SolutionsEnolization and enolate ions
Exercise 1
Rank the indicated hydrogens in order of increasing acidity: (a) CH3CO2Et; (b) CH3COCH2COCH3 (the CH2); (c) CH3COCH3; (d) CH3NO2. Justify the order.
(a) < (c) < (d) < (b), with approximate pKa values of 25, 19–20, 10 and 9. The ester is the least acidic because the alkoxy oxygen donates electrons to the carbonyl, which then stabilizes the charge less well; the ketone does better; the nitro group better still; and in acetylacetone the charge is shared by two carbonyls.
Exercise 2
Explain why LDA in THF at −78 °C, rather than sodium ethoxide in ethanol, is used to alkylate cyclohexanone with methyl iodide.
Ethoxide (pKa of ethanol ≈ 16) deprotonates only a small fraction of the ketone (pKa ≈ 19): the enolate coexists with plenty of ketone and gives the aldol condensation, and proton exchange leads to mixtures of mono- and polyalkylated products. LDA (pKa of diisopropylamine ≈ 36) converts all the ketone into the enolate irreversibly and does not attack the carbonyl because of its steric hindrance; moreover, ethoxide would also react with CH3I.
Exercise 3
The potassium enolate of cyclohexanone in HMPA reacts with dimethyl sulfate to give mainly 1-methoxycyclohexene; the lithium enolate in THF reacts with methyl iodide to give 2-methylcyclohexanone. Explain the difference.
The enolate is an ambident nucleophile. K+ in HMPA is strongly solvated and leaves the enolate “naked”, with the charge concentrated on the oxygen, and dimethyl sulfate is a hard electrophile: O-alkylation (enol ether) predominates. Li+ binds tightly to the oxygen and blocks it, and methyl iodide is a soft electrophile: C-alkylation predominates.