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

A double-barbed (full) arrowhead means an electron pair (2 electrons) moves. A fishhook (half arrowhead) means one electron moves. Here one electron moves ⇒ fishhook arrow. Rule of thumb: Full arrow = Full pair; Half arrow = Half the pair (one electron).

Both electrons to one atom = uneven split = heterolysis. One atom keeps the pair → becomes negative (an anion); the other loses its share → becomes positive (a cation). Products: two ions (a cation + an anion). Memory hook: HETERO = different → ions.

An arrow is moving electrons, so its tail must sit on something that actually holds electrons: a lone pair or a bond (σ or π). You can never start an arrow from an empty spot or from a bare positive charge — there is nothing there to move. This connects to Nucleophiles and Electrophiles: tails start at electron-rich sites (nucleophiles).

(a) The more electronegative atom grabs the pair (see Inductive effect and electronegativity). Cl is more electronegative ⇒ Cl keeps both electrons. (b) Arrow: tail on the H–Cl bond, head onto Cl. $H-Cl⟶H^{+}+:C{l}^{-}$ Cl now has 4 lone pairs (8 electrons) and charge $-1$; H lost its share and is bare ⇒ $+1$. (c) Reactant neutral $=0$. Products $=(+1)+(-1)=0$. ✓

A neutral carbon "owns" 4 bonding electrons (one from each of its 4 bonds). When the C–Br bond breaks heterolytically, Br walks off with that pair, so carbon is left sharing only 3 bonds = 6 valence electrons (short of the 8-electron octet). Missing one electron compared to neutral ⇒ charge $=+1$. This electron-poor carbon is a carbocation (see Carbocations — stability and structure) and behaves as an electrophile.

(a) Cl and Cl have identical electronegativity ⇒ the bond is non-polar; neither side "wants" both electrons, and the gas phase can't stabilise ions. So it splits evenly ⇒ homolysis, giving two $C{l}^{\bullet }$ radicals (two fishhooks). (b) C–Br is polar (Br more electronegative) and water stabilises ions by solvating them ⇒ heterolysis, giving $C^{+}$ and $:B{r}^{-}$ (one full arrow). Principle: polar bond + ion-stabilising surroundings → heterolysis; non-polar bond + energy, no stabiliser → homolysis.

Electrons flow rich → poor. The hydroxide's lone pair is electron-rich; the carbocation is electron-poor ($+$). Arrow: tail on a lone pair of $O{H}^{-}$, head onto the $C^{+}$. The arrowhead lands between O and C — and that new shared pair is the new C–O bond. Bond formation is just an arrow whose head stops between two atoms. Charge check: $(+1)+(-1)=0$ ⇒ neutral product. ✓

A full arrow moves the whole pair to one side — that would give $C{l}^{+}$ and $:C{l}^{-}$ (heterolysis), which is wrong for a non-polar bond. To split evenly, each Cl must take one electron. That means two fishhook arrows, one electron peeling off to each atom: $Cl\stackrel{\curvearrowright }{\underset{\curvearrowleft }{-}}Cl⟶C{l}^{\bullet }+C{l}^{\bullet }$ Each product is neutral with 7 valence electrons and one unpaired electron — a radical (see Free radical substitution (halogenation of alkanes)).

Arrow 1 (bond forms): tail on a lone pair of $H{O}^{-}$, head to the carbon of $C{H}_{3}Br$ — makes the new O–C bond. Arrow 2 (bond breaks): tail on the C–Br bond, head onto Br — heterolysis, Br leaves as $:B{r}^{-}$. Carbon never exceeds its octet: as the new O–C pair arrives, the old C–Br pair leaves simultaneously, keeping carbon at 8 electrons. Charge check: reactants $(-1)+0=-1$; products $0+(-1)=-1$. ✓

Arrow 1: tail on the electron-rich C=C π bond, head to the H of H–Br. The π electrons are the nucleophile; the H is ${\delta }^{+}$ because Br pulls electron density toward itself (Inductive effect and electronegativity). Arrow 2: tail on the H–Br bond, head onto Br (heterolysis). H must release its bonding pair to Br so that H stays neutral while it bonds to carbon. Result: a new C–H bond, one carbon now has only 3 bonds ⇒ $C{H}_3-C{H}_2^{+}$ (carbocation), plus $:B{r}^{-}$. Charge check: neutral + neutral $=0$; products $(+1)+(-1)=0$. ✓

Add H to the terminal CH${}_2$; the positive charge then sits on the middle carbon, giving a secondary ($2^{\circ }$) carbocation $C{H}_3-\stackrel{+}{C}H-C{H}_3$. Why: more-substituted carbocations are more stable — neighbouring alkyl groups donate electron density (+I effect) toward the electron-poor centre, spreading the positive charge (Carbocations — stability and structure). The reaction funnels through the lower-energy, more stable cation. The competing route makes a primary ($1^{\circ }$) cation (no alkyl stabilisation), which is higher in energy and disfavoured.

Initiation (homolysis, needs light): $Cl-Cl\stackrel{h\nu }{\to }2C{l}^{\bullet }$ Cl–Cl is non-polar ⇒ splits evenly ⇒ two fishhooks, one to each Cl. This creates the radicals.

Propagation (radicals consumed and regenerated — a chain): $C{l}^{\bullet }+C{H}_4⟶HCl+{}^{\bullet }C{H}_3$ ${}^{\bullet }C{H}_3+C{l}_2⟶C{H}_{3}Cl+C{l}^{\bullet }$ Each step needs fishhooks because a single unpaired electron is doing the work. The regenerated $C{l}^{\bullet }$ keeps the chain running.

Termination (two radicals meet, single electrons pair into a bond): $C{l}^{\bullet }+{}^{\bullet }C{H}_3⟶C{H}_{3}Cl{}^{\bullet }C{H}_3+{}^{\bullet }C{H}_3⟶C_2{H}_6$ Two fishhooks (one from each radical) meet head-to-head to form one shared pair = a new bond. This removes radicals, ending the chain.

Every step here is homolytic (radical) chemistry ⇒ all fishhooks, never full arrows. See Free radical substitution (halogenation of alkanes) and Bond dissociation energy (weaker Cl–Cl bond breaks first under UV).

Error 1 — wrong direction of the pair. Br is more electronegative, so the pair must go to Br, not away from it. The correct products are $(C{H}_3{)}_3{C}^{+}$ (carbocation) and $:B{r}^{-}$, not $C^{-}$ and $B{r}^{+}$. Error 2 — arrow drawn from the atom. The tail should sit on the C–Br bond (the electrons), head onto Br. Charge-check on the fixed version: $0\to (+1)+(-1)=0$. ✓ (The student's version also happens to sum to $0$, which is exactly why a charge-check alone isn't enough — the electronegativity rule catches this error.)