No, generally there are no rearrangements in free-radical chemistry analogous to the rearrangements of carbocations.

There are such things as intramolecular hydrogen atom transfers, which can sometimes result in more stable radicals, but these are generally outside the scope of what we cover here.

I changed it to, “no resonance stabilization”

The reason the finkelstein proceeds to completion is not because of the nucleophilicity of iodine versus other halides, it is that NaI is soluble in acetone whereas NaCl and NaBr are not. This drives the equilibrium toward formation of the alkyl iodide product.

Hope this helps

James

All the confusion you guys are having is because you think that the author said that I- is a strong nucleophile. (For conclusion/summary, read the last para)

No. He never said that I- is a strong nu- ; only that it becomes a ‘better’ nucleophile with protic solvents. In general, the foremost requirements for SN1 reactions are:-
1. A good leaving group attached to the carbon chain.
2. More steric hinderance, so reactivity of carbon follows the order 3° > 2° > 1°.
3. Polar protic solvent, so that the good leaving group has lots of protons flowing around to bond to.
4. A weak nucleophile : This one is more much of a common sense fact. Because we are already choosing a carbon chain containing a good/excellent leaving group then why would we need a strong nucleophile? And in fact, if we choose a strong nucleophile then it will bond to the protons of the protic solvent well before the leaving group could bond to it. So, an SN1 reaction would never occur.

And that’s why I- is a “better” nu- in protic solvents (not stronger) because in protic conditions, a weaker nu- is more favorable for the SN1 reaction.
And to clarify, I- is a weak nu- because HI is a strong acid, so its conjugate base (I-) would be very weak.

I have the same confusion. Of course whenever I look no one seems to be able to answer this problem clearly if anyone bothers to answer it at all besides just saying “I- is a strong nucleophile so SN2.”-gee thanks. However, the answer in your scenario where it’s polar protic would definitely go SN1. Why? Because the transition state of the SN1 reaction is stabilized by the high dielectric constant of the polar protic solvent. Stabilization is particularly pronounced in a protic one due to H bonding that develops as the LG becomes negatively charged. My problem is very similar to yours but slightly different. What do you do if you’re looking at a secondary substrate and an I- is the nucleophile but the solvent is not stated. What do you do in this case? I have a ton of examples where this is the case in my problem sets and they’re extremely confusing. According to logic, and this article, the I- is too stable to act as a good nucleophile if the the solvent is aprotic, and so one would assume the mechanism to be SN1 but I don’t know, this is my question. On the other hand if the solvent is polar protic the I- all of a sudden becomes a strong nucleophile so you’d think SN2, but because the solvent is polar protic the key feature is transition state stabilization making this scenario to be SN1 without a doubt. It almost makes me wonder why we are even bothered being taught the trend where nucleophilicity is inversely related to basicity going down a group on the periodic table. It seems like a waste of time and extremely confusing. Realistically what we should be taught is that nucleophilicity follows basicity, while also knowing that if you’re in a polar protic solvent, despite the reverse trend becoming the case, it doesn’t matter. Because polar protic will go SN1 in a secondary substrate. If anyone can PLEASE answer me that question where it’s secondary and no solvent is stated and your nucleophile is I-, I’d greatly appreciate it.
And sorry, the double bond for the beta part of your question I don’t know.

Hi!

Just wanted to point out that in this case, sterics explains the anti-Markonikov addition in hydroboration, NOT the syn stereochemistry. (Of course, the explanation provided in point #2 of the post is also a sensible explanation for anti-Markonikov addition!)

The concerted addition is what explains the syn stereochemistry.

Racemization does occur actually, since the concerted addition can occur either above or below the plane of C=C.