Apply the Hammond postulate to reaction mechanisms
Apply the Hammond postulate to one elementary step at a time, not to an overall equation with hidden intermediates.
Continue where you stopped
The core idea
On this page
Learning objectives
- Apply the Hammond postulate to reaction mechanisms
Locate the elementary step
Apply the Hammond postulate to one elementary step at a time, not to an overall equation with hidden intermediates.
On a reaction-coordinate profile, stable species are minima and transition states are maxima; compare a maximum with the adjacent minima.
Nearest in energy, not nearest on paper
A transition state resembles the stable species nearest to it in energy.
A strongly endothermic step has a late, product-like transition state; a strongly exothermic step has an early, reactant-like transition state.
Connect energetics to structure
When carbocation formation is endothermic, the transition state already has substantial C–leaving-group cleavage and developing positive charge.
A substituent that stabilises that carbocation can therefore also stabilise the product-like transition state and lower the activation barrier.
The shortcut is ‘nearer in energy means more alike in structure’. Apply it between one maximum and the minima directly beside it; horizontal distance along the reaction coordinate is not the evidence.
Ionisation of tert-butyl bromide
Consider the endothermic elementary cleavage (CH₃)₃C–Br → (CH₃)₃C⁺ + Br⁻.
The ionisation transition state is late and carbocation-like because the carbocation products of this step lie higher than the reactant.
Sketch this elementary-step profile and annotate the extent of C–Br cleavage and positive-charge development at the maximum.
Check your answer
A strong answer should include an uphill product minimum, a single intervening maximum, and a late transition-state sketch with a substantially broken C–Br bond and developing tertiary carbocation character.
Read an exothermic capture step
Supplied step: (CH₃)₃C⁺ + CN⁻ → (CH₃)₃C–CN, strongly downhill as the new C–C bond forms.
Apply Hammond: a transition state resembles the species nearest to it in energy. Decide which side that is here before placing the maximum.
Place the transition state for (CH₃)₃C⁺ + CN⁻ → (CH₃)₃C–CN on an exothermic profile and describe its structure.
Check your answer
Because the step is strongly exothermic, the transition state is early: the maximum lies closer in energy to the reactants, and the structure retains carbocation and cyanide character with only limited C–CN bond formation.
Compare substituent sensitivity
An activation barrier is the energy difference between a transition state and that step’s reactants. A substituent changes the barrier only insofar as it affects those two energies differently.
In endothermic ionisation, substantial positive charge develops between neutral R–Br and the late transition state. Alkyl stabilisation can therefore lower the transition state relative to the reactant.
An early capture transition state retains much of the carbocation reactant’s character. Stabilising both similarly changes their energy gap much less than stabilising the transition state alone.
Compare replacing methyl by hydrogen in endothermic R–Br ionisation and strongly exothermic capture of R⁺ by Br⁻. Assume the change mainly reduces stabilisation of carbon-centred positive charge; neglect changes in steric, solvent and neutral-substrate effects. Which activation barrier should change more, and why must each transition state be compared with its own reactants?
Check your answer
The ionisation barrier should rise more: reduced alkyl stabilisation raises its carbocation-like transition state relative to neutral R–Br. In capture, the carbocation reactant and early transition state lose similar stabilisation, so much of the change cancels in their energy difference. This is a qualitative matched-case argument, not a numerical barrier prediction or a rule for unmatched reactions.
Correct the time-order shortcut
The nearest stable species means nearest in energy, not whichever species is drawn immediately before or after the transition state.
Hammond predicts qualitative resemblance; it does not say that a transition state is an isolable intermediate or an exact copy of a reactant or product.
Correct the statement: ‘Every transition state is product-like because products occur after it.’
Check your answer
State that exothermic steps tend to have reactant-like transition states and endothermic steps product-like transition states because energy proximity governs resemblance.
Put the reasoning together
Complete the Hammond check questions before trying the unseen profile; retain the distinction between a transition state and an intermediate.
Later, try the different profile; compare how nucleophile identity and concentration affect substitution rate.
Build the answer in four links: state whether the elementary step is exothermic or endothermic, identify the nearer stable species, describe the transition-state bonding or charge, then explain the barrier change.
Attempt the Hammond final practice question, mark both the energy comparison and the structural resemblance, then try a fresh question later.
Check your answer
Continue when the profile direction and transition-state structure agree, then test nucleophile effects with controlled rate data.
Apply the Hammond postulate to reaction mechanisms scientific representation
Text alternative states the relative energies of both minima, locates each maximum, and describes the partial bond and charge at each transition state.
About 5 minutes
Endothermic R–Br ionisation
- R–Br
- late TS: Rδ+···Brδ−; long C–Br
- R⁺ + Br⁻
Exothermic R⁺ capture by CN⁻
- R⁺ + CN⁻
- early TS: R⁺···CN⁻; weak forming C–C
- R–CN
Text alternative: Text alternative states the relative energies of both minima, locates each maximum, and describes the partial bond and charge at each transition state.