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Homework Help: Lovastatin in dilute aqueous acid

  1. Mar 6, 2008 #1
    1. The problem statement, all variables and given/known data

    If Lovastatin were heated in dilute aqueous acid, what would the molecular weight of the product be? (The large of the two products. Assume that there is no hydration of the alkene functionalities.)

    2. Relevant equations


    3. The attempt at a solution


    I drew out the structure of Lovastatin and found there to be 2 ester linkages: 1 at the left separating the chain from the ring, and the 2nd ester linkage within the ring of the top right ring. So, based off this, I'm assuming that reacting the Lovastatin with dilute aqueous acid (adding water and acid) would cleave the ester linkages. One smaller product would be a separate carboxylic acid, 2-methylbutanoic acid, I think, and the other would be a larger product. At the point where the carboxylic acid separates, there would be an alcohol group on the larger product. However, I'm unsure if the same cleavage would occur in the ester within the top right ring. If the ring DOES get cleaved, resulting in a carboxylic acid and alcohol, then the molecular weight would be ~338.4 g/mol. If not, then the moleculare weight would be ~320.4 g/mol.

    At this point, I'm basically just stuck on this. Any help is appreciated.

    Thanks in advance.
  2. jcsd
  3. Mar 7, 2008 #2


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    The cyclic ester is of a special type known as a lactone. Under aqueous acid, this ring opens to form an alcohol and the carboxylic acid. It is a hydrolysis (hydrolysis = cleavage through the addition water).
  4. Mar 7, 2008 #3
    Thanks a bunch!

    I'm pretty sure I got it now.
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