What are the products in this SN1 reaction?
OK, so let's look at your starting material, 3-iodopentane: $\ce{(CH3CH2)2CH-I}$

So it's SN1, 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:


SN1 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.
SN2 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.



I just realized you probably meant why doesn't $\ce{I-}$ perform an SN2 on the protonated ethanol and have water leave. There are a couple of halfway decent chemical explanations for this:


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.)
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.
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.