Curly Arrows Revisited: Tracking Electron Flow

Introduction

A curly-arrow mechanism is a compact account of electron-pair movement. It should explain a product rather than decorate a memorised one. Every full arrow must start where electrons currently exist and end where that pair goes. If an arrow creates a bond without releasing space at an ordinary saturated carbon, or changes charge without accounting for electrons, the mechanism is incomplete.

Core explanation

Full-headed curved arrows move two electrons. The tail starts at a lone pair, a sigma bond or a pi bond. The head points to an atom when a pair becomes a lone pair, or between atoms when it forms a bond. In a substitution of CH₃Br by OH⁻, oxygen's lone pair points to methyl carbon, while C–Br bond electrons point to bromine. One C–O bond forms, one C–Br bond breaks, and formal charges move from OH⁻ to Br⁻. Without the second arrow, the drawing incorrectly gives carbon five bonds.

In nucleophilic carbonyl addition, an oxygen or carbon nucleophile points to the carbonyl carbon and the C=O pi bond points to oxygen. The immediate product is a tetrahedral alkoxide if the nucleophile was an anion; protonation may then give an alcohol. The electrophilic carbon does not simply accept a fifth bond while retaining C=O. Draw the intermediate and its charge before jumping to the final work-up product.

In acid-catalysed carbonyl chemistry, protonation can precede nucleophilic attack. An oxygen lone pair points to H of an acid, and the acid H–A bond pair moves to A. Protonated carbonyl oxygen increases carbonyl-carbon electrophilicity. Later deprotonation restores the catalyst. Catalysts appear in the mechanism but are regenerated overall; forgetting their charge-balance arrows can make a catalytic cycle seem to consume acid permanently.

Elimination uses a linked set of arrows. In E2, base lone pair points to a beta H; the beta C–H bond pair forms the alpha–beta pi bond; the alpha C–X bond pair goes to the leaving group. All three changes occur in one concerted step. An SN1 mechanism instead first moves C–X electrons to X, leaving a carbocation, then uses a separate nucleophile-attack step. Arrow timing therefore encodes mechanism, not just final connectivity.

Curved arrows in resonance drawings also move electron pairs, but resonance is not a chemical reaction between isolable structures. Atom nuclei and sigma-bond framework stay fixed while pi and lone-pair positions are redrawn. Use a resonance double-headed arrow between contributors, not an equilibrium arrow that implies two distinct interconverting species. Keto–enol tautomerism actually moves a proton and changes sigma connectivity, so it is an equilibrium between compounds and needs reaction steps.

Radical mechanisms use half-headed fishhook arrows for one-electron movement. Copying full pair arrows into a homolytic bond cleavage predicts the wrong electronic species. For example, photolytic Br–Br homolysis sends one electron to each Br atom, making two Br radicals. Heterolytic cleavage sends the bonding pair to one atom and gives ions. The two processes differ in both notation and products.

An arrow audit checks atoms, electrons and charge. Formal charges need not remain on the same atom, but total charge must be conserved for the complete step. A neutral nucleophile that forms a fourth bond on nitrogen may become positively charged until deprotonation. A negatively charged nucleophile that forms a new single bond may become neutral. Carbon's ordinary valence, oxygen's typical two bonds and nitrogen's typical three neutral bonds are quick error detectors, though valid charged intermediates can have different bond counts.

Use arrow logic to predict, not just confirm. Identify the highest-energy electrophilic site, accessible nucleophile and plausible leaving group. Draw the first elementary step, then let the resulting intermediate suggest the next. If the route requires an implausible leaving group such as unactivated OH⁻ under ordinary substitution conditions, consider protonation or conversion to a better leaving group before drawing departure.

Step-by-step reasoning

For each elementary step, write all charges and lone pairs needed for reasoning. Locate electron source and destination. Draw every concurrent arrow required to keep valence legal. Construct the immediate intermediate, recalculate formal charges and total charge, then decide whether proton transfer, leaving-group loss or another addition follows. Distinguish pair-electron polar steps from radical one-electron steps.

Visual explanation

Draw three panels. Panel one: OH⁻ + CH₃Br with two synchronised arrows, ending CH₃OH + Br⁻. Panel two: CN⁻ attack on acetone with CN-to-carbon and C=O-to-O arrows, ending an alkoxide. Panel three: Br–Br homolysis with two half-headed arrows to separate Br radicals. Colour arrow tails blue at the original electrons.

Real-world analogy

An arrow is an accounting entry recording where a pair of workers moves. One cannot begin the transfer from an empty desk. If workers occupy a new desk that was already full, someone else must move out. Valence and charge checks are the ledger reconciliation after each move.

Real-world example

A mechanism answer for hydroxide attack on acetone draws a new C–O bond but leaves C=O unchanged. The carbonyl carbon then appears to have five bond units. Adding the pi-bond-to-oxygen arrow yields a tetrahedral alkoxide, which can be protonated in a later step. The correction changes the intermediate and charge, not just the picture's neatness.

Why?

Why must arrow tails begin at electrons? A curved arrow describes electron movement, so its origin is a lone pair or bond, not an electron-deficient atom. Why separate protonation from attack when they occur in different elementary steps? The intermediate's charge and electrophilicity change, affecting which path is plausible next.

Common misconception

"An arrow from a positive carbon toward a nucleophile shows attraction." It reverses electron flow. A nucleophile supplies electrons toward the positive or partially positive carbon. Electrostatic attraction can motivate the encounter, but the curved arrow records electron-pair donation.

Worked example

Question: Show the electron-pair changes for CN⁻ addition to acetone before protonation.

Reasoning: The cyanide carbon's electron pair attacks acetone's carbonyl carbon. At the same time, the C=O pi pair shifts to oxygen. The former carbonyl carbon now bears CN and O⁻ in a tetrahedral intermediate.

Answer: Draw CN⁻ carbon → carbonyl carbon and C=O pi bond → O; the intermediate is (CH₃)₂C(O⁻)CN, followed later by protonation to the cyanohydrin.

Quick check

1. What arrowhead style represents movement of one electron in a radical step? Answer: A half-headed fishhook arrow, rather than a full two-electron curved arrow.

Exam focus

Use full arrows from bonds or lone pairs only. Draw all simultaneous arrows of a concerted step, then check valence and total charge. Show immediate intermediates before work-up. Use resonance arrows only for alternative electron contributors with the same atom positions, and fishhook arrows for radicals.

Advanced insight

Curved-arrow notation summarises orbital interactions without claiming electrons travel along literal curved paths. The arrows are still constrained by orbital availability, charge and conservation. A correctly drawn mechanism should predict both the product and characteristic intermediate or stereochemical behaviour, making it testable against reaction evidence.

Summary

Full curly arrows move electron pairs from real sources to destinations. Substitution couples bond formation and leaving-group loss; carbonyl addition shifts the pi pair to oxygen; E2 links proton removal, pi formation and leaving-group departure. Audit intermediates for valence and charge. Separate polar pair arrows, radical fishhooks, resonance contributors and true chemical equilibria.

Practice questions

1. Where does the C=O pi pair go when CN⁻ adds to acetone? Answer: Onto oxygen, forming an alkoxide intermediate. 2. Why does CH₃Br + OH⁻ SN2 need a C–Br-to-Br arrow? Answer: Bromide must leave with the old bonding pair as the new C–O bond forms, preserving carbon valence. 3. Do resonance contributors differ in atom positions? Answer: No. Only electron placement changes; nuclei and sigma framework stay fixed. 4. What kind of cleavage makes two bromine radicals from Br₂? Answer: Homolytic cleavage, represented with two one-electron fishhook arrows.