[Brook]

My answer differs from that in the book so I just wanted to make sure what I am doing is plausible.
Radical halogenation, elimination , hydroboration oxidation, forming an alkoxide ion by treatment with sodium hydride, and a williamson ether synthesis.  Thank you for your help.

[orgopete]

I am guessing the difference is in step 2. Ethanol, in the absence of base, will not guarantee elimination. Remember, HBr is the by-product. The butene can revert to an ethyl ether. NaOEt would be much better.

I like NaH, but your book may have used something different. That would be okay.

[Brook]

Thank you for reply. I will add the base to favor elimination. My book actually did this for the synthesis. I am not quite sure I understand it though. Is it just like an oxymercuration demercuration? If someone could please explain a bit , it will be greatly appreciated.

[spirochete]

Yes it is essentially the same as hydroboration/oxidation, but using a different nucleophile than water. Methanol and water have about the same nucleophilicity, so it makes sense that they could both attack a mercuinium ion.

The use of the weaker base trifluoroacetate rather than acetate probably enhances the electrophilicity of the Mercury(II) ion and the mercuinium ion.

[orgopete]

@Brook, your textbook has an error. Your hydroboration route is correct, the oxymercuration gives the wrong alcohol.

[Brook]

Thank you orgopete, but I performed the synthesis as the book outlined, and it does not yield an alcohol, but in fact an ether as in the book due t the fact that water is not used in the mercuration as a nucleophile, but CH3OH. When the CH3OH is deprotonated by the trifluoroacetate anion, it leaves an OCH3 attached instead of a hydroxyl group.

Thank you also to spirochete.

[orgopete]

Honestly, I didn't look at the solvent. I just knew this would give the wrong isomer.

[Brook]

Thanks again orgopete.