[laze]

I presume the product is trans-1,2-dimethoxycyclohexane upon opening of the oxiranium ring. My comments were not directed to providing a substitution product, but what one should reasonably expect. You would also get trans-1,2-dimethoxycyclohexane if the reaction were carried out in methanol without NaOMe. It would be easier to predict that as the product as those conditions are more easily understood as SN1 conditions. If the reaction is an SN1 mechanism, then NaOMe is not a kineticly important reagent nor a catalyst. It may well have been present, but in this case misleading to predicting the product.

So your saying that NaOME doesnt do any thing in this reaction? How do you get to the oxiranium ring to the final product?

[orgopete]

I am told to predict the major substitued product for a Williamson Ether Synthesis involving 1-methoxy-2-chloro-cyclohexane and sodium methoxide and to rationalize it.

I said that since the chlorine is attached to a 2ndary carbon it is a 2ndary haloalkane, and that with secondary/tertiary haloalkanes in a williamson ether synthesis it tends to be elimination rather than an SN2 reaction. I said it undergoes dehydrohalogenation and forms the product 1-methoxy-cyclohex-1-ene, which is incorrect.

I was then hinted to think about involvement of anchimeric assistance
in the formation of the major observed product.

I don't want to mislead you in this problem. I agreed with your analysis. My explanations were to further that my agreement.

Second issue, "What is the product and how might it form?" If I must exclude elimination and the hint is that anchimeric assistance occurs, then it must be a solvolysis with the OCH3 group forming a three membered ring (as shown by disco). It is unstable and will react with the solvent. I suggest you can think of it as similar to a bromonium ion. Methoxide can open the ring, but methanol should also be sufficient. That should lead to the product.

I also know this can be a confusing area. If KOtBu is used, it is generally acknowledged this is supposed to be an elimination reaction. If NaOEt/EtOH is used on a secondary halide, the majority of examples will show an elimination reaction. If the same reaction is presented without the NaOEt and especially with water present, then it will give the SN1 (and E1) product. When NaOEt is left out, then we expect that it is not playing a kinetically important role. I was simply arguing as you had, if NaOMe is present, it should cause a reaction. If the question stated the rate of the reaction was independent of the [NaOMe], then you would know it wasn't causing the reaction. You would get the same rate if it were not present. The only reasonable reaction for it to cause would be an elimination, as you originally proposed.