The following reaction is a Reformatsky reaction.

Reformatsky reaction of substituted 2-bromoacetate with a ketone. Source: https://en.wikipedia.org/wiki/Reformatsky_reaction

Why is the ester bond not attacked in the last step? Couldn’t $\ce{OR3}$ be replaced by $\ce{OH}$ or $\ce{O(molecule)}$?

  • $\begingroup$ You ask why the ester doesn't react with the H3O to form an actic acid and release an alcohol? $\endgroup$ – Avishai Barnoy Jan 12 '17 at 6:20
  • $\begingroup$ @AvishaiBarnoy yes , I want to know that $\endgroup$ – Icandoahandstand99 Jan 12 '17 at 7:53
  • $\begingroup$ It probably occurs as it's an equilibrium step.maybe the base adding is under cold condition and prevents further reaction. $\endgroup$ – Avishai Barnoy Jan 12 '17 at 11:15

Yes, technically you have a zinc alcoholate which could display the basicity and nucleophilicity of any other alcoholate and you know that you can transesterificate with e.g. sodium methanolate.

Aside from the oxyanion here being rather closely associated to the zinc cation which has a larger positive charge than e.g. sodium and thus exercises a greater attractive force, there is also the fact that the alcohol in question is tertiary and tertiary alcohols are bad nucleophiles. In most of the cases, $\ce{R3}$ will be primary or at most secondary, thus the transesterification reverse reaction will be more rapid than the forward reaction.

If, however, $\ce{R3}$ is tertiary, too, you would be dealing with a huge steric bulk on that carbonyl carbon, attempting to have two tertiary $\ce{OR}$ groups on the tetrahedric intermediate. That is unlikely to happen for steric reasons.

If you are talking about the absolute final step, i.e. the aquaeous workup, then you will likely be using ammonium chloride solution or something else mildly acidic. In mildly acidic or otherwise buffered solutions, there is not a high enough concentration of hydroxide ions to attack the ester carbonyl sufficiently for an observable reaction to occur.


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