EK1001 Ochem #476 Question

[shffl]

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The question was which alkane will react faster than (2R)-iodobutane with methanol to form 2-methoxybutane.

a. 1-iodobutane
b. 2-bromobutane
c. 2-iodopropane
d. 2-iodo-2-methylbutane

The way I thought of it was since methanol is weak nucleophile this reaction will most likely proceed with SN1 reaction. SN1 reactions favor highly substituted molecules and in this case D would be the correct answer (which it is). However I didn't pick D because I don't see how that compound can form 2-methoxybutane via reacting with methanol. I thought it would form 2-methoxy-2-methylbutane instead. Where did I go wrong here?

I was actually wrestling between B and C. Let's say choice D is something else that is wrong and it's down between B and C, which one would be the right answer? I ended up choosing C because I- is a better leaving group since it's more stable overall than Br-.

 

[thestrokes14]

The question was which alkane will react faster than (2R)-iodobutane with methanol to form 2-methoxybutane.

a. 1-iodobutane
b. 2-bromobutane
c. 2-iodopropane
d. 2-iodo-2-methylbutane

The way I thought of it was since methanol is weak nucleophile this reaction will most likely proceed with SN1 reaction. SN1 reactions favor highly substituted molecules and in this case D would be the correct answer (which it is). However I didn't pick D because I don't see how that compound can form 2-methoxybutane via reacting with methanol. I thought it would form 2-methoxy-2-methylbutane instead. Where did I go wrong here?

I was actually wrestling between B and C. Let's say choice D is something else that is wrong and it's down between B and C, which one would be the right answer? I ended up choosing C because I- is a better leaving group since it's more stable overall than Br-.

This seems odd. I agree that Methanol is a weak nucleophile, but if we are required to form the product that you mentioned, both C and D can't be used. Keep in mind, it says "propane" in C. But you are right about iodide being a better leaving group than bromide. Iodide is larger and thus can support a negative charge better than bromide can. I am curious as to what the explanation is for the solution. I have the solution, but not with me at the moment. I will check it out.

 

[shffl]

This seems odd. I agree that Methanol is a weak nucleophile, but if we are required to form the product that you mentioned, both C and D can't be used. Keep in mind, it says "propane" in C. But you are right about iodide being a better leaving group than bromide. Iodide is larger and thus can support a negative charge better than bromide can. I am curious as to what the explanation is for the solution. I have the solution, but not with me at the moment. I will check it out.

The solution states that since its SN1 reaction, the tertiary haloalkane will react faster due to a more stable cation. Which is correct but I'm just not sure how 2-iodo-2-methylbutane can react with methanol to form 2-methoxybutane.

 

[thestrokes14]

The solution states that since its SN1 reaction, the tertiary haloalkane will react faster due to a more stable cation. Which is correct but I'm just not sure how 2-iodo-2-methylbutane can react with methanol to form 2-methoxybutane.

The question is likely unclear. You don't need to form the product mentioned in the question. You are only comparing the reaction kinetics between methanol and the various compounds. That is the only thing that makes sense to me. You understand that tertiary compounds undergo SN1 reactions faster than secondary compounds, so I wouldn't get to hung up on this question.

 

[shffl]

The question is likely unclear. You don't need to form the product mentioned in the question. You are only comparing the reaction kinetics between methanol and the various compounds. That is the only thing that makes sense to me. You understand that tertiary compounds undergo SN1 reactions faster than secondary compounds, so I wouldn't get to hung up on this question.

I thought the same too. It's just I wasn't clear if there truly is a reaction that can lead to that product. Thanks 👍