The Hammett Equation — Substituent Effects as a Straight Line
Physical organic chemistry has one central question: what is actually happening at the transition state? You cannot photograph a transition state, so you probe it indirectly. The Hammett equation is the oldest and still the most used of those probes. You change a substituent on a benzene ring, measure how the rate or equilibrium changes, plot the results, and the slope of that straight line tells you whether positive or negative charge is building up in the rate-determining step. This is standard CSIR-NET and GATE organic material, and it is one of the few topics where a genuine least-squares fit is part of the chemistry.
The equation
What each symbol means:
- kX — rate constant for the compound carrying substituent X on the ring.
- kH — rate constant for the unsubstituted parent (X = H). This is the reference, so by definition σ(H) = 0 and log(kH/kH) = 0.
- σ (sigma) — the substituent constant. It belongs to the substituent and its position (meta or para), not to the reaction. Positive σ means electron-withdrawing; negative σ means electron-donating.
- ρ (rho) — the reaction constant. It belongs to the reaction, the solvent and the temperature, not to any one substituent. It measures how sensitive that reaction is to electronic change on the ring.
Both σ and ρ are dimensionless. Because log k is proportional to a free energy (ΔG‡ = −RT ln k + constant), a straight line in log k is a straight line in free energy — which is why this family of relationships is called a linear free energy relationship (LFER).
Where σ values come from
σ is not fitted to the reaction you are studying. It is defined once, from the ionisation of substituted benzoic acids in water at 25 °C, and that reference reaction is assigned ρ = 1 exactly:
Worked example 1 — deriving σp for the nitro group.
Benzoic acid has pKa ≈ 4.20; 4-nitrobenzoic acid has pKa ≈ 3.42.
σp(NO2) = 4.20 − 3.42 = +0.78
The positive sign says the nitro group withdraws electron density, stabilises the carboxylate anion, and therefore makes the acid stronger. Different compilations quote pKa values that differ in the second decimal place, so tabulated σp for NO2 is usually given as 0.78 but you will also see 0.81 — use one table consistently rather than mixing sources.
Some widely tabulated values, for orientation:
| Substituent | σmeta | σpara | Electronic character |
|---|---|---|---|
| −NO2 | +0.71 | +0.78 | Strongly withdrawing (−I, −M) |
| −CN | +0.56 | +0.66 | Withdrawing (−I, −M) |
| −Cl | +0.37 | +0.23 | Withdrawing by induction, weakly donating by resonance |
| −H | 0.00 | 0.00 | Reference |
| −CH3 | −0.07 | −0.17 | Weakly donating (+I, hyperconjugation) |
| −OCH3 | +0.12 | −0.27 | Withdrawing from meta, donating from para (+M) |
| −NH2 | −0.16 | −0.66 | Strongly donating by resonance |
Look carefully at −Cl and −OCH3. Their meta and para values have opposite signs or very different sizes, because resonance donation reaches the para position but not the meta position, while induction reaches both. That single observation is the reason two separate scales exist at all.
Worked example 2 — predicting a rate ratio
Problem: a reaction of substituted benzene derivatives is found to have ρ = +2.30. By what factor does a para-nitro group speed it up relative to the unsubstituted compound?
log (kX/kH) = σp × ρ = 0.78 × 2.30 = 1.794
kX/kH = 101.794
Break the power up: 101.794 = 101 × 100.794.
Since log 6.22 = 0.794, we get 100.794 = 6.22.
kX/kH = 10 × 6.22 = 62 times faster.
Now repeat for a para-methoxy group, σp = −0.27:
log (kX/kH) = (−0.27)(2.30) = −0.621
kX/kH = 10−0.621 = 0.239, i.e. about
4.2 times slower (1 ÷ 0.239 = 4.18).
So across just these two substituents the rate spans a factor of 62 ÷ 0.239 ≈ 260. A modest σ range produces a large rate range because the relationship is logarithmic.
Worked example 3 — fitting ρ by least squares
In practice you do the opposite: you measure rates, and the slope gives you ρ. Suppose a set of measurements gives these four points.
| Substituent | x = σ | y = log(kX/kH) |
|---|---|---|
| p-OCH3 | −0.27 | −0.62 |
| H | 0.00 | 0.00 |
| p-Cl | 0.23 | 0.53 |
| p-NO2 | 0.78 | 1.79 |
Step 1 — means.
Σx = −0.27 + 0 + 0.23 + 0.78 = 0.74, so x̄ = 0.74 ÷ 4 = 0.185
Σy = −0.62 + 0 + 0.53 + 1.79 = 1.70, so ȳ = 1.70 ÷ 4 = 0.425
Step 2 — deviations and their products.
(−0.455)(−1.045) = 0.475475
(−0.185)(−0.425) = 0.078625
(0.045)(0.105) = 0.004725
(0.595)(1.365) = 0.812175
Σ(x − x̄)(y − ȳ) = 1.371000
Step 3 — sum of squared x deviations.
(−0.455)² = 0.207025; (−0.185)² = 0.034225; (0.045)² = 0.002025; (0.595)² = 0.354025
Σ(x − x̄)² = 0.597300
Step 4 — slope and intercept.
ρ = 1.371000 ÷ 0.597300 = +2.30
intercept = ȳ − ρx̄ = 0.425 − (2.2953 × 0.185) = 0.425 − 0.4246 = 0.0004 ≈ 0
The intercept coming out at essentially zero is not luck — it is the built-in check on your working. Because σ(H) = 0 and log(kH/kH) = 0, a correct Hammett plot must pass through the origin. If your fitted intercept is far from zero, you have either used the wrong reference kH or mixed up meta and para σ values.
Reading the sign and size of ρ
| ρ value | What it means about the transition state | Typical situation |
|---|---|---|
| Large positive (ρ > +1) | Negative charge builds up near the ring; electron-withdrawing groups help | Nucleophilic attack on a carbonyl; ester saponification |
| Small positive | Slight negative charge build-up, or the reaction centre is far from the ring | Side-chain reactions several bonds away |
| Near zero | Little charge change, or opposing effects cancel | Radical reactions, some concerted processes |
| Negative (ρ < 0) | Positive charge builds up; electron-donating groups help | SN1 solvolysis of benzylic halides; electrophilic aromatic substitution |
A break or curvature in the plot is itself a result, not a failure. If electron-rich substituents lie on one line and electron-poor ones on another, the usual interpretation is a change of mechanism or a change in the rate-determining step across the series. Hunting for that break is often the whole point of running the study.
The σ+ and σ− scales
The ordinary σ scale is calibrated on benzoic acid ionisation, where the substituent and the reaction centre are separated by the carboxyl carbon. When the transition state puts charge in direct conjugation with the substituent, ordinary σ under-describes it and the plot goes crooked. Two extra scales exist for this:
- σ+ — for transition states with positive charge conjugated to the substituent, such as benzylic cations. Donors get much more negative values; σ+p(OCH3) is far more negative than σp(OCH3).
- σ− — for transition states with negative charge conjugated to the substituent, such as phenoxide ions. Withdrawing groups get larger positive values; σ−p(NO2) is larger than σp(NO2).
Choosing the scale that straightens the plot is itself mechanistic evidence: if only σ+ gives a good line, direct resonance stabilisation of a cationic centre is implicated. Related treatments you may meet by name include the Yukawa–Tsuno equation, which blends σ and σ+, and the Taft equation, which separates polar from steric effects for aliphatic systems.
Common mistakes that cost marks
- Using σpara for a meta substituent. They are genuinely different numbers, and for OCH3 they even have opposite signs. Always read the position from the structure first.
- Applying Hammett to ortho substituents. The standard scales do not cover ortho, because steric interference and direct field effects contaminate the purely electronic term. Ortho points are simply left out.
- Forcing the fit through a nonzero intercept. The origin is a theoretical requirement here, not a fitting choice.
- Reading ρ as a bond order or a charge. ρ measures sensitivity, not the number of electrons moved. A ρ of +2.3 does not mean "2.3 units of negative charge".
- Mixing σ and σ+ in the same plot. Pick one scale for the whole data set and say which you used.
- Comparing ρ across different solvents or temperatures. ρ is defined for a specific set of conditions; changing the solvent changes ρ.
- Using ln instead of log. The equation is defined with base-10 logarithms. Using natural logs inflates every value by a factor of 2.303.
Where this appears in exams
| Exam | Typical use |
|---|---|
| CSIR-NET | Sign and interpretation of ρ; deducing a mechanism from a Hammett plot; σ vs σ+ choice |
| GATE (Chemistry) | Numerical: calculate kX/kH from given σ and ρ, or find ρ from two data points |
| IIT-JAM | Qualitative substituent effects on acidity and reactivity, the conceptual base of this topic |
| M.Sc. / research work | Building a full plot from measured rate constants and reporting ρ with its fit quality |
Check the current official syllabus for whichever paper you are sitting — the depth expected on physical organic topics is revised from time to time.
Fit your own Hammett plot. The Linear Regression tool fits y = mx + c by least squares. Enter σ as x and log(kX/kH) as y, and the slope m it returns is ρ — with the intercept c as your built-in check that the line really does pass through the origin.
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