Organic Chemistry 7e by Wade

1. Chapter 4
©2010, Prentice Hall
Organic Chemistry, 7th
Edition
L. G. Wade, Jr.
The Study of
Chemical Reactions

2. Chapter 4 2
Introduction
• Overall reaction: reactants → products
• Mechanism: Step-by-step pathway.
• To learn more about a reaction:
 Thermodynamics
 Kinetics

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.
• Many molecules of product are formed from
absorption of only one photon of light (chain
reaction).

4. Chapter 4 4
The Free-Radical Chain
Reaction
• Initiation generates a radical 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: Formation of
Chlorine Atom
A chlorine molecule splits homolytically into
chlorine atoms (free radicals).

6. Chapter 4 6
Propagation Step: Carbon
Radical
The chlorine atom collides with a methane
molecule and abstracts (removes) an H,
forming another free radical and one of the
products (HCl).

7. Chapter 4 7
Propagation Step: Product
Formation
The methyl free radical collides with another
chlorine molecule, producing the organic
product (methyl chloride) and regenerating the
chlorine radical.

8. Chapter 4 8
Overall Reaction

9. Chapter 4 9
Termination Steps
• A reaction is classified as a termination step when
any two free radicals join together producing a
nonradical compound.
• Combination of free radical with contaminant or
collision with wall are also termination steps.

10. Chapter 4 10
More Termination Steps

11. Chapter 4 11
Lewis Structures of Free
Radicals
• Free radicals are unpaired electrons.
• Halogens have 7 valence electrons so one of them
will be unpaired (radical). We refer to the halides as
atoms not radicals.

12. Chapter 4 12
Equilibrium Constant
• Keq = [products]
[reactants]
• For CH4 + Cl2  CH3Cl + HCl
Keq = [CH3Cl][HCl] = 1.1 x 1019
[CH4][Cl2]
• Large value indicates reaction “goes to
completion.”

13. Chapter 4 13
Free Energy Change
 ∆G = (energy of products) - (energy of reactants)
 ∆G is the amount of energy available to do work.
 Negative values indicate spontaneity.
∆Go
= -RT(lnKeq) = -2.303 RT(log10Keq)
where R = 8.314 J/K-mol and T = temperature in
kelvins.

14. Chapter 4 14
Factors Determining ∆G°
Free energy change depends on:
 Enthalpy
 ∆H°= (enthalpy of products) - (enthalpy of reactants)
 Entropy
 ∆S° = (entropy of products) - (entropy of reactants)
∆G° = ∆H° - T∆S°

15. Chapter 4 15
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).

16. Chapter 4 16
Entropy
• ∆So
= change in randomness, disorder,
or 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.

17. Chapter 4 17
Calculate the value of ∆ G° for the chlorination of methane.
∆ G° = –2.303RT(log Keq)
Keq for the chlorination is 1.1 x 1019
, and log Keq = 19.04
At 25 °C (about 298 ° Kelvin), the value of RT is
RT = (8.314 J/kelvin-mol)(298 kelvins) = 2478 J/mol, or 2.48 kJ/mol
Substituting, we have
∆ G° = (–2.303)(2.478 kJ/mol)(19.04) = –108.7 kJ/mol (–25.9 kcal>mol)
This is a large negative value for ∆ G°, showing that this chlorination has a large driving force that
pushes it toward completion.
Solved Problem 1
Solution

18. Chapter 4 18
Bond-Dissociation Enthalpies
(BDE)
• Bond-dissociation requires energy (+BDE).
• Bond formation releases energy (-BDE).
• BDE can be used to estimate ∆H for a reaction.
• BDE for homolytic cleavage of bonds in a
gaseous molecule.
 Homolytic cleavage: When the bond breaks, each atom gets
one electron.
 Heterolytic cleavage: When the bond breaks, the most
electronegative atom gets both electrons.

19. Chapter 4 19
Homolytic and Heterolytic
Cleavages

20. Chapter 4 20
Enthalpy Changes in Chlorination
CH3-H + Cl-Cl  CH3-Cl + H-Cl
Bonds Broken ∆H° (per Mole) Bonds Formed ∆H° (per Mole)
Cl-Cl +242 kJ H-Cl -431 kJ
CH3-H +435 kJ CH3-Cl -351 kJ
TOTALS +677 kJ TOTAL -782 kJ
∆H° = +677 kJ + (-782 kJ) = -105 kJ/mol

21. Chapter 4 21
Kinetics
• Kinetics is the study of reaction rates.
• Rate of the reaction is a measure of how the
concentration of the products increase while
the concentration of the products decrease.
• A rate equation is also called the rate law and
it gives the relationship between the
concentration of the reactants and the
reaction rate observed.
• Rate law is experimentally determined.

22. Chapter 4 22
Rate Law
• For the reaction A + B → C + D,
rate = kr[A]a
[B]b
 a is the order with respect to A
 b is the order with respect to B
 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.

23. Chapter 4 23
Activation Energy
• The value of k depends on temperature as
given by Arrhenius:
RTEa
Aek
/
r
−
=
where A = constant (frequency factor)
Ea = activation energy
R = gas constant, 8.314 J/kelvin-mole
T = absolute temperature
Ea is the minimum kinetic energy needed to react.

24. Chapter 4 24
Activation Energy (Continued)
• At higher temperatures, more molecules have
the required energy to react.

25. Chapter 4 25
Energy Diagram of an Exothermic
Reaction
• The vertical axis in this graph represents the potential
energy.
• The transition state is the highest point on the graph,
and the activation energy is the energy difference
between the reactants and the transition state.

26. Chapter 4 26
Rate-Limiting 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.

27. Chapter 4 27
Energy Diagram for the
Chlorination of Methane

28. Chapter 4 28
Rate, Ea, and Temperature
X + CH4 HX + CH3
X Ea(per Mole) Rate at 27 °C Rate at 227 °C
F 5 140,000 300,000
Cl 17 1300 18,000
Br 75 9 x 10-8
0.015
I 140 2 x 10-19
2 x 10-9

29. Chapter 4 29
Consider the following reaction:
This reaction has an activation energy (Ea) of +17 kJ/mol (+4 kcal/mol) and a ∆ H° of +4 kJ/mol (+1
kcal/mol). Draw a reaction-energy diagram for this reaction.
We draw a diagram that shows the products to be 4 kJ higher in energy than the reactants. The barrier
is made to be 17 kJ higher in energy than the reactants.
Solved Problem 2
Solution

30. Chapter 4 30
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).

31. Chapter 4 31
Primary, Secondary, and Tertiary
Hydrogens

32. Chapter 4 32
Chlorination Mechanism

33. Chapter 4 33
Bond Dissociation Energies for
the Formation of Free Radicals

34. Chapter 4 34
Tertiary hydrogen atoms react with Cl• about 5.5 times as fast as primary ones. Predict the product
ratios for chlorination of isobutane.
There are nine primary hydrogens and one tertiary hydrogen in isobutane.
(9 primary hydrogens) x (reactivity 1.0) = 9.0 relative amount of reaction
(1 tertiary hydrogen) x (reactivity 5.5) = 5.5 relative amount of reaction
Solved Problem 3
Solution

35. Chapter 4 35
Even though the primary hydrogens are less reactive, there are so many of them that the primary
product is the major product. The product ratio will be 9.0:5.5, or about 1.6:1.
Solved Problem 3 (Continued)
Solution

36. Chapter 4 36
Stability of Free Radicals
• Free radicals are more stable if they are
highly substituted.

37. Chapter 4 37
Chlorination Energy Diagram
• Lower Ea, faster rate, so more stable
intermediate is formed faster.

38. Chapter 4 38
Rate of Substitution in the
Bromination of Propane

39. Chapter 4 39
Energy Diagram for the
Bromination of Propane

40. Chapter 4 40
Hammond Postulate
• Related species that are similar in
energy are also similar in structure.
• The structure of the transition state
resembles the structure of the closest
stable species.
• Endothermic reaction: Transition state
is product-like.
• Exothermic reaction: Transition state
is reactant-like.

41. Chapter 4 41
Energy Diagrams: Chlorination
Versus Bromination

42. Chapter 4 42
Endothermic and
Exothermic Diagrams

43. Chapter 4 43
Radical Inhibitors
• Often added to food to retard spoilage
by radical chain reactions.
• 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.

44. Chapter 4 44
Radical Inhibitors (Continued)
• A radical chain reaction is fast and has many
exothermic steps that create more reactive radicals.
• When an inhibitor reacts with the radical, it creates a
stable intermediate, and any further reactions will be
endothermic and slow.

45. Chapter 4 45
Carbon Reactive Intermediates

46. Chapter 4 46
Carbocation Structure
• Carbon has 6 electrons, positively charged.
• Carbon is sp2
hybridized with vacant p orbital.

47. Chapter 4 47
Carbocation Stability

48. Chapter 4 48
Carbocation Stability
(Continued)
• Stabilized by alkyl
substituents in two ways:
1. Inductive effect: Donation
of electron density along the
sigma bonds.
2. Hyperconjugation:
Overlap of sigma bonding
orbitals with empty p orbital.

49. Chapter 4 49
Free Radicals
• Also electron-deficient.
• Stabilized by alkyl substituents.
• Order of stability:
3° > 2° > 1° > methyl

50. Chapter 4 50
Stability of Carbon Radicals

51. Chapter 4 51
Carbanions
• Eight electrons on
carbon: 6 bonding
plus one lone pair.
• Carbon has a
negative charge.
• Destabilized by alkyl
substituents.
• Methyl >1° > 2 ° > 3 °

52. Chapter 4 52
Carbenes
• Carbon is neutral.
• Vacant p orbital, so can be electrophilic.
• Lone pair of electrons, so can be nucleophilic.

53. Chapter 4 53
Basicity of Carbanions
• A carbanion has a negative charge on its
carbon atom, making it a more powerful base
and a stronger nucleophile than an amine.
• A carbanion is sufficiently basic to remove a
proton from ammonia.

54. Chapter 4 54
Carbenes as Reaction
Intermediates
• A strong base can abstract a proton from tribromomethane
(CHBr3) to give an inductively stabilized carbanion.
• This carbanion expels bromide ion to give dibromocarbene.
The carbon atom is sp2
hybridized with trigonal geometry.
• A carbene has both a lone pair of electrons and an empty p
orbital, so it can react as a nucleophile or as an electrophile.