[MarxG]

I know the reaction below is E2 as I looked up the answer, but my question is why can't it be Sn1.
 
 
 
 (1-bromocyclohexyl)benzene     +      CH3CH2O-        ---EtOH----->         E2 rxn.
 
 
 
 
I realize Ethoxide Anion is a decent base, but as I was instructed Ethoxide has slightly higher nucelophlicity than basicity. I understand that the nucleophile is not part of the Sn1 rate law, but why would it not perform the reaction that involves ethoxide to be a nucelophile just because it can "act better as one" (so to speak).
 
Additionally, its tertiary, with a decent leaving group, and in a polar protic solvent---All suitable for Sn1.  I just don't understand why Sn1 is not possible? Any input would be appreciated so I can straighten this is my head.

[Altered State]

It will indeed give both products of elimination and substitution... In most cases you can only predict what will be in most yield, and sometimes not even that without running the reaction in the lab.

[spirochete]

I realize Ethoxide Anion is a decent base, but as I was instructed Ethoxide has slightly higher nucelophlicity than basicity. I understand that the nucleophile is not part of the Sn1 rate law, but why would it not perform the reaction that involves ethoxide to be a nucelophile just because it can "act better as one" (so to speak).
 
Additionally, its tertiary, with a decent leaving group, and in a polar protic solvent---All suitable for Sn1.  I just don't understand why Sn1 is not possible? Any input would be appreciated so I can straighten this is my head.

Strong bases like R-O^- are too reactive to "wait" for alkyl halides to ionize into carbocations. This means that even with tertiary structures you cannot get much Sn1 and/or E1 product.

You are correct that polar protic solvents favor ionization pathways, but the ethoxide base is too strong for that to happen.

I'm pretty sure a good leaving group favors all of the possibly pathways roughly equally, so leaving group ability is not a good way to narrow things down.

[orgopete]

This is how I look at problems like this. First of all, I believe the answer coud be E1/SN1, but probably is not. The kinetics will depend upon concentration of halide and ethoxide. If the reaction were to be diluted sufficiently, the amount of carbocation that may form would increase. In this case, since it is tertiary and benzylic, more solvolysis would occur than other examples. However, adding concentration to the question is too complicated. We should interpret that listing ethoxide as a reagent means it should have a kinetic effect upon the reaction, meaning it must be E2.

[MarxG]

Thank you all for helping me clear this up it was starting to bug me haha.