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Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: burntbread07 on February 25, 2008, 12:10:57 AM

Title: Products of Bromo octanol in the presence of aqueous NaOH
Post by: burntbread07 on February 25, 2008, 12:10:57 AM

"(3R,6R)-6-Bromo-3-octanol (A) is expected to form optically inactive B (C8H16O) upon exposure to aqueous NaOH. A stereoisomer of A, namely, C also forms B under the same conditions. Two other stereoisomers of A, namely D and E as a 50/50 mixture, form optically inactive F, a diastereoisomer of B. What are the structures of A-F? The structures D and E are not distinguishable. Explain and illustrate with mechanisms."

I hope I am not breaking any rules here, but apart from coming up with the structure for A, I can't get much further.  I still do not understand what happens to the OH- from the NaOH. 

Title: Re: Products of Bromo octanol in the presence of aqueous NaOH
Post by: Rico on February 25, 2008, 07:10:35 AM

Hey burntbread07

The stereoisomeres react via a ring closure reaction forming tetrahydrofuran-derivatives and the hydroxide ions deprotonates the oxonium-ion intermediate. I would recommend you to make a model, since this makes the stereochemical issues much easier to vizualise.

Try figuring it out, and if its not possible for you please let me know and I will be happy to post my solution to the problem.

Rico