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Princeton Review CBT 4150 error?

Discussion in 'MCAT Discussions' started by RPedigo, May 5, 2007.

  1. RPedigo

    Physician Faculty Moderator Emeritus 10+ Year Member

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    Hi everyone,

    I'm not sure if I'm missing something, or if the Princeton Review CBT diagnostic 4150 is distinctly in error.

    Problem 18 on the Biological Sciences section asks:

    "Based on the information in the passage, which of the following is the most likely product of the reaction between propanal and 2-propanone, in the presence of base?"


    My reasoning:

    2-propanone is acetone, the simplest methyl ketone. There is a methyl group on each side, and either of these methyl groups, if deprotonated by a base, will give an equivalent product. 2x the chance of reaction = good. Also, there is good resonance stabilization of the anion after deprotonation, and the anion is primary even without resonance stabilization. Thus, the anion generated by the 2-propanone can attack the aldehyde (propanal) and make the aldol condensation product. In addition, the aldehyde is less sterically hindered, and thus more open to nucleophilic attack by the 2-propanone carbanion.




    Princeton's solution says:

    This is the product of an aldol reaction, resulting from the enolate of propanal adding to the ketone (propanone).







    Why would propanal get deprotonated, when there is only one site, and it's secondary? 2-propanone has two equivalent primary sites that can be deprotonated, and the carbanion can then have nucleophilic attack on the wide-open aldehyde that's completely sterically unhindered to attack. Even if we made the anion on the propanal, it'd have to attack the 2-propanone, which isn't as sterically accessible.


    Am I just dumb? Thanks! =)
     
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  3. Surf Rx

    7+ Year Member

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    The alpha hydrogen on aldehydes have a pka of 17 and ketones have a pka of 19. Therefore, propanal is a better "acid" and would be deprotonated.

    Besides, I think that the correct answer choice to that question is wrong because it has one too many carbons.
     
  4. RPedigo

    Physician Faculty Moderator Emeritus 10+ Year Member

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    Yeah, I originally crossed that one out because it had too many carbons, and so I figured my answer was even more correct.

    Although the alpha hydrogens on the aldehyde are more acidic, there are only two of them, and the enolate ion has to attack a secondary carbon... the alpha hydrogens on the 2-propanone are less acidic, but there are 6 of them, and once formed, are very ready for nucleophilic attack.

    Ah well, either way the diagnostic is in error because their answer has 7 carbons. Thanks for the response.
     
  5. CmpE

    2+ Year Member

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    yeah, i thought something was fishy.
    i've been staring at that problem/solution for the past 20 minutes trying to see what i did wrong ;)

    nothing more frustrating than solutions that aren't right ;)
     

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