[xoannnnna]

We just got our test back and these were a few problems that I'm still confused about, can you clear it up for me?

1)Methyl halides react with sodium ethoxide in ethanol by this mechanism. (I think it's SN2 because its a methyl halide, and those undergoSN2 reactions. But isn't ethanol polar protic and will therefore make it an SN1 too?)

2) When cyclohexyl bromide is treated with sodium ethoxide in ethanol, the major product is formed by this mechanism. (I said E2 and I got it right but I don't really understand why. Why does it say "is treated with" compared to "reacts with" as in question 1? And again, isn't ethanol a polar protic solvent and will therefore make it an E1 mechanism? I would have put E1, but it wasn't one of the choices.)

3) Alkyl iodides react faster than alkyl bromides in reactions that proceed by these mechanisms. ( I said SN2, and my teacher wrote a " + ?" next to indicating that there were two choices. Again, this was a total guess. I thought that the leaving group, as in I and Br will have the same effect no matter what mechanism. I will be a better leaving group because it is bigger than bromine. Or am I wrong?)

Thanks, everything just keeps confusing me about these.

[dunno260]

1 goes by an Sn2 mechanism because primary carbocations are fairly unstable.  Sn2 reactions work best in polar aprotic solvents, but they will still occur in other solvent systems.

For 2, reacted with and treated with are somewhat interchangable.  The terminology treated with is slightly better, but no real different.  Again, the solvent can aid, but its not the first thing you should look for.  E2 is going to be favored with strong bases and alkoxides are relatively strong organic bases. 

For 3, Iodine is a better leaving group than Bromine.  Its going to enhance the rate of an E2 reaction as well.  With leaving groups, the ability of the group to accept the electron pair from the bond determines the rate of the reaction.  I- is more stable than Br- and is therefore a better leaving group.  This is why fluorides, alcohols, ethers, and amines don't react in Sn2 mechanisms generally.

[orgstudy]

Ka of ethanol is < 10^-8 so I don’t think it would really affect the rxn

Sn1 and E1 depends on the the carbocation stability but Sn2 and E2 depends on bases as "dunno" said.
You will easily differentiate between two after doing many questions... which happened in my case!!!
roup
If you have any trouble in understaing why I is better leaving group than Br then ...
As the size of Br- is smaller than I-, therefore it can hold its lone pair strongly making it a good target for protons, making him stronger base than I
and as you will know weak base is a good leaving group (conjugate acid base theory)