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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![**Which of these would be the best electrophile for an S<sub>N</sub>1 reaction?**
A) CH<sub>3</sub>Br
B) (CH<sub>3</sub>)<sub>3</sub>CBr
C) (CH<sub>3</sub>)<sub>3</sub>CO<sup>−</sup>
D) CH<sub>3</sub>O<sup>−</sup>
*Explanation:*
In an S<sub>N</sub>1 reaction, the best electrophile is one that can form a stable carbocation. Typically, tertiary carbocations are more stable than secondary or primary ones due to hyperconjugation and inductive effects.
- **Option A (CH<sub>3</sub>Br)**: Methyl bromide forms an unstable methyl cation.
- **Option B ((CH<sub>3</sub>)<sub>3</sub>CBr)**: Tertiary butyl bromide forms a stable tertiary carbocation.
- **Option C ((CH<sub>3</sub>)<sub>3</sub>CO<sup>−</sup>)**: Presents as a negative ion, not suitable as an electrophile.
- **Option D (CH<sub>3</sub>O<sup>−</sup>)**: Also a negative ion, unsuitable as an electrophile.
Thus, the best choice for an S<sub>N</sub>1 reaction is **Option B** - (CH<sub>3</sub>)<sub>3</sub>CBr.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff69c34af-4b69-4606-a1b5-5e3396e36f88%2Fa38a6677-e1be-4443-b593-eb671dee1ef3%2Fxkne47s_processed.png&w=3840&q=75)
Transcribed Image Text:**Which of these would be the best electrophile for an S<sub>N</sub>1 reaction?**
A) CH<sub>3</sub>Br
B) (CH<sub>3</sub>)<sub>3</sub>CBr
C) (CH<sub>3</sub>)<sub>3</sub>CO<sup>−</sup>
D) CH<sub>3</sub>O<sup>−</sup>
*Explanation:*
In an S<sub>N</sub>1 reaction, the best electrophile is one that can form a stable carbocation. Typically, tertiary carbocations are more stable than secondary or primary ones due to hyperconjugation and inductive effects.
- **Option A (CH<sub>3</sub>Br)**: Methyl bromide forms an unstable methyl cation.
- **Option B ((CH<sub>3</sub>)<sub>3</sub>CBr)**: Tertiary butyl bromide forms a stable tertiary carbocation.
- **Option C ((CH<sub>3</sub>)<sub>3</sub>CO<sup>−</sup>)**: Presents as a negative ion, not suitable as an electrophile.
- **Option D (CH<sub>3</sub>O<sup>−</sup>)**: Also a negative ion, unsuitable as an electrophile.
Thus, the best choice for an S<sub>N</sub>1 reaction is **Option B** - (CH<sub>3</sub>)<sub>3</sub>CBr.
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