--->This reaction takes place in presence of bases like NaOEt , NaOtBu etc......In this mechanism intermediate anion is stabilized by resonance, hence this is one of the driving force of reaction.
--->If two different types of esters taken then, the one which can capable of forming more stable anion stabilized through resonance will form carbane ion another one will be electron pair acceptor from carbane ion and that product will be more .There will be four different products.
--->Rate of reaction will be dependent of conc. of both ester and base, so we can consider this reaction as SN2 type.(order in the initiation step, total 3 in whole reaction. With respect to ester 2, and with respect to base1.)
--->This SN2 manner can be supported by the fact that, the use of stronger bases, e.g. sodium amide or sodium hydride instead of sodium ethoxide, often increases the yield.
--->Here the carbane ion formed acts as nucliophyl and substitution will be on the other ester's carbonyl group.
--->Simply it is an addition and elimination type reaction, elimination of small components like EtOH taking place hence this is a condensation reaction.
Example reaction(Formetion of aceto acetic ester):
Mechanism
--->If the reacting ester has two α H- atoms the intermediate product stabilized by enolizetion.
Hence reaction is more favorable.
--->Since this reaction final step is reversible removal of Ethyl alcohol increases the forward reaction rate leads to product.
Energy-Profile Diagram:
--->I1 if product does't have resonance stability, I2 is if product has resonence stability.
SOME MORE EXAMPLES
$ CH3CH2COOME +BASE ==>CH3CH(COC2H5)COOME
$PhOCOCH3+NaOME ==>CH3COCH2COOPh
ANYONE HAVING DOUBTS REGARDING ABOVE TOPICS MAIL ; pavan@chem.iitb.ac.in
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