What are the products in this SN1 reaction?

Question

I am asked to identify the organic/inorganic products of an S_N1 reaction between 3-iodopentane and ethanol. I have identified correctly that the substrate is the 3-iodopentane, the leaving group is the iodide, and that the nucleophile is the ethanol. My problem is that I do not understand what the products will be, and I am unsure why.

I decided that the products should be 3-pentanol and ethyl iodide (or iodoethane) because iodide is a good leaving group, and the oxygen takes two electrons and leaves ethane to bond with the pentane group where the iodide left off. Now it makes sense that the ethane and iodide react to create ethyl iodide.

Unfortunately, I am wrong in this case, and I just can not put my finger on why. What is the error in my thinking and what is the concept that I am missing?

I realize that my major organic product will be the ether compound which I am having trouble naming, and the other compounds are a protonated ethanol plus the iodide. The most confusing part now is the protonated ethanol and the iodide, why not just create the ethyl iodide?

Answer 1

OK, so let's look at your starting material, 3-iodopentane: $\ce{(CH3CH2)2CH-I}$

So it's S_N1, so the $\ce{I}$ is going to leave spontaneously and give you the carbocation: $\ce{(CH3CH2)2CH+}$

The oxygen is in the ethanol so that will attack the carbocation giving the oxonium ion: $\ce{(CH3CH2)2CH-O+} ~ \ce{H-CH2CH3}$

Which will then be deprotanated by something acting as a base. There's probably some water around, but I suppose saying protonated ethanol and iodide is fine. This gives the final product of: $\ce{(CH3CH2)2CH-O-CH2CH3}$

For ethyl iodide to form, one of two things would need to happen:

1. S_N1 mechanism: $\ce{OH-}$ would have to spontaneously leave forming a primary carbocation. It is SUPER DISFAVORED, so much so that it straight up won't happen.

2. S_N2 mechanism: $\ce{I-}$ would have to act as a nucleophile AND $\ce{OH-}$ would have to leave. Iodide is an OK nucleophile, but OH- is a pretty bad leaving group. Both things considered this pathway is unlikely.

This was a pretty weak question by your (I am assuming) professor. This reaction would be horribly slow, and probably would barely go. Now, if they used sodium ethoxide instead of ethanol you get a nice Williamson ether synthesis and form the same ether and NaI, much more favored thermodynamically.

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I just realized you probably meant why doesn't $\ce{I-}$ perform an S_N2 on the protonated ethanol and have water leave. There are a couple of halfway decent chemical explanations for this:

1. Assuming there is any water around (there is pretty much always water around in the real world (especially with ethanol), unless remarkably heroic measures are taken to avoid it), the protonated ethanol will quickly react with it and form hydronium, ethanol and $\ce{I-}$. (Protonated ethanol has a $\mathrm pK_\mathrm a$ of about -2.4 and hydronium has a $\mathrm pK_\mathrm a$ of about -1.7.)

2. Assuming there is not water around, you are right it will probably react and form small amounts of ethyl iodide, however this will generate water which will stop this from continuing for the above reason.

3. You could also write the mechanism where the iodide deprotonates to form the ether (and avoid this whole fuss). BUT $\ce{HI}$ has a $\mathrm pK_\mathrm a\approx10$ where as oxonium proton $\mathrm pK_\mathrm a$s are usually between -7 and -2, so this would NOT be an accurate way to write the mechanism (in my opinion), but some professors get a little hand wavy here and allow it.

I know those are kind of sketchy workaround reasons, but honestly it won't be formed in any real quantity. There is probably another (probably better) argument from a thermodynamic stand point, but I don't know the bond energies off hand.