pKa values of estrone, 17 beta-estradiol and 2-methoxyestrone

An understanding of the solvating effect helps rationalize the relation between solvation and acidity of the compound. In addition, this also explains the relative stability of conjugate bases for compounds with different pKa values. This lesson details, in-depth, the principle of solvating effects. The strength of an acid and the stability of its corresponding conjugate base are determined using pKa values. This observed relationship is a consequence of solvation, which is the interaction...

SN2 substitutions and E2 eliminations of alkyl halides proceed via a concerted pathway. While the nucleophile attacks the alpha carbon in SN2 reactions, it functions as a strong base and abstracts a beta hydrogen in the E2 mechanism. The rate-limiting transition state in E2 elimination reactions is characterized by partially broken carbon–hydrogen and carbon–halogen bonds and a partially formed pi bond between the alpha and beta carbons. The beta hydrogen and halide are eliminated...

Elimination reactions of alkyl halides can yield one or more alkenes depending on the specific regiochemical and stereochemical considerations. While the regiochemistry of the reaction governs the location of the double bond in the product, the stereochemical requirements often influence the geometry.
When a substrate with two different β hydrogens undergoes an E2 elimination, the presence of a strong base can yield two regioisomeric alkenes. The more-substituted alkene is the major product and...

One of the critical aspects of the E1 reaction mechanism, as also observed in E2, is the regiochemistry, with multiple regioisomers obtained as products. In the example discussed, the presence of water as a weak base favors elimination over substitution to generate two alkenes. Given that alkenes’ stability increases with the number of alkyl groups across the double bond, typically, E1 reactions lead to the Zaitsev product, for this is more substituted and stable than the Hofmann product.

Aldehydes and ketones form enols, although only about 1% of the enol is present at the equilibrium for simple monocarbonyl compounds. The enol form is undetectable for acetaldehyde, present as only 1.5 × 10−4 % of acetone, and present as only 1.2% of cyclohexanone. Two kinds of regioisomeric enols are possible for unsymmetrical ketones, and their net composition is 1% at equilibrium. This instability is due to the lower bond energy of C=C than the C=O group. The additional instability of enols...