Reaction mechanisms

Two arrow types, two meanings. A mechanism is drawn using exactly two kinds of curved arrow, and mixing them up is the single most common scoring error at this level.

• A full-headed arrow (two barbs, like a normal arrowhead) always represents a pair of electrons moving together — from a lone pair, a $\pi$ bond, or a $\sigma$ bond, to wherever the new bond or negative charge ends up. This is the arrow used for heterolytic processes.
• A half-headed, "fish-hook" arrow (a single barb) represents a single electron moving alone, the notation reserved for homolytic, free-radical processes.

Where an arrow may start and end. An arrow's tail must sit on an actual source of electron density — a lone pair, or an existing bond — never on a positive charge or an empty orbital, because you cannot push electrons from nothing. Its head points to where those electrons end up: a new bond forming, or an atom that keeps the pair and becomes negatively charged (or neutral, if it started anionic).

The general recipe for writing any mechanism step. First, sort the two reacting species into the electron-rich one (the nucleophile) and the electron-poor one (the electrophile) — often by spotting a lone pair or a negative charge on one side and an atom pulled electron-deficient by an electronegative neighbour or a good leaving group on the other. Second, draw one arrow from the nucleophile's electron source to the electrophilic atom, showing the new bond appearing. Third, if that electrophilic atom cannot simply gain a fifth bond, draw a second arrow showing an existing bond at that atom break, sending its electrons off with the departing leaving group. A full mechanism is just this two-arrow pattern repeated, step after step, until a stable product remains.

Three named categories, one shared grammar. Substitution swaps a leaving group for an incoming nucleophile in a single two-arrow step. Addition opens a $\pi$ bond across two new $\sigma$ bonds, usually via a carbocation intermediate, so it needs two such steps chained together. Elimination runs the addition logic in reverse — a leaving group departs as a neighbouring C–H bond's electrons swing in to build a new $\pi$ bond, with a base carrying the hydrogen away. All three are read the same way: find the electron source, draw where it goes, and account for every bond that appears or disappears.

Curved-arrow mechanisms are not decoration — they are a bookkeeping system that keeps every electron accounted for at every step, and that accounting is exactly why the notation is trustworthy enough to predict real products, not just describe known ones.

Step 1. Every atom in a stable organic molecule already has a full outer shell — carbon with four bonds, oxygen with two bonds and two lone pairs, and so on. A mechanism step can only rearrange existing electrons; it cannot create or destroy them. That constraint is why a curved arrow is never allowed to start from an empty orbital or a bare positive charge — there is nothing there to move. It must start from a genuine electron source: a lone pair, a $\sigma$ bond, or a $\pi$ bond.

Step 2. When a nucleophile's lone pair moves to form a new bond, the atom that supplied it goes from "owning" that pair alone to "sharing" it in a bond — its formal charge rises by one (an anion becomes neutral, for instance). Simultaneously, if the electrophilic carbon is already at its full valence of four bonds, it cannot simply gain a fifth; something must leave at the same instant. This is why a substitution step is drawn with two arrows, not one — the incoming and outgoing electron pairs are a single, simultaneous event, and showing only the first arrow would (incorrectly) predict a five-bonded carbon.

Step 3. The same electron-counting logic explains why addition to a $\pi$ bond needs two separate steps rather than one. The alkene's $\pi$ electrons attack the electrophile in step one, but that only builds one new bond; the carbon that supplied the other end of the $\pi$ bond is left with just three bonds and a positive charge — a carbocation. It takes a second nucleophilic attack, in a second step, to complete its fourth bond. Reading the arrows this way lets you predict the carbocation intermediate before ever being told it forms.

Step 4. Because the notation is really just "electrons flow from rich to poor, and every bond that appears is balanced by a bond that either forms fresh or already existed," the same three rules — identify nucleophile and electrophile, draw the bond-forming arrow, draw any necessary bond-breaking arrow — reproduce substitution, addition, and (run in reverse) elimination without needing a different set of rules for each. That is the payoff of learning the notation properly now: one grammar, three reaction types.

Question. Using curved-arrow notation reasoning, describe the mechanism by which hydroxide ion, $\text{OH}^-$, converts bromomethane, $\text{CH}_3\text{Br}$, into methanol, $\text{CH}_3\text{OH}$, and state what kind of reaction this is.

Step 1. Identify the electrophile and the nucleophile. Bromine's greater electronegativity pulls electron density away from the carbon in $\text{CH}_3\text{Br}$, leaving that carbon electron-deficient — the electrophilic site. Hydroxide carries a full negative charge and a lone pair on oxygen, making it the nucleophile.

Step 2. Draw the bond-forming arrow. A curved arrow starts at a lone pair on the oxygen of $\text{OH}^-$ and points to the electrophilic carbon, showing the new O–C $\sigma$ bond beginning to form.

Step 3. Draw the bond-breaking arrow. Because carbon cannot exceed four bonds, a second curved arrow starts at the existing C–Br bond and points to bromine, showing that bond breaking heterolytically as bromine departs with both electrons, becoming $\text{Br}^-$.

Step 4. Assemble the product. The carbon keeps its three C–H bonds and gains the new bond to oxygen, giving $\text{CH}_3\text{OH}$, while $\text{Br}^-$ leaves as a separate, stable ion.

Answer: simultaneous nucleophilic attack by $\text{OH}^-$ and heterolytic departure of $\text{Br}^-$ convert $\text{CH}_3\text{Br}$ into $\text{CH}_3\text{OH}$ — a substitution reaction, captured completely by the two-arrow pattern.

Competitive stretch

The same grammar, run as two linked steps, explains the addition of $\text{HBr}$ to ethene, $\text{CH}_2=\text{CH}_2$. In step one, the electron-rich $\pi$ bond attacks the partially positive hydrogen of polarised $\text{H}-\text{Br}$, forming a new C–H bond while the H–Br bond breaks and releases $\text{Br}^-$; this leaves a carbocation intermediate, $\text{CH}_3-\text{CH}_2^+$. In step two, the released $\text{Br}^-$ acts as the nucleophile, its lone pair attacking the positively charged carbon to form the new C–Br bond. The overall product is bromoethane, $\text{CH}_3-\text{CH}_2-\text{Br}$, produced by an addition reaction that proceeds through a carbocation intermediate — the same "form a bond, break a bond" bookkeeping as the substitution above, just spread across two steps instead of one.

• Starting a curved arrow from a positive charge or an empty orbital — an arrow's tail must sit on a real electron source (a lone pair, a $\sigma$ bond, or a $\pi$ bond); there are no electrons to push from a cation.
• Drawing only one arrow in a substitution step — the bond forming and the bond breaking happen together, and showing just one leaves the electron count unbalanced (and implies an impossible five-bonded carbon).
• Using a full two-barb arrow for a single-electron, homolytic process, or a single-barb fish-hook arrow for a heterolytic, electron-pair process — the barb count must match how many electrons actually move.
• Reaching for the formal labels SN1, SN2, E1 or E2 at this stage — that kinetic classification, and the rate laws behind it, belong to Class 12; here the goal is correct electron-flow reasoning only.

Ask the tutor: Try sketching the curved-arrow mechanism for a nucleophile attacking a different alkyl halide, or the elimination step that reverses today's addition example, and have the tutor check your arrows.