Organic chemistry question

[doitfortheppl2016]

Advertisement - Members don't see this ad

I have a question regarding a problem in the organic odyssey chapter 11 # 38. I attached a picture of the problem to this thread. I thought that the answer would be B because after protonating the OH you would have a strong nucleophile, Br-. So, I thought that it would undergo SN2. However, the answer is A indicating that it undergoes SN1. Could someone please explain why it goes Sn1 and not SN2? Is it because the solvent is overall protic?

Thank you

(copyrighted image removed by moderator staff at request of copyright holder)

 

[clutchfans]

First Br- isn't a very good nucleophile. If I remember correctly, conjucated bases of strong acids are weak nucleophiles.
H2O is a very good leaving group. It falls off making a secondary carbocation.
There is a 1,2 hydride shift from C2 to C3 resulting in a tertiary carboation on C2. Then the Br attackes the tertiary carbocation resulting in 2 bromo iso pentane

Edit: Check that Br- is a weak base because it's the conjugate base of a strong acid. Weak bases make good nucleophiles. Br is a good nucleophile.

 

[C10H15N]

Here are some of my notes from my last semester of ochem. It really helped me keep it organized and clear in my head.

Because it is a secondary alcohol it would be an Sn1 reaction. And thus it would occur in multiple steps and a carbocation would form. And then there would be a hydride shift resulting in the most stable conformation which is with the tertiary carbon. The ranking of stability is 3>2>1. If it was a primary Alcohol rather than a secondary it would result in an Sn2 reaction and there would be no hydride shift.

 

[doitfortheppl2016]

Here are some of my notes from my last semester of ochem. It really helped me keep it organized and clear in my head.

Because it is a secondary alcohol it would be an Sn1 reaction. And thus it would occur in multiple steps and a carbocation would form. And then there would be a hydride shift resulting in the most stable conformation which is with the tertiary carbon. The ranking of stability is 3>2>1. If it was a primary Alcohol rather than a secondary it would result in an Sn2 reaction and there would be no hydride shift.

This is really helpful. Thanks for the response!

 

[O Gaúcho]

Thanks!

 

[DAT Destroyer]

I have a question regarding a problem in the organic odyssey chapter 11 # 38. I attached a picture of the problem to this thread. I thought that the answer would be B because after protonating the OH you would have a strong nucleophile, Br-. So, I thought that it would undergo SN2. However, the answer is A indicating that it undergoes SN1. Could someone please explain why it goes Sn1 and not SN2? Is it because the solvent is overall protic?

Thank you

(copyrighted image removed by moderator staff at request of copyright holder)

Not a chance. You have a branched secondary alcohol. Protonation would form a secondary carbocation that can easily rearrange into a highly stable tertiary carbocation. This rearrangement is so rapid, it would surely beat any direct attack. As a general rule, secondary and tertiary alcohol react by Sn1 or E1 processes .

Hope this helps.

Dr. Romano