Chemistry · Lesson

Substitution, elimination and competing mechanisms

Compare concerted and stepwise substitution with elimination under stated kinetic and geometric assumptions. Use rate units, excess-reagent limits and intermediate branching without treating one rate law as a unique mechanism.

Back to the Chemistry map

By the end, you can

Distinguish selected SN2, SN1 and E2 model features.
Calculate a second-order rate and a bounded pseudo-first-order approximation.
Keep initial ionization and later intermediate capture separate.
Read inversion, ion-pair effects and periplanar geometry with appropriate scope.

A selected SN2 event couples entry and departure

In the elementary SN2 model, the entering nucleophile forms a bond while the leaving-group bond breaks in one concerted event. The transition-state geometry is not an ordinary stable carbon intermediate with five complete single bonds. For a suitable configured tetrahedral reaction center, backside displacement gives geometric inversion. A change of R/S label additionally depends on the product’s priority ranking, so descriptor comparison is not a substitute for tracking geometry.

For the simplest homogeneous bimolecular model, v = k[RX][Nu], with k in L mol⁻¹ s⁻¹ when concentrations use mol L⁻¹. Substrate sterics, nucleophile, leaving group and solvent affect kinetics. The observed form alone does not prove a uniquely elementary pathway; different mechanisms and approximations can produce the same dependence.

A second-order process can look first order in one reactant

If Nu stays approximately constant in large excess, define k_obs = k[Nu] and estimate [RX](t) = [RX]₀ exp(−k_obs t). This is a pseudo-first-order limit of the stated model rather than automatic evidence for SN1. Check how much Nu could actually be consumed before using the constant approximation.

The invented case k=0.250 L mol⁻¹ s⁻¹, [Nu]₀=0.500 mol L⁻¹ and [RX]₀=0.000500 mol L⁻¹ has maximum one-for-one Nu depletion only 0.100%. Its approximate k_obs is 0.125 s⁻¹ and half-life ln 2/k_obs≈ 5.54518 s. Real side consumption, changing volume or activities would require additional balances.

Initial ionization and product capture can have different sensitivities

A selected SN1 model first ionizes substrate, then captures a cationic intermediate. An idealized rate-limiting ionization can give substrate-first-order kinetics, but solvent, ionic partners, reversibility and other steps affect the full observed response. A rate coefficient for one step does not fix every product branch.

A planar free-ion idealization allows capture at two faces, but a persistent leaving-group ion pair or other asymmetric environment can bias them. Therefore complete racemization is not an unconditional SN1 rule. Compare the stated ion-pair and stereochemical evidence rather than assigning an exact face ratio merely from a name.

A cation can leave by capture or elimination

In an invented quasi-steady intermediate model, ionization has k_ion=0.00100 s⁻¹, nucleophile capture has k_cap=0.400 L mol⁻¹ s⁻¹ and a separate effective elimination rate is k_elim=0.0500 s⁻¹. Nu and the elimination conditions are held approximately constant. Capture’s effective rate is k_cap[Nu], so its branching fraction is k_cap[Nu]/(k_cap[Nu]+k_elim).

At [Nu]=0.500 M this fraction is 0.800; at 1.00 M it is 0.888889. The quasi-steady cation/substrate ratio is k_ion/(k_cap[Nu]+k_elim), about 0.00400 or 0.002222 respectively. These stipulated model values do not predict a real substrate’s competition, reverse recombination or yield.

Changing capture conditions can change a later branch without changing the stipulated ionization coefficient

Stipulated [Nu] / mol L⁻¹	Capture effective rate / s⁻¹	Capture fraction	Elimination fraction	Quasi-steady I/RX
0.500	0.200	0.800	0.200	0.00400
1.00	0.400	0.888889	0.111111	0.002222

Exact arithmetic for the invented irreversible capture/elimination quasi-steady cation model. k_ion=0.00100 s⁻¹, k_cap=0.400 L mol⁻¹ s⁻¹, k_elim=0.0500 s⁻¹ and the selected reservoirs are clamped. Fractions describe departure from that intermediate, not isolated yield or a measured SN1 product distribution.

An E2 proposal must retain the beta hydrogen and geometry

In an elementary E2 ledger, a base–H bond forms, a beta C–H pair contributes to C=C and the leaving-group bond pair goes to the departing group in one concerted event. The beta H is on a carbon adjacent to the leaving-group-bearing alpha carbon. A geometry that aligns the participating bonds is part of the proposal, not an optional detail.

Anti-periplanar arrangements commonly favour the required orbital alignment; syn-periplanar elimination is possible in appropriate cases, so anti is not a universal prohibition of every other pathway. A symbolic reference with alpha→beta as +x, leaving-group direction +y and beta-H direction −y has projected dihedral 180°; beta-H +y instead gives 0°. These ideal reference coordinates are not measured carbon bond angles or a real ring conformation.

Worked examples

Keep second-order units with the two concentrations

For the stipulated bimolecular model, use k=0.250 L mol⁻¹ s⁻¹, [RX]=0.0200 M and [Nu]=0.0300 M.

v = 0.250 × 0.0200 × 0.0300 = 0.000150 mol L⁻¹ s⁻¹.
Doubling one concentration gives 0.000300 M s⁻¹.
Doubling both gives 0.000600 M s⁻¹.

Initial rate 1.50× 10⁻⁴ M s⁻¹ under the stated model; the dependence itself does not prove a unique pathway.

Bound the pseudo-first-order reservoir change

Use the separate excess case [Nu]₀=0.500 M, [RX]₀=0.000500 M and one Nu consumed per RX.

Complete consumption of RX would remove at most 0.000500 M Nu.
Relative maximum Nu depletion is 0.000500 / 0.500 = 0.00100, or 0.100%.
k_obs = 0.250 × 0.500 = 0.125 s⁻¹.
Approximate half-life is ln 2 / 0.125 ≈ 5.54518 s.

A bounded 0.100% reservoir change supports this stated approximation; its exponential shape is not by itself an SN1 diagnosis.

Calculate the departure fractions at one clamped concentration

For the quasi-steady cation model at [Nu]=0.500 M, find capture fraction and I/RX.

Effective capture = 0.400 × 0.500 = 0.200 s⁻¹.
Total departure = 0.200 + 0.0500 = 0.250 s⁻¹.
Capture fraction = 0.200 / 0.250 = 0.800.
I/RX = 0.00100 / 0.250 = 0.00400.

Capture accounts for 80.0% of modeled departures, with a quasi-steady I/RX ratio 0.00400. No practical isolated yield was specified.

Try it yourself

What are k’s units in v=k[RX][Nu] when v uses mol L⁻¹ s⁻¹ and both concentrations mol L⁻¹?
Show the answer and reasoning
L mol⁻¹ s⁻¹.
The coefficient removes one of the two concentration factors to give concentration per time.
Does a pseudo-first-order exponential prove unimolecular substrate ionization?
Show the answer and reasoning
No.
A bimolecular model with an approximately constant excess partner can give that shape.
Does an SN1 name require exact 50:50 capture at both faces?
Show the answer and reasoning
No.
Ion pairing and other asymmetric conditions can bias capture; the free-ion symmetric idealization has a defined scope.
Is syn-periplanar E2 geometry universally forbidden?
Show the answer and reasoning
No.
Anti is commonly favoured, but compatible syn pathways can occur in appropriate cases.

What to carry forward

Couple all bond changes in a concerted event.
Retain rate units and concentration assumptions.
Bound an excess-reagent approximation quantitatively.
Separate initial ionization from later branching.
Use actual geometry and evidence for stereochemical claims.

References

Teaching resources and reference material supporting this lesson’s explanations and examples.