Organic Chemistry

1. Chapter 4
The Study of
Chemical Reactions
Organic Chemistry

2. Chapter 4 2
Tools for Study
• To determine a reaction’s
mechanism, look at:
Equilibrium constant
Free energy change
Enthalpy
Entropy
Bond dissociation energy
Kinetics
Activation energy =>

3. Chapter 4 3
Chlorination of Methane
• Requires heat or light for initiation.
• The most effective wavelength is blue, which
is absorbed by chlorine gas.
• Lots of product formed from absorption of
only one photon of light (chain reaction).
=>
C
H
H
H
H + Cl2
heat or light
C
H
H
H
Cl + HCl

4. Chapter 4 4
Free-Radical Chain Reaction
• Initiation generates a reactive intermediate.
• Propagation: the intermediate reacts with a
stable molecule to produce another reactive
intermediate (and a product molecule).
• Termination: side reactions that destroy the
reactive intermediate.
=>

5. Chapter 4 5
Initiation Step
A chlorine molecule splits homolytically
into chlorine atoms (free radicals)
=>
Cl Cl + photon (hν) Cl + Cl

6. Chapter 4 6
Propagation Step (1)
The chlorine atom collides with a
methane molecule and abstracts
(removes) a H, forming another free
radical and one of the products (HCl).
C
H
H
H
H Cl+ C
H
H
H
+ H Cl
=>

7. Chapter 4 7
Propagation Step (2)
The methyl free radical collides with
another chlorine molecule, producing
the other product (methyl chloride) and
regenerating the chlorine radical.
C
H
H
H
+ Cl Cl C
H
H
H
Cl + Cl
=>

8. Chapter 4 8
Overall Reaction
C
H
H
H
H Cl+ C
H
H
H
+ H Cl
C
H
H
H
+ Cl Cl C
H
H
H
Cl + Cl
C
H
H
H
H + Cl Cl C
H
H
H
Cl + H Cl =>
Cl Cl + photon (hν) Cl + Cl

9. Chapter 4 9
Termination Steps
• Collision of any two free radicals
• Combination of free radical with
contaminant or collision with wall.
C
H
H
H
Cl+ C
H
H
H
Cl
Can you suggest others?
=>

10. Chapter 4 10
Equilibrium constant
• Keq = [products]
[reactants]
• For chlorination Keq = 1.1 x 1019
• Large value indicates reaction “goes to
completion.”
=>

11. Chapter 4 11
Free Energy Change
• ∆G = free energy of (products -
reactants), amount of energy available to
do work.
• Negative values indicate spontaneity.
• ∆Go
= -RT(lnKeq)
where R = 1.987 cal/K-mol
and T = temperature in kelvins
• Since chlorination has a large Keq, the free
energy change is large and negative.
=>

12. Chapter 4 12
Problem
• Given that -X is -OH, the energy difference for
the following reaction is -1.0 kcal/mol.
• What percentage of cyclohexanol molecules
will be in the equatorial conformer at
equilibrium at 25°C?
=>

13. Chapter 4 13
Factors Determining ∆G°
• Free energy change depends on
enthalpy
entropy
∀∆H° = (enthalpy of products) - (enthalpy
of reactants)
∀∆S° = (entropy of products) - (entropy of
reactants)
∀∆G° = ∆H° - T∆S° =>

14. Chapter 4 14
Enthalpy
• ∆Ho
= heat released or absorbed during
a chemical reaction at standard
conditions.
• Exothermic, (-∆H), heat is released.
• Endothermic, (+∆H), heat is absorbed.
• Reactions favor products with lowest
enthalpy (strongest bonds).
=>

15. Chapter 4 15
Entropy
• ∆So
= change in randomness, disorder,
freedom of movement.
• Increasing heat, volume, or number of
particles increases entropy.
• Spontaneous reactions maximize
disorder and minimize enthalpy.
• In the equation ∆Go
= ∆Ho
- T∆So
the
entropy value is often small.
=>

16. Chapter 4 16
Bond Dissociation Energy
• Bond breaking requires energy (+BDE)
• Bond formation releases energy (-BDE)
• Table 4.2 gives BDE for homolytic
cleavage of bonds in a gaseous molecule.
A B A + B
We can use BDE to estimate ∆H for a reaction.
=>

17. Chapter 4 17
Which is more likely?
Estimate ∆H for each step using BDE.
CH4 HCl+ +Cl CH3
CH3 + Cl2 CH3Cl + Cl
or
Cl+CH4 CH3Cl + H
H Cl2+ HCl Cl+
104 103
58 84
=>
104 84
58 103

18. Chapter 4 18
Kinetics
• Answers question, “How fast?”
• Rate is proportional to the concentration
of reactants raised to a power.
• Rate law is experimentally determined.
=>

19. Chapter 4 19
Reaction Order
• For A + B → C + D, rate = k[A]a
[B]b
a is the order with respect to A
a + b is the overall order
• Order is the number of molecules of that
reactant which is present in the rate-
determining step of the mechanism.
• The value of k depends on temperature as
given by Arrhenius: ln k = -Ea + lnA
RT
=>

20. Chapter 4 20
Activation Energy
• Minimum energy required to reach
the transition state.
• At higher temperatures, more
molecules have the required energy.
=>
C
H
H
H
H Cl

21. Chapter 4 21
Reaction-Energy Diagrams
• For a one-step reaction:
reactants → transition state → products
• A catalyst lowers the energy of the
transition state.
=>

22. Chapter 4 22
Energy Diagram for a
Two-Step Reaction
• Reactants → transition state → intermediate
• Intermediate → transition state → product
=>

23. Chapter 4 23
Rate-Determining Step
• Reaction intermediates are stable as long
as they don’t collide with another molecule
or atom, but they are very reactive.
• Transition states are at energy maximums.
• Intermediates are at energy minimums.
• The reaction step with highest Ea will be the
slowest, therefore rate-determining for the
entire reaction. =>

24. Chapter 4 24
Rate, Ea, and Temperature
X + CH4 HX + CH3
X E a Rate @ 300K Rate @ 500K
F 1.2 kcal 140,000 300,000
Cl 4 kcal 1300 18,000
Br 18 kcal 9 x 10-8
0.015
I 34 kcal 2 x 10-19
2 x 10-9
=>

25. Chapter 4 25
Conclusions
• With increasing Ea, rate decreases.
• With increasing temperature, rate
increases.
• Fluorine reacts explosively.
• Chlorine reacts at a moderate rate.
• Bromine must be heated to react.
• Iodine does not react (detectably).
=>

26. Chapter 4 26
Chlorination of Propane
• There are six 1° H’s and two 2° Η’s. We
expect 3:1 product mix, or 75% 1-
chloropropane and 25% 2-chloropropane.
• Typical product mix: 40% 1-chloropropane
and 60% 2-chloropropane.
• Therefore, not all H’s are equally reactive.
=>
1° C
2° C
CH3 CH2 CH3 + Cl2
hν CH2
Cl
CH2 CH3 + CH3 CH
Cl
CH3

27. Chapter 4 27
Reactivity of Hydrogens
• To compare hydrogen reactivity, find
amount of product formed per hydrogen:
40% 1-chloropropane from 6 hydrogens
and 60% 2-chloropropane from 2
hydrogens.
• 40% ÷ 6 = 6.67% per primary H and
60% ÷ 2 = 30% per secondary H
• Secondary H’s are 30% ÷ 6.67% = 4.5
times more reactive toward chlorination
than primary H’s. =>

28. Chapter 4 28
Predict the Product Mix
Given that secondary H’s are 4.5 times as
reactive as primary H’s, predict the
percentage of each monochlorinated
product of n-butane + chlorine.
=>

29. Chapter 4 29
Free Radical Stabilities
• Energy required to break a C-H bond
decreases as substitution on the carbon
increases.
• Stability: 3° > 2° > 1° > methyl
∆H(kcal) 91, 95, 98, 104
=>

30. Chapter 4 30
Chlorination Energy Diagram
Lower Ea, faster rate, so more stable
intermediate is formed faster.
=>

31. Chapter 4 31
• There are six 1° H’s and two 2° Η’s. We
expect 3:1 product mix, or 75% 1-
bromopropane and 25% 2-bromopropane.
• Typical product mix: 3% 1-bromopropane
and 97% 2-bromopropane !!!
• Bromination is more selective than
chlorination. =>
1° C
2° C
CH3 CH2 CH3 + CH2
Br
CH2 CH3 +Br2
heat
CH3 CH
Br
CH3
Bromination of Propane

32. Chapter 4 32
• To compare hydrogen reactivity, find amount of
product formed per hydrogen: 3% 1-
bromopropane from 6 hydrogens and 97% 2-
bromopropane from 2 hydrogens.
• 3% ÷ 6 = 0.5% per primary H and
97% ÷ 2 = 48.5% per secondary H
• Secondary H’s are 48.5% ÷ 0.5% = 97 times
more reactive toward bromination than primary
H’s.
=>
Reactivity of Hydrogens

33. Chapter 4 33
Bromination Energy Diagram
• Note larger difference in Ea
• Why endothermic?
=>

34. Chapter 4 34
Bromination vs. Chlorination
=>

35. Chapter 4 35
Endothermic and
Exothermic Diagrams
=>

36. Chapter 4 36
Hammond Postulate
• Related species that are similar in energy are
also similar in structure. The structure of a
transition state resembles the structure of the
closest stable species.
• Transition state structure for endothermic
reactions resemble the product.
• Transition state structure for exothermic
reactions resemble the reactants.
=>

37. Chapter 4 37
Radical Inhibitors
• Often added to food to retard spoilage.
• Without an inhibitor, each initiation step
will cause a chain reaction so that many
molecules will react.
• An inhibitor combines with the free
radical to form a stable molecule.
• Vitamin E and vitamin C are thought to
protect living cells from free radicals.
=>

38. Chapter 4 38
Reactive Intermediates
• Carbocations (or carbonium ions)
• Free radicals
• Carbanions
• Carbene
=>

39. Chapter 4 39
Carbocation Structure
• Carbon has 6 electrons,
positive charge.
• Carbon is sp2
hybridized
with vacant p orbital.
=>

40. Chapter 4 40
Carbocation Stability
• Stabilized by alkyl
substituents 2 ways:
• (1) Inductive effect:
donation of electron
density along the
sigma bonds.
• (2) Hyperconjugation:
overlap of sigma
bonding orbitals with
empty p orbital.
=>

41. Chapter 4 41
Free Radicals
• Also electron-
deficient
• Stabilized by alkyl
substituents
• Order of stability:
3° > 2° > 1° > methyl
=>

42. Chapter 4 42
Carbanions
• Eight electrons on C:
6 bonding + lone pair
• Carbon has a negative
charge.
• Destabilized by alkyl
substituents.
• Methyl >1° > 2 ° > 3 °
=>

43. Chapter 4 43
Carbenes
• Carbon is neutral.
• Vacant p orbital, so
can be electrophilic.
• Lone pair of
electrons, so can be
nucleophilic.
=>

44. Chapter 4 44
End of Chapter 4