1. Chapter 3: Radical Halogenation
O O
O
O
N

2. The Ozone Hole

3. Radical Halogenation and Bond
Strength
Reactions require bond breaking and bond making
Bond strengths: Homolytic cleavage
radicals
∆H = DHº = Bond dissociation energy (kcal mol-1)
This process contrasts with heterolytic cleavage
A B A B
+ -
+
A B A· B+
e.g. H OH, DHº = +119,
yet: H2O + H2O H3O + OH easy
+ -
·

4. H H H H+
DHº = +104 kcal mol-1
To functionalize alkanes, we need to break C H
A 1-minute problem: DHº = ?
The simplest bond dissociation:
A. Same as H–H
B. Larger
C. Smaller

5. We expect C H to be less than for H2.
But: Are all C–H bonds the same ?
DHº s decrease along the series:
CH4 > Rprim―H > Rsec―H > Rtert―H
 No!
Primary
Secondary
Tertiary

6. CH3 H DHº = +105 kcal mol-1

8. Calculate feasibility of a reaction:
CH3–OH + H–I CH3–I + H–OH ΔH° = ??

9. CH3–OH + H–I CH3–I + H–OH
93 71 57 119
164 – 176 = –12 kcal mol-1

11. Remember BH3!
Why do we see such a trend?
R is sp2-hybridized.
Substitution stabilizes the radical. How?

12. Hyperconjugation
p-Orbital (with single e) overlaps with
bonding molecular orbital of
neighboring C-H (or any other) bond.
C C
H
C C
Hyperconjugation

13. Hyperconjugation
Potential Energy Diagram
C H
Hyperconjugation
E
C H bond
p-Orbital
MO picture of
sp3
1s(H)
Net stabilization: 3e
Bonding MO
Antibonding MO

14. Hyperconjugation

15. Radical Halogenation:
Methane and Chlorine (kcal mol-1)
CH3 H + Cl Cl CH3 Cl + H Cl
Exothermic, but needs heat (∆) and/or light to start.
105 58 ∆Hº = -25 85 103
hv, ∆
CCl4
Mechanism
1. Initiation: Cl2 2 Cl ∆Hº = +58
“light the match”
hv or ∆

16. How does the Cl–Cl bond break?
Thermally: Vibrational
energy gets
sufficiently large to
cause bond breaking.
Photochemically:
Absorption of
photon causes
excitation of
bonding electron
to antibonding
molecular orbital.

17. 2. Propagation (“fire”): A radical chain mechanism
a. CH4 + Cl  CH3 + HCl ∆Hº = +2
b. CH3 + Cl2  CH3Cl + Cl ∆Hº = -27
up!
down!
[a. + b.]: CH4 + Cl2  CH3Cl + HCl ∆Hº = -25
3. Termination: 2Cl  Cl2
CH3 + Cl  CH3Cl
CH3 + CH3  CH3 CH3
Kills
propagation
Mech
105 103
58 85
Note: Initiation step does not enter into equation. Only a
few Cl∙ needed to convert all of the starting material.

18. Orbital Picture of H·
Abstraction
Fast!
Partial
radical
character
δ∙
resembles
products

19. Potential energy diagram of propagation steps
gives picture of the energetic “ups and downs”.

20. Lipshutz
Trenet

21. Other Halogenations of Methane
Compare important DH º values:
Reactivity: F2 > Cl2 ~ Br2 > I2 won’t go!
F2 Cl2 Br2 I2 HF HCl HBr HI F Cl Br I
38 58 46 36 136 103 87 71 110 85 70 57
explodes good !
CH3 X
Initiation OK for all
Why?

22. F2 Cl2 Br2 I2 HF HCl HBr HI F Cl Br I
38 58 46 36 136 103 87 71 110 85 70 57
Won’t go!
Endothermic.
Remember: CH3--H 105 kcal mol-1
CH3―X

23. Reactivity determined in step a. by DHº = H―X. Let’s
compare the position of the transition states along
reaction coordinate. Hammond Postulate.
not far, small δ∙ on C
far larger δ∙ on C
Early TS  fast , exothermic step ( F).
Late TS  slow , endothermic step ( Br, I).

24. Selectivity for Differing C-H Bonds
CH3CH2CH3 CH3CH2CH2Cl + CH3CHCH3
Statistical (expected) 3 : 1
Reactivity (expected) Less (prim) More (sec)
Found (25 ºC) : 43 : 57
Reactivity per H: 43/6 = 7.2 57/2 = 28.5
1 : 4
CH3
CH2
CH3 CH3
CH3
CH3C
H
prim, sec, tert
Cl
-HCl
Cl2, hv

25. Transition states radical-like; reflect relative stabilities of products
For alkanes other
than CH4, 1st pro-
pagation step is
exothermic

26. Tertiary?
Statistical (expected) 9 : 1
Reactivity (expected) Less (prim) More (tert)
Found (25 ºC) 64 : 36
Normalized per H: 64/9 = 7 36/1 = 36
1 : 5
Result: Relative reactivity (selectivity) in
chlorinations at 25ºC: Tert : Sec : Prim = ~ 5 : 4 : 1
CH3
CH3
CH3
ClCH2
CH3
CH3
CH3
CH3
CH3C C ClHC H
-HCl
Cl2, hv
+

27. Selectivity and Other Halogens
(CH3)3CH + F2  FCH2CH + (CH3)3CF 9:1
(CH3)3CH + Br2  (CH3)3CBr only !
statistical !CH3
CH3
“Early TS”
(CH3)3CH + F2  FCH2CH + (CH3)3CF 9:1
(CH3)3CH + Br2  (CH3)3CBr only !
“Late TS”

28. Just to get a feel for the numbers……..
Selectivities vary extensively with the reagent employed, e.g., ICl, ROCl, R

29. Problem
Rank H3C∙, H2N∙, and HO∙ in the order of increasing
reactivity (diminishing selectivity):
a. H3C∙, H2N∙, HO∙
b. H2N∙, H3C∙, HO∙
c. H2N∙, HO∙, H3C∙