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

Before any example, two pieces of vocabulary we will re-use everywhere.

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The scenario matrix

Every bond-breaking / bond-forming event this topic can throw at you falls into one of these cells. Read the table and the map-diagram beneath it: the diagram shows the two questions you ask (Is the bond polar? Are we breaking or forming?) that route you to each cell.

Cell	Case class	What decides it	Example
A	Heterolysis → carbocation (C keeps no electrons)	C bonded to a more electronegative atom	Ex 1
B	Heterolysis → carbanion (C keeps both)	C bonded to a less electronegative (electropositive) atom	Ex 2
C	Bond formation — lone-pair nucleophile → cation	electron-rich meets electron-poor	Ex 3
D	Homolysis (bond breaking, even) — non-polar bond, external energy	equal electronegativity + $h\nu$/heat	Ex 4
E	π-bond as nucleophile + concerted heterolysis	polar reagent (${\delta }^{+}/{\delta }^{-}$) meets alkene	Ex 5
F	Zero / degenerate case — perfectly non-polar bond that refuses to heterolyse (only homolyses under force)	$\Delta$(electronegativity) $=0$, no ion stabiliser	Ex 6
G	Limiting / borderline — which way does a bond of intermediate polarity split?	tug-of-war decided by ion stability + solvent	Ex 7
H	Real-world word problem	translate English → arrows	Ex 8
I	Exam twist — spot the illegal arrow (octet violation)	conservation & octet rules	Ex 9
J	Homolytic bond formation — two radicals couple	two fishhooks meet, one pair forms	Ex 10

The decision map (which two questions route you to which cell). Note that a non-polar bond has only one option left — heterolysis is off the table, so both the "will-it-homolyse" cell D and the "refuses-to-heterolyse" cell F land on the same homolysis branch; F is simply the extreme where heterolysis is impossible:

Prerequisites you'll lean on: Inductive effect and electronegativity, Nucleophiles and Electrophiles, Carbocations — stability and structure, Bond dissociation energy, Resonance and arrow pushing.

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Worked examples

Ex 1 — Cell A: heterolysis giving a carbocation

1. Compare electronegativity. Cl $\approx 3.0$, C $\approx 2.5$. Chlorine pulls harder. Why this step? The greedier atom (higher electronegativity, see Inductive effect and electronegativity) grabs the shared pair — this is the whole reason the split is heterolytic and not even.
2. Draw one double-barbed (full) arrow — the pair-mover from the legend above: tail on the C–Cl bond, head onto Cl. Why this step? A full arrow = a full pair moving; both electrons go to Cl. (See figure s01: the yellow full arrow leaves the bond and curls onto the red Cl.)
3. Compute formal charge on Cl. Free Cl: $V=7$. Now: 3 original lone pairs + the 2 it just captured $=8$ lone-pair electrons, $0$ bonds. ${\text{FC}}_{Cl}=7-8-0=-1.$ Why this step? We must confirm Cl really became a full anion — the whole claim of heterolysis rests on it walking off with the complete pair.
4. Compute formal charge on central C. Free C: $V=4$. It now has $0$ lone pairs and 3 remaining C–C bonds $=6$ bonding electrons. ${\text{FC}}_C=4-0-\frac{1}{2}(6)=+1.$ Why this step? To prove carbon is genuinely electron-deficient (only 6 electrons), which is what makes it a carbocation and an electrophile.

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Ex 2 — Cell B: heterolysis giving a carbanion

1. Compare electronegativity. C $\approx 2.5$, Mg $\approx 1.3$. Carbon is the greedier one now. Why this step? Mg is a metal, electropositive. So the shared pair goes to carbon, the reverse of Ex 1.
2. Full arrow: tail on the C–Mg bond, head onto carbon. Why this step? Same rule, opposite direction, because the electronegativity order flipped.
3. Formal charge on C. $V=4$; carbon now has one lone pair (the captured 2 electrons) $=2$ lone-pair electrons, and 3 C–H bonds $=6$ bonding electrons. ${\text{FC}}_C=4-2-\frac{1}{2}(6)=-1.$ Why this step? To prove carbon ends up electron-rich (a full extra pair), the defining feature of a carbanion.
4. Formal charge on the Mg end. Take the two atoms of the leaving fragment together. Why this step? Mg didn't leave alone — it kept its $Br$ ligand, so we must charge-count the whole $\text{MgBr}$ group, not Mg in isolation. Bridging the numbers slowly: the Mg atom lost its share of the C–Mg pair, so on its own ${\text{FC}}_{Mg}=2-0-0=+2$. But hanging off Mg is one bromide ligand, $B{r}^{-}$, which carries a charge of $-1$. Adding them: $\underset{\text{Mg}}{\underbrace{(+2)}}+\underset{{\text{Br}}^{-}\text{ ligand}}{\underbrace{(-1)}}=+1\text{for the whole }{\text{MgBr}}^{+}\text{ fragment.}$

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Ex 3 — Cell C: bond formation, lone pair attacks a cation

1. Identify source and sink. $O{H}^{-}$ has lone pairs (rich); $C^{+}$ is empty (poor). Why this step? Golden rule — electrons flow rich → poor (see Nucleophiles and Electrophiles).
2. Full arrow: tail on an O lone pair, head between O and C. Why this step? The arrowhead landing between the two atoms is the new C–O shared pair — that arrowhead literally draws the new bond.

Figure s02 (below): the tert-butyl carbon carries a blue $+$; oxygen (red) carries a yellow lone pair. A green full arrow curls from that lone pair into the gap between O and C — that curl is the new C–O bond being drawn. Read left-to-right: empty carbon, generous oxygen, electrons crossing to heal the deficit.

3. Formal charge on O in the product. $V=6$; after bonding O has 2 lone pairs $=4$ lone-pair electrons and 2 bonds (to C and to H) $=4$ bonding electrons. ${\text{FC}}_O=6-4-\frac{1}{2}(4)=0.$ Why this step? Oxygen gave away a slice of a lone pair into a shared bond — we check it lands back at neutral, not still $-1$.
4. Formal charge on C. Back to 4 bonds → ${\text{FC}}_C=4-0-\frac{1}{2}(8)=0$. Why this step? To confirm the electron-poor carbon is fully "healed" (8 electrons, neutral).

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Ex 4 — Cell D: homolysis of a non-polar bond

1. Check polarity. Both atoms are Cl → identical electronegativity → $\Delta =0$. Why this step? When neither atom is greedier, an uneven split would dump charge nobody can stabilise (gas phase, no solvent). So the bond splits evenly.
2. Draw TWO fishhook arrows, each carrying one electron of the bond to one Cl. Why this step? One electron per arrowhead → each Cl keeps exactly one → homolysis (see Free radical substitution (halogenation of alkanes)).
3. Formal charge on each Cl. $V=7$; each now has 3 lone pairs (6 e⁻) + 1 unpaired electron $=7$ non-bonding electrons, $0$ bonds. ${\text{FC}}_{Cl}=7-7-0=0.$ Why this step? To prove neither fragment is an ion — the whole point of an even split is that no charge is created.

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Ex 5 — Cell E: π-bond nucleophile + concerted heterolysis

1. Spot the electron-rich region. The C=C π bond sits above/below the plane, loosely held → a nucleophile. Why this step? π electrons are the most available electrons in the molecule (see Electrophilic addition to alkenes).
2. Spot the electrophilic atom in H–Br. Br is electronegative → H carries ${\delta }^{+}$ (recall: ${\delta }^{+}$ = "slightly positive," a fraction of a charge, from the definition above). Why this step? The π pair goes for the partially positive H, not Br.
3. Arrow 1 (full): tail on the C=C π bond, head to H. Arrow 2 (full, concerted): tail on the H–Br bond, head onto Br. Why this step? As H forms a new bond to carbon, its old bond must break heterolytically so H stays neutral — one push, one release, in the same instant.

Figure s03 (below): the alkene's blue π bond sits under the two carbons; H–Br on the right shows H tagged ${\delta }^{+}$ (yellow) and Br tagged ${\delta }^{-}$ (red). A green full arrow leaps from the π cloud to the ${\delta }^{+}$ H (forming the new C–H bond); a yellow full arrow curls from the H–Br bond onto Br (breaking it heterolytically). Two arrows, one concerted motion.

4. Formal charge on the carbon that GAINS the H. This carbon started with 2 C–H bonds and a share of the π bond; it now has 3 C–H bonds + 1 C–C bond = 4 bonds, no lone pair. ${\text{FC}}_{C(\text{got H})}=4-0-\frac{1}{2}(8)=0.$ Why this step? The bond-forming half of the mechanism must be checked too — we prove the carbon that accepted H is neutral with a full octet, not accidentally charged.
5. Formal charge on the OTHER carbon (the one left short). It kept only 3 bonds (2 C–H + 1 C–C) and no lone pair. ${\text{FC}}_{C(\text{cation})}=4-0-\frac{1}{2}(6)=+1.$ Why this step? To locate exactly where the new positive charge lands now that the partial $\delta$ charges have resolved into real ions.
6. Formal charge on Br (leaves with its pair). ${\text{FC}}_{Br}=7-8-0=-1.$ Why this step? To confirm the heterolysis half produced a genuine bromide anion.

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Ex 6 — Cell F: the zero / degenerate case

1. Compute polarity. Both atoms are carbon → $\Delta =2.5-2.5=0$. Perfectly non-polar. Why this step? Zero difference means neither carbon is greedier — the two heterolysis outcomes ("left C keeps the pair" vs "right C keeps the pair") are degenerate, so neither is favoured, and each would put a raw $+$ and $-$ with nothing to stabilise it. Energetically hopeless → heterolysis is off the table entirely (this is exactly what makes it Cell F, not Cell D's "which way").
2. Conclusion — no heterolysis. If you do supply enough energy (very high heat, gas phase), it snaps homolytically into two ${\text{CH}}_3^{\bullet }$ methyl radicals. Why this step? This is the mirror image of Ex 4: identical atoms → even split is the only legal breaking option.
3. Formal charge on each methyl carbon (radical). $V=4$; 3 C–H bonds (6 e⁻) + 1 unpaired electron. ${\text{FC}}_C=4-1-\frac{1}{2}(6)=0.$ Why this step? To confirm homolysis makes two neutral radicals, not ions — consistent with there being no charge stabiliser.

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Ex 7 — Cell G: the limiting / borderline case

1. Name the tug-of-war. Two forces decide heterolysis: (i) bond polarity, (ii) how well the leaving ion is stabilised (large soft ions + weak bond). Why this step? Polarity alone is a trap — it's only half the story.
2. Compare bond strengths. C–I bond dissociation energy $\approx 240\text{kJ/mol}$; C–F $\approx 485\text{kJ/mol}$ (see Bond dissociation energy). Why this step? The weaker C–I bond snaps far more readily; the huge, polarisable $I^{-}$ is a stable, "happy" anion, whereas tiny $F^{-}$ holds its charge tightly and clings to carbon.
3. Take the ratio — and see why a ratio settles it. $\frac{E(\text{C–F})}{E(\text{C–I})}=\frac{485}{240}\approx \mathrm{2.02.}$ Why this step? Reaction rate depends exponentially on the energy barrier: a bond that costs about twice the energy to break is not "twice as slow" but slower by an enormous factor. Putting the two energies as a ratio strips away units and shows at a glance that the C–F barrier is in a completely different league — decisive enough that the small polarity advantage of C–F is irrelevant. The mechanism follows the cheaper bond every time.

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Ex 8 — Cell H: real-world word problem

1. Find the weakest, most non-polar bond. Cl–Cl (BDE $\approx 242\text{kJ/mol}$) is weaker and less polar than C–H (BDE $\approx 413\text{kJ/mol}$). Why this step? UV energy goes to the bond that is cheapest to break and symmetric — Cl–Cl (see Free radical substitution (halogenation of alkanes)).
2. Choose cleavage type. Non-polar, gas phase, light-driven → homolysis. Why this step? Same reasoning as Ex 4.
3. Count arrows. Homolysis needs two fishhooks → two ${\text{Cl}}^{\bullet }$ radicals, the chain initiation step.
4. Numeric sanity — is UV enough? We use the Planck relation $E=\frac{hc}{\lambda }$. This is just the $E=h\nu$ from the definition above rewritten: a wave's frequency $\nu$ equals $c/\lambda$ (speed of light over wavelength), so $E=h\nu =hc/\lambda$. Here $h$ is Planck's constant $=6.626\times 10^{-34}\text{J⋅s}$, $c$ is the speed of light $=3.0\times 10^8\text{m/s}$, and $\lambda$ is the light's wavelength in metres. For $\lambda =300\text{nm}=300\times 10^{-9}\text{m}$: $E=\frac{hc}{\lambda }=\frac{(6.626\times 10^{-34})(3.0\times 10^8)}{300\times 10^{-9}}\approx 6.6\times 10^{-19}\text{J/photon}.$ Why this step? One photon breaks one molecule, so we must check a single photon carries at least the bond's dissociation energy. Multiplying by Avogadro's number $N_A=6.022\times 10^{23}$ converts per photon to per mole: $E_{\text{mol}}=6.6\times 10^{-19}\times 6.022\times 10^{23}\text{J/mol}\approx 399\text{kJ/mol}.$

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Ex 9 — Cell I: the exam twist (spot the illegal arrow)

1. Count carbon's bonds after the illegal arrow. Methane already has 4 C–H bonds (8 electrons). Adding a C–O bond makes 5 bonds = 10 electrons around carbon. Why this step? Period-2 atoms (C, N, O, F) cannot exceed the octet — 8 electrons is the hard ceiling.
2. Octet check. $\text{electrons around C}=2\times 5=10>8.\text{ILLEGAL.}$ Why this step? A quick electron count is the fastest way to catch an octet violation before you commit to a wrong product.
3. The fix. For the nucleophile to bond to carbon, an existing group must leave with its pair (a leaving group) so carbon never exceeds 8. Methane has no good leaving group → the reaction simply does not go. Why this step? Every new bond to a "full" carbon must be paired with a bond broken — conservation of the octet.

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Ex 10 — Cell J: homolytic bond formation (radical coupling)

1. Locate the two single electrons. Each radical carries one unpaired electron on carbon. Why this step? You can only form a bond from electrons that exist; each radical brings exactly one to the new shared pair.
2. Draw TWO fishhook arrows, each pointing from a radical's lone electron into the gap between the two carbons. Why this step? This is the reverse of Ex 4: one electron from each side meets in the middle to make the new shared pair. Two half-arrows build one full pair (see Free radical substitution (halogenation of alkanes) — this is a chain-termination step).
3. Formal charge on each carbon in ethane. $V=4$; now 4 bonds (3 C–H + 1 C–C) $=8$ bonding electrons, $0$ lone pairs, $0$ unpaired. ${\text{FC}}_C=4-0-\frac{1}{2}(8)=0.$ Why this step? To prove both carbons reached a full, neutral octet — the unpaired electrons are gone, so we're no longer dealing with radicals.

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Active recall

Recall Test yourself on the matrix

In C–Mg, which atom keeps the electron pair? ::: Carbon (it is more electronegative than Mg → carbanion). Why won't ethane's C–C bond heterolyse in water? ::: Zero electronegativity difference — the two outcomes are degenerate, so no atom is greedier; it only homolyses under strong energy. C–I breaks more easily than C–F despite being less polar. Why? ::: The C–I bond is far weaker (240 vs 485 kJ/mol) and $I^{-}$ is a large, well-stabilised leaving ion. How many arrows for a homolysis, and for a radical coupling? ::: Two fishhook arrows each — breaking splits a pair, coupling merges two singles. What do ${\delta }^{+}$ and ${\delta }^{-}$ mean? ::: Partial (fractional) positive and negative charge from unequal electron sharing — not full ions. What does the symbol $h\nu$ stand for? ::: One photon of light; energy $E=h\nu$ where $h$ is Planck's constant and $\nu$ is the light's frequency. Why can't hydroxide add a 5th bond to methane's carbon? ::: It would give carbon 10 electrons, violating the octet — a bond must break for a bond to form.