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7-thermochemistry(U).pptxHow do we know if a reaction can occur?
And – if a reaction can occur what do we know about the reaction?
Information we want to know:
How much heat is generated?
How fast is the reaction?
Are any intermediates generated?
(What is the reaction mechanism?)
All of this information is included in an Energy Diagram
Potential energy
Reaction Coordinate
Potential energy
Reaction Coordinate
Starting material
If we know the “shape” of the reaction coordinate, then all questions about the mechanism can be answered (thermodynamics and kinetics) 172  
Equilibrium Constants
Equilibrium constants (Keq) indicate thermodynamically whether the reaction is
favored in the forward or reverse direction and the magnitude of this preference
Keq = ([C][D]) / ([A][B]) = ([products]) /([starting material])
Gibb’s Free Energy
The Keq is used to determine the Gibb’s free energy
ΔG = (free energy of products) – (free energy of starting materials)
If we use standard free energy then ΔG (25C and 1 atm)
Keq = e(-ΔG/RT)
ΔG = -RT(ln Keq) = -2.303 RT(log10 Keq)
A favored reaction thus has a negative value of ΔG (energy is released)
The free energy term has two contributions: enthalpy and entropy
Enthalpy (ΔH): heat of a reaction (due to bond strength) Exothermic reaction: heat is given off by the reaction (-ΔG)
Endothermic reaction: heat is consumed by the reaction (+ΔG)
Entropy (ΔS): a measure of the freedom of motion - Reactions (and nature) always prefer more freedom of motion
Organic reactions are usually controlled by the enthalpy
Bond Dissociation Energies
The free energy of organic reactions is often controlled by the enthalpic term
-The enthalpic term in organic reactions is often controlled by the energy of the bonds being formed minus the energy of the bonds being broken
The energies of bonds is called the Bond Dissociation Energy
Many types of bonds have been recorded (both experimentally and computationally) we can therefore predict the equilibrium of a reaction by knowing these BDE’s
A second important feature is the RATE of a reaction
The rate is not determined by Keq, But instead by the energy of activation
Knowing the Ea of a reaction tells us how fast a reaction will occur
Reaction Coordinate
Rate therefore depends on the structure of the transition state along the rate determining step
While both the thermodynamics and kinetics depend on the structure of the starting material, the thermodynamics depends on product structure
while rate depends on transition state structure 177  
Rate Equation
The rate of a reaction can be written in an equation that relates the rate to the concentration of various reactants
Rate = kr [A]a[B]b
The exponents are determined by the number of species involved for the reaction step - The exponents also indicate the “order” of the reaction with respect to A and B
Overall order of the reaction is a summation of the order for each individual reactant
Arrhenius Equation
Referring back to our energy diagram the rate can be related to the energy of activation (Ea)
kr = Ae(-Ea/RT)
A is the Arrhenius “preexponential” factor
Ea is the minimum kinetic energy required to cause the reaction to proceed
As a general guide, the rate of a reaction generally will double every ~10C increase in temperature
(as the temperature of a reaction increases, there are more molecules with the minimum energy required to cause a reaction to occur)
The Arrhenius equation is an empirical fit to observed data
In an attempt to correlate the rate characteristics (which are not known exactly, primarily due to the structure of the transition state is not determined exactly)
to the well established thermodynamic equations (Gibb’s equation is an exact equation) transition state theory was developed
Instead of calling the energy barrier as an “energy of activation” (Ea), the barrier will be called the ΔG‡
A B k1
If A goes to B it passes a transition state A‡, once reaction reaches A‡ it can either return to A or proceed to product B
A A‡ B
An equilibrium constant for the transition state versus the starting material can be written
K‡ A‡
A =
The equations can be solved in a similar method as Gibb’s for the equilibrium to obtain:
k1 = (T/h) e (-ΔG‡/RT) Eyring equation
k1 = (T/h) e (-ΔH‡/RT) e (ΔS‡/R) 181  
kr = Ae(-Ea/RT)
k1 = (T/h) e (-ΔH‡/RT) e (ΔS‡/R)
Arrhenius plot
Eyring plot
While the Arrhenius and Eyring equations are widely used to determine either energy of activation (Ea) or ΔH‡ for reactions,
they are both merely empirical equations that are not quantum mechanically correct
Main problem is that unlike Gibb’s equation where the points of the starting materials and products along the reaction coordinate are stable structures, the transition state is a high
energy point along the coordinate
The equations consider the pathway to the transition state as isolated from other pathways (and the subsequent pathway to the product) while in reality the pathways are linked
While it is convenient to draw a reaction coordinate as a one-dimensional curve, the actual surface is multidimensional
Reaction Coordinate Reaction Coordinate 183  
One consequence of this trajectory along a multidimensional surface has been proposed by Carpenter as a “conservation of momentum”
Consider a reaction that generates a diradical intermediate that can continue to generate two equal energy products
By symmetry these two potential energy barriers are identical, therefore the rate of inversion product must be identical to rate of retention product
Experimentally determine that ki/kr = 5.0
T.H. Peterson, B.K. Carpenter, J. Am. Chem. Soc., 1992, 114, 766-767 184  
In order to predict the equilibrium for a reaction need to know the relative energies of starting material and product
Since starting materials and products are stable structures, the energies can be accurately determined either experimentally or computationally
In a molecular mechanics calculation, the energy is predicted by comparison to known compounds already included in the database
Another way to determine the energy is similar to a molecular mechanics calculation, but use group additives to estimate the energy
In this approach, a group in one molecule will be identical in energy to the same group in another molecule
For n-Butane:
Have two groups with a C-(C)(H)3
Have two groups with a C-(C)2(H)2
Add the value of these groups to determine enthalpy of butane 185  
These groups values were initially tabulated by a chemist, Sidney William Benson (and are called Benson group equivalents)
Group   ΔH298     (kcal/mol)  
Group   ΔH298     (kcal/mol)  
C-­(C)(H)3   -­10.20   Cd-­(CB)(C)   8.64  
C-­(C)2(H)2   -­4.93   C-­(CB)(C)(H)2   -­4.86  
C-­(C)3(H)1   -­1.90   C-­(CB)(C)2(H)   -­0.98  
C-­(C)4   0.50   Ct-­(H)   26.93  
Cd-­(H)2   6.26   Ct-­(C)   27.55  
Cd-­(C)(H)   8.59   Ct-­(Cd)   29.20  
Cd-­(C)2   10.34   CB-­(H)   3.30  
Cd-­(Cd)(H)   6.78   CB-­(C)   5.51  
Cd-­(Cd)(C)   8.88   CB-­(Cd)   5.68  
Cd-­(CB)(H)   6.78  
S.W. Benson, F.R. Cruickshank, D.. Golden, G.R. Haugen, H.E. O’Neal, A.S. Rodgers, R. Shaw, R. Walsh, Chem. Rev., 1969, 69, 279-324 186  
Using these values, the enthalpy for any hydrocarbon can be determined
Have two groups with a C-(C)2(H)2
Add the value of these groups to determine enthalpy of butane
ΔH = 2(-10.20) + 2(-4.93) = -30.26 kcal/mol Experimentally determined value is -30.15 kcal/mol
For Benzene:
ΔH = 6(3.30) = 19.80 kcal/mol Experimentally determined value is 19.82 kcal/mol
2 C-(CB)(H)3
2(-10.20) 2(-10.20)
Values would be identical, how to account for steric strain? Need to apply corrections
Group ΔH298 Ring ΔH298 gauche 0.80 cyclopropane 27.6 cis 1.00 cyclobutane 26.2 ortho 0.57 cyclopentane 6.3 cis (1=tbutyl) 4.00 cyclohexane 0
cycloheptane 6.4
Value for trans isomer
Can therefore calculate the ΔH value for any hydrocarbon by adding the appropriate steric corrections (E/Z, ortho in aromatic, or ring effects)
There are also Benson group tables for oxygen, sulfur, nitrogen, and halogen containing compounds
We can therefore determine, or predict, energy of stable compounds (and thus predict the thermodynamics of a reaction)
*How do we determine energy of transition states?
Transition state energies are directly related to the rates of reactions
Transition states only have a transitory existance and by definition they cannot be isolated
Reaction Coordinate 190  
Hammond Postulate
In an ENDOTHERMIC reaction, the transition state is closer to the PRODUCTS in energy and structure. In an EXOTHERMIC reaction, the transition state is closer to the
REACTANTS in energy and structure
The rate of a reaction is dependent upon the energy difference between the starting material and transition state in the rate determining step
While the structure of the starting material and products can be determined, the structure of the transition state is difficult to determine experimentally
because it is at an energy maximum and cannot be isolated
As an aide in predicting rates, a generalization was made that is now referred to as the “Hammond Postulate”
Reaction Coordinate Reaction Coordinate
Exothermic reaction, transition state structure resembles starting material 191  
Hammond Postulate
Reaction Coordinate
Another consequence of the Hammond postulate is how the transition state character changes when modifications of a reaction occur
Consider a reaction that generates a radical intermediate during the reaction
A 3 radical is more stable than a 2 radical which is more stable than a 1 radical
Due to the 1 radical intermediate being more endothermic than the 2 or 3 radical, the transition state will more closely resemble the intermediate (therefore more radical character)
Increasing radical character
Curtin-Hammett Principle
Consider a reaction that equilibrates between two low energy states (often two conformers), and each can react to a different product
state 1 state 2
product 1
Product 2
Reaction Coordinate
The Curtin-Hammett principle states that the only thing that determines the ratio of products 1 & 2 is the energy of activation difference (1‡ versus 2‡)
The ratio of states 1 and 2 do not influence formation of P1 or P2 (which is more stable does not affect which product is obtained, the more stable transition state is important)
Ph Br
CH3 base
Whether the E or Z isomer is favored is not due to energy difference of conformers
not important
Kinetic versus Thermodynamic Control What forms faster (kinetic product) and what is more stable (thermodynamic product)
need not be the same
Consider the addition to conjugated dienes (as example react with H+/H2O)
Generate allylic cation in first step Allylic cation can have water react at two sites
Thus the kinetic product has water reacting at 2 site
(more substituted double bond)

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