Difficulty: Intermediate | Prerequisites: Part 1 of these notes (acid-base fundamentals, pKa, Keq), plus familiarity with resonance structures, orbital hybridisation and electronegativity from earlier Chem 210 lectures.
The pKa table is not something you memorise line by line. It is a reference. What you do need is the ability to look at a molecule that is not in the table and estimate where it falls, or to explain why one compound is more acidic than another. That reasoning comes down to five structural factors, all of which affect how well the conjugate base can stabilise the negative charge left behind after deprotonation. This set of notes covers those five factors, the logic behind each, and how they appear in exam problems (including the CN bond-length and conjugate-base-stability questions from the Chem 210 practice set).
Five structural features determine how stable a conjugate base is, and therefore how strong its parent acid is: electronegativity, charge, atomic size, hybridisation (s-character) and resonance (delocalisation). A more stable conjugate base means a lower pKa and a stronger acid. When the pKa table does not list your compound, use these five factors to reason by analogy.
Electronegativity
The tendency of an atom to attract electron density toward itself. More electronegative atoms stabilise negative charge better, making the conjugate base more stable and the parent acid stronger.
Charge stabilisation
A neutral conjugate base is inherently more stable than a negatively charged one. If losing a proton converts a positively charged species into a neutral one, that acid will be stronger than one whose conjugate base carries a full negative charge.
Atomic size
Larger atoms spread charge over a bigger volume. A conjugate base centred on a larger atom (e.g. sulfur vs. oxygen, or iodide vs. fluoride) is more stable, corresponding to a lower pKa.
Hybridisation / s-character
Electrons in orbitals with more s-character sit closer to the nucleus and are held more tightly. An sp-hybridised carbon (50% s) stabilises a negative charge better than an sp3 carbon (25% s). This is why terminal alkynes (pKa ~25) are far more acidic than alkanes (pKa ~50).
Resonance / delocalisation
When the negative charge on a conjugate base can be spread across two or more atoms through resonance, the base is more stable. Carboxylate anions (charge shared over two oxygens) are much more stable than alkoxide anions (charge on one oxygen), which is why carboxylic acids (pKa ~4 to 5) are far more acidic than alcohols (pKa ~15 to 17).
Conjugate base stability
The central concept tying all five factors together. The more stable the conjugate base, the more the equilibrium favours deprotonation, and the lower the pKa.
1. Electronegativity
More electronegative atoms stabilise negative charge on the conjugate base.
Example: HF (pKa 3.2) is a stronger acid than H2O (pKa 15.7) because fluorine is more electronegative than oxygen and better stabilises the negative charge.
2. Charge
A neutral conjugate base is more stable than one carrying a negative charge.
Example: NH4+ (pKa ~10) is a stronger acid than NH3 (pKa ~36) because losing a proton from NH4+ gives neutral NH3, while losing a proton from NH3 gives the charged NH2-.
3. Atomic size
Larger atoms distribute charge over a greater volume, stabilising the conjugate base.
Example: HI (pKa -9) is a stronger acid than HF (pKa 3.2). Iodide is much larger than fluoride, so I- is more stable.
This factor dominates when comparing atoms in different rows of the periodic table.
4. Hybridisation (s-character)
More s-character means the orbital holding the lone pair is closer to the nucleus, stabilising the charge.
sp (50% s) > sp2 (33% s) > sp3 (25% s) in ability to stabilise a negative charge.
Example from the pKa table: HCN (pKa 9.1, sp carbon) vs. terminal alkyne C-H (pKa 25, sp carbon but with less electronegative neighbours) vs. alkane C-H (pKa 49, sp3 carbon).
5. Resonance / delocalisation
Spreading the negative charge across multiple atoms through resonance structures stabilises the conjugate base.
Example: acetic acid (pKa 4.8) vs. ethanol (pKa ~17). The acetate anion delocalises its charge over two oxygens; ethoxide cannot.
This factor produces some of the largest pKa differences.
The lecture's practice problem asks you to rank CN bond lengths in three structurally related ammonium compounds with different pKa values (M: 5.29, N: 4.6, O: 2.35).
The reasoning works backward from pKa:
Lower pKa means the conjugate base (the free amine) is more stable.
A more stable conjugate base has more electron density on nitrogen (less donation into the ring), which means less C-N double-bond character.
Wait: this lecture shows the opposite. The conjugate base of compound O (pKa 2.35, strongest acid, most stable conjugate base) has the most resonance donation of the nitrogen lone pair into the adjacent carbonyl. That resonance gives partial double-bond character to the C-N bond, shortening it.
Result: O has the shortest C-N bond, N is intermediate, M has the longest. The pKa values track how much the nitrogen lone pair is delocalised, which simultaneously stabilises the conjugate base and shortens the C-N bond.
The practice exam asks why alcohol group X is more acidic than alcohol group Y in artochamin F.
The conjugate base from deprotonating X can form an intramolecular hydrogen bond that stabilises the alkoxide.
The conjugate base from deprotonating Y cannot form the same interaction because the rigid ring geometry holds the donor and acceptor too far apart.
A stabilised conjugate base means a lower pKa (stronger acid), so X is deprotonated more readily.
pKa reflects conjugate base stability: more stable conjugate base = lower pKa = stronger acid.
The five factors are not a formula in the mathematical sense, but a checklist applied in order of relevance. When two factors conflict (e.g. electronegativity says one thing, resonance says another), the factor that produces the larger effect in the specific comparison usually wins. Resonance and hybridisation tend to produce the largest pKa shifts.
The structural reasoning behind pKa prediction is how medicinal chemists decide which nitrogen or oxygen in a drug molecule will be protonated at blood pH. That protonation state controls solubility, membrane permeability and binding to the target protein. The same logic applies in materials science (choosing monomers for polymers that resist acidic or basic environments) and environmental chemistry (predicting how pollutants ionise in water at different pH levels).
Students often assume electronegativity always wins. It does not. When comparing atoms in different rows of the periodic table, atomic size usually dominates (this is why HI is a stronger acid than HF despite fluorine being more electronegative).
Confusing the stability of the acid with the stability of the conjugate base. Acid strength is about how stable the conjugate base is after the proton leaves, not about how stable the starting acid is.
Thinking resonance stabilisation only matters when the charge is on oxygen. Nitrogen lone pairs delocalise into adjacent pi systems too, as the CN bond-length problem demonstrates.
Applying the five factors when the pKa table already has the answer. The lecture says it plainly: defer to the pKa table when you can. Use structural reasoning only for compounds not in the table or when an exam asks you to explain a trend.
\u26a0\ufe0f "Rank these compounds by acidity and explain" is a staple question. You must name which structural factor(s) drive the ranking and connect them to conjugate base stability.
\u26a0\ufe0f The CN bond-length problem (compounds M, N, O) tests whether you can link pKa to resonance, and resonance to bond order. Expect variations on this theme.
\u26a0\ufe0f Intramolecular hydrogen bonding as a stabilising interaction for conjugate bases (artochamin F) is a less obvious factor. The exam explicitly tests whether you can draw the interaction and explain why geometry matters.
\u26a0\ufe0f When estimating pKa by analogy, you are warned to avoid comparing structures that differ in hybridisation or delocalisation, because those differences introduce the largest errors.
True or false: Atomic size is the dominant factor when comparing acid strengths across a row of the periodic table. (False. Across a row, electronegativity is more important. Atomic size dominates down a column.)
Fill in the blank: An sp-hybridised carbon has ______% s-character. (50%.)
True or false: Resonance stabilisation of the conjugate base lowers the pKa of the parent acid. (True.)
Fill in the blank: A neutral conjugate base is ______ stable than a negatively charged conjugate base. (More.)
True or false: You should always use the five structural factors rather than the pKa table. (False. Defer to the table when the compound is listed.)
Q: Compounds M (pKa 5.29), N (pKa 4.6) and O (pKa 2.35) are structurally related amines with different substituents. Rank their conjugate bases by C-N bond length, shortest to longest, and explain.
A: O has the shortest C-N bond, N is intermediate, M has the longest. Lower pKa means the conjugate base is more stable, which in these compounds corresponds to greater resonance donation of the nitrogen lone pair into the adjacent pi system. Greater resonance donation gives the C-N bond more double-bond character, shortening it.
Q: In the artochamin F problem, the conjugate base from deprotonating alcohol X is more stable than the one from deprotonating alcohol Y. What interaction stabilises the X conjugate base, and why can the Y conjugate base not benefit from it?
A: The X conjugate base is stabilised by an intramolecular hydrogen bond between the alkoxide and a nearby OH donor. The Y conjugate base cannot form this hydrogen bond because the rigid ring system holds the potential donor and acceptor in a geometry that is too far apart or at the wrong angle.
Q: Explain why acetic acid (pKa 4.8) is a much stronger acid than ethanol (pKa ~17), using the five structural factors.
A: The dominant factor is resonance. The acetate conjugate base delocalises its negative charge equally over two oxygen atoms through resonance. Ethoxide has no such delocalisation; the charge sits entirely on one oxygen. This massive difference in conjugate base stability accounts for the roughly 12 pKa units between them.
Q: A base B- has a conjugate acid with pKa 43. Will it completely deprotonate an alkyne with pKa 25?
A: Keq = 10^(43 - 25) = 10^18. This is enormously greater than 10^3, so B- will completely deprotonate the alkyne.
The five structural factors for conjugate base stability reappear when you study nucleophilicity and leaving-group ability in substitution reactions (SN1/SN2). A good leaving group is, by definition, a stable conjugate base.
Resonance and hybridisation arguments from this unit carry forward to carbonyl chemistry (Chem 210's second half), where the acidity of alpha-hydrogens depends on enolate stabilisation through resonance.
The intramolecular hydrogen-bonding example connects to conformational analysis: molecular geometry determines which non-covalent interactions are possible.
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