrm{S}_{\mathrm{N}} 1 \) reaction has occurred due to carbocation formation e) an \( \mathrm{S}_{\mathrm{N}} 1 \) reaction followed by an \( \mathrm{S}_{\mathrm{N}} 2 \) "backside" attack A B C D Previous

Ask by Hobbs Howell. in the United States
Mar 18,2025

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The reaction resulted in a racemic mixture because it proceeded via an \( \mathrm{S}_{\mathrm{N}}1 \) mechanism, leading to inversion of configuration.

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Bonus Knowledge

In the reaction of (R)-2-chloro-4-methylhexane with excess Nal in acetone, the conditions favor \( \mathrm{S}_{\mathrm{N}} 1 \) mechanism due to the formation of a carbocation. This initially results in racemization of the chiral center, leading to a mixture of enantiomers. Options D and E are both feasible because D emphasizes the results of an \( \mathrm{S}_{\mathrm{N}} 1 \) reaction while E includes the possibility of an \( \mathrm{S}_{\mathrm{N}} 2 \) attack after carbocation formation. If you’re ever stuck on these stereochemical reactions, remember that the environment (like using a protic or aprotic solvent) can significantly influence the reaction pathway! Keep an eye out for carbocation stability as well, since that can direct you toward \( \mathrm{S}_{\mathrm{N}} 1 \) mechanisms. For further exploration, consider diving into "Advanced Organic Chemistry" by Francis A. Carey and Richard J. Sundberg. It’s packed with detailed explanations of reaction mechanisms, stereochemistry, and practical examples to enhance your understanding of these fundamental organic principles!

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