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A reaction takes place between $\ce {CH3COCH2COOC2H5}$ with 4-chlorophenyl magnesium bromide, followed by acidic work-up. What will be major product of the reaction? Would the product be the same as the original $\beta$-keto ester due to the active methylene group, at which the acidic doubly-$\alpha$ $\ce {H}$ gets deprotonated?

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    $\begingroup$ It would be helpful if you could provide illustration of these molecules. Also, could you also provide your thoughts on the matter? $\endgroup$ – Tan Yong Boon Nov 26 '18 at 9:11
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    $\begingroup$ You will get the ethyl acetoacetate back. It is sufficiently acidic (pKa ~11) to quench the Grignard and form the Mg salt of acetoacetate. $\endgroup$ – Waylander Nov 26 '18 at 14:17
  • $\begingroup$ @Waylander What counts as sufficiently acidic? For example, why isn't the alpha H of a ketone deprotonated by grignard? $\endgroup$ – Tan Yong Boon Nov 26 '18 at 15:24
  • $\begingroup$ I don't know precisely where to draw the line, somewhere around 18/19 perhaps. EtOH (pKa 15.9) and tBuOH (17.9) quench Grignards. Acetone (pKa 19.2) certainly undergoes Grignard addition readily enough. $\endgroup$ – Waylander Nov 26 '18 at 15:46
  • $\begingroup$ @Tan Yong Boon: Plenty of examples exist where enolization of ketones competes with addition upon treatment with Grignard reagents and organolithium reagents. $\endgroup$ – user55119 Nov 26 '18 at 16:53