CHEM 532, FALL 2013

Answer Key Problem Set 4 A.)

=slope=+1.90. Note - can also be used. The large positive rho value shows that the rate

limiting step involves breaking of the S-O bond.B.)

Y  k (10^8)  +  log(kx/kh) p‐CH3O  488,000 ‐0.12 ‐0.78 2.779398968 p‐CH3  16,400 ‐0.14 ‐0.31 1.305822994 H  811p‐Cl  318 0.24 0.11 ‐0.406593734 p‐NO2  1.44 0.81 0.79 ‐2.750658362 

y=1.9025x+0.0024R²=0.9929

‐0.5

0

0.5

1

1.5

2

‐0.2 0 0.2 0.4 0.6 0.8 1

log(KX/K

H)

m

The large (and negative) value for suggests that (+) charge is built up in the transition state, and there is more charge build-up than for chlorodimethylphenylmethane ionization (the reference point for the + substituent parameter). Also, note that the + parameter provides a better fit than p, as is expected for a directly conjugated cation.

C (a).

ratek2[anion]

K1[anion][R3NH ]

[olefin][R3N ]

[anion]K1[olefin][R3N ]

[R3NH ]

rateK1k2[olefin][R3N ]

[R3NH ]

ratekobv

[olefin][R3N ]

[R3NH ]

kobvK1k2

Note that the concentration of the protonated amine is still in the rate equation. This is OK because we have a pre-equilibrium (i.e. there is a measurable amount of this species in the solution). Also note that a pre-equilibrium does not necessarily mean that you can

use the SSA. For instance, if K1 = 1 then the anion and ammonium species are not consumed faster than they are formed. C (b).

d tBuOH dt

k2[tBu ][H2O]

d olefin dt

k3[tBu ]

k2, k3 k1 Steady State Approx.

d[tBu ]

dt 0 k1[SM ] k2[tBu ][H2O] k3[tBu ]

[tBu ] k1[SM ]

k2[H2O] k3

ratetBuOH k1k2[SM ][H2O]

k2[H2O] k3

&rateolefin k1k3[SM ]

k2[H2O] k3

D.) (a) This reaction shows a normal secondary isotope effect (kH/kD (Obsd.) = 1.111).

These hyperconjugative resonance structures will contribute less than in the case of a cation since a hydride ion is not very stable. Hence we will get a very small normal (sp3 sp2) K.I.E. due to combination of 6 small isotope effects (6 protons/deuteriums).

H3C CH2 CH CH2 CH3

HC

HCH3C CH2 CH3

H

H3C CH2HC

HC CH3

H

* *

*

* *

*

Hyperconjugatively stabilised intermediate

* *

SM = Starting Material

H*

H*

H*

NO2

H

H*

H*H*

(bond elongated for clarity)

OH- H*

H*

H*

NO2H*

H*H*

H*

H*H*

O2N H*

H*H*

etc.

(b) In the cyclic transition state the isotopically labeled CH2 changes from sp2 to sp3. This would lead to an inverse secondary kinetic isotope effect (kH/kD < 1).

(c) In the transition state of the r.d.s. the quaternary carbon gains sp2 character, an

example of a normal secondary kinetic isotope effect (kH/kD > 1). (d) The C-H/D bond is cleaved during the course of the reaction. A normal primary

kinetic isotope effect would be observed. Possibly a large value (kH/kD > 2). (e) Similar to (d). Normal primary kinetic isotope effect observed. Again expect large

effect (kH/kD > 2). (f) Similar to (d+e) Normal primary kinetic isotope effect observed. Again expect large

effect (kH/kD > 2). E.) The first approach: Mechanism I:

55Mconstant is ]O[H solution, aqueous In

][][

][][][][

]][[]][[][

2

221

221

221

AkOHkk

AkOHkazodt

azod

AazokOHazokdt

azod

obv

Thus the rate of phenyldiazonium consumption is 1st order dependent on concentration of both phenyldiazonium ion and nucleophilic anion. NOTE: when you write a rate equation, always indicate what you mean by rate. For instance in this case the rate could be for consumption of starting material or for production of phenol, or for production of the nucleophilic substitution product. Mechanism II: The first step is slow and the second two steps are fast, therefore the cation will never build up and the steady-state approximation will apply. We do not need this, though to calculate the rate of consumption of starting material since the first step for this is rate determining. (NOTE if you want to use the rates for production of either one of the two products you have to use the SS-approx. and the ratio of the two products is dependent on the individual concentrations of A- and H2O)

][][

1 azokdt

azod

The rate of phenyldiazonium consumption is NOT dependent on the concentration of the nucleophilic anion. Thus your experiments to distinguish between both mechanisms would involve experimentally determining the kinetic order in the concentration of the anion by varying its concentration. It would be 1st order in mechanism I and 0th order in mechanism II. The second approach: You could place substituents on the phenyl ring of the aryldiazonium salts. Mechanism I involves nucleophilic attack of A- or H2O on the ipso carbon. This should be accelerated by electron withdrawing groups. Mechanism II involves the generation of a positive charge. This reaction rate should be decreased by electron withdrawing groups. *HOWEVER the positive charge cannot be directly resonance stabilized and so the values of + would not provide a linear relationship, while the values would work fine. The positive charge is localized in a orbital that is orthogonal to the system. So the stabilization by electron donating substituents would be field and inductive effects, not resonance stabilization. Additional possibilities that can be used but are less definitive: Entropic considerations: Mechanism I predicts negative S of activation, mechanism

II positive S of activation. e.g. Do reaction at different temperatures different rate constants calculate G# from temperature dependence of rate constant. plot G# v/s T, then slope = -S# )

Solvent isotope effects Solvent polarity effects Change in product ratio with different para-substituents: an EWG would increase

amount of product formed by nucleophilic substitution by the anion for mechanism I by increasing the rate.

F.) Derivation of the rate equation (to show you how to use the conservation of mass for a catalyst as an additional piece of information; I did not ask for this, but it is good to have seen it) rate k2[A '][vinylSnR3]

rate kobv[vinylSnR3]

kobv k2[A ']

K [L][A ']

[A]

K[A] [L][A ']

[A] [L][A ']

K

[totalPd] [A][A ']

[A] [totalPd][A ']

[L][A ']

K [totalPd][A ']

[L][A '] K[totalPd][A ']

[L][A '] K[totalPd]K[A ']

[L][A ']K[A '] K[totalPd]

[A '] [L]K K[totalPd]

[A '] K[totalPd]

[L]K

kobv Kk2[totalPd]

[L]K 1

kobv

[L]K

Kk2[totalPd]

1

kobv

K

Kk2[totalPd] [L]

Kk2[totalPd]

1

kobv

1

Kk2[totalPd][L] 1

k2[totalPd]

(a)

If you were to measure the rate of a series of reactions where the total Pd and SnR3 concentrations were held constant, but with varying ligand concentration, a plot of

Note: We know neither [A] nor [A’]

Substitute into kobv= k2[A’]

This form of the equation is most useful as we can easily use it to obtain K and k2 (see (b))

1/kobv versus [L] should be linear. The slope of this line would be 1/Kk2[totalPd], and the intercept would be 1/k2[totalPd]. Since [totalPd] is known, the values of K and k2 can be obtained. The rate of the reaction would also be proportionally related to [Pd], but this could be explained by other mechanistic models, and therefore is less diagnostic for Scheme 2. (b) If Scheme 2 is correct, then the two parameters that govern the overall observed rate are K and k2. Both parameters could be influenced by the electronics of the ligand L. More electron donating ligands could dissociate more slowly from complex A (i.e. lower K). This would lower the concentration of A’ at a given time, and thereby decrease the observed rate of the reaction. This is observed in the data given. One could argue that the decrease in K could be offset by an increase in the conversion of A’ into B, that is if k2were to increase with more electron donating phosphines. The second column, however, shows the relative inhibition behavior for the same ligands. This inhibition by excess ligand (more than 2 equiv.) is consistent with the shifting of the equilibrium between A and A’ towards A, decreasing the overall rate of the reaction. If the rates were determined by the effect of the bound ligand in A’ on it’s transformation into B (i.e., effect on k2), then excess ligand would not be expected to influence the overall rate. Therefore both columns in the table support Scheme 2.