Kinetics as a tool of the trade

Collum, Angew Chem, 2007, 46, 3002Espenson, Chemical Kinetics & Reaction Mechanisms,

McGraw-Hill 1995Atwood, Inorganic and Organometallic Reaction Mechanisms,

2nd Ed, Wiley VCH 1997

Illustrative example from Collum LDA Chemistry

NiPr

iPr Li

Li

NiPr

iPr

• The rate law provides the stoichiometry of the rate-limiting transition structure,relative to the reactants.

• The rate law reveals changes in aggregation and solvation numbers required toreach the rate-limiting TS.

• Let A = LiNiPr2 and S = solvent

NiPr

iPr Li

Li

NiPr

iPr

(1) A2S2 + substrate + S ➝ product(2) d[product]/dt ➝ kobs[substrate](3) kobs ➝ k[A2S2]a[S]b

Entry a b kobs [stoichiometry]‡

1 1/2 -1 k[A2S2]1/2[S]-1 [A(substrate)]‡

2 1 -1 k[A2S2]1[S]-1 [A2S(substrate)]‡

3 1/2 1 k[A2S2]1/2[S]1 [AS2(substrate)]‡

4 1/2 0 k[A2S2]1/2[S]0 [AS(substrate)]‡

5 1/2 2 k[A2S2]1/2[S]2 [AS3(substrate)]‡

Ok. So what do these TS stoichiometries mean chemically?

[AS(substrate)]‡

iPr2N Li(solvent)

R Br

[AS2(substrate)]‡

iPr2N Li(solvent)2

R Br

[AS3(substrate)]‡

iPr2N Li(solvent)3

R Br

Br

H

H

H

NLi(THF)3

iPr2N

iPr2N

• Changes in solvent conc. and/or polarity can effect a mechanism. For the case of LDA,for any fixed solvent there is always a competition between a monomer & dimer pathway.

• Pseudo 1st order rate constants (kobs) can be established by setting the substrateas the limiting reagent.

[substrate]

t

ln[substrate]

t

x

xx

xx x

xx

xx

xx

slope = kobs

• Of more value is to plot kobs vs organolithium & solv concs.The dependence reveals the rxn orders & the aggregation state & degree of solvation

in the rate-limiting TS structure

• Ex: 1st order solvent dependence of an LDA monomer mech. indicates assoc. of1 solvent per monomer in the TS:

a = ½, b = 1 ⇨ kobs = k[A2S2]1/2[S]1 ⇨ [AS2(substrate)]‡ kobs

[s]

b = 1

b = 0

b = -1inverse order requires

dissociation!

Multiple pathways are often common. Even the simplest rxns can have competing,

kinetically competent pathways. ⟹⟹⟹ rate laws are then additive combinations.

Ex: kobs = k’[A2S2]1[S]0 + k”[A2S2]

1/2[S]1

• One can detect parallel pathways by plotting kobs vs [solv]

kobs

[s]

b = 1

b = 0

observed curvature indicates1st order dependence dominates

at higher [S]

• So rxn orders in LDA at low/high conc can differ, implying a change in aggregation.

• Similar observations are made in varying [L] in examining assoc subst rxnsin inorganic/organometallic chemistry

• Note: Fleeting intermediates are of NO KINETIC CONSEQUENCE

fleeting intermediates

SM

TS

Product

∆G‡

• Saturation often reveals a change in reagent structure.Leveling of kobs can indicate:

(i) change in rate-limiting step(ii) change in the observable form of the reactant

• Fortunately, unless you happen to be studying Cu-mediated chemistry,speciation of complexes is commonly simpler in organometallic chemistry than for

organolithium chemistry (but beware, copper chemists…)

• In principle, strong covalency helps to simplify rxn mechanisms (less overall lability).

• 1st order rxns: A k!→! P

v = −d[A]dt

=d[P]dt

= k[A]

k dt0

t

∫ = −d[A][A][A]0

[A]

∫

kt = − ln [A][A]0

⇒ ln[A]− ln[A]0 = kt

⇒

ln[A]= ln[A]0 − ktor

[A]= [A]0e−kt

differential rate law expression

⇒

ln[A]= ln[A]0 − ktor

[A]= [A]0e−kt

[A]

t

ln[A]

t

x

x

xx

x x

xx

xx

xx

slope = - k

[A]0

ln[A]0

• Half-life for a 1st order rxn: [A]= 12[A]0

ln [A][A]0

= −kt⇒ ln12 [A]0[A]0

#

$%

&

'(= −kt

ln 12= − ln2 = −kt

ln2k

= t1/2

0.693k

= t1/2

• Consistent w/1st order kinetics would be:(i) ln[A] vs t ⇨ straight line over ~ 3 half-lives(ii) Show that k is constant over a range of [A]0

• 2nd order kinetics: v = k[A]2

rates are proportional to the square of a single reagent. This is rather common forintermediates that undergo bimolecular decay.

−d[A]dt

= k[A]2

d[A][A]2

= −k dt = −kt⇒⇒⇒−1[A]

+1[A]0

= −kt0

t

∫[A]0

[A]

∫

1[A]

=1[A]0

+ kt

Mo Mo

Cp

CpOC

CO

OCOC

COCO

532 nmMo

COOCOC

2

Cp

OC CO

CO

Mo

Cp

COOC

OCkfast kslow

Mo Mo

Cp

CpOC

CO

OCOC

COCO

microseconds milliseconds

gauche

C2hC2h

2nd order integrated rate law

1[A]

=1[A]0

+ kt

• A plot of 1/[A] vs t should give a straight line for 2nd order kinetics

1/[A]

t

slope = k

intercept = 1/[A]0

should be positive if we’re monitoringdecay of SM

this plot spreads out, and emphasizes leastaccurate pts at the ends owing to thereciprocal function.

• Half-life for 2nd order reaction depends on initial concentrations.

[A]= 12[A]0 ⇒

112[A]0

= kt + 1[A]0

⇒2[A]0

= kt + 1[A]0

⇒ 2 = kt[A]0 +11= kt[A]0

⇒ t1/2 =1

k[A]0

� Example: Assume you are trying to study a rxn that is 2nd order that proceeds at 1000 M-1 s-1

What is a reasonable starting [A]0 to use if you’d like t1/2 ~ 1 hr?

3600 s = 1/{103 M-1 s-1�[A]0} ⇨⇨⇨ 3.6 x 106 M-1 = 1/[A]0 ⇨ [A]0 ~ 2.7 x 10-7 M

not very practical!

� Example: What if you want to work at 0.01 M (NMR conc)?

What sort of k is possible to ballpark for t1/2 = 1 hr?

k = 1/{3600 s * 0.01 M} = 36 M-1 s-1

v = k[A][B][A]+[B]→ [P]

• 2nd order kinetics where: this is a common scenario.

note: if [A] = [B] then previous case covers it.

If we assume [B] to be in excess: For [B] > [A] the stoichiometry requires that:

[B]t = [B]o −[A]0 +[A]t = Δ0 +[A]tsubstituting& rearranging :

−d[A][A](Δ0 +[A]t )

= kdt

noting dx(x)(ax + b)

= −b−1 ln[ax + bx

]∫

Integrating within the limits: ln([B]t[A]t

) = ln([B]0[A]0

)+ kΔ0t where Δ0 = [B]0 −[A]0k∝ 1

Δ0closer initial concentrationcreates higher uncertainty in kOR

1Δ0ln([B]t[A]0[A]t[B]0

) = kt

• A plot of ln([A]0[B]/[A][B]0) wrt time will be linear, with a slope = k∆0 & intercept = ln([B]0/[A]0)

• A plot of ln([A]0[B]/[A][B]0) wrt time will be linear, with a slope = k∆0.

ln([A]0[B]/[A][B]0)

t

k∆0

• Alternatively, we can equivalently express as follows instead:.

ln([A]t[B]t

) = ln([A]0[B]0

)+ k([A]0 −[B]0 )t

with a slope now of

k([A]0 −[B]0 )

and an intercept of

ln([A]0[B]0

)

• Principle of microscopic reversibility:

If a reaction coordinate finds a low energy pathway for the forward reaction, this pathwaymust also be lowest energy for the reverse reaction.

REVIEWING LIGAND SUBSTITUTIONS & KINETICS

MLx + L’ ➝ MLx-1L’ + L

• Lets have a look at Associative Substitutions.

MLx + L’ ⇋ MLxL’ ⇋ MLx-1L’ + L

• The above rxn should be 1st order in [MLx].• The above rxn should be 1st order in [L’].• The above rxn is 2nd order overall

• Experimentally if we add a large excess of L’ we can assume [L’] ≈ constant

rate ≅ kobs[MLx ] kobs = k[L ']where

rate ≅ kobs[MLx ] kobs = k[L ']where

rate = − d[MLx ]dt

= kobs[MLx ]

d[MLx ][MLx ]

=[MLx ]0

[MLx ]

∫ − kobs dt0

t

∫

therefore

ln[MLx ]− ln[MLx ]0 = −kobst

ln[MLx]

t1/2

ln[MLx]0

t

slope = kobs = ln(2)/t1/2

kobs

[L’]

kobs = k[L’]back-out k from the slopex

x

x

• As already shown

[L’] small

[L’] large

rate = d[cis−PtX2L2 ]dt

= k1[cis−PtX2L2 ][Li ]

• In many cases, this intercept is non-zero and indicates solvolysis is a competing pathway,leading to an additive rate law expression:

• Associative Substitution

• Lets have a closer look at Dissociative Substitutions.

(i) MLx ⇌ MLx-1 + L

(ii) MLx-1 + L’ ➝ MLx-1L’

• Kinetic analysis

k1

k-1

k2

• Pre-equilibrium can be much faster than k2. In such a case:

rate = k2[MLx−1][L ']………[MLx−1]=Κ1[MLx ][L]

rate = −d[MLx ]dt

=d[MLx−1L ']

dt= k2Κ1

[MLx ][L '][L] CASE 1

suggests a build-upof MLx-1

• But it is more typical not to see the lower coordinate intermediate build-up.This happens when k1 is slower or comparable to the two second order steps,

k-1[MLx-1] & k2[MLx-1][L’]

CASE 2: No build-up of [MLx-1]: k2 is fast

• It’s concentration remains undetectably small and we say that:

d[MLx−1]dt

≈ 0 = k1[MLx ]− k−1[MLx−1][L]− k2[MLx−1][L ']

math (steady-state approximation)

rate = k2[MLx−1][L ']=k1k2[MLx ][L ']k−1[L]+ k2[L ']

CASE 2

• Note: If k-1 >> k2, in other words a rapid pre-equilibrium situation, then:

rate = k2k1k−1

[MLx ][L '][L]

= k2Κ1[MLx ][L '][L]

(i) MLx ⇌ MLx-1 + L

(ii) MLx-1 + L’ ➝ MLx-1L’

k1

k-1

k2

Crabtree 6th, pp 116

• Note: If we make k2[L’] >> k-1[L], perhaps by adding large [L’] relative to [L], then:

rate ≈ k1[MLx] this suggests the first step is rate-limiting.

•Experimentally, as before we can add large [L’] (& also [L]):

rate ≈ kobs[MLx]; and all the other terms drop out!

kobs =k1k2[L ']

k−1[L]+ k2[L ']

kobs

[L’]

1st order region;MLx-1 + L’ ➝ product, nearly every time

k1

2nd order region;MLx ⇋ MLx-1 + L occurs many times before

MLx-1 + L’ ➝ product

rate = k2[MLx−1][L ']=k1k2[MLx ][L ']k−1[L]+ k2[L ']

CASE 2

• If the above rate law is to provide a successful test of a Dissoc Mech, then conditions shouldbe met such that: k-1[L] ≈ k2[L’]

• Then, if [L] is held constant while [L’] is varied, kobs will change as shown in thesaturation plot above. This type of “saturation” behavior shows that:

kobs approaches k1 when k2[L’] >> [k-1][L]

• Also, if L’ is varied using different ligands, k1 should not change, unless of course themechanism changes!

• So long as the order in [L’] drops below 1, suggested when the curve “bends over”,we can extract k1, and (k-1/k2): 1

kobs=k−1[L]k1k2[L ']

+1k1

1kobs

slope = k−1k1k2

intercept = 1k1

[L][L ']

• Thus, a plot of kobs-1 vs [L]/[L’]

should be linear andk1 and k-1/k2 may be determined.

“double-reciprocal plot”

• Dissociative Substitution

� Note: the reverse of the dissociation step is oftenkinetically competitive.

rate = k2[MLx−1][L ']=k1k2[MLx ][L ']k−1[L]+ k2[L ']

� Destabilization of the starting complex is commonly accomplished by adding steric bulk to its ligands. Naturally, dissociation relieves steric congestion in the starting complex. Chelation has the opposite effect, and tends to steel the starting complex against dissociation.

• Tolman Cone Angle Parameter

Credits to George Stanley, the true “King”

So what should dissociate? Trans Influence and Trans Effect

PEt3

Pt

Cl

Et3P Cl

2.42 Å

2.25 ÅCl

Pt

Cl

Et3P PEt3

2.32 Å

2.31 ÅPEt3

Pt

Cl

Et3P CH3

1JPtP = 1719 Hz

1JPtP = 4197 Hz

Trans Influence: Extent to which that ligand weakens the bond trans to it.It is a ground-state effect.

here there is mores-character in the Pt-C

bond, ‘stealing it’ from Pt-P, observedin the attenuated coupling

Crude ordering of some common ligands wrt trans influence.

R2B-, R3Si-, H-, CH3- ~ CN- ~ olefin > CO > PR3 ~ NO2 > I > Br > Cl > NH3 > H2O

Cl

Pt

Ltrans

Et3P PEt3 +

N

EtOH

k2, 25 °C

py

Pt

Ltrans

Et3P PEt3

Cl-

Trans Effect: Effect of that ligand on the rate of ligand displacement at the positiontrans to it. This is a ground-state and transition-state effect.

Ltrans k2rel

Cl 1Ph 400

CH3 1700PEt3, H > 104

Transition-State Theory in OM Chemistry

• In 1887 Arrhenius proposed based on empirical data that rate constants vary exponentiallywith inverse T:

k = Ae−EaRT

lnk = lnA− Ea

RT

Arrhenius Eqn:

� k = rate constant�A = pre-exponential factor, which can be thought of as

as a steric change or entropy change associated withapproaching the “activated complex”. This pre-factoris T-independent w/same units as k.*Reacting molecules must become “activated”*Kinetic Energy (KE) ➝ Potential Energy (PE) as moleculesbecome activated.

� R = 1.987 cal mol-1 K-1

• Note: −Ea = RT ln(kobsA)

• This should remind us of: ∆G˚ = RTln(Keq)so it’s natural to think of Ea as energy (like ∆H˚)

Arrhenius Plot:

T1 > T2

Note: now assumingThe same Ea in one reaction coordinate,and probing the k rate at two differ T’s.

General

• It’s convenient to think of a rxn proceeding along a potential energy (PE) surface, w/motionof the atoms characterizing the chemical rxn:

H

C

HH

HO Br

OH- + CH3Br

Br- + CH3OH

Ea1Ea2

∆E˚

q

• TS theory was developed for elementary single rxns. Eyring extended it toensembles, developing a theory of absolute rates based on postulatedequilibria of the activated complex (molecules at the top of the TS)with other molecules.

Eyring Eqn:

Eyring Eqn: Assume a unimolecular decay of A according to the following scheme.

AK

A*k*

products

k1 A

A*

products[A*] = K‡[A], rate = k*[A*] = k*K‡[A]

but since conventional rate expression is: −d[A]dt

=d[products]

dt= k1[A]

∴

k1 = k*K ‡

k* is analyzed by statistical mechanics to be:kBTh

K‡ is related to ∆G‡ in the usual way…∆G‡ = ∆H‡ - T∆S‡ = -RTlnK‡

• Eyring obtained an expression of the same form as the Arrhenius relationship:

k = kBThe−ΔG‡

RT =kBThexp(−ΔH

‡

RT+ΔS‡

R)

∆S‡ = “entropy of activation”, provides idea of how ordered the TS is.if ∆S‡ is very negative (maybe -20 to -35 e.u.), step is likely bimolecular.if ∆S‡ is very positive (maybe > 15 e.u.), it is suggestive of a late TS fordecay of one molecule to several fragments.

• Rearranging (taking the log): ln( kT) = −ΔH

‡

RT+ ln(kB

h)+ ΔS

‡

R

23.76

xx

xx

xx

slope = -(∆H‡/R)

23.76 + (∆S‡/R)

ln( kT)

1T