The following reaction proceeds in two parts. The first part produces an intermediate compound A and…
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Question “The following reaction proceeds in two parts. The first part produces an intermediate compound A and…”
The following reaction proceeds in two parts. The first part
produces an intermediate compound A and byproduct HCOO–. The
product of the second part is a bicyclic ?,?-unsaturated carbonyl
compound.
Answer
This reaction is between an aldehyde (or an \alpha {\rm{,}}\;\beta).
Robinson’s annulation refers to a reaction that involves two reactions: a Michael addition and aldol condensation.
The first step in Michael addition is deprotonation of a base, which results in the formation an enolate. Next, the enolate is formed and attacked the carbonyl moiety by 1,4 conjugate adding. Protonation is the final step in Michael addition.
Finally, the aldol condensation occurs to produce the final product.
The intermediate A is:
This is the first step.
The second step of the process is as follows:
The third step is as follows:
The fourth step of the process is as follows:
The fifth step of the mechanism is as follows:
Ans:
This is the mechanism:
Conclusion
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