[SodiumBicarbonate]

Hello,

I don't understand the solution of McMurry 8e, problem 11.11.

The problem is to rank the following substances in order of their expected S_N1 reactivity:

CH₃CH₂Br
H₂C=CHCH(Br)CH₃
H₂C=CHBr
CH₃CH(Br)CH₃

And the answer is:

H₂C=CHCH(Br)CH₃ > CH₃CH(Br)CH₃ > CH₃CH₂Br > H₂C=CHBr

But why?

It's a matter of carbocation stability, of course, and it obviously relates to the attached image, which appears in the book right before this problem. So I do understand why CH₃CH(Br)CH₃ > CH₃CH₂Br (secondary vs. primary) and why H₂C=CHCH(Br)CH₃ > CH₃CH₂Br (allylic vs. primary), but I don't understand why H₂C=CHCH(Br)CH₃ > CH₃CH(Br)CH₃ or why H₂C=CHBr is the least stable.

I admit I'm more concerned about having missed something in the book, than simply not knowing organic chemistry well enough... I mean, there will always be things I won't know, but if I should have known how to solve this problem and I don't, then that's a problem.

Please help.

Thanks!

[Rutherford]

H₂C=CHCH(Br)CH₃ dissociation of bromine, will produce a carbocation that is both secondary and in alylic position in regards to the double bond while the carbocation of CH₃CH(Br)CH₃ is only secondary (no resonance with a double bond). Also, there is no resonance stabilization in vinyl carbocations. You should review the resonance topic.

[SodiumBicarbonate]

Thanks, Raderford, I get what you're saying about H₂C=CHCH(Br)CH₃.
But I still don't understand why H₂C=CHBr is the least stable. I mean, right, there's no resonance, but there's no resonance stabilizing the carbocation of CH₃CH₂Br too, so it can't be the whole explanation.

[Rutherford]

You can look at it this way: the formed carbocation will be sp hybridized, there is a 50% s orbital character of the sp hybrid orbital where the positive charge will be accommodated. As s orbital is closer to the nucleus, and positive repels positive the positive charge on a sp C atom will cause higher energy to that carbocation, than the positive charge accommodated on the sp² C atom of the ethyl carbocation.

[SodiumBicarbonate]

What do you mean by "sp hybrid orbital where the positive charge will be accommodated"? The only positively charged thing is the nucleus, isn't it?

If I get what's going on, both sp orbitals will be fully occupied, as well as one of the the two p orbitals (the one forming the π bond). The remaining p orbital will remain empty. Why is it less stable than three fully occupied sp² orbitals?

[Rutherford]

I tried to make analogy with carboanion stabilization regarding hybridization, but I see it doesn't make too much sense.

What can you conclude from the comparison of the hyperconjugation stabilization in the two carbocations?

[SodiumBicarbonate]

Oh, I think I get it. With three σ bonds we have more electrons interacting with the empty p orbital, thus stabilizing it?

[Rutherford]

Right.

[SodiumBicarbonate]

Great, I really appreciate your help, thanks!