Question 15

[eo]

I thought C explanation would lead to product 2.... Wouldn't you need 1-butanol to form a carbocation intermediate in order for it to have a hydride shift? (similar to experiment 1)

[matthew]

picked A, too. I was told that Sn1 has rearrangement for sure. But since these are primary halides they will not be Sn1.

So for Sn2, I thought rearrangements could not happen. Is there an exception? I am at a tough spot b/c there is no strong nucleophile, being Cl(-1) are weak bases, strong conjugate acids.

I think this problem would be so much better if it was tertiary alcohol than everything would make sense.

Been reading and found that Sn1 and 2 work with nucleophiles but the Sn2 is more dependent on the strength of the nucleophile. Even if this would have been secondary alcohol answer choice A would make sense.

I am not sure what's going on.

[matthew]

Forgot, even though Cl -1 is a weak base just read that halogens make nice nulceophiles b/c their anions are large. (cations being small, like cats)


So just need to know if Sn2 in this problem can do arrangement.

[admin]

Answer C actually explained the mechanism for Product 2 as opposed to Product 1. Product 1 would occur as follows:

Protonation of a hydroxyl group, followed by the dissociation of the protonated hydroxyl group (water is a great leaving group), hydride shift (to convert a primary carbocation into a more stable secondary carbocation) then attack by a chloride ion.

[matthew]

So just to make sure, for MCAT purposes, Sn2 can do rearrangements? (hydride, methyl shift?)

[admin]

Yes, but, pardon the pun, the AAMC will never by shifty in the way that you would have to apply this concept. In other words, there simply won't be any other reasonable explanation for the occurrence or the molecule or intermediate which they show. Because it won't be ambiguous, as long as you are comfortable with the possibility, then you have nothing to worry about.

[admin]

bump