Part 8: MECHANISMS

Note that no other reagents are needed in order to complete any of these sequences, you should only be using what is there.

Part A
This reaction is the hydrolysis of an acetal.


In this scheme, the base, B:, could be R-O-R, R-OH, H2O or the conjugate base of the acid catalyst

ii This is a Friedel-Crafts alkylation of an aromatic via an electrophilic aromatic substitution involving a carbocation rearrangement.

Friedel Crafts alkylation


iii This is an acid catalysed transesterification.... the -OH of the molecule displaces the methanol to give the cyclic ester.

In this scheme, the base, B:, could be C=O, R-OH, or the conjugate base of the acid catalyst


Part B
i  
LiN(iPr)2 is LDA, a strong base (pKa = 35), which will form an enolate of the ester.  This enolate will then react with the ketone (similar to the aldol reaction). The product of this step contains an alkoxide and there is a good leaving group (the chlorine) on the adjacent carbon so an intramolecular Williamson type reaction - just SN2 - happens to form a cyclic ether, the epoxide (note : this last step is also like using a halohydrin to form an epoxide)

Darzen's condensation

ii 
The starting material has a ketone and an epoxide.  The product has neither of these, but on careful examination has a cyclic acetal.  The key steps are therefore (a) opening of the epoxide to give the 1,2-diol then (b) reaction of the diol with the ketone to give the cyclic acetal.  Numbering helps you to see how the structures relate to each other.

brevicomin

epoxide to diol and diol to cyclic acetal in brevicomin
In this scheme, the base, B:, could be C=O, R-O-R, R-OH, H2O or the conjugate base of the acid catalyst




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