tiny529 Posted March 17, 2011 Share Posted March 17, 2011 Hello again! I'm back with another Orgo 2 question. My prof has set up some online homework questions that require a registration code that came in the shrink wrap of the textbook. Not knowing that I'd need to keep the stupid shrink wrap, I tossed it and am forced to do a whole semester's worth of homework in 4 days as I'm on a "free trial". That being said, I might have a bunch of questions between now and the weekend. I've done 4 chapter's worth of the homework so far, but I'm stuck on the last problem of the 5th chapter. Haven't even looked at the other chapters yet. *mutter* So here's my question... First off, I'm not 100% sure of the name of the starting compound. My best guess is 3,3-dimethyl-2,4-dioxane-ethanoic acid. Anyway, that needs to be treated with H2SO4 in water. I'm supposed to draw the resulting product. I have no idea how to go about doing this. My 1st guess was just the starting product, figuring that the acid would hydrate the carboxylic acid, but that it would lose water and end up back where it started. That was, obviously, wrong, but it did tell me that I need to do hydrolysis of a ketal. When I work that out on paper, I get 2 products, but this is only asking for "the expected product". Can anyone lend me some insight??? Thanks in advance... =) Here's a picture of the problem... Link to comment Share on other sites More sharing options...
John Cuthber Posted March 17, 2011 Share Posted March 17, 2011 "When I work that out on paper, I get 2 products," Me too. But I bet they are not interested in acetone. 1 Link to comment Share on other sites More sharing options...
Horza2002 Posted March 17, 2011 Share Posted March 17, 2011 (edited) In you molecule, the R-O-C(Me)2-O-R is the ketal. Remember than oxygen has two lone pairs of electrons (i.e. they are slightly basic in nature). Once you know that, along with having water in thre reaction, you could be able to give a mechanism. Post your ideas on here if you need some other guidence. As John Cuthber has said, they wont be interested in the acetone byproduct (HINT!). This type of functional group is often used as a base stable protecting group for ___. Seeing as ___ is the anwser to your problem, I won't tell you just yet! Edited March 17, 2011 by Horza2002 Link to comment Share on other sites More sharing options...
tiny529 Posted March 17, 2011 Author Share Posted March 17, 2011 Yeah, I got acetone and a diol. If I give the diol as my answer, it tells me that I'm on the right track, but how can it react further. The annoying thing is that I asked my prof today and he said that my answer was right. He said nothing about reacting it further. I tried reacting it further to lose water and give a cyclic ether, but it didn't like that answer either. Okay... Just tried it yet again and got a lactone. It seemed to like that answer. Link to comment Share on other sites More sharing options...
kristina221 Posted March 17, 2011 Share Posted March 17, 2011 I believe that the first step would be deprotection of the keto-group and that the products would be acetone and a diol. The next step would be formation of a cyclic ester (lactone). There are two possible products and I would go with the one that has less strained ring. Link to comment Share on other sites More sharing options...
tiny529 Posted March 17, 2011 Author Share Posted March 17, 2011 I believe that the first step would be deprotection of the keto-group and that the products would be acetone and a diol. The next step would be formation of a cyclic ester (lactone). There are two possible products and I would go with the one that has less strained ring. Yep, that's what I finally ended up with. =) Now I've got a new question, which I'll put in a new post... =) Link to comment Share on other sites More sharing options...
Horza2002 Posted March 17, 2011 Share Posted March 17, 2011 Good good. Yes a ketal like that is a common protecting group for 1,3-diols as you have in your product. Under the reaction conditions, an internal esterification will occur because you will get a 6-membered lactone. These are very stable compounds indeed. Link to comment Share on other sites More sharing options...
Dan_Ny Posted March 18, 2011 Share Posted March 18, 2011 (edited) Good good. Yes a ketal like that is a common protecting group for 1,3-diols as you have in your product. Under the reaction conditions, an internal esterification will occur because you will get a 6-membered lactone. These are very stable compounds indeed. Hm, that's interesting. What conditions does it take to open up a 6-membered lactone? (Or, to which conditions are 6-membered lactones in a substrate stable?) Edited March 18, 2011 by Dan_Ny Link to comment Share on other sites More sharing options...
Horza2002 Posted March 18, 2011 Share Posted March 18, 2011 (edited) 6-membered lactones are VERY stable. I have tried to hydrolyse the lactame by boiling them in 5M NaOH over the weekend....it basically just laughed at me!! It did nothing to it. Ester are fairly stable anyway and couple that with being in a stable 6-membered ring, it all makes a suprisingly stable compound. If you add groups around the ring especially on the alpha position, then it makes it even more stable! You sterically protect the carbonyl group from the hydroxide ion. With that said, lactones can be reduced with LiAlH4... to the corrosponding 1,6-diol (you end up reducing the carboxyl section as well) Edited March 18, 2011 by Horza2002 Link to comment Share on other sites More sharing options...
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