Organic Chemistry · Part 9 of 9

Determining a Mechanism — The Physical Organic Toolkit

Reaction Mechanisms, Part 9 · 12 sections · about 18,355 words · CSIR-NET Chemical Sciences, GATE Chemistry & IIT-JAM

The previous eight parts told you what the mechanisms are. This one tells you how anybody knows. Every mechanism in this book was established by experiment, and the exam frequently asks you to reason in that direction — here is the evidence, what does it prove? This part covers the standard tools, then works through real cases where the evidence decided between competing proposals, and closes with the master index to all nine parts. Two layers on every section: a slow, hand-held beginner path and a research-grade advanced/reference path.

The 12 sections in Part 9

  • 1Rate laws — what the order in each component does and does not tell you Free below
  • 2Catalysis — specific, general, nucleophilic and intramolecular
  • 3Kinetic isotope effects — locating the atom that moves
  • 4The Hammett equation — substituents as a probe of charge
  • 5The Taft treatment — separating sterics from electronics
  • 6Solvent effects and activation parameters
  • 7Stereochemistry, crossover experiments and isotopic labelling
  • 8Trapping and direct observation of intermediates
  • 9Case studies — four disputes the evidence settled
  • 10The nine parts at a glance
  • 11Section-by-section index, A.1 to J.9
  • 12What this book has built

Rate laws — what the order in each component does and does not tell you

Free extract

Section J.1 of Part 9, reproduced in full from the book — figures and all. No sign-in, no paywall on this section.

Beginner layer

The rate law is a measurement. The mechanism is a story that has to be consistent with it. Keeping those two things apart is most of the skill.

A rate law is an experimental result: an equation, found by measuring concentrations against time, that expresses the rate of a reaction in terms of the concentrations of the species present. It is not derived, it is measured. For the hydrolysis of a bromoalkane you might find

rate = k[RBr][OH⁻]

and for a different bromoalkane, under the same conditions,

rate = k[RBr]

Both are facts. Neither is a mechanism. What they do is constrain the mechanisms you are allowed to propose, and that constraint is sharp:

The fundamental rule of kinetics: the rate law of a multi-step reaction contains the concentrations of every species that appears in the rate-determining transition state, together with everything consumed in any equilibrium before it. It contains no information whatsoever about anything that happens after the rate-determining step.

Order and molecularity — the distinction everything else rests on

These two words are constantly confused and the confusion is fatal.

OrderMolecularity
What it isAn experimental number: the exponent on a concentration in the measured rate lawA theoretical number: how many species come together in one elementary step of a proposed mechanism
Applies toThe overall reaction, however many steps it hasOne elementary step only
Possible valuesAny number — 0, 1, 2, 3, ½, 1.5, −1, and it can change with conditionsOnly 1, 2 or (rarely) 3 — you cannot have half a molecule collide
Known before or after experimentOnly afterProposed before, tested after
They coincide in exactly one situation: for a single-step reaction, order equals molecularity. For everything else, treating them as the same word is the error that generates most wrong answers in this area.
Note: The names SN1 and SN2 are the origin of half the trouble. The digits are molecularities of the rate-determining step, not orders. SN1 solvolysis in water is kinetically first order because water is the solvent and its concentration cannot change; but SN2 attack of water on a primary halide in water is also kinetically first order, for exactly the same reason. The observed order does not separate them — which is why Part 4 needed stereochemistry, salt effects and isotope effects to do it.

Reading a rate law backwards

Take each term in turn and ask what it licenses you to say.

ObservationWhat it meansWhat it does not mean
First order in substrate, zero order in nucleophileThe nucleophile is absent from the rate-determining transition state; the substrate ionises or fragments firstThat the nucleophile is unimportant — it still determines the product, just not the rate
First order in each of two reactantsBoth are present in the rate-determining transition state, one molecule of eachThat the reaction is one step. A fast pre-equilibrium followed by a slow unimolecular step gives the same law
Fractional order, e.g. ½Almost always a dissociative pre-equilibrium: a dimer breaking into monomers, or an initiator into two radicals, before the rate-determining stepAnything about the molecularity of the rate-determining step directly
Order changes from 1 to 0 in one reactant as its concentration rises (saturation)That reactant is involved in a binding pre-equilibrium that becomes complete; classic for enzymes, for micellar catalysis and for reactions through a complexed intermediateThat there are two competing mechanisms — though there may be
The third and fourth rows are the ones that separate a good candidate from an average one. A fractional or saturating order is a gift: it is the kinetics telling you explicitly that there is a pre-equilibrium.

A substitution shows rate = k[RX] and the rate is unchanged when the concentration of azide ion is tripled — but the product changes from almost pure alcohol to almost pure alkyl azide. Explain, and state what the data exclude. Easy

The kinetics. Zero order in azide means azide is not present in the rate-determining transition state. A concerted bimolecular displacement is therefore excluded: any SN2 pathway must show first order in the nucleophile.
The product change. The nucleophile clearly does react with something. Since it is not in the rate-determining step, it must attack an intermediate formed after that step — here a carbocation.
The picture that fits both. Slow, rate-determining ionisation to R⁺; then fast, product-determining capture of R⁺ by whichever nucleophile is present. This is the SN1 mechanism of D.4.
The general lesson. The rate-determining step and the product-determining step need not be the same step. Whenever a change of nucleophile alters the product but not the rate, you are looking at a common intermediate. This is the single most reliable piece of kinetic reasoning in the subject.
What is still not proved. Nothing here shows the cation is free rather than an ion pair, and nothing here shows there is only one intermediate. Stereochemistry (J.7) and the common-ion test (D.5) are needed for that.
Advanced / reference layer

The two approximations, derived rather than quoted; the real meaning of “rate-determining”; and kinetic ambiguity, which is the reason kinetics alone is never enough.

The general two-step scheme

Almost every mechanism in this book reduces, kinetically, to one scheme. Write it once and derive everything from it:

A + B    I  (k₁ forward, k−₁ back)    I + C → P  (k₂)

I is an intermediate at a local energy minimum — a carbocation, a tetrahedral intermediate, a Meisenheimer complex, an arenium ion. It is never present in more than trace amounts, and that fact is what makes the algebra tractable. Two approximations handle essentially all cases, and they are limits of the same expression.

The steady-state approximation, derived

The assumption is that after a short induction period the concentration of the reactive intermediate stops changing appreciably, because it is destroyed as fast as it is made:

d[I]/dt ≈ 0

This is justified when [I] is always very small compared with [A] and [B] — which is the definition of a reactive intermediate. Writing every process that makes and destroys I:

d[I]/dt = k₁[A][B] − k−₁[I] − k₂[I][C] = 0

Solve for the unmeasurable quantity [I]:

[I] = k₁[A][B] / (k−₁ + k₂[C])

and substitute into the rate of product formation, which is the thing you can measure:

rate = k₂[I][C] = k₁k₂[A][B][C] / (k−₁ + k₂[C])

That single expression contains both limiting mechanisms, and moving between them is just a question of which term in the denominator dominates.

LimitConditionRate lawPhysical meaning
First step rate-determiningk₂[C] ≫ k−₁rate = k₁[A][B]Every I formed goes on to product. The intermediate never returns to reactants; the rate is simply the rate of making it. Zero order in C — the classic signature.
Pre-equilibriumk−₁ ≫ k₂[C]rate = (k₁/k−₁)k₂[A][B][C] = K₁k₂[A][B][C]I reverts to reactants many times for each passage to product, so the first step is at equilibrium. Third order overall, and the observed constant is a product of an equilibrium constant and a rate constant.
Neitherk−₁ ≈ k₂[C]the full expressionOrder in C is between 0 and 1 and changes with [C]. Saturation kinetics. This is not a nuisance — it is the most informative case, because a plot of 1/rate against 1/[C] is linear: the intercept gives k₁, and the slope-to-intercept ratio gives k−₁/k₂. The individual constants k−₁ and k₂ are not separable by kinetics alone.
The double-reciprocal plot in the third row is the same manoeuvre as the Lineweaver–Burk plot of enzyme kinetics, and for the same reason: both describe a reaction proceeding through a reversibly formed intermediate.

Writing K₁ = [I]/[A][B] directly gives the same pre-equilibrium result without any steady state, which shows that pre-equilibrium is a special case of the steady state, not a rival approximation. Use the steady state whenever you are unsure.

Two limits of the same two-step schemeFree energy G →reaction coordinate →blue = first step rate-determining | red = pre-equilibrium, second step rate-determining
Which barrier is higher decides which rate law you measure. Blue: the first barrier is the higher one, so k₂[C] ≫ k−₁, the intermediate never comes back, and the rate law is zero order in C. Red: the second barrier is the higher one, so the intermediate reverts many times before it goes on, the first step is at equilibrium, and C appears in the rate law. Note that the thermodynamics is identical in the two cases — same reactants, same intermediate depth, same products. Only the relative barrier heights differ, and that is enough to change the measured order. Schematic.

What “rate-determining step” actually means

The textbook phrase “the slowest step” is loose and occasionally wrong. A step with a small rate constant is not necessarily rate-determining if its reactant is abundant, and in a chain reaction the propagation steps are individually very fast yet one of them controls the overall rate. The precise statement is:

Rate-determining (rate-limiting) step: the step whose transition state is the highest point on the free-energy profile relative to the reactants, and hence the step whose rate constant most strongly controls the overall rate. More rigorously, the step with the largest degree of rate control: the fractional change in the overall rate produced by a fractional change in that step’s rate constant, all others held fixed.

Three consequences worth stating explicitly, because each is an exam trap:

  • It is the highest transition state, not the deepest well. A very stable intermediate does not slow a reaction down unless the barrier out of it is high. Compare the tetrahedral intermediate of G.4, which is a genuine minimum yet rarely rate-determining.
  • It can change with conditions. Raising [C] in the scheme above pushes k₂[C] past k−₁ and moves the rate-determining step from the second to the first. Changing pH does the same in imine formation (G.3): below the optimum, dehydration is rate-determining; above it, addition is. That is exactly why the pH-rate profile has a maximum.
  • It need not be a single step at all. “The” rate-determining step is an idealisation that is only clean when one barrier stands well above the rest. When two barriers lie within roughly 4 kJ mol⁻¹ of each other — a factor of about five in rate at room temperature — rate control is shared between them, and both appear in the observed rate constant. The partial isotope effect in the sulfonation of benzene (J.3, H.1) is exactly this situation caught in the act: kH/kD ≈ 1.7 is neither the 1 of clean first-step control nor the 7 of clean second-step control.

Kinetic ambiguity — why kinetics is never the last word

Here is the deepest limitation of the whole method, and the one most likely to be examined at Part C level. A rate law does not determine a mechanism, because different mechanisms can give identical rate laws. Two mechanisms whose rate laws are algebraically indistinguishable are called kinetically equivalent, and no amount of kinetic work will separate them.

The standard example. Suppose a reaction shows

rate = k[S][HA][B]

where HA is a general acid and B a base. Two readings:

  • Mechanism I: the substrate S reacts with HA and B together in one transition state — concerted general acid–base catalysis.
  • Mechanism II: S is protonated in a fast pre-equilibrium by HA to give SH⁺ plus A⁻, and SH⁺ then reacts with B. Since [SH⁺] ∝ [S][HA]/[A⁻], and A⁻ is fixed by the buffer, the rate law is the same.

More generally: for any acid-base pair, the transition state composition is what kinetics reports, and a transition state of composition {S + HA + B} is reached identically from S + HA + B and from SH⁺ + A⁻ + B. Kinetics sees the atomic composition and charge of the transition state — nothing about how the atoms are arranged in it or the order in which they arrived.

Note: How ambiguity is broken. Never by more kinetics. It is broken by an isotope effect (is the proton in flight at the transition state, J.3?), by a Brønsted exponent (how far has it moved?), by trapping (does the protonated intermediate exist long enough to catch, J.8?), or by a structural change that affects one formulation and not the other. This is the whole argument of J.9: mechanisms are settled by converging independent lines, not by one decisive measurement.

A reaction A + B → P shows the following initial rates at 25 °C. Deduce the rate law and propose a mechanism consistent with it.
(i) [A]=0.10, [B]=0.10, rate = 2.0 × 10⁻⁴; (ii) [A]=0.20, [B]=0.10, rate = 4.0 × 10⁻⁴; (iii) [A]=0.20, [B]=0.20, rate = 5.4 × 10⁻⁴; (iv) [A]=0.20, [B]=0.40, rate = 6.4 × 10⁻⁴; (v) [A]=0.20, [B]=0.80, rate = 7.0 × 10⁻⁴ (all concentrations mol dm⁻³, rates mol dm⁻³ s⁻¹). Hard

Order in A. Compare (i) and (ii): doubling [A] at fixed [B] doubles the rate. First order in A, cleanly.
Order in B. Compare (ii)–(v) at fixed [A] = 0.20. Doubling [B] from 0.10 to 0.20 raises the rate by only 1.35×; the next doubling by 1.19×; the next by 1.09×. The order in B is less than one and falling towards zero. This is saturation, not experimental error — the trend is monotonic and smooth.
Fit the general expression. Saturation in B is exactly the third row of the table above with B playing the role of C: rate = k₁k₂[A][B] / (k−₁ + k₂[B]). At low [B] this is first order in B; at high [B] the k₂[B] term dominates the denominator, cancels, and gives rate = k₁[A], zero order in B.
The mechanism. A reacts reversibly with B to give an intermediate; that intermediate goes on to product in a step that competes with its reversion. At high [B] the forward channel wins and formation of the intermediate from A becomes rate-determining. Note that A stays first order throughout, so A is consumed in the first step and not regenerated.
What you have not shown. Nothing here identifies the intermediate, and nothing excludes the kinetically equivalent formulation in which A binds B in a rapid complexation equilibrium and the complex fragments. Both give this algebra. Say so — examiners give credit for the caveat.

⚠ Common mistakes & exam traps

  • Reading order as molecularity. A second-order rate law does not mean a bimolecular one-step mechanism, and a first-order rate law does not mean a unimolecular mechanism. Solvolysis in water is first order whatever the mechanism, because the solvent concentration cannot be varied. This is the most examined error in the whole part.
  • Assuming the rate-determining step is the product-determining step. In every SN1 reaction it is not. Whenever a change of nucleophile changes the product but not the rate, the two steps are different and there is a common intermediate.
  • Forgetting that the rate law is blind to everything after the highest barrier. No kinetic experiment can distinguish mechanisms that differ only in fast steps following the rate-determining one. Do not propose kinetics as evidence for such a difference.
  • Treating a fractional or drifting order as bad data. An order of ½ means a dissociative pre-equilibrium; a falling order means saturation. Both are information. Rounding them to the nearest integer throws away the most interesting part of the result.
  • Believing that a rate law settles a mechanism. Kinetically equivalent mechanisms give identical rate laws by construction. Kinetics narrows the field; it does not choose the winner.

Read the rest of Part 9

The remaining 11 sections of this part — Catalysis — specific, general, nucleophilic and intramolecular, Kinetic isotope effects — locating the atom that moves, The Hammett equation — substituents as a probe of charge, The Taft… — and all nine parts of Reaction Mechanisms are part of ChemVidya Full Access, along with the other books, 55 Study Notes and 6,000+ practice questions.

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