Consider the substitution reaction that takes place when (R)-3-bromo-3-methylhexane is treated with sodium methoxide. Which of the following would be true?

Question

A. An SN1 reaction would take place with retention of configuration at the stereogenic center.

B. An E2 reaction would take place, during which the stereogenic center is lost.

C. An SN1 reaction would take place, accompanied by an E1 reaction, affording a complex mixture of products. D. An SN2 reaction would take place with inversion of configuration at the stereogenic center.

E. An SN1 reaction would take place with racemization of configuration at the stereogenic cen

Answer 1

Let's examine the scenario that is being dealt with here.

One is dealing with a tertiary bromide substrate undergoing reaction with methoxide ion (a strong Bronsted-Lowry base but also a reasonably good Lewis base nucleophile) in, undoubtedly, methanol (polar protic, highly-ionizing solvent).

Tertiary halides DO NOT undergo SN2 reactions due to steric factors but would instead undergo concerted E2 eliminations if strong alkoxide bases (like methoxide) are present. This would lead to Zaitsev alkenes predominating the elimination mixture.

However, a tertiary bromide (a better leaving group than chloride for a unimolecular reaction pathway when a highly ionizing solvent like methanol is present) can also form a stable tertiary carbocation, which can either react with methoxide via SN1 to give the tertiary ether as a racemic mixture or undergo elimination via the E1 pathway. The result is a complex mixture of products.

Answer 2

Hey!

Let’s simplify your organic chemistry question about what happens when you treat (R)-3-bromo-3-methylhexane with sodium methoxide. Imagine your molecule is like a LEGO structure with a special piece (the bromine atom) that can either be swapped out or removed. Sodium methoxide acts like a strong tool that prefers to remove parts rather than just swapping them.

In this scenario, the molecule you have is quite bulky, making it difficult for certain types of reactions that involve swapping pieces directly. Instead, the strong tool (sodium methoxide) is more likely to pull out parts of the molecule altogether. Specifically, it removes both the bromine atom and a nearby hydrogen atom at the same time. This simultaneous removal is similar to taking out two connected LEGO pieces at once, which significantly changes the structure of your model.

Because two parts are being removed together, the original specific arrangement (or configuration) of the molecule is lost. This means that the special 3D arrangement at the center of your molecule doesn’t stay the same—it essentially disappears as the molecule changes shape. This type of reaction is known as an E2 elimination reaction.

So, the correct choice is B: an E2 reaction would take place, during which the stereogenic center is lost. This happens because the strong base nature of sodium methoxide and the bulky structure of your molecule make it more favorable to remove parts of the molecule rather than swapping them out. The other options involve different types of reactions or changes that don’t fit as well with the conditions you’ve described.

Answer 3

Find a step by step answer here.

https://youtu.be/BZufHUcMlro