Homolytic and Heterolytic Cleavage

Introduction

Breaking a covalent bond does not specify who receives its two electrons. If each fragment takes one, the process is homolytic and radicals result. If one fragment takes both, it is heterolytic and ions result. These alternatives can lead to very different mechanisms even when the same initial bond is broken.

Core explanation

Consider Cl–Cl. In a homolytic representation, Cl₂ → 2Cl·. Each chlorine receives one electron from the bond, producing two neutral chlorine radicals with unpaired electrons. Draw two single-headed fishhook arrows from the Cl–Cl bond, one toward each atom. Light can initiate chlorine-radical chemistry under appropriate conditions, but the mere presence of light does not mean every bond cleaves homolytically or that the resulting radicals are long-lived.

For H–Cl heterolysis in a suitable polar environment, both bonding electrons go to chlorine, producing H⁺ and Cl⁻ in the idealised equation. Draw one full-headed curved arrow from the H–Cl bond to Cl. In water, a free bare H⁺ is not the best physical picture; proton transfer to water produces hydronium and more extensive hydration. The simple heterolysis notation is useful electron bookkeeping but should not be confused with an isolated gas-phase proton freely roaming in solution.

For an alkyl halide R–Br, heterolytic cleavage of C–Br can give R⁺ and Br⁻ if a pathway and environment can stabilise the resulting ions. A tertiary carbocation may be better stabilised than an analogous primary one in a simple alkyl series, but solvent and leaving-group effects determine whether the ionisation pathway is feasible. A primary haloalkane may instead undergo concerted nucleophilic substitution without a free carbocation intermediate. Writing R⁺ + Br⁻ as a possibility does not prove it is the actual mechanism.

Homolysis of a C–H bond gives a carbon radical and H· in an idealised cleavage. Its bond dissociation enthalpy is defined for gas-phase homolysis under specified conditions. Heterolysis of that same C–H bond would produce ionic fragments with different energies and solvent requirements. Thus a quoted “bond strength” must name the cleavage process; homolytic dissociation data cannot automatically predict how readily a bond ionises in water.

Electron count is the reliable distinction. In homolysis, each atom gains one electron from the former bonding pair. In heterolysis, one atom gains both and the other none. The total electrons and total charge are conserved in both cases. For a bond between different atoms, which fragment gets the pair in heterolysis is often guided by electronegativity and product stability, but do not apply an electronegativity slogan without considering the actual reaction and medium.

Radicals can be neutral or charged in more advanced chemistry; the elementary definition is an unpaired electron, not “a neutral atom.” Conversely, an ion need not contain an unpaired electron. A carbocation is typically electron deficient and positively charged; a carbanion has a lone pair and negative charge; a carbon radical has one unpaired electron. These intermediates must be drawn with their electronic features explicitly.

Step-by-step reasoning

1. Identify the bond being cleaved and its electron pair. 2. Decide whether electrons split one each or both go to one fragment. 3. Use two fishhooks for homolysis or one pair arrow for heterolysis. 4. Draw every fragment, unpaired electron and formal charge. 5. Check total atoms, electrons and net charge before and after.

Visual explanation

Split one A–B bond into two rows. Top: A· + ·B with two single-headed arrows. Bottom: A⁺ + B⁻ with one full-headed arrow to B. Write “one each” and “both to B” beside the rows.

Real-world analogy

Two partners can split a pair of tokens one each, or one partner can keep both. The bookkeeping of the tokens resembles homolytic and heterolytic bond-electron allocation, although real electrons obey quantum mechanics.

Real-world example

Photochemical chlorination of methane proceeds through radical-chain steps initiated by Cl–Cl homolysis under suitable light. Acid dissociation in water instead involves heterolytic proton-transfer accounting and solvated ionic products.

Why?

Why do two fishhook arrows represent Cl₂ homolysis? The bonding pair has two electrons, and each chlorine radical receives one. A full-headed arrow to one chlorine would instead show both electrons moving together.

Common misconception

“A C–Br bond breaking must first form a carbocation.” Concerted substitution can break C–Br as a new bond forms, with no discrete free carbocation. Cleavage notation must match the proposed mechanism and conditions.

Worked example

For CH₃–Br, compare hypothetical cleavage modes. Homolysis gives CH₃· and Br·, one bond electron to each. Heterolysis with both electrons to Br gives CH₃⁺ and Br⁻. Both equations conserve atoms and overall zero charge, but the products have different electron configurations and energy requirements. A real reaction pathway cannot be chosen from the formula alone.

Quick check

1. Which cleavage of Cl–Cl produces two chlorine radicals with one unpaired electron each? Answer: Homolytic cleavage.

Exam focus

Use the correct arrowhead and write charges or radical dots on products. Separate hypothetical bond cleavage from evidence that a particular reaction actually follows that pathway.

Advanced insight

Solvent polarity can stabilise separated ions and thereby change the energetics of heterolytic processes. Radical stability and bond dissociation energy follow different thermochemical comparisons, so gas-phase and solution trends should not be casually merged.

Summary

Homolysis splits a bonding pair one electron per fragment and forms radicals. Heterolysis sends both electrons to one fragment and forms ions. Arrow notation, product charge and reaction conditions distinguish the pathways.

Practice questions

1. What arrowheads represent homolysis? Answer: Two single-headed fishhooks, one for each electron. 2. What products result from idealised H–Cl heterolysis with both electrons to Cl? Answer: H⁺ and Cl⁻, with the proton solvated or transferred in water. 3. Does every radical have zero charge? Answer: No. A radical is defined by an unpaired electron; charged radicals can exist. 4. Why does C–Br cleavage notation not by itself prove an S N1 mechanism? Answer: A concerted pathway may break C–Br as another bond forms without a free carbocation.