Reaction mechanisms — curved-arrow notation, bond formation - breaking (heterolysis vs homolysis): Visual walkthrough

We derive the parent's central result — heterolysis vs homolysis, tracked by curved arrows — one picture at a time. Prerequisites we lean on: Nucleophiles and Electrophiles, Inductive effect and electronegativity, Bond dissociation energy, and Carbocations — stability and structure. This is the visual companion to the parent note.

Step 1 — Draw a bond as two dots between two letters

WHAT. We start with the most basic object in chemistry: a covalent bond. Two atoms — call them $A$ and $B$ — are stuck together because they share two electrons. We draw each electron as a dot. So the bond is literally:

$A ∣ \cdot\cdot B = A : B$

Read the symbols:

$A$ , $B$ — the two atoms (think two people).
The pair of dots $:$ sitting between them — that is the bond, the two shared electrons.

WHY draw dots at all? Because the whole story is "where do these two dots end up?" If we hide the electrons and only draw a line " $A - B$ ", we can't track them. The dots are the thing that moves. Everything else on this page is bookkeeping for these two dots.

PICTURE. Two black circles ( $A$ and $B$ ), two red electron dots exactly halfway between them. Red = the object we will follow the whole page.

Figure — Reaction mechanisms — curved-arrow notation, bond formation - breaking (heterolysis vs homolysis)

Step 2 — When the bond breaks, there are ONLY two destinations

WHAT. Snap the bond. The two dots are now homeless. Ask: where can two dots go? There are exactly two ways to split a pair:

Both dots go to the same atom (2 for one, 0 for the other).
One dot goes to each atom (1 for each).

That's it. There is no third option — you cannot split "two things" any other way. This is why the parent note says there are only two kinds of bond breaking. It's not chemistry yet; it's just counting.

WHY only two? Two electrons is an even number sliced between two atoms. The only integer splits of $2$ are $2 + 0$ and $1 + 1$ . (A $\frac{1}{2} + \frac{3}{2}$ split is impossible — electrons don't cut in half.) So counting alone gives us the two named processes.

PICTURE. A fork in the road: the bonded pair at the top, then two branches. Left branch (red) = both dots slide to one atom. Right branch = the dots separate, one each.

Step 3 — The "both dots to one atom" branch → heterolysis makes ions

WHAT. Follow the left branch. Say both dots go to $B$ . Then:

$A : B ⟶ A^{+} + : B^{-}$

Term by term:

$A^{+}$ — atom $A$ lost its share of the pair. It had 1 of the 2 "owed" to it; now it has 0. Losing an electron's worth of negative charge leaves it $+ 1$ : a cation.
$: B^{-}$ — atom $B$ kept both dots (that's the $:$ drawn on it now). It gained an extra electron's worth of negative charge: an anion, charge $- 1$ .

WHY does this make charges? Because the electrons split unevenly — one atom is now electron-rich, the other electron-poor. Uneven electrons = unequal charge = ions. The name says it: hetero = "different", the two products are different in charge.

The single curved arrow. We record "both dots move to $B$ " with one double-barbed arrow: tail on the bond (the source of the pair), head pointing at $B$ (the destination). One arrow = one pair moving.

PICTURE. The red pair drawn already sitting on $B$ ; a red double-barbed (full head) arrow curving from where the bond was onto $B$ . $A$ is now marked $+$ , $B$ marked $-$ .

Step 4 — The "one dot each" branch → homolysis makes radicals

WHAT. Now the right branch. Each atom keeps one dot:

$A : B ⟶ A^{∙} + B^{∙}$

Term by term:

$A^{∙}$ — atom $A$ keeps exactly the one electron that was "its half" of the shared pair. It neither gained nor lost overall → neutral. The $∙$ is that lonely unpaired electron.
$B^{∙}$ — same story for $B$ : one electron, neutral, one unpaired.

WHY no charges here? Because the split was even — each side keeps precisely what it "brought." Even electrons = no net gain or loss = no charge. But each atom now has an odd number of electrons (one is unpaired), and a neutral atom with an unpaired electron is a free radical. Homo = "same/even".

Two fishhook arrows. One electron moving needs a half-headed (fishhook) arrow. Two electrons are moving (one to each atom), so we draw two fishhooks — one dot to $A$ , one dot to $B$ .

PICTURE. The red pair splitting: two red fishhook arrows fanning outward, one dot landing on each atom. Both atoms labelled neutral with a single $∙$ .

Step 5 — WHICH branch does nature take? Electronegativity decides

WHAT. Counting gave two options; chemistry picks between them. The tiebreaker is electronegativity — how strongly an atom pulls shared electrons toward itself (see Inductive effect and electronegativity).

If $A$ and $B$ are very different in electronegativity (a polar bond, big $Δ$ EN), the greedier atom grabs both dots → heterolysis (Step 3). A polar solvent that can hug the resulting ions helps too.
If $A$ and $B$ are similar / identical in electronegativity (a non-polar bond, $Δ$ EN $\approx 0$ ), neither wants both dots, so they split evenly → homolysis (Step 4), usually kick-started by UV light or heat.

WHY. Nature takes the path that costs the least energy. Making ions is only worth it if something stabilizes the charge (an electronegative atom eager to hold the pair, plus a friendly solvent). With no such help and a symmetric bond, forcing charges apart is expensive — splitting evenly into radicals is cheaper.

PICTURE. A horizontal "electronegativity gap" axis. Small gap (left, red) → homolysis road. Big gap (right, red) → heterolysis road.

Step 6 — The arrowhead that forms a bond is the same act, reversed

WHAT. So far arrows broke bonds. Turn the idea around: an arrowhead that lands between two atoms creates a bond there. Take a lone pair on a nucleophile $Nu$ attacking an electron-poor carbon (a carbocation from Step 3):

$Nu :^{-} + C^{+} ⟶ Nu-C$

Tail on the lone pair of $Nu$ (that's the electron source — a nucleophile, see Nucleophiles and Electrophiles).
Head lands between $Nu$ and $C$ → those two electrons become the new shared pair = the new bond.

WHY. Bond-breaking and bond-making are literally the two ends of one arrow. Breaking: pair leaves the space between atoms. Making: pair arrives in the space between atoms. Same picture, watched from the other end. That's why one grammar (curved arrows) covers all of organic mechanism.

PICTURE. Red lone pair on $Nu^{-}$ curving with a double-barbed arrow into the gap next to $C^{+}$ ; the new bond drawn as the freshly-arrived red pair.

The one-picture summary

Everything on one canvas: the shared pair at the center, the fork, and the two fates — ions vs radicals — with the arrow type that produces each, and electronegativity as the switch.

Recall Feynman retelling — the whole walkthrough in plain words

Two friends hold hands using two shared candies (the bond = two dots). Something pulls them apart. Two dots can only be split two ways. Way 1 — one friend grabs both candies (heterolysis). Happens when one friend is greedy (more electronegative). That friend walks off rich = negative (an anion); the empty-handed one is positive (a cation). We record this with one full-headed arrow — a whole pair moving to the greedy friend. Way 2 — each takes one candy (homolysis). Happens when the friends are equal (non-polar bond) and the Sun (UV) pushes them apart. Each leaves with one candy = neutral but lonely = a radical. We record this with two fishhook arrows — one candy each. Turn any arrow around and instead of breaking a bond it makes one: an electron-rich thing (a nucleophile's lone pair) reaches into an electron-poor spot (a cation) and the arrowhead landing between them is the new bond. Count the charges before and after — they always match. That's the entire grammar of organic reactions.

Related deeper dives: Free radical substitution (halogenation of alkanes) (homolysis in action), Electrophilic addition to alkenes and Resonance and arrow pushing (heterolysis + bond-forming arrows).