Comparing E2 E1 Sn2 Sn1 Reactions | Summary and Q&A

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September 18, 2010
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Khan Academy
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Comparing E2 E1 Sn2 Sn1 Reactions

TL;DR

A comprehensive analysis of the possible reactions - Sn2 and E2 - that can occur in the given solvent and with the provided reactants.

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Questions & Answers

Q: How does the nature of the solvent affect the reaction possibilities?

The aprotic nature of the dimethylformamide solvent favors Sn2 and E2 reactions. Being aprotic means that it does not have readily available protons, which would hinder the strong nucleophile or base required for Sn2 and E2 reactions.

Q: Is the methoxide ion a strong nucleophile or base?

The methoxide ion is both a strong nucleophile and a strong base. It is a strong nucleophile because it can attack the substrate carbon. It is also a strong base because it can remove a hydrogen from a beta carbon in an E2 reaction.

Q: Why is the carbon with the leaving group considered neutral?

The carbon with the leaving group is a secondary carbon bonded to two other carbons. This makes it neutral in the reaction, allowing for both Sn2 and E2 reactions. A tertiary carbon would favor Sn1 or E1 reactions, while a primary carbon would favor Sn2 exclusively.

Q: Can both Sn2 and E2 reactions occur simultaneously?

Yes, both Sn2 and E2 reactions can occur simultaneously. The environmental factors, such as the aprotic solvent and the strong nucleophile/base, enable both reactions to take place.

Summary & Key Takeaways

  • In the video, the content discusses the reactions that can occur when bromocyclopentane is dissolved in dimethylformamide solvent with methoxide ion.

  • The solvent being aprotic suggests favorability for Sn2 or E2 reactions.

  • The methoxide ion acts as both a strong nucleophile and a strong base, indicating the potential for Sn2 and E2 reactions.

  • The carbon attached to the leaving group is a secondary carbon, making it neutral and allowing for both Sn2 and E2 reactions.

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