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Aldol Condensation

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Hey,

 

Im doing an assignment from my org.chem. class, we are using the OWL cengage website. Im supposed to show the second step in aldol condensation, but i cant get my arrows to go where i want. the second step is when the 'negative' C goes to the 'positive' C and the double bond goes up to the O right? if im explaining it in a weird way, sorry.. but i cant figure it out and its driving me insane. or maybe ive misunderstood the second step... anyone? :)

So far as I understand your question, you've deprotonated the acidic site of one of your carbonyls and your up to the point in the mechanism where the newly formed carbanion (negative C, in your words) is attacking the electropositive carbon of the other carbonyl, so I'll answer you based on that. Yes, the electrons in the pi bond of the C=O will move up onto the oxygen following attack at the carbon position. There aren't really any other viable options for it to go as a first point of call. Do you know what happens next?

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