Alkyne Synthesis Reaction Problem | Summary and Q&A

TL;DR

The reaction between propine, sodium amide, and ethyl bromide forms 2-pentyne through a series of steps.

Key Insights

• 💁 Propine reacts with sodium amide and ethyl bromide to form 2-pentyne through a deprotonation and SN2 reaction.
• 💁 Sodium amide acts as a strong base, while the alkaline ion formed acts as a nucleophile.
• 👶 The formation of 2-pentyne involves the creation of a new carbon-carbon bond.
• 🧘 The naming of the compound is based on the position of the triple bond, resulting in the name 2-pentyne.
• 🈸 The reaction demonstrates the application of organic chemistry principles in understanding reaction mechanisms.
• ❓ The significance of correct answer selection in organic chemistry problems.
• 🎮 Additional resources and exams reviews are available through the YouTube membership program and video playlists.

Transcript

number nine which of the following is the major product that is formed in the reaction of propine with sodium amide followed by ethyl bromide so let's draw it propine is basically a three carbon alkyne we're going to react it with sodium amide nanh2 followed by ethyl bromide ch3ch2 br sodium has a positive charge so the amide ion is going to have a... Read More

Questions & Answers

Q: What is propine and how does it react with sodium amide and ethyl bromide?

Propine is a three-carbon alkyne. It reacts with sodium amide, which acts as a strong base and deprotonates the alkyne, forming an alkaline ion. This alkaline ion then undergoes an SN2 reaction with ethyl bromide, resulting in the formation of 2-pentyne.

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

Sodium amide acts as a strong base in the reaction. It deprotonates the alkyne CH hydrogen, forming an alkaline ion. This alkaline ion then acts as a nucleophile in the subsequent SN2 reaction.

Q: What is the final product of the reaction?

The final product of the reaction is 2-pentyne. It is formed through the SN2 reaction between the alkaline ion and the primary carbon of ethyl bromide.

Q: Why is the answer choice B, 2-pentyne?

The reaction of propine with sodium amide followed by ethyl bromide leads to the formation of a five-carbon system with a triple bond between carbons two and three. Therefore, the correct answer is 2-pentyne.

Summary & Key Takeaways

• Propine, a three-carbon alkyne, reacts with sodium amide nanh2 followed by ethyl bromide ch3ch2 br.

• Sodium amide acts as a strong base, deprotonates the alkyne CH hydrogen, and forms an alkaline ion.

• The resulting alkaline ion behaves as a nucleophile and undergoes an SN2 reaction with the primary carbon of ethyl bromide, leading to the formation of 2-pentyne.