O-Chem

The Hammond Postulate: Transition States, Energy Diagrams & Reaction Selectivity

One of the most broadly applicable principles in all of organic chemistry — once you truly understand it, you can predict the selectivity of reactions you have never seen before simply by knowing whether the key step is exothermic or endothermic.

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SectionTopic
1What is a Transition State? Revisiting the Reaction Coordinate
2The Hammond Postulate: Statement and Logic
3Energy Diagrams: Exothermic Steps and Early Transition States
4Energy Diagrams: Endothermic Steps and Late Transition States
5Application: Chlorination vs Bromination Selectivity
6Application: Why More Stable Intermediates Form Faster (SN1, Markovnikov)
7Hammond Across the Course: A Reference Map
1

Before the Hammond Postulate makes sense, you need a clear mental picture of what a transition state is and how it sits on an energy diagram.

The Reaction Coordinate Diagram

Every elementary reaction step can be plotted on a reaction coordinate diagram: energy on the y-axis, progress of bond breaking and forming on the x-axis. The x-axis is not time — it is how far along the chemical transformation has gone, from reactants on the left to products on the right.

Reaction Coordinate → Energy → Reactants Transition State [‡] Products Ea ΔH
Generic reaction coordinate diagram — the TS sits at the energy peak; reactants and products sit at energy minima.

Transition State vs Intermediate — A Critical Distinction

FeatureTransition State (TS)Intermediate
Energy positionPeak of energy curveValley between two peaks
StabilityUnstable; collapses instantlyCan persist briefly
BondsPartially formed / brokenFull bonds
Isolable?No — neverSometimes (in principle)
Notation[A···B]‡Written as normal structure
Common exam mistake: confusing a TS with an intermediate. Intermediates sit in energy valleys and can (in principle) be trapped. Transition states are fleeting saddle points that exist for femtoseconds and cannot be isolated under any conditions.
2

George Hammond published his postulate in 1955 to address a practical problem: transition states cannot be observed directly — so how can we reason about their structure and energy?

"If two states, as for example a transition state and an unstable intermediate, occur consecutively during a reaction process and have nearly the same energy content, their interconversion will involve only a small reorganization of the molecular assemblage."

— G. S. Hammond, J. Am. Chem. Soc. 1955, 77, 334.

Plain-Language Version

The transition state of a reaction step resembles whichever species is closest to it in energy — the reactants (if the step is exothermic) or the products (if the step is endothermic).

The Logic

Think about the geometry of the energy curve. The TS sits at the peak. If the products are much lower in energy (strongly exothermic), the peak comes early on the x-axis and is close to the reactant energy level. If the products are much higher (strongly endothermic), the peak comes late and is close to the product energy level.

Step ThermodynamicsTS PositionTS Resembles
Strongly exothermicEarlyReactants
Slightly exothermicSlightly earlyMostly reactants
ThermoneutralMiddleBoth equally
Slightly endothermicSlightly lateMostly products
Strongly endothermicLateProducts
3

Consider chlorine radical H-abstraction from an alkane — an exothermic step:

R–H + Cl• → R• + HCl    ΔH ≈ −2 to −5 kcal/mol

Reaction Coordinate → Energy → R–H + Cl• Early TS ‡ R• + HCl Ea (small) TS ≈ reactants
Exothermic H-abstraction by Cl•. The TS appears early — close in energy to the reactants. Very little radical character has developed at the TS.

What This Means for Selectivity

Because little radical character has developed at the early TS, the stability difference between a 1°, 2°, or 3° radical barely registers in the TS energy. The energy curves for 1° and 3° abstraction are nearly identical — giving low selectivity (~5:1 per H for 3° vs 1°). Chlorination gives a statistical mixture of products.

4

Now contrast with bromine radical H-abstraction — an endothermic step:

R–H + Br• → R• + HBr    ΔH ≈ +5 to +18 kcal/mol

Reaction Coordinate → Energy → R–H + Br• Late TS ‡ R• + HBr Ea (large) TS ≈ products
Endothermic H-abstraction by Br•. The TS appears late — close in energy to the products (R• + HBr). Significant radical character has developed at the TS.

What This Means for Selectivity

Because the late TS closely resembles the product radical R•, the stability of R• directly controls the TS energy. The 3° radical is ~4–5 kcal/mol more stable than a 1° radical — and because the TS is late, this full difference appears in the TS energy gap. A 4 kcal/mol difference in Ea translates to ~1000-fold selectivity at room temperature via the Arrhenius equation. Bromination is essentially completely selective for the most substituted position (~1600:1 per H for 3° vs 1°).

The key insight: The energy difference between radical types is the same for chlorination and bromination (~4–5 kcal/mol). What changes is how much of that difference shows up at the TS. Early TS = barely any shows up. Late TS = nearly all of it shows up. This is the entire mechanistic basis of the selectivity difference.
5
FeatureChlorination (Cl• + R–H)Bromination (Br• + R–H)
ΔH of H-abstraction~−2 to −5 kcal/mol (exothermic)+5 to +18 kcal/mol (endothermic)
TS positionEarly (reactant-like)Late (product-like)
C–H breaking at TSMinimal (<25%)Substantial (>75%)
Radical character at TSSmallLarge
3° vs 1° ΔEa~0.5–1 kcal/mol~4–5 kcal/mol
3° vs 1° selectivity per H~5:1~1600:1
2° vs 1° selectivity per H~4.5:1~82:1
Practical outcomeMixture of productsHighly predictable single major product

The Reactivity–Selectivity Principle

A highly reactive reagent is less selective. A less reactive reagent is more selective. High reactivity means low Ea → early TS → intermediate stability barely registers in the TS energy. Cl• is faster but unselective. Br• is slower but exquisitely selective. The cost of selectivity is rate. This trade-off appears throughout organic chemistry.
6

The Hammond Postulate is the mechanistic foundation for several rules you already know. In each case the logic is identical: an endothermic step has a late TS that closely resembles the intermediate, so the intermediate's stability controls the rate.

SN1 Reactions — Carbocation Formation

The rate-determining step of SN1 is ionization of the C–X bond to form a carbocation. This step is endothermic. By Hammond, the TS is late and resembles the carbocation. Therefore, the stability of the carbocation directly controls the activation energy.

Markovnikov's Rule — HX Addition to Alkenes

When HX adds to an unsymmetrical alkene, the step that forms the carbocation is endothermic. By Hammond, the TS is late and resembles the carbocation. The more stable carbocation forms faster.

Markovnikov's rule is not an empirical coincidence — it is a direct consequence of endothermic carbocation formation having a late TS.

Carbocation Rearrangements

When a less-stable carbocation forms in an endothermic step, the activation energy for 1,2-hydride or 1,2-alkyl shifts to a more stable carbocation is low — because the TS for these shifts also resembles the more stable cation being formed. Rearrangements proceed in the direction of greater carbocation stability because that direction has the lower TS energy.

7
Reaction / ConceptHow Hammond Applies
Radical halogenation: Br₂ vs Cl₂Endothermic H-abstraction (Br•) → late TS → radical stability controls selectivity
SN1 substrate requirementsEndothermic ionization → late TS → stable carbocation = lower Ea = faster rate
Markovnikov's rule (HX + alkene)Endothermic protonation → late TS → more stable carbocation forms faster
E1 and Zaitsev's ruleEndothermic carbocation formation → same logic as SN1
Carbocation rearrangementsLate TS for rearrangement steps → more stable cation = lower TS = faster shift
Allylic/benzylic radical selectivityEndothermic H-abstraction by Br• → resonance-stabilized radical lowers TS dramatically
Acid-base: endothermic proton transferLate TS → stability of conjugate base controls activation energy and rate
The core insight: Whenever a step is endothermic, the TS resembles the products, and the stability of the product intermediate directly controls the rate and selectivity. Whenever a step is strongly exothermic, the TS resembles the reactants and intermediate stability barely matters.

Quick Recap

ConceptKey Point
Transition stateEnergy peak; partial bonds; cannot be isolated; drawn as [A···B]‡
IntermediateEnergy valley; full bonds; can persist briefly; can sometimes be trapped
Exothermic stepProducts lower; TS early; resembles reactants; intermediate stability ≈ invisible
Endothermic stepProducts higher; TS late; resembles products; intermediate stability controls rate
Hammond PostulateTS resembles the species closest to it in energy on the reaction coordinate
Reactivity–selectivityMore reactive reagent → lower Ea → earlier TS → less selective
Bromination selectivityEndothermic H-abstraction → late TS → radical stability matters → ~1600:1
SN1/MarkovnikovEndothermic cation formation → stable cation = lower TS = faster = major product

Final Checklist — Don't Lose Points on Exams

  • TS is always at the energy peak — never in a valley
  • Exothermic step → early TS → intermediate stability barely matters
  • Endothermic step → late TS → intermediate stability controls rate and selectivity
  • Bromination: ~1600:1 selectivity for 3° over 1° per H
  • Chlorination: only ~5:1 selectivity for 3° over 1° per H
  • Hammond explains Markovnikov's rule, SN1 substrate requirements, and carbocation rearrangements
  • Don't confuse TS (energy peak) with intermediate (energy valley)
  • Don't expect high selectivity from strongly exothermic steps — the early TS erases stability differences
The Hammond Postulate — Complete Study Guide Full guide with energy diagrams, worked examples, and reference map — free to download and print
Download PDF

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