Acid Strength Trends and Conjugate Base Stability

Problem # 745

The molecule below has five different types of hydrogens (A through E). Rank each in order of decreasing acidity.

(1 = most acidic). Explain your reasoning.

Hydrogens D and E are alpha to two carbonyls and so will be the more acidic than C, which is only alpha to one carbonyl. This is because the enolates that arise from deprotonate at D and E have more resonance forms than the enolate that arrises at C.

Because ketones are more electron withdrawing than esters, D will be more acidic than E. So far we have:

D > E > C > (other)

The carbanions that would arrise from deprotonation of carbons A and B both do not have any resonance forms, so we don't expect either to be acidic. But because A is next to a fluorine atom, which is an electron withdrawing group, A will be more acidic than B. So the overall order is:

(most acidic) D > E > C > A > B (least acidic)

Problem Details

MendelSet practice problem # 745 submitted by Matt on July 27, 2011.

Subject: Organic Chemistry

Subject Topic(s): Acid Strength Trends and Conjugate Base Stability, Carbonyl Alpha-Substitution Reactions (Enols and Enolates, Malonic Ester Syn, Stork)

Question Type: Concept (explain why this is so..)

Keyword(s): acid strength

Problem # 540

Rank the following anions in order of decreasing stability (1 = most stable)

Solution

Each of these ions has a charge of -1, and none of them has any resonance forms. So the two things to consider are electronegativity and size. Electronegativity is relevant as we go from left to right across the periodic table. But here we are going up to down on the periodic table, so size is relavant. Larger ions are more stable than smaller ions (due to a smaller charge:size ratio), so I^- is the most stable of the group.