Difficulty: Intermediate to Advanced | Prerequisites: Parts 1 and 2 of these notes, plus comfort with curved-arrow notation and Lewis structures from earlier Chem 210 lectures.
This set of notes covers the applied, problem-solving side of the acid-base unit: drawing curved-arrow mechanisms for proton transfer and Lewis acid-base complexation, identifying the most acidic proton in a multi-functional molecule, and determining the major species at a specific pH. These are the question types that appear on Exam 1. The lecture's practice problems (Topic 12) walk through each of them step by step, and the notes below reconstruct that reasoning so you can replicate it on your own.
Proton transfer and complexation are the two reaction types in this unit. For proton transfer, draw the curved arrow from the base's lone pair to the acidic proton, then from the bond to the acid's conjugate base. For complexation, draw the arrow from the Lewis base's lone pair to the Lewis acid's empty orbital. Always show formal charges, draw complete Lewis structures, and pick the most significant resonance contributor for each product.
Proton transfer reaction
A Br\u00f8nsted acid-base reaction in which a proton (H+) moves from one molecule to another. Both bonds (old and new) to the proton are shown with curved arrows.
Complexation (Lewis acid-base reaction without proton transfer)
A reaction in which a Lewis base donates an electron pair directly to a Lewis acid, forming a new covalent bond. No proton moves. Example: NH3 donating into the empty p-orbital of BF3.
Curved arrow mechanism
The notation system where arrows show the movement of electron pairs. The arrow tail starts at the electron source (lone pair or bond), and the head points to the electron destination (atom or bond). In proton transfer: two arrows. In complexation: one arrow.
Most significant resonance contributor
When a product can be drawn as multiple resonance structures, the one that best satisfies the octet rule, minimises formal charges, places negative charges on electronegative atoms and avoids charge separation is the "most significant" contributor. The exam requires you to draw this form.
Major species
The dominant form of a molecule at a given pH, determined by comparing each ionisable group's pKa to the solution pH. Groups with pKa below the pH are deprotonated; groups with pKa above the pH remain protonated.
Every proton transfer requires exactly two curved arrows:
Arrow 1: from the lone pair on the base to the acidic proton (forms the new bond).
Arrow 2: from the bond between the acid atom and the proton to the acid atom itself (breaks the old bond, generating the conjugate base).
After drawing the arrows, write out the products with complete Lewis structures. Show all formal charges. If the product has resonance structures, draw the most significant contributor.
Example from lecture: cyclohexane carbanion (from NaH deprotonation, pKa ~49 for cyclohexane) reacting with a ketone as a proton source. The carbanion's lone pair attacks the acidic proton; the C-H bond breaks and the electrons go to the carbon of the ketone.
Complexation is simpler: one curved arrow, from the lone pair on the Lewis base to the Lewis acid.
The Lewis acid gains an electron pair and a formal negative shift.
The Lewis base loses a lone pair and gains a formal positive shift (or loses its negative charge).
Example from lecture: AlI3 (Lewis acid, empty p-orbital on Al) reacting with a diethyl ether (Lewis base, lone pair on oxygen). One arrow from the oxygen lone pair to aluminium. Product: an aluminium-oxygen dative bond, with a positive formal charge on oxygen and a negative formal charge on aluminium.
When a molecule has multiple ionisable protons (e.g. an amino acid with a carboxylic acid, an ammonium and a thiol), rank the sites by their pKa values from the table:
The site with the lowest pKa loses its proton first.
Work through each ionisable group systematically. For a compound with an amide NH (pKa ~15), a carboxylic acid OH (pKa ~4.8) and a thiol SH (pKa ~10.5), the carboxylic acid proton is the most acidic.
The aminophenylalanine problem from lecture is the model for this question type. The molecule has three measured pKa values: 2.15 (carboxylic acid), 6.10 (alkylammonium) and 10.60 (arylammonium).
Process for any pH:
Compare the pH to each pKa, one at a time.
If pH > pKa for a group, that group is deprotonated (A- form dominates).
If pH < pKa for a group, that group is protonated (HA form dominates).
If pH = pKa, the two forms are present in a 1:1 ratio.
Worked examples from lecture:
pH 7.10: pH > 2.15 (COO-), pH > 6.10 (free amine NH2), pH < 10.60 (ArNH3+ stays protonated). Major species: deprotonated carboxylate, free alkylamine, protonated arylamine.
pH 10.60: this equals the arylammonium pKa, so that group is 1:1 (ArNH3+ and ArNH2 in equal amounts). The other two groups are deprotonated (pH >> their pKa values).
pH 5.45: pH > 2.15 (COO-), pH < 6.10 (NH3+ stays), pH < 10.60 (ArNH3+ stays). Major species: deprotonated carboxylate, both amines protonated.
pH 1.30: pH < 2.15 (COOH stays), pH < 6.10 (NH3+ stays), pH < 10.60 (ArNH3+ stays). Major species: everything protonated.
The exam gives you incomplete reactions and asks you to supply either the curved-arrow mechanism, the products, or the missing starting materials. The instructions always specify that these are proton transfer or complexation reactions.
Checklist for each problem:
Identify whether it is proton transfer (a proton moves) or complexation (an electron pair is donated to an empty orbital).
Draw the curved arrow(s).
Write complete Lewis structures for all products.
Show all formal charges.
Account for all atoms and charges (the reaction must balance).
If more than one resonance structure is possible, draw the most significant contributor.
No new formulas in this section beyond those in Part 1. The key relationship to keep to hand:
Keq = 10^(pKa of product acid - pKa of reactant acid)
For the "will this base completely deprotonate this acid" question type, compute Keq. If Keq >= 10^3 (i.e. the pKa difference is 3 or more in favour of products), the reaction goes to completion.
Proton transfer mechanisms are how enzymes catalyse reactions in your body. An enzyme active site positions a basic residue to abstract a proton from a substrate, lowering the activation energy. The curved-arrow logic is identical to what you draw on paper.
Lewis acid-base complexation is the basis of coordination chemistry and catalysis. Industrial processes (Friedel-Crafts alkylation, Ziegler-Natta polymerisation) rely on Lewis acid catalysts such as AlCl3 accepting electron pairs from substrates.
Drawing three arrows for a proton transfer instead of two. Only two arrows are needed: lone pair to proton, and bond to conjugate base atom.
Forgetting formal charges on products. Every proton transfer or complexation changes the formal charge on at least two atoms. If your product has no formal charges and the starting materials had none either, double-check: the base that gained a proton should carry a positive charge (or have lost a negative one), and the acid that lost a proton should carry a negative charge (or have lost a positive one).
Confusing proton transfer with complexation. If a proton moves, it is proton transfer. If no proton moves but a lone pair forms a new bond to an electron-deficient atom, it is complexation. The exam specifies the reaction type.
Drawing the wrong resonance contributor. When the exam says "most significant contributor," it wants the structure with the fewest formal charges, complete octets, and any remaining negative charges on the most electronegative atoms.
\u26a0\ufe0f The exam will give you incomplete reactions and ask you to supply curved arrows, products, or missing starting materials. Practise all three directions.
\u26a0\ufe0f "Draw the major species at pH = X" with a multi-pKa molecule is a near-certain exam question. The aminophenylalanine example is the template.
\u26a0\ufe0f "Which base can completely deprotonate this acid?" requires you to calculate Keq and check that it exceeds 10^3. Know how to work this quickly.
\u26a0\ufe0f Always draw the most significant resonance contributor when more than one is possible. Points are deducted for a minor contributor.
\u26a0\ufe0f Complete Lewis structures with all lone pairs and formal charges are required for full marks. Partial structures lose points even if the chemistry is correct.
How many curved arrows are needed for a proton transfer reaction? (Two.)
True or false: In a Lewis acid-base complexation, a proton always moves. (False. No proton moves in complexation.)
Fill in the blank: At pH 3, a carboxylic acid group with pKa 4.8 is mostly in its ______ form. (Protonated, HA. Because pH < pKa.)
True or false: The most significant resonance contributor is the one with the most formal charges. (False. It is the one with the fewest formal charges and complete octets.)
Fill in the blank: For a reaction to go to completion, Keq should be at least ______. (10^3, or 1000.)
Q: You are given a reaction between a cyclopentadienyl anion (pKa of conjugate acid = 19) and sodium amide (NaNH2, conjugate acid NH3 with pKa = 36). Provide the curved arrow mechanism and predict whether products are favoured.
A: NH2- (the base) donates its lone pair to a proton on cyclopentadiene. The C-H bond breaks, electrons go to carbon (forming the cyclopentadienyl anion). Products: cyclopentadienyl anion + NH3. Keq = 10^(36 - 19) = 10^17. Products are overwhelmingly favoured.
Q: AlI3 reacts with diethyl ether. Is this proton transfer or complexation? Draw the curved arrow mechanism.
A: Complexation. No proton moves. One curved arrow from the oxygen lone pair on diethyl ether to the aluminium of AlI3. Product: an Al-O dative bond with formal positive charge on O and formal negative charge on Al.
Q: A compound has ionisable groups with pKa 4.8 (carboxylic acid) and 10.5 (thiol). What is the major species at pH 8?
A: pH 8 > 4.8, so the carboxylic acid is deprotonated (COO-). pH 8 < 10.5, so the thiol is protonated (SH). Major species: deprotonated carboxylate, protonated thiol.
Q: An amino acid has pKa values of 2.15, 6.10 and 10.60. At pH 14, what is the major species?
A: pH 14 is above all three pKa values. Every ionisable group is deprotonated. The carboxylate is COO-, the alkylammonium has lost its proton to become NH2, and the arylammonium has lost its proton to become ArNH2. Net charge is -1 (one negative from the carboxylate, both amines neutral).
Q: For the reaction of NaH with acetone: identify the acid, the base, draw the curved arrows, and write the products with formal charges.
A: NaH provides H- (the base, though here H- acts by donating its pair to a proton). Acetone's alpha-C-H is the acid (pKa ~19). Curved arrow 1: from the C-H bonding electrons... actually, NaH is a strong base, and H- abstracts a proton from the alpha carbon. Arrow 1: lone pair on H- to the alpha hydrogen. Arrow 2: C-H bond electrons go to carbon. Products: the enolate (carbanion stabilised by resonance with the carbonyl, draw the resonance contributor with the negative charge on oxygen as the most significant contributor) + H2. Na+ is a spectator ion.
Curved-arrow mechanisms for proton transfer are the same notation used for every reaction mechanism in Chem 210, from substitution to elimination to addition. Mastering the two-arrow proton transfer now saves time later.
The major-species-at-a-given-pH analysis returns in biochemistry for amino acids, peptides and proteins (isoelectric point calculations, buffer capacity, enzyme kinetics).
Lewis acid-base complexation is the entry point to organometallic chemistry and catalysis, which some Chem 210 sections introduce toward the end of the term.
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