Questions: Acidity of Organic Compounds and pKa Trends

Question 1

A molecule contains three different C–H bonds: one on a terminal alkyne (sp carbon), one alpha to a ketone (sp³ carbon adjacent to C=O), and one on a simple alkyl chain (sp³ carbon). Rank these from most to least acidic.

• A. Alkyl C–H > alpha C–H > alkyne C–H
• B. Alkyne C–H > alpha C–H > alkyl C–H
• C. Alpha C–H > alkyne C–H > alkyl C–H
• D. All three have similar acidity because they are all C–H bonds

Answer: (C)

The alpha C–H (pKa ~20) is most acidic because its conjugate base (an enolate) is resonance-stabilized — the negative charge delocalizes onto the electronegative carbonyl oxygen. The alkyne C–H (pKa ~25) comes next because sp hybridization gives more s-character, holding the resulting anion's electrons closer to the nucleus for stabilization. The alkyl C–H (pKa ~50) is least acidic because the resulting carbanion receives no resonance or hybridization stabilization. Option D is the classic misconception — all C–H bonds are not equivalent; they span roughly 30 pKa units in acidity.

Question 2

Carboxylic acids (pKa ~5) are far more acidic than alcohols (pKa ~16), even though both compounds lose an O–H proton. What best explains this large difference?

• A. Carboxylic acids have two oxygen atoms, so the molecule is simply more polar
• B. The carboxylate anion is resonance-stabilized, spreading negative charge over two oxygens, while the alkoxide anion localizes charge on one oxygen
• C. The carbonyl oxygen is more electronegative than a hydroxyl oxygen
• D. Carboxylic acids are stronger acids because they form hydrogen bonds more easily

Answer: (B)

Both compounds lose an O–H proton to the same oxygen, so electronegativity differences between the oxygens themselves are minimal. The decisive factor is what happens to the resulting anion. The carboxylate anion has two equivalent resonance structures, delocalizing the negative charge symmetrically over two oxygens — effectively halving the charge density. The alkoxide anion concentrates its full negative charge on a single oxygen. More stabilized conjugate base = stronger acid. This illustrates the core principle: acidity is about conjugate base stability, not just the acidity of the parent compound.

Question 3

An sp-hybridized C–H bond (e.g., in a terminal alkyne) is more acidic than an sp³ C–H bond because the sp orbital has greater s-character, which stabilizes the resulting carbanion.

• T. True
• F. False

Answer: True

sp orbitals are 50% s-character (vs. 33% for sp² and 25% for sp³). Electrons in orbitals with more s-character are held closer to the nucleus and experience stronger nuclear attraction, making them lower in energy. When the C–H bond breaks heterolytically, the carbanion's electrons reside in this high-s-character orbital — which stabilizes the negative charge more effectively than a pure sp³ orbital would. This hybridization effect raises the acidity of terminal alkynes to pKa ~25, compared to ~50 for alkane C–H bonds.

Question 4

Adding electron-withdrawing groups (like fluorine atoms) near the acidic site of a compound decreases its acidity because they make the molecule more electronegative and harder to deprotonate.

• T. True
• F. False

Answer: False

Electron-withdrawing groups increase acidity by stabilizing the conjugate base through inductive effects. They pull electron density away from the negatively charged site, reducing its charge density and making the anion more stable. Trifluoroacetic acid (pKa ~0) is thousands of times more acidic than acetic acid (pKa ~4.8) precisely because three fluorine atoms inductively stabilize the carboxylate anion. The confusion arises from conflating the charge on the acid (neutral) with the charge on the conjugate base (negative) — electron-withdrawal destabilizes the acid only marginally while strongly stabilizing the conjugate base.

Question 5

Why is the alpha-hydrogen of a ketone (pKa ~20) roughly 10³⁰ times more acidic than a regular sp³ C–H bond on an alkyl chain (pKa ~50)?

Model answer: Removing the alpha-hydrogen generates an enolate anion, where the negative charge is delocalized by resonance between the alpha carbon and the electronegative carbonyl oxygen. This resonance stabilization dramatically lowers the energy of the conjugate base compared to a simple carbanion, which cannot delocalize its charge. The more stable the conjugate base, the lower the pKa (stronger acid). A plain sp³ C–H on an alkyl chain produces an unstabilized carbanion with no resonance or hybridization advantage to lower the energy.

The key is that acidity reflects conjugate base stability. The enolate from a ketone alpha-H gets electron density spread over two atoms (C and O), reducing charge concentration. An alkyl carbanion concentrates all its negative charge on one carbon with no way to delocalize it. This difference in conjugate base stability — roughly 30 pKa units — corresponds to a factor of 10³⁰ in acid equilibrium constants, one of the largest effects in organic chemistry.