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 Post subject: Organic Chemistry II help!!!! I am so lost....?
PostPosted: Wed Apr 16, 2014 2:25 pm 
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I had to post the question on craigslist so whoever wants to help can see the question.

http://tulsa.craigslist.org/com/2288910518.html

I didn"t know how else to post this question so you can see what I am looking at


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 Post subject: Organic Chemistry II help!!!! I am so lost....?
PostPosted: Thu Mar 02, 2017 10:53 am 
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The product of the KMnO4 oxidation followed by acidification is a keto acid, 3,3-dimethyl-5-oxohexanoic acid. NaBH4 reduction of it will convert the ketone to an alcohol, the carboxylate in basic NaBH4 will not react. After acidification, it will cyclize to the lactone, 4,4,6-trimethyltetrahydro-2H-pyran-2-one (a cyclic ester).


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 Post subject: Organic Chemistry II help!!!! I am so lost....?
PostPosted: Thu Apr 20, 2017 12:14 pm 
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The sequence is set up to suggest an oxidative olefin cleavage. My only comment on the first reaction is that basic conditions tend to favor the diol. Replacing -OH with MgSO4 will give you the dialdehyde H3C-CO-CH2-C(CH3)2-CH2-CHO. Reduction with NaBH4 would give the diol: 3,3-dimethyl-1,5-hexanediol.



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