The SN1 Reaction: Substitution Following Solvolysis

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The SN1 Reaction: Substitution Following Solvolysis

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Practice Resources:

Guided Worksheet(s) + Solutions:

- Substitution/Elimination Practice | Solution

Worksheet Solutions Walkthrough(s):

- Substitution/Elimination Practice

Series Study Guide:

- Substitution/Elimination Study Guide

Description

https://joechem.io/videos/49 for video on jOeCHEM and attached worksheet + solution (below video on jOeCHEM aka the link). In this video, we look at the SN1 reaction, go through the mechanism, and talk about ideal/necessary conditions for SN1. We also talk about how racemic mixtures can be produced via SN1.

Tags: SN1 solvolysis carbocation polar protic racemic

Comments:

Anu Jayachandran • 6 years ago

Is there no preference in product for the last sample problem that you do? Does steric hindrance with the methyl group on the adjacent carbon not affect product preference?

REPLY
- Joe Del Nano • 6 years ago
  
  Hey Anu! While it might provide a tiny (tiny!) bit of hindrance, that wouldn't be a scenario on a test/quiz that you would be expected to recognize that and say that the AcOH nucleophile would add as a dash more than a wedge. If the methyl group was a t-butyl group, then I would agree with you: A wedged t-butyl group in place of that wedged methyl group would add steric hindrance and force a higher percentage of the dashed product (and not the typical racemic mixture that is characteristic of SN1 rxns). But that measly little methyl group doesn't bother anyone in this problem :) does that make sense?
  
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  Anu Jayachandran • 6 years ago
  
  Yup! I hadn't been sure if the methyl was something significant enough to differentiate between a major and minor product in this scenario. Is it a matter of relativity? Like if the nucleophile had been closer in size to the methyl, would it be more affected? Or would it be that there's actually so much more space that any steric hindrance is even less than with the methyl and AcOH nucleophile? Thanks for all of your help by the way!
  
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  Joe Del Nano • 6 years ago
  
  Of course! I'm glad you're using the site (and hope its helpful!). So really it comes down to being branched = being bulky = causing hindrance. If the methyl group had been an isopropyl group, there would probably be some hindrance, but while Ochem teachers are thought of as being devious (which they obviously aren't!) someone will make it painfully obvious to you if steric hindrance is a factor in the problem and include the classic t-butyl group.
  
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