SolutionsEnamines, imines and hydrazones
Exercise 1
Write the products of the sequence: cyclohexanone + morpholine (TsOH, −H2O); then ethyl bromoacetate; finally H3O+.
The morpholine enamine forms (1-morpholinocyclohexene); its β carbon attacks the CH2 of the bromoacetate (a very reactive electrophile) to give the iminium salt; hydrolysis releases ethyl 2-oxocyclohexylacetate (2-(ethoxycarbonylmethyl)cyclohexanone) and morpholine.
Exercise 2
The pyrrolidine enamine of 2-methylcyclohexanone has its double bond towards C6, not towards C2. Explain why and deduce which ketone is obtained on alkylation with methyl iodide and hydrolysis.
For the nitrogen lone pair to conjugate with the double bond, the pyrrolidine ring must lie in the plane of the alkene; if the double bond were C1=C2, the C2 methyl would clash with a CH2 of the pyrrolidine (A1,3 allylic strain). The less substituted enamine forms and alkylation occurs at C6: 2,6-dimethylcyclohexanone is obtained.
Exercise 3
What is the advantage of a metalated imine (aza-enolate) over an enamine for alkylating a ketone with 2-bromopropane? And over the lithium enolate of the ketone?
Compared with the enamine: the aza-enolate is a much stronger nucleophile and reacts with secondary halides, with which the enamine hardly reacts or undergoes N-alkylation. Compared with the enolate: it is less basic, so it favours substitution over elimination, and there is no free ketone for self-condensation.