Question

14. Reaction of the following diester with NaOEt followed by mild Ht makes product G, not product H. Explain why there is a preference for forming product Gover H. a. NaOEt OEt not "OEib mild H* 15. Give the curved arrow mechanism for the decarboxylation and keto-enol tautomerism steps in this reaction (from 3 to K). مه-حل-ا مل a. NaOEt b. H, H30, heat

Answer 1

O No (14) Ans: - Intramolecular Cyclisation of diester impresence of hoort le sodium ethoxide which acts as base followed by mildt Makes Product ce and not t which is! Mentioned in the question, because sodium ethoxide ie Naget accept the proton from sterically less crowded position. Eto -sterically more crowded Sterically a proton (more steric less crowded Hinderance) prolon due to presence a Methyl group. In the above substrate Nadet (base) prefer to abstract the proten from less substituted active methylene proten! ie from sterically free casson Formation of product a over product H is due to abstraction of proten by Hadel from less hindered carport is faster than

sterically hindered position. 2 Eto a Arto T Carbanian Intramolecular I cyclization - ETOH an i hot i Nadet Easy to abstruct the propen for base Nao Et from that active methylene carbon Where steric hindrance is minimum ar null and hence formation of Product "a" is taster than product H. product ( an 0) I finally the main reason or cause behind formation of product G is only sterically free active methylene group where it's very convinient or easy for base ie Naoet to abstruct the prodiect and then followed by nucleophilic intou molecular cyclization to form product"a" and Not H ! de