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I followed a procedure in my organic chemistry lab, but I don't understand what product the procedure created. Can someone tell me what chemicals reacts with which to create the product?

First $\ce{NaOH}$, water, and ethanol were mixed together and cooled. Then acetone was added, followed by benzaldehyde (the reaction mixture then turned yellow/orange). The reaction was stirred for a while. The product was washed with water and ethanol before being recrystallized from ethyl acetate. After crystal formation the product was reduced then cooled.

If I understand correctly the acetone and benzaldehyde reacted together and the ethyl acetate was only used as a solvent?

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  • $\begingroup$ NaOH is intended to act as base. The presence of ethanol, beside water, helps to accomodate both inorganic matter (NaOH) and organic matter (acetone, benzaldehyde) in solution. Where, by the way, acetone may be deprotonated? And, in its then different form could it react with ease on the subsquently added benzaldehyde? $\endgroup$ – Buttonwood Apr 8 '17 at 16:40
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Straightforward aldol between benzaldehyde and acetone. Enolate of acetone adds to benzaldehyde (which cannot enolise) then eliminates water to give the enone. Product recrystallised from EtOAc.

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  • $\begingroup$ @JQQ You just need to figure out whether it's single aldol or double aldol. $\endgroup$ – Zhe Apr 8 '17 at 17:24
  • $\begingroup$ I think I understand it now, but correct me if I'm wrong. I can take a mp of the product and see whether it's closer to the mp of the single or double aldol to determine what the major product is. If I used a 2:1 ratio of benzaldehyde to acetone, then the double aldol would be the expected product. The only thing left to figure out would be which conformation it has. If sufficient heat was used then I should get the thermo-favored product (E,E). $\endgroup$ – JQQ Apr 8 '17 at 18:13
  • $\begingroup$ True. The proportions of reagents used may give you a clue. If xs acetone is used then that would favour the single addition product, if xs benzaldehyde then you get the double $\endgroup$ – Waylander Apr 8 '17 at 18:15

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