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Predict the products of the controlled oxidation of 2-methylpentan-2-ol.

2-methylpentan-2-ol

From what I know, since 2-methylpentan-2-ol is a tertiary alcohol, there should be no reaction. However, my textbook states that the product includes 2-methylpentanone.

Edit: In response to the comments, my textbook neither specifies the isomer nor mentions any given conditions. I hate Nelson

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    $\begingroup$ What conditions are given for the oxidation? $\endgroup$ – NotEvans. Jul 18 '16 at 22:10
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    $\begingroup$ Nothing will happen under "controlled" conditions with any of the common oxidizing reagents. Under forcing conditions elimination of water to form olefins can occur along with epoxide formation and general molecular fragmentation. BTW, which isomer of 2-methylpentanone does your textbook claim is formed? $\endgroup$ – ron Jul 18 '16 at 22:26
  • $\begingroup$ The "reactant includes 2-methylpentanone"? Or the product? And what molecule is meant by "2-methylpentanone"? 4-methyl-2-pentanone? Or 3-methyl-2-pentanone? Or 2-methyl-3-pentanone? It's hard to see how any of those could be produced in high yield through the oxidation of this alcohol. $\endgroup$ – Curt F. Jul 18 '16 at 22:31
  • $\begingroup$ I suppose under dehydrating conditions the preferred product would be 2-methyl-2-pentene? Could that alkene be oxidized to 2-methyl-3-pentanone? $\endgroup$ – Curt F. Jul 18 '16 at 22:34
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    $\begingroup$ Given the level that book is aimed at, this seems far more like a typo than a genuine question. As @Ron pointed out, the most likely thing to happen is elimination.... anything after that is getting into significantly non 12th chemistry $\endgroup$ – NotEvans. Jul 18 '16 at 22:41
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Tertiary alcohols are resistant to oxidation in neutral or alkaline solution, but are readily oxidized by acidified oxidizing agents.

In your case two products will be acetone and propionic acid.

http://chemistry.tutorcircle.com/organic-chemistry/tertiary-alcohol.html

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    $\begingroup$ This simply isn't true. Even the Jones reagent (18M sulfuric acid and stochiometric chromic anhydride ) doesn't touch tertiary alcohols. There are indeed methods but reasily oxidised is simply nonsense $\endgroup$ – NotEvans. Jul 20 '16 at 7:07
  • $\begingroup$ @NotNicolaou If you are unfamiliar with the topic study it before commenting/downvoting pubs.acs.org/doi/abs/10.1021/ja01190a040 You cannot oxidize it using a school chemistry textbook, but you are allowed to oxidize them once you go to a university. $\endgroup$ – sixtytrees Jul 20 '16 at 13:30
  • $\begingroup$ Look at the limited substrate scope of that paper. This is not a general trivial method. I've also found that textbooks don't make very good oxidants.... $\endgroup$ – NotEvans. Jul 20 '16 at 13:45
  • $\begingroup$ @NotNicolaou Look at the link in my post. Scroll to the very bottom. chemistry.tutorcircle.com/organic-chemistry/… This is the reaction we are talking about. $\endgroup$ – sixtytrees Jul 20 '16 at 13:45
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    $\begingroup$ I see a lot of hostility in this comment section and I would like to ask everyone to be nice to each other. $\endgroup$ – Martin - マーチン Jul 21 '16 at 3:01

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