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IB DP Chemistry – R3.4.3 Heterolytic fission – Study Notes

IB DP Chemistry - R3.4.3 Heterolytic fission - Study Notes - New Syllabus - 2026, 2027 & 2028

IB DP Chemistry – R3.4.3 Heterolytic fission – Study Notes – New Syllabus

IITian Academy excellent Introduction to the Proton transfer reactions – Study Notes and effective strategies will help you prepare for your IB DP Chemistry  exam.

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Reactivity 3.4.3 — Heterolytic Fission

Reactivity 3.4.3 — Heterolytic Fission

Heterolytic fission is the breaking of a covalent bond in such a way that both of the shared electrons go to one of the atoms. This results in the formation of two ions—typically a cation and an anion.

It is contrasted with homolytic fission, where the electrons are split evenly, forming radicals.

Representation:

Heterolytic fission is represented using curly arrows (⟶) to show the movement of electron pairs. These arrows always originate from a lone pair or bond and point to where the electrons are going.

General Equation:

\( \text{A–B} \rightarrow \text{A}^+ + \text{B}^- \)

Here, B takes both bonding electrons, forming an anion, while A becomes a cation.

Common Conditions:

  • Polar bonds where one atom is more electronegative than the other.
  • Usually in substitution or elimination mechanisms in organic chemistry.

Mechanism Example 1: Heterolytic fission of hydrogen chloride (HCl)

In aqueous solution, HCl dissociates into ions by heterolytic fission:

\( \text{H–Cl} \rightarrow \text{H}^+ + \text{Cl}^- \)

Mechanism Example 2: Nucleophilic substitution of bromoethane

In the SN1 mechanism, the first step involves heterolytic fission of the C–Br bond:

\( \text{CH}_3\text{CH}_2\text{Br} \rightarrow \text{CH}_3\text{CH}_2^+ + \text{Br}^- \)

Mechanism Example 3: Heterolytic fission during acid dissociation

Ethanoic acid undergoes heterolytic fission when releasing a proton:

\( \text{CH}_3\text{COOH} \rightarrow \text{CH}_3\text{COO}^- + \text{H}^+ \)

This explains why acids act as proton donors in Brønsted–Lowry theory—the O–H bond breaks heterolytically.

Mechanism Example 4: Heterolytic cleavage in elimination reactions

In the E1 elimination of tert-butyl bromide:

\( (\text{CH}_3)_3\text{CBr} \rightarrow (\text{CH}_3)_3\text{C}^+ + \text{Br}^- \)

This forms a stable carbocation as the intermediate. A base then removes a proton from a β-carbon, leading to alkene formation.

Key Concepts for IBDP:

  • Curly arrows must start from electron pairs or bonds and point to where the electrons move.
  • Electronegativity differences often drive heterolytic fission.
  • Formation of ions is critical in many reaction mechanisms—especially substitution and acid–base reactions.

Example

Ethyl bromide undergoes hydrolysis in aqueous solution. Predict the ions formed in the first step of the mechanism, and indicate the movement of electrons.

▶️Answer/Explanation

The carbon–bromine bond breaks heterolytically because bromine is more electronegative:

\( \text{CH}_3\text{CH}_2\text{Br} \rightarrow \text{CH}_3\text{CH}_2^+ + \text{Br}^- \)

A curly arrow goes from the C–Br bond toward the Br atom, indicating both bonding electrons move to Br. This forms an ethyl carbocation and bromide ion. These ions participate in subsequent steps, such as reaction with water or hydroxide.

Example

When ethyl ethanoate is hydrolyzed under acidic conditions, one step involves the heterolytic fission of the C–O bond. Draw this step using curly arrows and identify the species formed.

▶️Answer/Explanation

In one step of acid-catalyzed hydrolysis, the protonated ester undergoes cleavage of the C–O bond:

\( \text{CH}_3\text{COOCH}_2\text{CH}_3 + \text{H}^+ \rightarrow \text{CH}_3\text{COOH} + \text{CH}_3\text{CH}_2^+ \)

A curly arrow is drawn from the bond between the ester oxygen and ethyl group toward the oxygen atom. This shows heterolytic fission where both electrons move to the oxygen, resulting in the formation of a neutral molecule of ethanoic acid and a positively charged ethyl carbocation.

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