CHEM 73/8311 CS3 Structure Effects on Stability and Reactivity.docx
2012Jan18
Mechanisms
Mechanism = description of rearrangement of reactants as they proceed to products
Never can prove a mechanism, only disprove it
covalent bonds are usually broken or formed during a reaction
classifications:
1. heterolytic - electrons on bond remain with one fragment
2. homolytic - each fragment retains one electron
3. pericyclic - cyclic transition state, no intermediates
4. electron transfer – bond making/breaking usually follows
Predicting products: Thermodynamic Requirements
1. free energy must be negative
2. G = H - TS, G = -RTLnK
3. entropy effect is often small compared to enthalpy except in certain cases
(a) phase change involving gases: gases have higher entropy than liquids and solids
(b) molecularity change
(c) high temperatures
(d) ring formation and breaking
4. enthalpy can be calculated directly from enthalpies of formation of reactions and
products
Hrxn = Hf(products) - Hf(reactants)
hydrogenation of butene: – 30.15 (n-butane) – 0 (H2) + 2.67 (t-2-butene) = – 27.48
experimental heat is –27.5 kcal/mol
5. Reaction enthlapies can be calculated from broken and formed bond energies
H = bond dissociation energies formed – bond dissociation energies broken
only applies if actual values are known
hydrogenation of ethene
Bonds formed
Bonds broken
H-CH2CH3
2 x 98.2
H-H
103
CH3-CH3
88
C=C
171
sum
284.4
274
H =-10.4 kcal/mol (7.5 from Hf)
Actual process does not correspond to bond energies used
171 kcal/mol + H C
2
CH2
H
103 kcal/mol + H2
H
H2C
CH2
CH3
CH3
CH3
H
CH2CH3
CH3
CH2
H2C
H
+?
CH2CH3
+?
CH3CH3 + 88 kcal/mol
CH3CH3 + 98 kcal/mol
1
CHEM 73/8311 CS3 Structure Effects on Stability and Reactivity.docx
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Methods of Determining a Mechanism
Identification of Products: mechanism
must account for products
Identify intermediates
1. stop reaction in progress and
isolate intermediate
2. observe intermediate during
reaction (hydroxyl amine)
3. trap intermediate with reactive
reagent (carbene with olefin)
4. may not be intermediate
5. independently create
intermediate under identical
conditions (common intermediate)
Isotopic Labeling - mechanism
requires a label to appear at specific
sites
ester hydrolysis with H2O18, example:
O
Me
Me
Me
I
Me
NO2
I
HNO3
O2N
I
+
+
I
NO2
+
H3C
CH CH
CH3
-
HCl
OH
HCl
CH3
CH3
H
CH2
H3C
CH3
CH3
Cl
Cl
OH2
H3C
Cl CH
3
-
Cl
CH2 C
CH3
CH CH
H3C
CH3
+
CH3
CH CH
H3C
CH3
Cl
CH3 etc.
CH CH
H3C
CH3
Br
O18
+ H2O18
+ HOR
O18H
OR
Stereochemistry - mechanism requires
stereochemistry to be preserved or lost
Br2 addition to cyclopentene: explain anti? Syn?
mixture products? 1,3 addition?
+
Br2
Br
-
Br
Br
X
Br
Br
kinetic evidence
Rate law - indicates which component contributes to RDS
1. follow appearance of product or disappearance of reactant
2. determine dependence on reactants, catalyst, product
3. rate = d[product]/dt, units are mol/Ls
[H+]
[A]
[B]
4. example: rate = ko + k[H+]1/2[A][B]2
0.001
0.1
0.1
order of reaction is 3 1/2, 1/2 in [H+], 1st in [A] and 2nd
0.001
0.2
0.1
[B] (sum of exponents)
0.001
0.1
0.2
5. compare rate law with proposed mechanism
0.004
0.1
0.1
inital rates (< 10%) can be determined for complex
reaction to avoid change in concentrations
common rate laws
k1
1. first-order rate law, A 
B, d[ A] / dt  d[B] / dt  k1[ A] (units of k1?)
k1
2. 2nd-order rate law, A  B 
 C, d[ A] / dt  d[B] / dt  d[C] / dt  k1[ A][B]
3. 2nd/3rd-order law
Br Br
rate
1 x 10-4
2 x 10-4
4 x 10-4
2 x 10-4
CHEM 73/8311 CS3 Structure Effects on Stability and Reactivity.docx
d[E] / dt  k 2 [C][D], C may not be detectable
d[C] / dt  0 , steady state assumption
d[C] / dt  k1[ A ][B]  k 1[C]  k 2 [C][D]  0
2012Jan18
A+B
3
k1
k-1
k2
C+D
k1[ A ][B]
k1k 2 [ A ][B][D]
[ C] 
 d[E] / dt 
k 1  k 2 [D]
k 1  k 2 [D]
k1k 2
kk
for k 1  k 2 [D]  d[E] / dt 
[ A][B][D], kobs = 1 2 , 3rd order overall
k 1
k 1
for k 2 [D]  k 1  d[E] / dt  k 1[A ][B], kobs = k 1 2nd order overall
for [B]  [A], [B] does not change and k obs  k1[B], pseudo -1st order
a useful rate law is expressed in terms of observable species (reactants)
Integrated rate expression
k1
1st-order in A, A 
 product
[ A]t
t
[ A]0
 d [ A] / dt  k1 [ A]   d [ A] /[ A]    k1 dt  ln[ A]t  ln[ A]0  k1t  ln
 k1t
[ A]t
[ A]0
0
plot ln[Ao/At] versus t, slope is k1, units are s-1
2nd-order in A
k1
 d[ A ] / dt  2d[P] / dt
A  A 
product :
[ A]t
C
E
d[P] / dt  k 1[A ]2
t
d [ A]
 d [ A] / dt  2k1 [ A]  
   2k1 dt  1 /[ A] |[[ AA]]t0  2k1t  1 /[ A]t  1 /[ A]0  2k1t
2
[ A]0 [ A]
0
2
plot 1/[A] - 1/[A]o versus t
k is the rate constant of the event not the rate of molecules disppearing (units are M -1 s-1)
mixed second order
use pseudo-1st order conditions
k1
A  B 
products ,[B]  [ A],k obs  k1[B] , units = s-1
plot ln([A]o/[A]t) versus t, slope = k1[B]
Reaction Half-Life
1. most useful for 1st-order and pseudo 1st-order processes, independent of
concentrations
[ A ]0
[A]t1/2 = 1/2 [A]0 ,ln
 k1t , ln[2] = 0.693 = k1t1/2
[ A ]t
t1/2 = 0.693/k1  1/k1 , e. g. 10-9 s  109 s-1 or 10 hr  1/10 hr-1 (t1/2  1/k1)
note no concentration dependence
2. higher order processes dependent on initial concentration
1 /[ A]t1 / 2  1 /[ A]0  2k1t1 / 2  1 /( 0.5[ A]0 )  1 /[ A]0  2k1t1 / 2  t1 / 2  1 /( 2k1 [ A]0 )
i.e. increasing A decreases lifetime
Rates of Molecular processes
1. unimolecular  frequency of vibration (velocity of fragments)  1012 - 1013 s-1
2. bimolecular  limited by diffusion, reactant must "meet" diffusion rate  109 - 1011 s-1 in
solution, depends on viscosity of media and molecular size
3. termolecular is seldom encountered
Eyring Equation
CHEM 73/8311 CS3 Structure Effects on Stability and Reactivity.docx
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1. theory of transition state decomposition – show rxn coordinate for A+B  TS  product
2. rate of transition state breakdown = kBT/h = 6 x 1012 @ 25 oC
 =transmission coefficient =1, kB = Boltzman, T = temp in Kelvin, h = Planck's
rate= [TS]kBT/h
[TS]
k BT / h

A  B 
TS 

 product , K* 
 [TS]  K * [A][B] , rate = K*[A][B]kBT/h
[A][B]
 G*
RT
G* = -RTlnK*, = free energy difference between reactants and TS, K*  e
k T G*
k T  G*
k
rate = B e RT [ A ][B] , rate constant = k  B e RT for A  B 

product
h
h
3. log k = 13 - G*/1.4 @ 25 oC (G* = kcal/mol)
4.2
14
24
G*
4. G* =H* - TS* substituting,
10
3
k
10
10
10-4
kB T S*/R  H*/RT
t1/2
k
e
e
10-10
10-3
104
h
100 ps 1 ms
3 hr
5. ln k/T = ln kB/h + S*/R - H*/(RT), plot Ln k/T
assumes B in excess at 1 M.
versus 1/(RT)
slope = H*, calculate S* from intercept
6. H* is almost always positive, larger the more bond breaking in TS
7. S* is negative if # of components decrease in TS, positive if increased
solvation can make interpretation tricky, e.g. SN1 in polar solvents is negative
Arrhenius equation
k  AeEa /RT , log k = log A - Ea/(2.303RT)
k T
A  B e( S*R)/R ,Ea  H * RT
h
28
10-7
107
4 mo
Linear Free-Energy Relations
A. Hammett observed the rate of base hydrolysis of aromatic esters and reactions of other
aromatic compounds correlated with the acid dissociation constants of related aromatic acids.
x indicates substituent on aromatic ring
B. log kx =  log Kx + C
C. substituents that increase Kx also increase kx: NO2 increases as acidity (stabilizes anion)
and rate of hydrolysis (increases the positive charge density at the carbonyl).
D. X = H: log kH =  log KH + C
O
log kx - log kH = log Kx -  log KH
log kx/kH =  log Kx/KH
OMe
X
x = log Kx/KH is a measure of electronic effects of X, x
 Hammett substituent constant
- electron withdrawing groups have positive x and
electron donating groups have negative x
- other rates and equilibria can be correlated with x
O
kx
O
X
O
OH
Kx
-
+ MeO-
O
O
-
+ H3O+
X
X
CHEM 73/8311 CS3 Structure Effects on Stability and Reactivity.docx
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- the  values indicate sensitivity to substituent changes relative to ionization of benzoic acid
derivatives.
x =(log Kx/KH)benzoic acids = (-Gx/RT + GH/RT)benzoic acids
for some reaction sensitive to electronic effects we have
log Kx/KH = (log Kx/KH)benzoic acids = (-Gx/RT + GH/RT) = (-Gx/RT + GH/RT)benzoic acids
similarly
log kx/kH = (log Kx/KH)benzoic acids = (-G*x/RT + G*H/RT) = (-Gx/RT + GH/RT)benzoic acids
E. sign and magnitude of 
1. ionization of benzoic acid derivatives by definition have  =1.
2. reactions with positive  have increase in charge density in TS (or product)
3. reactions with negative  have decrease in charge density in TS (or product)
4.  is negative for the protonation of anilines
5. | > 1 if reaction is more sensitive to substitutent
effects that ionization of benzoic acids (reverse is also
true, e.g. 2-phenylethanoic acid derivative)
Compare x
Deviation from LFER indicates change in mechanism (see
figure above) or electronic effect different than reference
reaction.
X
X
AcO-
X
OBs
OAc
X
AcO-
group
p
m
O-
-0.81
-0.47
OH
-0.38
0.13
OMe
-0.28
0.10
H
0
0
F
0.15
0.34
Cl
0.24
0.37
Br
0.26
0.37
I
0.28
0.34
NH3+
0.60
0.86
N2+
1.93
1.65
Taft Ingold equation
Ingold noted that acid and base hydrolysis of aliphatic
esters are nearly equal in sensitivity to steric effects
and that acid hydrolysis was insensitive to electronic
effects
O
R'
thus log(kx/kH)A = Es (Es is steric parameter, 
indicates sensitivity to steric effects)
and I = 0.181[log(kx/kH)B -log(kx/kH)A] (electonic
parameter)
OH
H+
OR
R'
H2O
OR
OH
R'
OR
OH2
O
O
HO
R'
R'
OR
OR
OH
CHEM 73/8311 CS3 Structure Effects on Stability and Reactivity.docx
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in general for alkyl systems: log kx/kH = II + Es
the more negative Es the greater the steric bulk of X
consider thiol adition to aldehydes versus ketones.
Isotope Effects
1. rate can change if reactant has isotopic substitution (chlorination of ketone)
2. deuterium for hydrogen most popular, kH/kD
3. rate is usually slower due to difference in 0-point vibrational energy
4. position in potential energy for vibration determined by reduced mass
5. maximum is 7 for CH, 11 for OH and 6 for NH+
6. for CH, kH/kD = 2-7 for primary isotope effect, kH/kD = 1-2 for secondary
primary: bond broken in TS, timing effects magnitude
secondary: broken before or change in bonding (e.g. hybridization)
7. iodination of phenol
OH
OH
k1k 2 [P][ I 2 ][ B]
(a) rate =
k1
k 1[I  ]  k 2 [B]
+ I+ I2
k-1
(b) k2 [B] >> k-1[I ] then secondary isotope effect
(kH/kD = 1-2)
I H(D)
H(D)
(c) k-1[I-] >> k2[B] , second step is RDS, then
primary isotope effect
(d) observed kH/kD  4
8. hydrolysis of benzyl chloride
rds
PhCH 2Cl 
PhCH 2  H 2O  , kH/kD  1.4 ( secondary isotope effect)
rds
R 2CHCR 2 X 
R 2CHCR 2  , kH/kD  1.1 ( secondary isotope effect)
9. inverse isotope effect - greater bonding in TS
10. in following hydrogen transfer from Mn, kH/kD  0.4
R2C=CH2 + HMn(CO)4  Mn(CO)4 + R2CCH3
CH bond forming in TS, C-H is stronger than Mn-H
11. rigorous treatment requires consideration of all
bond vibration changes
kH
kD
 e
 h (  H *  D *  H  D )
2 k BT
= exp[0.1865/T], for CH bonds,  in cm-1
Kinetic versus Thermodynamic control
1. low temperatures favor product with lowest activation
energy even if not most stable
2. high temperature favors most stable product even if
barrier is higher, higher temperature allows access over
higher barrier
H
H2SO4
H
D
H
D
OH
k2[B]
I
H
SO3H
SO3H
SO3H
SO3H
CHEM 73/8311 CS3 Structure Effects on Stability and Reactivity.docx
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Hammond Postulate
"If two states, as for example, a transition state and an unstable intermediate, occur
consecutively during a reaction process and have nearly the same energy content, their
interconversion will involve only a small reorganization of the molecular structures.”
“In highly exothermic steps it will be expected that the transition states will resemble
reactants closely and in endothermic steps the products will provide the best models for the
transition states.'
Curtin-Hammett Postulate (Move)
absolute height difference between transition states determines
relative rates for products formed
Microscopic reversibility (Move)
reaction always follows lowest energy path so path is same in
reverse as it is in forward direction. Critique the following
mechanism.
Br
Br
Br
Br
reactants
product
product
ACIDS and BASES
A. many reactions are acid and/or base catalyzed
B. reactive intermediates are often acids and bases
2. Ka defined by ability to protonate water
[H  ][A  ]


HA  H2O 
A

H
0
Ka

3

[HA ]
H
O
HBr
H
NBr
NBr
O
O
O
O
O
Br
HBr
Br
NH + Br2
N
NBr
H
O
NH
N
O
O
O
+ Br2
Br
C. extent of formation of intermediates can be determined
from pKas
1. strength of acid or base related to pKa of acid or
conjugate acid
O
O
O
O
3. pKeq = pKa = pKa1 - pKa2, a1 = acid, a2 = conjugate acid, pKa = -logKa
4. strong acid (a1) + strong base  weak acid (a2) + weak base
5 strongest acid in water is hydronium and strongest base is hydroxide
6. pKas are determined by the relative stability of acid and conjugate base
electronegativity: CH4 (48) NH3 (38) H2O (18) HF (5)
charge/field: NH3 (38) vs. NH4+ (9) CH3CO2H (5) vs. CF3CO2H (0)
delocalization: CH3CO2H (5) vs. CH3CH2OH (16)
hybridization: HCCH (25) H2C=CH2 (44) CH3CH3 (50)
overlap: HI > HBr > HCl > HF
solvation: H2O v t-BuOH in solution and gas-phase
see list of pKa's
D. proton transfer is diffusion controlled if pKa is  2 for transfer between heteroatoms
CHEM 73/8311 CS3 Structure Effects on Stability and Reactivity.docx
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1. hydrogen bond is formed first
2. proton transter to and from carbon are almost always slower than diffusion control
because of poor hydrogen bonding (why?) and reorganization
E. determination of pKa for very strong acids (conjugate acids of very weak bases)
1. Two problems:
a. Typical determination of Ka for strong acid does not work, no HA: Ka 
[H  ][A  ]
[HA ]
b. Need acidic medium (conc. H2SO4) to inhibit dissociation: incomplete solvation
H+, non-ideal behavior, concentrated acids or bases have greater activity if not
completely solvated
2. create acidity function HO equivalent to pH, determined using indicators of known pKa in
water
a. in a concentrated acid solution (e.g. 2 M H2SO4), effective concentration of acid will
be greater than the apparent concentration
Ka 
Ka[HI]
[HI]
[HI]
[H ][I ]
, [H ] 
, pH  pKa  log  , H0  pKaw  log 

[I ]
[I ]
[I ]
[HI]
from known Ka and a measurement of [HI]/[I-], a calculation of [H+] is greater than added
acid!!!!, define new measure of acid strength (H0) to avoid confusion
b. H0 is defined for specific solvent and concentration of acid
c. pKa of stronger acid can be determined using new H0
d. series of H0 are determined with increasingly acidic indicators
e. similar functions can be obtained for basic solutions
F. acid/base catalysis
1. specific acid/base - rate dependent directly on pH, independent of acid or base used to
adjust pH
2. general acid/base - rate depends on added acid or base even at constant pH
3. Bronsted acid catalysis law – GA or GB rate constant is proportional to acidity of acid
log k =  log Ka + C,  can be 0-1, indicates extent of proton transfer in TS?
4. if reaction is general acid/base catalyzed then proton transfer occurs in RDS
CHEM 73/8311 CS3 Structure Effects on Stability and Reactivity.docx
SA:
SB: H3N +
H3O+ + X-
HX +H2O
O H
O
H3O+ +
O-
O
+
SR
H2O
H3N
OH2
O H
O
GA:
H4N + HO
H2O
O H
O H
H2O +
+
HX
+
9
HO
RSH
+
2012Jan18
GB:
O
H
RS
+
ONH4
+
SR
X-
OH2
Separation of catalytic terms
HA + H 2O
H3O + S
HA + S
k1
k-1
H 3O + A
kH+
kobs  k0  k H  [ H  ]  k HA[ HA]
plot kobs v. [HA] at constant [H+]
kHX
slope = kHA (assuming not GB), intercept = k0 + kH+[H+]
repeat at other [H+]
ko
S
rate expression for scheme at left
d [ S ] / dt  k0 [ S ]  k H  [ H  ][ S ]  k HA[ HA][ S ]
plot intercepts versus [H+]
slope = kH+, intercept = k0
what if kobs  k0  k H  [ H  ]  k HA[ HA]  k A [ A ]  k0  k H  [ H  ]  k HA[ HA]  k A K a

then kobs  k0  k H  [ H  ]  k HA 


slope = k HA 

[ HA]
[H  ]
k A K a 
[ HA] and plot kobs v. [HA] at constant [H+]
 
[H ] 
k A K a 
and intercept = k0 + kH+[H+], determine k0 + kH+ as above
[H  ] 
plot slope v. 1/[H+], intercept is kHA and slope is kA-Ka
GA equivalent ot SA + GB and GB is equivalent to SB + GA
CHEM 73/8311 CS3 Structure Effects on Stability and Reactivity.docx
GA:
O H
O
+ HX
NH2
H2O +
SA + GB:
H3O+ +
rds
NH2
k
+ X
k1
NH2
+ NH3
OH
10
O
+
+ H
OH
OH
H+
H3O+ + X-
O H
OH+
k-1
OH+
NH3+
OH2+
H2O + HX
O
O H
-H+/+H+
2012Jan18
+ H2O + X
NH2
k2
NH2
rds
OH
comparison of kinetic expressions:
GA : Rate  k[ HX ][ RCONH 2 ][ H 2O] , SA  GB : rate  k2[ X  ][ AH  ][ H 2O] , A = amide
d [ AH  ]
 k1[ H  ][ A]  k1[ AH  ][ H 2O]  k2 [ X  ][ AH  ][ H 2O]
dt
k1[ H  ][ A]
[ AH ] 
k1[ H 2O]  k2 [ X  ][ H 2O]

rate 
k2 k1[ H  ][ A][ X  ][ H 2O] k2 k1[ H  ][ A][ X  ]

k1[ H 2O]  k2 [ X  ][ H 2O]
k1  k2 [ X  ]
if water addition is rds then k-1 >>
where ka 
k2[X-]
k2 k1[ H  ][ A][ X  ]
and rate 
k1
k [ HX ]
[ H 3O  ][ X  ]
, then rearrange to [ X  ]  a  and substituting for X[ H 3O ]
[ HX ]
k2 k1[ H  ][ A]ka [ HX ] /[ H  ] k2 k1ka

[ A][ HX ] which is kinetically indestinguishable from the
k1
k1
GA rate expression: Rate  k[ HX ][ RCONH 2 ][ H2O]
rate 
Lewis Acids and bases
A. Lewis acid - vacant or electron deficient orbital
B. Lewis base - pair of electrons (lone-pair,  and -electrons)
C. Hard Soft Acid Base Theory
1. soft acids and bases have low electronegativity and high polarizability
2. hard acids and bases have high electronegativity and low polarizability
3. hard acids prefer to bond to hard bases and soft acids prefer to bond to softs
bases
4. hard acids and bases tend to form ionic bonds and soft acids and bases tend to form
covalent bonds
CHEM 73/8311 CS3 Structure Effects on Stability and Reactivity.docx
2012Jan18
Marcus Theory (for single step processes)
activation energy is composed of two parts
1.kinetic part - intrinsic free energy of activation G*int: barrier that would occur if G of
reaction was zero
2. thermodynamic part - results from G of reaction, wR and wP (work required to bring
together reactant or products
3. for OH  MeBr  HOMe  Br 
(a) G* for OH- + MeOH (41.8 kcal/mol) and Br- + MeBr (23.7 kcal/mol)
(b) approximate symmetrical potential energy surface with parabolas facing up
-point of intersection corresponds to activation energies
-use average parabolas to create intersection at averaged barrier (32.3 kcal/mol)
-adjust vertically to reflect free energy of reaction (G = -23.4 kcal)
( Go  w R  w P )2
*
 ( Go  w R  w P ) / 2 
(c) G*  Gint
*
16( Gint
 wR )
(d) assume work terms are small
( Go )2
*
G*  Gint
 Go / 2 
*
16 Gint
*
*
*
 ( GMeOH
 GMeBr
)/2
(e) Gint
Kurtz, J. L. J. Org Chem. 1983, 48, 5117.
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