Reaction types and intermediates

Bond fission — two ways to break a bond. In homolytic fission (homolysis), the shared electron pair splits equally, one electron staying with each fragment. It needs an energy push — heat or light, written $h\nu$ or $\Delta$ — and is favoured when there is no polar solvent to stabilise ions, e.g. $\text{Cl}-\text{Cl} \xrightarrow{h\nu} \text{Cl}^{\bullet} + \text{Cl}^{\bullet}$. In heterolytic fission (heterolysis), the pair goes entirely to one fragment, favoured in polar bonds or polar solvents, producing ions directly: $R_3\text{C}-X \rightarrow R_3\text{C}^+ + X^-$.

The three intermediates fission produces. A free radical is carbon with an unpaired electron and no formal charge — neutral but violently reactive. A carbocation is carbon with only three bonds and a positive charge; it is $sp^2$ hybridised, flat (trigonal planar), with an empty $p$ orbital, and its stability rises with alkyl substitution because alkyl groups feed electron density into the empty orbital inductively and by hyperconjugation:

where: $3^\circ$, $2^\circ$, $1^\circ$ label a carbocation by how many carbon groups are attached to the positively charged carbon. A carbanion is carbon with three bonds, a lone pair and a negative charge — roughly $sp^3$, pyramidal — and its stability runs the opposite way, because alkyl groups only intensify an already electron-rich centre:

Electrophiles and nucleophiles — who attacks whom. An electrophile ("electron-loving") is electron-deficient and seeks electron-rich targets: $\text{H}^+$, $\text{NO}_2^+$, $\text{BF}_3$, and carbocations all qualify. A nucleophile ("nucleus-loving") is electron-rich and donates a lone pair to an electron-poor centre: $\text{OH}^-$, $\text{CN}^-$, $\text{NH}_3$, and carbanions all qualify. Neither label depends on formal charge — $\text{BF}_3$ is a neutral electrophile (boron has an empty orbital) and $\text{NH}_3$ is a neutral nucleophile (nitrogen has a lone pair).

The four reaction categories. Substitution swaps one atom or group for another, e.g. $\text{CH}_4 + \text{Cl}_2 \xrightarrow{h\nu} \text{CH}_3\text{Cl} + \text{HCl}$. Addition combines two reactants across a multiple bond, e.g. $\text{CH}_2=\text{CH}_2 + \text{H}_2 \xrightarrow{\text{catalyst}} \text{CH}_3-\text{CH}_3$. Elimination removes atoms from adjacent carbons to create a multiple bond, e.g. $\text{CH}_3\text{CH}_2\text{Br} \xrightarrow{\text{alc. KOH}} \text{CH}_2=\text{CH}_2 + \text{HBr}$. Rearrangement shuffles atoms within one molecule to give a structural isomer, with no change in molecular formula.

This topic is a classification scheme, not a numerical formula, so the goal is to reason out why each intermediate has the properties it does, rather than to derive an equation. Everything traces back to one idea: how the electrons of the broken bond are distributed decides the charge, the shape, and the reactivity of what remains.

Step 1. Count the electrons carbon keeps. In homolysis carbon keeps exactly one electron from the shared pair — no charge, but an unpaired electron makes the species a highly energetic free radical that reacts by pairing that lone electron with another. In heterolysis carbon either keeps both electrons (three bonds plus a lone pair, so a full negative charge, a carbanion) or keeps neither (three bonds and an empty orbital, so a full positive charge, a carbocation). The formal charge is nothing more than bookkeeping of where the two bonding electrons ended up.

Step 2. Read hybridisation from that electron count. A carbocation has only three bonding pairs around carbon and an empty $p$ orbital, so it adopts the geometry that keeps three groups as far apart as possible: trigonal planar, $sp^2$. A carbanion has three bonding pairs plus a lone pair — four electron domains — so it adopts the roughly tetrahedral, pyramidal $sp^3$ geometry that a lone pair forces, exactly like ammonia.

Step 3. Explain the opposite stability trends from that same electron count. A carbocation is short of electrons, so any neighbouring alkyl group that can push electron density toward it (through the sigma bond, inductively, and by hyperconjugation, donating C–H bond electrons into the empty $p$ orbital) makes it more stable — hence $3^\circ > 2^\circ > 1^\circ$. A carbanion already has surplus electron density, so an alkyl group pushing more electrons at it only worsens the crowding — hence the trend reverses to methyl $> 1^\circ > 2^\circ > 3^\circ$. The same inductive effect stabilises one species and destabilises the other because the two species have opposite electron needs.

Step 4. Connect the intermediate to the reaction category. A carbocation is electron-poor, so it is attacked by an electron-rich nucleophile — the resulting new bond replaces the group that left, a substitution, or it can lose a proton from an adjacent carbon to form a double bond, an elimination. A carbanion is electron-rich, so it behaves as the nucleophile itself, attacking an electrophilic carbon elsewhere. Free radicals propagate by pairing their odd electron with another radical or by abstracting an atom to generate a new radical, which is why radical reactions (like methane's chlorination) proceed as a self-sustaining chain rather than a single step.

Question: Classify the fission and the reaction type, and name the intermediate, for: (a) $\text{Cl}-\text{Cl} \xrightarrow{h\nu} 2\text{Cl}^{\bullet}$; (b) $(\text{CH}_3)_3\text{C}-\text{Br} \rightarrow (\text{CH}_3)_3\text{C}^+ + \text{Br}^-$; and (c) $\text{CH}_3\text{Br} + \text{OH}^- \rightarrow \text{CH}_3\text{OH} + \text{Br}^-$, stating whether $\text{OH}^-$ is acting as an electrophile or a nucleophile.

Step 1. Look at bond (a). Both chlorine atoms come away with one electron each from the original shared pair — the split is equal, so this is homolytic fission, driven by light ($h\nu$), and it produces two chlorine free radicals ($\text{Cl}^{\bullet}$).

Step 2. Look at bond (b). Bromine leaves with both bonding electrons, becoming $\text{Br}^-$, while carbon is left with only three bonds and a positive charge. The split is unequal, so this is heterolytic fission, and because carbon here carries three methyl groups it produces a stabilised tertiary (3°) carbocation, $(\text{CH}_3)_3\text{C}^+$.

Step 3. Classify reaction (c). The bromine atom on $\text{CH}_3\text{Br}$ is replaced by an $-\text{OH}$ group — one group exchanged for another on the same carbon — so this is a substitution reaction. $\text{OH}^-$ carries a lone pair and a negative charge, and it is the species donating its electron pair to the electron-deficient carbon (which bears the leaving bromine), so $\text{OH}^-$ is acting as a nucleophile, and the carbon it attacks is behaving as an electrophilic centre.

Answer: (a) homolytic fission, two chlorine free radicals; (b) heterolytic fission, a tertiary carbocation; (c) substitution, with $\text{OH}^-$ acting as a nucleophile.

Competitive stretch

Ethene reacting with bromine, $\text{CH}_2=\text{CH}_2 + \text{Br}_2 \rightarrow \text{CH}_2\text{Br}-\text{CH}_2\text{Br}$, is an addition reaction — the double bond opens and two new C–Br bonds form without anything leaving, so the degree of unsaturation drops. The reverse-looking reaction, ethanol losing water under conc. $\text{H}_2\text{SO}_4$ and heat, $\text{CH}_3\text{CH}_2\text{OH} \xrightarrow{\Delta} \text{CH}_2=\text{CH}_2 + \text{H}_2\text{O}$, is an elimination — atoms leave from two adjacent carbons ($-\text{OH}$ from one, $-\text{H}$ from the next) to create a double bond, raising the degree of unsaturation. Spotting whether unsaturation goes up or down is the fastest way to tell addition from elimination on sight.

• Reversing the two stability orders — carbocations are stabilised by more alkyl substitution ($3^\circ > 2^\circ > 1^\circ$), carbanions are destabilised by it (methyl $> 1^\circ > 2^\circ > 3^\circ$); they are opposite because the two ions have opposite electron needs.
• Assuming electrophile always means positively charged and nucleophile always means negatively charged — $\text{BF}_3$ is a neutral electrophile and $\text{NH}_3$ is a neutral nucleophile; the test is electron deficiency versus electron richness, not formal charge.
• Confusing addition with elimination as simple opposites without checking unsaturation — addition removes a multiple bond (adds atoms across it), elimination creates one (removes atoms from adjacent carbons).
• Treating "free radical," "carbocation" and "carbanion" as synonyms for "reactive intermediate" — each has its own charge, hybridisation and formation pathway (homolysis for radicals, heterolysis for the two ions), and mixing them up misidentifies the whole mechanism.

Ask the tutor: Want to try classifying a few more bond-fission and reaction examples together to lock this in before you move to full mechanisms?