E2 reactions | Substitution and elimination reactions | Organic chemistry | Khan Academy

TL;DR

Sodium methoxide reacts with 2-chlorobutane in a methanol solution, resulting in the elimination of the chloro group and formation of but-2-ene.

Transcript

Let's try to come up with the reaction of when we have this molecule right here reacting with sodium methoxide in a methanol solution, or with methanol as the solvent. Just so we get a little practice with naming, let's see, this is one, two, three, four carbons. So it has but- as a prefix and no double bond or triple bonds, so it's butane. And we ... Read More

Key Insights

• 💪 Sodium methoxide in a methanol solution acts as a strong base and participates in an E2 reaction with 2-chlorobutane.
• 💁 The E2 reaction involves the elimination of a leaving group and the formation of a double bond.
• 👥 The chloro group acts as the leaving group in this reaction.
• ☠️ The E2 reaction follows a one-step mechanism, with both reactants participating in the rate-determining step.
• ⚾ The E2 reaction is favored when the substrate and the base are both suitable for elimination.
• 🏑 The E2 reaction is part of the broader field of organic chemistry and has similarities to the Sn2 reaction.

Questions & Answers

Q: What is the role of sodium methoxide in the reaction?

Sodium methoxide acts as a strong base, abstracting a proton from 2-chlorobutane and facilitating the elimination of the chloro group.

Q: How does the E2 reaction differ from the Sn2 reaction?

The E2 reaction involves the elimination of a leaving group, while the Sn2 reaction involves the substitution of a leaving group by a nucleophile.

Q: Why is the E2 reaction the most likely reaction to occur in this case?

The E2 reaction occurs because sodium methoxide is a strong base and 2-chlorobutane is a suitable substrate for elimination.

Q: What is the significance of the rate-determining step in the E2 reaction?

The rate-determining step in the E2 reaction involves the simultaneous participation of both reactants, leading to the formation of but-2-ene and the elimination of the chloro group.

Summary & Key Takeaways

• Sodium methoxide, a strong base, reacts with 2-chlorobutane in a methanol solution.

• The reaction involves the elimination of the chloro group and the formation of a double bond, resulting in the formation of but-2-ene.

• This reaction follows the E2 mechanism, where both reactants participate in the rate-determining step.