Homolytic and Heterolytic Bond Cleavage

Introduction

Breaking a covalent bond does not always produce the same kinds of fragments. In homolysis, each bonded atom receives one electron and radical species result. In heterolysis, both bonding electrons go to one fragment, creating ions or changing charge distribution. A mechanism's conditions and arrow notation must match the cleavage type. Confusing the two leads to the wrong intermediates and usually the wrong predicted products.

Core explanation

For a bond A–B containing an electron pair, homolytic cleavage can be represented A–B → A· + B· when one electron goes to each atom. The dot represents an unpaired electron. A chlorine molecule under suitable light or heat can cleave homolytically into two chlorine radicals, initiating alkane radical chlorination. Use two single-headed fishhook arrows from the Cl–Cl bond, one to each chlorine. One full-headed arrow would send both electrons to one atom and depict a different process.

Heterolytic cleavage places the bond electron pair on one fragment. For tert-butyl bromide in a polar ionising context, an arrow from C–Br bond to Br gives a tertiary carbocation and Br⁻. The bromine takes both bonding electrons. The relative stability of the ionic fragments, solvent solvation, and bond polarisation affect feasibility. It would be a mistake to draw a free tert-butyl radical and bromine radical after using a full-headed arrow toward bromine.

The same elemental bond can cleave in different ways under different conditions. Photolysis and radical initiators can promote homolytic pathways, while polar solvents and good leaving groups can facilitate heterolytic pathways. However, “light means every bond homolyses” and “polar solvent means every bond heterolyses” are oversimplifications. The substrate and reaction network determine which bond is activated and whether electron transfer or concerted substitution happens instead of a free cleavage step.

Bond dissociation energy commonly refers to homolytic bond breaking in a specified phase, while heterolytic cleavage energy includes formation and stabilisation of charged fragments. Gas-phase bond strengths cannot alone predict an ionic reaction rate in solution because solvent can greatly stabilise ions and transition states. A polarised C–Br bond may favour electron pair departure to bromine, but whether a discrete carbocation forms depends on carbon stabilisation and solvent, not polarity alone.

Some reactions combine radical and ionic events. A radical can transfer an electron to form an ion, or a radical intermediate can react with a charged species. Such changes require explicit electron bookkeeping. Mechanistic labels should be assigned to each elementary step rather than imposed on the whole reaction without checking. This matters in photoredox and electrochemical organic chemistry, where single-electron transfer can lead to radical ions.

Step-by-step reasoning

1. Locate the bond proposed to break and its electron pair. 2. Decide whether electrons split evenly or travel together. 3. Use two fishhooks for homolysis or one full-headed arrow for heterolysis. 4. Assign radical dots or formal charges to the resulting fragments. 5. Check whether reaction conditions stabilise the proposed intermediates.

Visual explanation

Draw C–Br twice. In one path, send one fishhook electron to carbon and one to bromine; in the other, send the full bond pair to bromine and label C⁺ and Br⁻.

Real-world analogy

Two partners can divide a shared pair of tickets one each or one partner can take both. The allocation determines what each partner carries after separating.

Real-world example

UV light can initiate chlorination by splitting Cl₂ into radicals. By contrast, a tertiary alkyl bromide can ionise in a suitable polar solvent to a carbocation and bromide.

Why?

Why does heterolytic C–Br cleavage give Br⁻ rather than neutral bromine? The entire bond electron pair moves to bromine, adding one electron relative to neutral bonded bromine's share.

Common misconception

“Every broken C–X bond makes a carbocation.” Radical conditions can yield a carbon radical, while concerted SN2 can break C–X without a free carbocation.

Worked example

Compare initiation of methane chlorination with a hypothetical SN1 ionisation of tert-butyl bromide. In the first, Cl₂ under light gives Cl· + Cl· by homolysis; each chlorine receives one electron, and fishhooks represent the split. In the second, (CH₃)₃C–Br can heterolyse under suitable ionising conditions to (CH₃)₃C⁺ + Br⁻; a full-headed arrow sends C–Br electrons to Br. The radical and cation have different electron counts, charge, and subsequent chemistry.

Quick check

1. What species result from homolytic cleavage of a neutral Cl–Cl bond? Answer: Two neutral chlorine radicals, each with one unpaired electron.

Exam focus

Match arrowhead style to electron count and product charges. Do not infer a free intermediate from overall product formula alone.

Advanced insight

Bond cleavage can be coupled to bond formation in one transition state. SN2 is heterolytic in electron allocation but does not require isolated C–X ionisation.

Summary

Homolysis splits a bond electron pair into radicals; heterolysis sends both electrons to one fragment and creates ionic charge changes. Conditions and arrow notation must agree.

Practice questions

1. Which arrows represent two single electrons leaving a bond separately? Answer: Two single-headed fishhook arrows. 2. What charge does bromine acquire after receiving both electrons from C–Br heterolysis? Answer: It becomes bromide, Br⁻. 3. Does concerted SN2 require an isolated carbocation? Answer: No. Nucleophile attack and heterolytic C–X bond departure occur together.