[la_tiburtina]

Okay…here's another question that I'm getting wrong.

The question is, which of the following reacts with HBr at the fastest rate? (see attached image)

I _think_ this question is basically asking about the relative stabilities of the compounds and the preferred equatorial position of the -OH

Here's my rationale (I could be completely wrong):

The idea is to find the compound that is most likely to have its hydroxyl group in the equatorial position.

Compound 1. although there's a penalty for flipping the -OH into the axial position, the penalty is less than that of compound # 2. so the answer can't be # 1.

Compound 2. I believe there's a greater penalty for having the methyl group in the axial position, so the -OH in this compound would be favored in the axial position, and we're looking for a compound where the -OH prefers to be in the equatorial. So, it can't be # 2.

Compound 3. both the methyl and -OH are equatorial. ring flipping would put both in the axial position which would be severely disfavored. I think this is the correct answer

Compound 4. putting the -OH in the equatorial position forces the methyl to "go axial" which is not favored.  Compound 4 is not the answer

Compound 5. ring flipping the hydroxyl here has no net effect on the 2 methyls--one will always be equatorial and one axial…so I'm guessing that while the -OH will favor the equatorial position, it will more frequently sample the axial position than the hydroxyl in compound 3. I.e., the hydroxyl in compound 5 can become axial more easily than the hydroxyl in compound 3. therefore compound 5 isn't it.

However, the book says compound 5 reacts fastest, so my logic isn't quite right.

Thoughts on where I'm going wrong?

TKS

[orgopete]

I am sorry, but I don't know that I agree with your textbook (or that I don't know the principle your textbook is trying to illustrate with this question). If this were to illustrate an SN2 reaction, then 3 or 4 should be the fastest for the conformational reasons given. I believe it actually should be from the axial OH, 4. 5 (and 1) gives neither preference.

When I first saw the question, I imagined an SN1 reaction in which the slow step is formation of a carbocation. The tertiary alcohol 2 should give the most stable carbocation, hence the fastest of the possible rate limiting steps. Therefore I thought 2 would be fastest.

[la_tiburtina]

I'm sorry...my bad...the correct answer is listed as # 2, so you're correct.

I'm still not sure I understand though...The _axial_ -OH is more reactive than equatorial?  Is there an explanation for this or is it just empirically determined?

[orgopete]

Let's not mix two different reactions. 2 can be either conformer, axial or equatorial. The faster rate of reaction is due to it gives a tertiary carbocation. That enables an SN1 reaction to take place.

Re: axial v equatorial
Since I was trying to guess what reaction your textbook was referring to, I was also reasoning the rates of an SN2 reaction. I have not tried to identify factors preferring one over the other. This is from A Handbook of Organic Chemistry Mechanisms,

An SN2 reaction of a ditosylate with one equivalent of cyanide gives a mono-nitrile. Note, one equivalent signifies the mole to mole ratio. Axial substituents react faster than equatorial ones is a general trend. While a di-nitrile could be produced with excess cyanide, in this example, the axial tosylate reacts faster and leads to the major product.©Curved Arrow Press

[SVXX]

I concur with what orgopete says... let's go over the question again.
What it asks : Which compound reacts fastest with HBr.
Reagents and Functional Groups involved : HBr and -OH, and some methyl groups in a few places.
Reaction Occurring(or most likely to occur) : S_N1, as HBr is a good acid(proton donator) and -OH is an acid-labile group.
For S_N1 reactions, it is well known that the rate-determining step is the formation of the carbocation. If you can speed up that step, you can speed up the entire reaction. A carbocation forms easily if stabilizing factors are involved.
Here that would be the formation of a tertiary carbocation, which happens in compound #2. Hence that is your answer. Axial and equitorial considerations would rather affect S_N2, I think.