DAT Organic Chemistry: Proton Transfer Favorability

DAT High-Yield Concept

Equilibrium Direction: Proton transfer equilibria always lie toward the side with the weaker acid (higher $\text{p}K_a$ ) and weaker base.
Equilibrium Constant Equation: Calculate $K_{\text{eq}}$ quantitatively using $K_{\text{eq}} = 10^{\text{p}K_a(\text{conjugate acid}) - \text{p}K_a(\text{starting acid})}$ .
Thermodynamic Favorability: A proton transfer is product-favored ( $K_{\text{eq}} > 1$ ) when the starting acid is stronger (lower $\text{p}K_a$ ) than the conjugate acid formed.
Site Selectivity: When a molecule contains multiple acidic sites, a base preferentially removes the proton attached to the site with the lowest $\text{p}K_a$ .

Determining Proton Transfer Favorability

In a Brønsted-Lowry acid-base reaction, a proton is transferred from an acid (proton donor) to a base (proton acceptor). The direction of equilibrium is determined by comparing the relative acid strengths of the reactants and products. Acids with lower $\text{p}K_a$ values donate protons more readily, whereas weaker acids (higher $\text{p}K_a$ ) retain protons more tightly. As a result, proton transfer reactions always proceed spontaneously toward the formation of the weaker acid and weaker base.

Scheme of the dissociation of acetic acid to acetate, with structural formulae and curly arrows indicating the mechanism of proton transfer. A water molecule is protonated to form a hydronium ion in the process. (Source: Ben Mills · Public domain · Wikimedia Commons)

For instance, when acetic acid ( $\text{p}K_a = 4.75$ ) transfers a proton to water, forming hydronium ( $\text{p}K_a = -1.7$ ), the reaction is reactant-favored because hydronium is a stronger acid than acetic acid:

$\ce{CH3COOH + H2O <=> CH3COO^- + H3O^+}$

Deprotonation of acetic acid by a hydroxide ion (Source: loltim · CC BY-SA 4.0 · Wikimedia Commons)

In contrast, when acetic acid reacts with hydroxide ion, it forms acetate and neutral water ( $\text{p}K_a = 15.7$ ). Because water is a much weaker acid than acetic acid, this reaction is strongly product-favored:

$\ce{CH3COOH + OH^- <=> CH3COO^- + H2O}$

Key Principle: Proton transfer runs downhill toward the weaker acid (higher

\text{p}K_a

) and weaker base. A lower

\text{p}K_a

value signifies a stronger acid.

Worked Example 1: Predicting Equilibrium Favorability

Problem: Is the reaction between phenol ( $\text{p}K_a = 10.0$ ) and ammonia ( $\ce{NH3}$ ) to form phenoxide and ammonium ( $\ce{NH4^+}$ , $\text{p}K_a = 9.25$ ) product-favored or reactant-favored?

$\ce{C6H5OH + NH3 <=> C6H5O^- + NH4^+}$

Reasoning: Identify the starting acid ( $\ce{C6H5OH}$ , $\text{p}K_a = 10.0$ ) and conjugate acid ( $\ce{NH4^+}$ , $\text{p}K_a = 9.25$ ). Comparing the two acids, ammonium is stronger because it has a lower $\text{p}K_a$ . Because equilibrium favors the weaker acid, the position lies on the reactant side.

Answer: The reaction is reactant-favored ( $K_{\text{eq}} < 1$ ).

Exam Trap: Students often confuse $\text{p}K_a$ values with acid strength. Remember that lower $\text{p}K_a$ means a stronger acid, so equilibrium always favors the side with the higher $\text{p}K_a$ .

Equilibrium Position from a $\text{p}K_a$ Difference

The exact equilibrium constant ( $K_{\text{eq}}$ ) for a proton transfer reaction can be calculated directly from the difference in $\text{p}K_a$ between the conjugate acid (product acid) and the starting acid (reactant acid):

$K_{\text{eq}} = 10^{\text{p}K_a(\text{conjugate acid}) - \text{p}K_a(\text{starting acid})}$

When $\Delta \text{p}K_a = \text{p}K_a(\text{conjugate acid}) - \text{p}K_a(\text{starting acid}) > 0$ , the equilibrium is product-favored ( $K_{\text{eq}} > 1$ ). Every unit of difference in $\text{p}K_a$ corresponds to a factor of 10 in the equilibrium constant.

Proton transfer always runs downhill toward the weaker acid — A proton transfer is decided entirely by which side holds the weaker acid, and the pKa table tells you that at a glance. Here ammonium at 9.25 is a weaker acid than acetic acid at 4.76, so the products are the comfortable side and the equilibrium sits there. The arithmetic makes it quantitative: every pKa unit of difference is one power of ten in Keq, so a gap of about 4.5 units is roughly thirty thousand to one. As a working rule, a difference of three or more pKa units means the reaction goes essentially to completion.

Worked Example 2: Calculating $K_{\text{eq}}$ from $\text{p}K_a$ Values

Problem: Calculate $K_{\text{eq}}$ for the deprotonation of ethanol ( $\ce{CH3CH2OH}$ , $\text{p}K_a = 16.0$ ) by amide ion ( $\ce{NH2^-}$ ) to yield ethoxide ( $\ce{CH3CH2O^-}$ ) and ammonia ( $\ce{NH3}$ , $\text{p}K_a = 38.0$ ).

$\ce{CH3CH2OH + NH2^- <=> CH3CH2O^- + NH3}$

Reasoning: The starting acid is ethanol ( $\text{p}K_a = 16.0$ ) and the conjugate acid formed is ammonia ( $\text{p}K_a = 38.0$ ). Substitute into the exponent formula:

$K_{\text{eq}} = 10^{38.0 - 16.0} = 10^{22.0}$

Answer: $K_{\text{eq}} = 10^{22}$ . The equilibrium is overwhelmingly product-favored.

Exam Trap: Subtracting in reverse (starting acid minus conjugate acid) gives $10^{-22}$ , incorrectly predicting a reactant-favored system. Always use $\text{p}K_a(\text{conjugate acid}) - \text{p}K_a(\text{starting acid})$ .

Site Selectivity: Which Proton Is Removed by a Base?

When a single compound contains multiple acidic functional groups, treatment with one equivalent of base leads to selective deprotonation at the most acidic position (the proton with the lowest $\text{p}K_a$ ). Structural factors like resonance delocalization, electronegativity, and induction govern these relative acidity differences.

DAT Mnemonic: LOW pKa GOES FIRST — The proton attached to the site with the Lowest

\text{p}K_a

is Overwhelmingly Withdrawn first by any base.

Functional Group / Species	Approximate $\text{p}K_a$	Conjugate Base	Relative Acidity
Alkane ( $\ce{CH4}$ )	50	$\ce{CH3^-}$	Extremely Weak
Amine ( $\ce{NH3}$ )	38	$\ce{NH2^-}$	Very Weak
Terminal Alkyne ( $\ce{HC#CH}$ )	25	$\ce{HC#C^-}$	Weak
Alcohol ( $\ce{CH3OH}$ )	16	$\ce{CH3O^-}$	Moderate
Water ( $\ce{H2O}$ )	15.7	$\ce{OH^-}$	Moderate
Phenol ( $\ce{C6H5OH}$ )	10	$\ce{C6H5O^-}$	Strong Organic Acid
Carboxylic Acid ( $\ce{CH3COOH}$ )	4.75	$\ce{CH3COO^-}$	Very Strong Organic Acid
Hydronium ( $\ce{H3O^+}$ )	-1.7	$\ce{H2O}$	Strong Mineral Acid

Worked Example 3: Predicting Selective Deprotonation

Problem: Predict the major organic product formed when 4-hydroxybenzoic acid is treated with 1.0 equivalent of $\ce{NaOH}$ .

$\ce{HO-C6H4-COOH + OH^- -> HO-C6H4-COO^- + H2O}$

Reasoning: 4-hydroxybenzoic acid possesses two distinct acidic protons: a carboxylic acid O-H proton ( $\text{p}K_a \approx 4.5$ ) and a phenolic O-H proton ( $\text{p}K_a \approx 10.0$ ). Because the carboxyl proton has a much lower $\text{p}K_a$ , it is significantly more acidic. Hydroxide selectively deprotonates the carboxylic acid to yield 4-hydroxybenzoate and water.

Answer: 4-hydroxybenzoate ( $\ce{HO-C6H4-COO^-}$ ).

Exam Trap: Deprotonating the phenolic OH group assuming aromaticity makes it more reactive. Although phenol is acidic ( $\text{p}K_a = 10$ ), carboxylic acids ( $\text{p}K_a \approx 4.5$ ) are vastly more acidic due to resonance delocalization across two electronegative oxygen atoms.

Critical Thinking Questions

If acetic acid ( $\text{p}K_a = 4.75$ ) reacts with hydrogen cyanide ( $\ce{HCN}$ , $\text{p}K_a = 9.2$ ), which direction does the equilibrium favor, and what is the approximate value of $K_{\text{eq}}$ ?
Why is the deprotonation of a terminal alkyne ( $\text{p}K_a = 25$ ) using hydroxide ( $\ce{OH^-}$ , conjugate acid $\text{p}K_a = 15.7$ ) unfavorable, whereas using amide ion ( $\ce{NH2^-}$ , conjugate acid $\text{p}K_a = 38$ ) is highly favorable?
A compound contains both an aliphatic amine proton ( $\text{p}K_a \approx 38$ ) and a thiol proton ( $\text{p}K_a \approx 10.5$ ). Which site is deprotonated first when treated with sodium hydride ( $\ce{NaH}$ )?

References

Sources returned by Google Search grounding during lesson verification.

Predicting Acid-Base Reactions from pKa Values - Chemistry LibreTexts — chem.libretexts.org
Calculating the Equilibrium Constant (Keq) from Ka and pKa Values - Chemistry Steps — chemistrysteps.com

Proton Transfer Favorability

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