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Aldol Condensation rxn- NaOH + acetone, then add benzaldehyde = ?

  1. Nov 10, 2013 #1
    1. "Students were instructed to mix the benzaldehyde and acetone starting materials in their conical vials before adding the ethanolic sodium hydroxide solution.
    Why was this essential to the success of the reaction?
    What would have been the most likely product formed if the sodium hydroxide solution were added to the vial first, followed by acetone and then waiting a few minutes to add the benzaldehyde?
    Provide a balanced equation showing this reaction:"


    2. From the reaction of NaOH with acetone, I thought the product would be
    4-methyl-3-penten-3-one + 2H2O + NaOH.

    Then I'm confused about what happens once benzaldehyde is added...

    I'm wondering if the NaOH deprotonates the alpha carbons of 4-methyl-3-penten-3-one again, which would cause the benzaldehydes to add...

    If another deprotonation occurs, I get this as the final product:
    bigmoleculepic_zps2e7fad34.png

    I thought the steric strain from the 2 hydroxyls might cause a dehydration reaction, which results in the ether bond...???

    Not sure... This seems like too weird of a product to be right... Please help?!
    Thanks! :)
     
  2. jcsd
  3. Jan 28, 2017 #2

    TeethWhitener

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    Gold Member

    You've got the right idea: acetone will undergo an aldol self-condensation in the presence of strong base. But your product is wrong. The alpha carbon of an acetone enolate species will attack the carbonyl carbon of a neutral acetone to give 4-hydroxy-4-methyl-2-pentanone (praying to the IUPAC gods that I got that name right). It's basically an acetone with an isopropyl alcohol hanging off the alpha carbon.

    This one is a little trickier and, honestly, you'll probably get an ugly mixture of stuff. Again, you have the kernel of the right idea: NaOH could give you the enolate of the 2-pentanone product above. Then you'd expect an aldol condensation with the benzaldehyde. But there's a wrinkle, because the hydroxyl group can lose a proton as well. I'm not entirely sure where the equilibrium will lie in this situation, so I can't say whether the enolate or the alkoxide will dominate. If it's the enolate, the condensation will proceed as normal, but if it's the alkoxide, the reaction will either be slow or won't go at all.
     
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