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Our topic was about ketones and aldehydes (carbonyl group) including hemiketals and hemiacetals and he asked us to do a homework. I cannot find an answer to this one.Please explain me the reaction and the product formed.

PS: The figure at bottom is my best effort to do this question.

  • $\begingroup$ Have you tried to write a mechanism for this reaction? $\endgroup$
    – jerepierre
    Commented Mar 24, 2016 at 15:53
  • $\begingroup$ How did you end up at the structure on the bottom? $\endgroup$ Commented Mar 24, 2016 at 16:15
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    $\begingroup$ Would it help to know that $\ce{R'-C(OR)3}$ is an orthoester? $\endgroup$ Commented Mar 24, 2016 at 16:49
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    $\begingroup$ There are 5 oxygens at the bottom and three at the top. The only way you get more oxygens in these types of reactions is if you lose a double bond. Therefore, something is amiss at the bottom. $\endgroup$
    – Lighthart
    Commented Mar 25, 2016 at 4:20

2 Answers 2


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But.....6 member lactones are super stable so I wouldn't be surprised if only 1 ring opened/ some recyclized

  • $\begingroup$ I'm not sure I would say "super stable" necessarily. Delta-valerolactone always (for some aggravating reason) comes as liquid lactone over polymerized material. See impurities listed here. $\endgroup$ Commented Mar 25, 2016 at 7:56
  • $\begingroup$ Let me rephrase since 'super stable' is not a technical term (though probably assessable and appropriate for those with beginning organic chemistry homework problems). Molecules like this with 2 nucleophilic atoms which if they attacked the carbonyl would make a 6 membered ring would likely react in acidic media to some extent. Please post the solvent, acid used, concentration, and temperature of your homework such that we can guess at the product distribution. $\endgroup$
    – StevieD
    Commented Mar 25, 2016 at 21:52
  • $\begingroup$ Also of note; 4 membered transition states of proton transfer are unlikely in reality, but often used in these sorts of mechanisms. The proton is likely transferred first to a water molecule, a solvent molecule, or some other base in solution before it ends up on the indicated position. Chemists do this (and take it with a grain of NaCl) since it doesn't overly complicate the real mechanism in question, and could be any or all bases present. $\endgroup$
    – StevieD
    Commented Mar 25, 2016 at 22:06

You mentioned ketals, acetals, hemiketals and hemiacetals.
Let's have a look at some structures (1 to 5). acetals_and_ketals

What do they have in common?

Apparently, 1 to 5 each have one carbon atom with two single bonds to oxygen atoms. In 1, this central carbon atom bears a hydrogen atom (omitted in the drawing). Do you recognize that 1 is a "masked" aldehyde? 2 and 3 are apparently different. The central carbon atom, the one with the two oxygen atoms, bears single bonds to two other carbon atoms. 2 is made from cyclopentanone, 3 is made from cyclohexanone. When the protecting groups are split off in 1 to 3, they are gone and no longer a part of the molecule with the rebuild $\ce{C=O}$ group. But it mustn't always be like that. Take a closer look at 4 and 5 and figure out what will happen to them under acidic conditions. Will there be $\ce{C=O}$ and $\ce{OH}$-groups in the products?

Now have a look at the structures 6 and 7.


Are these different from 1-5?

Apparently, they are! Each has one carbon atom with three single bonds to oxygen atoms! You will conclude that these are not the acetals that we worked on above. Under aqueous acidic conditions, the central carbon will not be converted to an aldehyde or ketone, but to a […]. On the other hand, 4 and 7 do have something in common when it comes to the question whether the "protecting groups will still be found in the product ;-)

  • $\begingroup$ actually what reaction is this?(i.e the one mentioned in the question). $\endgroup$ Commented Mar 26, 2016 at 15:39

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