Difficulty: Intermediate | Prerequisites: Organic Chemistry I acid-base fundamentals, familiarity with pKa tables.
This topic sits at the heart of organic reactivity. Before you can predict whether a reaction will proceed, you need to know which way the proton moves, and K_EQ tells you exactly that. Section 3.2 builds on the Bronsted-Lowry framework from Orgo I and applies it to organic structures where the acidic proton is not always obvious. If you are comfortable identifying Lewis structures and conjugate acid-base pairs, you are ready for this.
TL;DR
Acid-base reactions favour the side with the weaker acid and weaker base (more stable species). To predict whether K_EQ > 1, compare the pKa of the acid with the pKa of the conjugate acid of the base. If the starting acid is stronger (lower pKa), equilibrium favours products, and K_EQ > 1.
Bronsted-Lowry acid
A species that donates a proton (H+) to another species. In simple terms, it is anything that gives up a hydrogen ion.
Bronsted-Lowry base
A species that accepts a proton (H+) from another species. Think of it as the molecule with a lone pair ready to grab a hydrogen.
Conjugate acid
The species formed when a base accepts a proton. If B is the base, BH+ is its conjugate acid.
Conjugate base
The species formed when an acid donates a proton. If HA is the acid, A- is its conjugate base.
pKa
The negative log of the acid dissociation constant (Ka). A lower pKa means a stronger acid. In simple terms, pKa is a number on a scale: the smaller it is, the more readily the molecule gives up its proton.
K_EQ (equilibrium constant)
The ratio of product concentrations to reactant concentrations at equilibrium. For an acid-base reaction, K_EQ > 1 means products are favoured; K_EQ < 1 means reactants are favoured.
Acid-base equilibrium
The state at which the forward and reverse proton-transfer reactions proceed at equal rates. The position of this equilibrium is what K_EQ describes.
The central principle: equilibrium favours the formation of the weaker acid and weaker base.
Compare the pKa of the reacting acid (on the left) to the pKa of the conjugate acid (on the right).
If the reacting acid has a lower pKa (stronger acid) than the conjugate acid, K_EQ > 1 and products are favoured.
If the reacting acid has a higher pKa (weaker acid) than the conjugate acid, K_EQ < 1 and reactants are favoured.
The reaction always moves the proton from the stronger acid to the stronger base, ending with the weaker pair.
For a proton transfer reaction: HA + B- goes to A- + HB
K_EQ = Ka(HA) / Ka(HB)
In log form: pK_EQ = pKa(HB) - pKa(HA)
Then K_EQ = 10^(pKa(HB) - pKa(HA))
Example: if HA has pKa = 5 and HB has pKa = 16, then K_EQ = 10^(16-5) = 10^11, heavily favouring products.
Look for protons on electronegative atoms (O-H, N-H, S-H) as potential acids.
Look for lone pairs on electronegative atoms as potential bases (especially nitrogen, oxygen, and charged carbanions).
C-H bonds can be acidic when adjacent to electron-withdrawing groups or when the conjugate base is stabilised by resonance (e.g. terminal alkynes, pKa around 25).
Charged species are often key players: a negatively charged base (e.g. an alkoxide, RO-) is a much stronger base than its neutral counterpart (e.g. an alcohol, ROH).
When given multiple structures across several parts of a question:
Identify every species that could act as an acid (has a transferable proton).
Identify every species that could act as a base (has an available lone pair or negative charge).
Pick the strongest acid you can find from one part and the strongest base from a different part.
The larger the pKa gap between the acid (low pKa) and the conjugate acid of the base (high pKa), the more confidently K_EQ > 1.
A safe approach: pair a strong organic acid (e.g. a carboxylic acid, pKa around 4-5) with a strong organic base (e.g. an alkoxide or amide ion) and you will get K_EQ well above 1.
General acid-base equilibrium:
HA + B^- \rightleftharpoons A^- + HBEquilibrium constant from pKa values:
K_{EQ} = 10^{(pK_a(\text{conjugate acid}) - pK_a(\text{reacting acid}))}Relationship between Ka and pKa:
pK_a = -\log(K_a)K_EQ from Ka values directly:
K_{EQ} = \frac{K_a(\text{reacting acid})}{K_a(\text{conjugate acid})}Predicting acid-base equilibria is how chemists choose reagents for reactions. If you need to deprotonate a substrate, you pick a base whose conjugate acid has a higher pKa than the substrate, ensuring K_EQ > 1 so the deprotonation actually goes to completion. This logic underpins everything from pharmaceutical synthesis (choosing the right base to form an enolate) to industrial processes (neutralisation reactions in water treatment).
Students often think a higher pKa means a stronger acid. It does not. A higher pKa means a weaker acid. The scale runs in the opposite direction to acid strength.
Students frequently confuse K_EQ > 1 with the reaction going to 100% completion. K_EQ > 1 means products are favoured at equilibrium, but both reactants and products are still present. Only very large K_EQ values (say 10^10 or more) mean the reaction is essentially complete.
Students sometimes forget that the base's strength is assessed via the pKa of its conjugate acid, not the base itself. You always compare two pKa values on the same scale.
A common error on exam problems like POD 9: selecting an acid and a base from the same part of the question rather than from different parts. The question specifically asks for cross-part pairing.
⚠️ Predicting which direction an acid-base equilibrium favours is tested repeatedly throughout Orgo II. It is foundational for understanding reaction mechanisms.
⚠️ You will be expected to justify your K_EQ prediction with pKa values. A bare answer without reasoning will not earn full marks.
⚠️ The POD 9 question format (cross-selecting from multiple parts) tests whether you can scan a set of structures and rank their relative acidities and basicities, not just memorise individual pKa values.
⚠️ The concept returns in nucleophilic substitution and elimination topics: leaving group ability tracks conjugate base stability, which is the same pKa reasoning.
True or False: A reaction with K_EQ = 10^(-3) favours product formation.
False. K_EQ < 1 means reactants are favoured.
Fill in the blank: Equilibrium favours the side with the ______ acid and the ______ base.
Weaker, weaker.
True or False: If the pKa of the reacting acid is 4 and the pKa of the conjugate acid is 15, K_EQ > 1.
True. The stronger acid (lower pKa) donates to the stronger base, favouring products.
Fill in the blank: To find K_EQ from pKa values, calculate 10 to the power of (pKa of ______ minus pKa of ______).
Conjugate acid, reacting acid.
True or False: A negatively charged species is always a stronger base than a neutral species.
Generally true in comparable structures, but not universally. A resonance-stabilised anion (e.g. acetate) can be a weaker base than a neutral amine.
Q: A carboxylic acid (pKa approximately 5) is mixed with sodium hydride (NaH), where the conjugate acid H2 has a pKa of approximately 35. Will K_EQ be greater or less than 1? Explain.
A: K_EQ will be far greater than 1. The carboxylic acid (pKa 5) is much stronger than H2 (pKa 35), so the proton transfers from the carboxylic acid to the hydride ion. K_EQ = 10^(35-5) = 10^30.
Q: An alcohol (pKa approximately 16) reacts with an amide ion, NH2- (conjugate acid NH3, pKa approximately 38). Predict K_EQ and state which side is favoured.
A: K_EQ = 10^(38-16) = 10^22. Products are heavily favoured. The alcohol is the acid and the amide ion is the base; the proton transfers to give an alkoxide and ammonia.
Q: Explain why you cannot simply look at charge to determine which species in a reaction is the acid and which is the base.
A: Charge helps but is not definitive. A neutral molecule with an O-H bond can be a strong acid (e.g. a carboxylic acid), while a negatively charged species can be a poor base if resonance delocalises the charge extensively. You must consider pKa values, which account for resonance, electronegativity, and induction effects together.
Q: You are given five organic structures and asked to select an acid from one part and a base from a different part such that K_EQ > 1. Describe the strategy you would use.
A: First, identify every possible acidic proton across all five structures and estimate or look up their pKa values. Then identify every species that could act as a base and note the pKa of each one's conjugate acid. Pick the acid with the lowest pKa from one part and the base whose conjugate acid has the highest pKa from a different part. The larger the gap, the more confidently K_EQ exceeds 1. State both pKa values in your defence.
This connects directly to nucleophilic substitution (SN1/SN2) because leaving group ability is the flip side of base strength: a weak base makes a good leaving group, and you assess that through the same pKa framework. It also feeds into elimination reactions (E1/E2), where the choice of base determines whether substitution or elimination dominates. Later in the course, enolate chemistry relies entirely on choosing a base strong enough to deprotonate at the alpha-carbon (pKa around 20), which is the same K_EQ > 1 logic applied to C-H bonds.
Bronsted-Lowry acid-base, proton transfer, pKa, Ka, equilibrium constant, K_EQ, conjugate acid, conjugate base, acid dissociation constant, acid strength, base strength, organic acids, organic bases, carboxylic acid pKa, alcohol pKa, amine pKa, alkoxide, amide ion, deprotonation, protonation, CHEM 210, Organic Chemistry II, University of Michigan, Section 3.2, POD 9, Question 3.07