SolutionsRegioselectivity and enantioselectivity in ketone alkylation
Exercise 1
Give the main product of 2-methylcyclohexanone with: (a) LDA, THF, −78 °C, then allyl bromide; (b) Me3SiCl, Et3N, refluxing DMF, then MeLi and finally allyl bromide.
(a) Kinetic control: the enolate on the less substituted side (C6) forms, giving 2-allyl-6-methylcyclohexanone. (b) The thermodynamic silyl enol ether forms (C1=C2 double bond); MeLi regenerates that enolate and allylation gives 2-allyl-2-methylcyclohexanone.
Exercise 2
Why is the thermodynamic enolate the more substituted one? Which conditions allow the two enolates to equilibrate?
As with alkenes, a more substituted C=C double bond is more stable (hyperconjugation). Equilibrium is reached if the enolates can exchange protons: with a slight excess of ketone, with a base whose conjugate acid is acidic enough (an alkoxide in its alcohol), at room temperature or above and with counterions that form less covalent bonds (Na+, K+).
Exercise 3
In the Evans alkylation, which three roles does the chiral auxiliary play? How is it removed at the end, and why is the configuration of the new stereocentre not lost?
(1) It makes the enolate form with a defined Z geometry; (2) it chelates the metal between the two carbonyl oxygens and fixes a rigid conformation; (3) its benzyl group blocks one face of the enolate, so the electrophile enters from the other. It is removed by hydrolysis with LiOH/H2O2 (hydroperoxide selectively attacks the exocyclic carbonyl): the reaction does not go through the enolate, so the α stereocentre does not epimerize, and the oxazolidinone is recovered.