Full recap on Bsc. Level first year organic chemistry

1. RECAP OF FIRST YEAR
ORGANIC CHEMISTRY
Ch. 6 - 1

2. Ch. 6 - 2
ALKYL HALIDES

3. Ch. 6 - 3
1. Organic Halides
❖ Halogens are more electronegative
than carbon
C X
+
−
X = Cl, Br, I

4. Ch. 6 - 4
C F
H
H
H C Cl
H
H
H C Br
H
H
H C I
H
H
H
C–X Bond
Length (Å)
1.39 1.78 1.93 2.14
C–X Bond
Strength
(kJ/mol)
472 350 293 239
Carbon-Halogen Bond Lengths
and Bond Strength
increase
decrease

5. Ch. 6 - 5
Chloride
-23.8
13.1
78.4
68
69
51
Group
Me
Et
Bu
s
Bu
i
Bu
t
Bu
Fluoride
-78.4
-37.7
32
-
-
12
Bromide
3.6
38.4
101
91.2
91
73.3
Iodide
42.5
72
130
120
119
100(dec)
1A. Physical Properties of Organic Halides:
Boiling Point (bp/oC)

6. Ch. 6 - 6
Different Types of Organic Halides
❖ Alkyl halides (haloalkanes)
Cl Br I
a 1o chloride a 2o bromide a 3o iodide
Attached to
1 carbon atom
C
Attached to
2 carbon atoms
C
C
Attached to
3 carbon atoms
C
C
C
sp3
C X

7. Ch. 6 - 7
❖ Vinyl halides (Alkenyl halides)
❖ Aryl halides
❖ Acetylenic halides (Alkynyl halides)
sp2
X
sp2
X
benzene or aromatic ring
sp
X

8. Ch. 6 - 8
C X
sp3
+
−
Alkyl halides
Prone to undergo
Nucleophilic Substitutions
(SN) and Elimination
Reactions (E)
sp2
X X X
sp2
sp
❖ Different reactivity than alkyl halides,
and do not undergo SN or E reactions

9. Ch. 6 - 9
Nu + C X
+
−
C
Nu + X
(nucleophile) (substrate) (product) (leaving
group)
The Nu⊖
donates
an e⊖ pair
to the
substrate
The bond
between
C and LG
breaks,
giving both
e⊖ from the
bond to LG
The Nu⊖ uses
its e⊖ pair to
form a new
covalent bond
with the
substrate C
The LG
gains the
pair of e⊖
originally
bonded
in the
substrate
2. Nucleophilic Substitution Reactions

10. Ch. 6 - 10
❖ Two types of mechanisms
● 1st type: SN2 (concerted mechanism)
R
C Br
R
R
+
−
HO
R
C
R
R
Br
HO
−
−
transition state (T.S.)
+ Br-
R
C
HO
R
R
Timing of The Bond Breaking & Bond
Making Process

11. Ch. 6 - 11
R
C Br
R
R
R
C
R
R
Br
(k1)
Step (1):
+
slow
r.d.s.
H2O
(k3)
R
C
R
R
O
H
H R
C
R
R
OH H3O+
Step (3)
+ +
R
C
R
R
H2O
(k2) R
C
R
R
O
H
H
Step (2)
+
k1 << k2 and k3
fast
fast
● 2nd type: SN1 (stepwise mechanism)

12. Ch. 6 - 12
C X
+
−
❖ A reagent that seeks a positive center
❖ Has an unshared pair of e⊖
e.g.: HO , CH3O , H2N (negative charge)
H2O, NH3 (neutral)
This is the positive
center that the
Nu⊖ seeks
3. Nucleophiles

13. Ch. 6 - 13
❖ Examples:
HO + C
Cl
Cl
H H
CH3
(substrate) (product) (L.G.)
(Nu )
C
OH
H H
CH3
+
O + C
Cl
Cl
H H
CH3
(substrate) (L.G.)
(Nu )
C
O
H H
CH3
+
H H H
H
C
OH
H H
CH3
H3O
+
(product)

14. Ch. 6 - 14
❖ To be a good leaving group, the substituent
must be able to leave as a relatively stable,
weakly basic molecule or ion
e.g.: I⊖, Br⊖, Cl⊖, TsO⊖, MsO⊖, H2O, NH3
OMs =
OTs = O S
O
O
CH3
O S
O
O
CH3
(Tosylate)
(Mesylate)
4. Leaving Groups

15. Ch. 6 - 15
❖ The rate of the substitution reaction is
linearly dependent on the
concentration of OH⊖ and CH3Br
❖ Overall, a second-order reaction 
bimolecular
HO + CH3 Br HO CH3 +
Rate = k[CH3Br][OH
-
]
Br
5. Kinetics of a Nucleophilic Substitution
Reaction: An SN2 Reaction

16. Ch. 6 - 16
❖ The rate of reaction can be measured by
● The consumption of the reactants
(HO⊖ or CH3Cl) or
● The appearance of the products
(CH3OH or Cl⊖) over time
HO + C Cl HO C +
(Nu )
Cl
H
H
H
H
H
H
(substrate) (product)
(leaving
group)
e.g.:
5A. How Do We Measure the Rate of
This Reaction?

17. Ch. 6 - 17
Time, t
Concentration,
M
[CH3Cl] ↓
[CH3OH] ↑
Graphically…
Rate =
Δ[CH3Cl]
Δt
= −
[CH3Cl]t=t − [CH3Cl]t=0
Time in seconds
[CH3Cl] ↓
[CH3OH] ↑

18. Ch. 6 - 18
Time, t
Concentration,
M
[CH3Cl]
Initial Rate
[CH3Cl]t=0
[CH3Cl]t=t
Initial Rate
(from slope)
= −
[CH3Cl]t=t − [CH3Cl]t=0
Δt

19. Ch. 6 - 19
❖ Example:
HO + Cl CH3 HO CH3 + Cl
60o
C
H2O
[OH⊖]t=0 [CH3Cl]t=0
Initial rate
mole L-1, s-1 Result
1.0 M 0.0010 M 4.9 × 10-7
1.0 M 0.0020 M 9.8 × 10-7 Doubled
2.0 M 0.0010 M 9.8 × 10-7 Doubled
2.0 M 0.0020 M 19.6 × 10-7 Quadrupled

20. Ch. 6 - 20
❖ Conclusion:
HO + Cl CH3 HO CH3 + Cl
60o
C
H2O
● The rate of reaction is directly
proportional to the concentration of
either reactant.
● When the concentration of either
reactant is doubled, the rate of
reaction doubles.

21. Ch. 6 - 21
H
C Br
H
H
+
−
HO
H
C
H
H
Br
HO
−
−
transition state (T.S.)
+ Br-
H
C
HO
H
H
negative OH⊖
brings an e⊖
pair to δ+ C; δ–
Br begins to
move away with
an e⊖ pair
O–C bond
partially formed;
C–Br bond
partially broken.
Configuration of
C begins to invert
O–C bond
formed; Br⊖
departed.
Configuration
of C inverted
6. A Mechanism for the SN2 Reaction

22. Ch. 6 - 22
(R)
(S)
CH3
C Br
CH2CH3
H
HO +
+ Br
CH3
C
HO
CH2CH3
H
(inversion)
❖ Inversion of configuration
8.The Stereochemistry of SN2 Reactions

23. Ch. 6 - 23
CH3 OCH3 + I
❖ Example:
CH3 I
+ OCH3
Nu⊖ attacks from the TOP face.
(inversion of configuration)

24. Ch. 6 - 24
+ Br
CN
❖ Example:
+ CN
Br
Nu⊖ attacks from the BACK face.
(inversion of
configuration)

25. Ch. 6 - 25
CH3
C Br
CH3
CH3
H2O
CH3
C OH
CH3
CH3
HBr
+ +
9. The Reaction of tert-Butyl Chloride
with Hydroxide Ion: An SN1 Reaction
⚫ The rate of SN1 reactions depends only on
concentration of the alkyl halide and is
independent on concentration of the Nu⊖
Rate = k[RX]
In other words, it is a first-order reaction
 unimolecular nucleophilic substitution

26. Ch. 6 - 26
⚫ In a multistep reaction, the rate of the
overall reaction is the same as the rate
of the SLOWEST step, known as the
rate-determining step (r.d.s)
⚫ For example:
k1 << k2 or k3
9A. Multistep Reactions & the Rate-
Determining Step

27. Ch. 6 - 27
❖ The opening A is
much smaller than
openings B and C
❖ The overall rate at
which sand reaches
to the bottom of
the hourglass is
limited by the rate
at which sand falls
through opening A
❖ Opening A is
analogous to the
rate-determining
step of a multistep
reaction
A
B
C

28. Ch. 6 - 28
❖ A multistep process
CH3
C Br
CH3
CH3
CH3
C
CH3
CH3
Br
(k1)
Step (1):
+
(ionization
of alkyl
halide)
slow
r.d. step
10. A Mechanism for the SN1 Reaction

29. Ch. 6 - 29
CH3
C
CH3
CH3
H2O
(k2) CH3
C
CH3
CH3
O
H
H
Step (2)
+
fast

30. Ch. 6 - 30
H2O
(k3)
CH3
C
CH3
CH3
O
H
H CH3
C
CH3
CH3
OH
H3O+
Step (3)
+
+
CH3
C
CH3
CH3
H2O
(k2) CH3
C
CH3
CH3
O
H
H
Step (2)
+
fast
fast

31. Ch. 6 - 31
H2O
(k3)
CH3
C
CH3
CH3
O
H
H CH3
C
CH3
CH3
OH
H3O+
Step (3)
+
+
CH3
C
CH3
CH3
H2O
(k2) CH3
C
CH3
CH3
O
H
H
Step (2)
+
k1 << k2 and k3
fast
fast

32. Ch. 6 - 32
❖ Carbocations are
trigonal planar
❖ The central carbon
atom in a carbocation
is electron deficient; it
has only six e⊖ in its
valence shell
❖ The p orbital of a
carbocation contains
no electrons, but it can
accept an electron pair
when the carbocation
undergoes further
reaction
11A. The Structure of Carbocations
11. Carbocations
C
H3C
H3C
CH3
sp2-sp3 p bond

33. Ch. 6 - 33
❖ General order of reactivity (towards
SN1 reaction)
● 3o > 2o >> 1o > methyl
❖ The more stable the carbocation
formed, the faster the SN1 reaction
11B. The Relative Stabilities of
Carbocations

34. Ch. 6 - 34
❖ Stability of cations
R
C
R R
R
C
R H
R
C
H H
H
C
H H
> > >
most stable (positive inductive effect)
❖ Resonance stabilization of allylic and
benzylic cations
CH2 CH2
etc.

35. Ch. 6 - 35
Ph
Br
CH2CH3
CH3
CH3OH
(S)
Ph
C
CH2CH3
CH3
(trigonal planar)
CH3OH
attack from left
CH3OH
attack from right
CH3OH
Ph
C
CH3 OCH3
CH2CH3
(R) and (S)
racemic mixture
50:50
chance
12. The Stereochemistry of SN1 Reactions
Ph
CH3O
CH2CH3
CH3
(R)
Ph
OCH3
CH2CH3
CH3
(S)
(1 : 1)

36. Ch. 6 - 36
❖ Example:
Br
(R) H2O
(SN1)
slow
r.d.s.
(one enantiomer)
(carbocation)
H2O
attack from
TOP face
H2O attack from
BOTTOM face
O
H H
O
H H
OH
(R)
OH
(S)
+
H2O
H2O
racemic mixture
( 1 : 1 )

37. Ch. 6 - 37
t
Bu
CH3


t
Bu
O
Me
Me H
t
Bu
Me
O
H
Me
t
Bu
I
Me MeOH
❖ Example:
slow
r.d.s.
MeOH
MeOH
trigonal planar
t
Bu
OMe
Me t
Bu
Me
OMe
+
MeOH
MeOH

38. Ch. 6 - 38
❖ The structure of the substrate
❖ The concentration and reactivity of the
nucleophile (for SN2 reactions only)
❖ The effect of the solvent
13. Factors Affecting the Rates of
SN1 and SN2 Reactions

39. Ch. 6 - 39
13A. The Effect of the Structure of the Substrate
❖ General order of reactivity (towards
SN2 reaction)
● Methyl > 1o > 2o >> 3o > vinyl or aryl
DO NOT
undergo
SN2 reactions

40. Ch. 6 - 40
Relative Rate (towards SN2)
methyl 1o
2o
neopentyl 3o
2  106
4  104
500 1 < 1
Most
reactive
Least
reactive
CH3 Br CH3CH2 Br CH3CH Br
CH3
C CH2Br
CH3
CH3
CH3 C Br
CH3
CH3
CH3
R Br HO
+ R OH Br
+
❖ For example:

41. Ch. 6 - 41
H
C Br
CH3
CH3
+
−
H
C Br
H
H
+
−
❖ Compare
HO + Br
H
C
HO
H
H
faster
HO + Br
H
C
HO
CH3
CH3
slower
HO

42. Ch. 6 - 42
+ Br
CH3
C
HO
CH3
CH3
extremely
slow
+ Br
H
C
HO
CH3
t
Bu
very
slow
H
C Br
H
t
Bu
+
−
CH3
C Br
CH3
CH3
+
−
HO
HO
HO
HO

43. Ch. 6 - 43
❖ Note NO SN2 reaction on sp2 or sp
carbons
e.g.
H
H
H
I
+ Nu No reaction
No reaction
+ Nu
I
I No reaction
+ Nu
sp2
sp2
sp

44. Ch. 6 - 44
❖ General order of reactivity (towards SN1
reaction)
● 3o > 2o >> 1o > methyl
❖ The more stable the carbocation
formed, the faster the SN1 reaction
Reactivity of the Substrate in SN1
Reactions

45. Ch. 6 - 45
❖ Stability of cations
R
C
R R
R
C
R H
R
C
H H
H
C
H H
> > >
most stable (positive inductive effect)
❖ Allylic halides and benzylic halides also
undergo SN1 reactions at reasonable
rates I
Br
an allylic bromide a benzylic iodide

46. Ch. 6 - 46
❖ Resonance stabilization for allylic and
benzylic cations
CH2 CH2
etc.

47. Ch. 6 - 47
❖ For SN1 reaction
Recall: Rate = k[RX]
● The Nu⊖ does NOT participate in
the r.d.s.
● Rate of SN1 reactions are NOT
affected by either the
concentration or the identity of
the Nu⊖
13B. The Effect of the Concentration
& Strength of the Nucleophile

48. Ch. 6 - 48
❖ For SN2 reaction
Recall: Rate = k[RX][Nu]
● The rate of SN2 reactions depends
on both the concentration and
the identity of the attacking Nu⊖

49. Ch. 6 - 49
❖ Identity of the Nu⊖
● The relative strength of a Nu⊖ (its
nucleophilicity) is measured in
terms of the relative rate of its SN2
reaction with a given substrate
rapid
CH3O + CH3I CH3OCH3 + I
Good Nu⊖
Very
slow
CH3OH + CH3I CH3OCH3 + I
Poor Nu⊖

50. Ch. 6 - 50
❖ The relative strength of a Nu⊖ can be
correlated with 3 structural features
● A negatively charged Nu⊖ is always a
more reactive Nu⊖ than its conjugated
acid
⧫ e.g. HO⊖ is a better Nu⊖ than H2O
and RO⊖ is better than ROH
● In a group of Nu⊖s in which the
nucleophilic atom is the same,
nucleophilicities parallel basicities
⧫ e.g. for O compounds,
RO⊖ > HO⊖ >> RCO2
⊖ > ROH > H2O

51. Ch. 6 - 51
● When the nucleophilic atoms are
different, then nucleophilicities may
not parallel basicities
⧫ e.g. in protic solvents HS⊖, CN⊖,
and I⊖ are all weaker bases
than HO⊖, yet they are stronger
Nu⊖s than HO⊖
HS⊖ > CN⊖ > I⊖ > HO⊖

52. Ch. 6 - 52
Solvents
Non-polar solvents
(e.g. hexane, benzene)
Polar
solvents
Polar protic solvents
(e.g. H2O, MeOH)
Polar aprotic solvents
(e.g. DMSO, HMPA)
❖ Classification of solvents
13C. Solvent Effects on SN2 Reactions:
Polar Protic & Aprotic Solvents

53. Ch. 6 - 53
❖ SN2 Reactions in Polar Aprotic Solvents
● The best solvents for SN2 reactions
are
⧫ Polar aprotic solvents, which
have strong dipoles but do not
have OH or NH groups
⧫ Examples
O
S
CH3 CH3
O
H N
CH3
CH3
O
P
Me2N NMe2
NMe2
CH3CN
(DMSO) (DMF) (HMPA) (Acetonitrile)

54. Ch. 6 - 54
⧫ Polar aprotic solvents tend to
solvate metal cations rather than
nucleophilic anions, and this
results in “naked” anions of the
Nu⊖ and makes the e⊖ pair of
the Nu⊖ more available
Na
DMSO
+ DMSO Na
"naked anion"
CH3O CH3O

55. Ch. 6 - 55
CH3Br + NaI CH3I + NaBr
Solvent Relative Rate
MeOH 1
DMF 106
⧫ Tremendous acceleration in SN2
reactions with polar aprotic
solvent

56. Ch. 6 - 56
H
Nu H
H
H
OR
OR
OR
RO
❖ SN2 Reactions in Polar Protic Solvents
● In polar protic solvents, the Nu⊖
anion is solvated by the surrounding
protic solvent which makes the e⊖
pair of the Nu⊖ less available and
thus less reactive in SN2 reactions

57. Ch. 6 - 57
❖ Halide Nucleophilicity in Protic Solvents
● I⊖ > Br⊖ > Cl⊖ > F⊖
 Thus, I⊖ is a stronger Nu⊖ in protic
solvents, as its e⊖ pair is more available
to attack the substrate in the SN2 reaction.
H
H
H
H
H
H
OR
OR
OR
RO
RO
RO
(strongly solvated)
+
+
+
+
+
+
F-
H
H
H
RO
OR
OR
(weakly solvated)
I-

58. Ch. 6 - 58
❖ Halide Nucleophilicity in Polar Aprotic
Solvents (e.g. in DMSO)
● F⊖ > Cl⊖ > Br⊖ > I⊖
⧫ Polar aprotic solvents do not solvate
anions but solvate the cations
⧫ The “naked” anions act as the Nu⊖
⧫ Since F⊖ is smaller in size and the
charge per surface area is larger
than I⊖, the nucleophilicity of F⊖ in
this environment is greater than I⊖

59. Ch. 6 - 59
❖ Solvent plays an important role in SN1
reactions but the reasons are different
from those in SN2 reactions
❖ Solvent effects in SN1 reactions are due
largely to stabilization or destabilization
of the transition state
13D. Solvent Effects on SN1 Reactions:
The Ionizing Ability of the Solvent

60. Ch. 6 - 60
❖ Polar protic solvents stabilize the
development of the polar transition
state and thus accelerate this rate-
determining step (r.d.s.):
CH3
C
CH3 Cl
CH3
slow
r.d.s.
C Cl
+
−
H
H
OR
OR
O
R
H
+
−
+
CH2
C
CH3
CH3
H3C
CH3
CH3
+
Cl
-

61. Ch. 6 - 61
14. Organic Synthesis: Functional Group
Transformation Using SN2 Reactions
HO
Br
OH
MeO
Me
HS
SH
MeS
SMe
CN
CN

62. Ch. 6 - 62
Br
C C
Me
Me
MeCOO
O Me
O
Me3N
NMe3 Br
N3
N3
I
I

63. Ch. 6 - 63
❖ Examples:
Br O
??
NaOEt, DMSO
I SMe
??
NaSMe, DMSO

64. Ch. 6 - 64
Br
C C
Me
Me
MeCOO
O Me
O
Me3N
NMe3 Br
N3
N3
I
I

65. Ch. 6 - 65
❖ Examples:
I CN
(optically active, chiral) (optically active, chiral)
??
● Need SN2 reactions to control
stereochemistry
● But SN2 reactions give the inversion of
configurations, so how do you get the
“retention” of configuration here??
● Solution:
“double inversion”  “retention”

66. Ch. 6 - 66
Br
I CN
(optically active, chiral) (optically active, chiral)
??
(Note: Br⊖ is a stronger Nu than
I⊖ in polar aprotic solvent.)
NaBr
DMSO
NaCN
DMSO
(SN2 with
inversion)
(SN2 with
inversion)

67. Ch. 6 - 67
❖ Substitution
15. Elimination Reactions of Alkyl
Halides
❖ Elimination
C C
H
Br
Br-
C C
H OCH3
+
(acts as a
Nu )
OCH3
C C
H
H
Br
C C CH3OH Br-
(acts as a
base)
+ +
OCH3

68. Ch. 6 - 68
❖ Substitution reaction (SN) and
elimination reaction (E) are processes
in competition with each other
t
BuOK
t
BuOH
e.g.
I +
O
t
Bu
SN2: 15% E2: 85%

69. Ch. 6 - 69
15A. Dehydrohalogenation
C
H
C
X halide as LG
β carbon
β hydrogen
α carbon
H
Br
t
BuOK
t
BuOH, 60o
C
+ KBr +
t
BuOH
α
β
LG
β hydrogen
⊖OtBu

70. Ch. 6 - 70
❖ Conjugate base of alcohols is often used
as the base in dehydrohalogenations
15B. Bases Used in Dehydrohalogenation
R−O⊖ + Na⊕ + H2
R−O−H
R−O⊖ + Na⊕ + H2
Na
NaH
EtO Na
t
BuO K
sodium ethoxide potassium tert-butoxide
e.g.

71. Ch. 6 - 71
❖ Rate = k[CH3CHBrCH3][EtO⊖]
❖ Rate determining step involves both
the alkyl halide and the alkoxide anion
❖ A bimolecular reaction
16. The E2 Reaction
Br
H
EtO + + EtOH + Br

72. Ch. 6 - 72
Mechanism for an E2 Reaction
Et O
C C
Br
H
H
H
CH3
H
Et O
C C
Br
H
H
H
CH3
H
−
−
C C
H
H
H
CH3
Et OH + Br
+
β
α
EtO⊖ removes
a b proton;
C−H breaks;
new p bond
forms and Br
begins to
depart
Partial bonds in
the transition
state: C−H and
C−Br bonds
break, new p
C−C bond forms
C=C is fully
formed and
the other
products are
EtOH and Br⊖

73. Ch. 6 - 73
❖ E1: Unimolecular elimination
C Cl
CH3
CH3
CH3 H2O
slow
r.d.s
C
CH3
CH3
CH3
H2O as
nucleophile
(major (SN1))
C OH
CH3
CH3
CH3
H2O as
base
+CH2 C
CH3
CH3
(minor (E1))
17. The E1 Reaction

74. Ch. 6 - 74
Mechanism of an E1 Reaction
Cl
H2O
H
H2O
slow
r.d.s.
α carbon
β hydrogen
+ H3O
(E1 product)
fast
H2O
fast
O
H
H H2O
OH + H3O
(SN1 product)

75. Ch. 6 - 75
CH3
C Cl
CH3
CH3
CH3
C
CH3
CH3
Cl
(k1)
Step (1):
+
H2O
slow
r.d. step
Aided by the
polar solvent, a
chlorine departs
with the e⊖ pair
that bonded it to
the carbon
Produces relatively
stable 3o carbocation
and a Cl⊖. The ions
are solvated (and
stabilized) by
surrounding H2O
molecules

76. Ch. 6 - 76
H3C
C
H3C
H2O
(k2)
H O
H
H
Step (2)
+
C H CH2
H3C
H3C
+
H
H
fast
H2O molecule removes one of
the b hydrogens which are
acidic due to the adjacent
positive charge. An e⊖ pair
moves in to form a double
bond between the b and a
carbon atoms
Produces alkene and
hydronium ion

77. Ch. 6 - 77
18. How To Determine Whether
Substitution or Elimination Is Favoured
❖ All nucleophiles are potential bases and
all bases are potential nucleophiles
❖ Substitution reactions are always in
competition with elimination reactions
❖ Different factors can affect which type
of reaction is favoured

78. Ch. 6 - 78
C
Nu
C X
H
E2
(b)
(a)
SN2
(b)
(a)
C
C
Nu
H
+ X
C
+ X
C
+ Nu H
18A. SN2 vs. E2

79. Ch. 6 - 79
❖ With a strong base, e.g. EtO⊖
● Favor SN2
Primary Substrate
Br
NaOEt
EtOH
OEt
+
E2: (10%)
SN2: 90%

80. Ch. 6 - 80
❖ With a strong base, e.g. EtO⊖
● Favor E2
Secondary Substrate
Br
NaOEt
EtOH
OEt
+
+
E2: 80%
SN2: 20%

81. Ch. 6 - 81
❖ With a strong base, e.g. EtO⊖
● E2 is highly favored
Tertiary Substrate
Br OEt
NaOEt
EtOH
+
E2: 91% SN1: 9%

82. Ch. 6 - 82
❖ Unhindered “small” base/Nu⊖
Base/Nu⊖: Small vs. Bulky
❖ Hindered “bulky” base/Nu⊖
Br
KOt
Bu
t
BuOH
Ot
Bu
+
E2: 85%
SN2: 15%
Br
NaOMe
MeOH OMe
+
E2: 1%
SN2: 99%

83. Ch. 6 - 83
Basicity vs. Polarizability
Br
O CH3
O
+
O
C
CH3 O
(weak base)
E2: 0%
SN2: 100%
EtO
(strong base)
OEt
+
E2: 80%
SN2: 20%

84. Ch. 6 - 84
Tertiary Halides: SN1 vs. E1 & E2
EtO
Br
OEt
+
SN1: 0%
E2: 100%
(strong
base)
EtOH
heat
OEt
+
SN1: 80%
E1 + E2: 20%

85. Ch. 6 - 85
Review Problems
Br
DMF, 25
o
C
t
Bu
Na CN
(1)
CN
t
Bu
NaH
Et2O
I O
H
(2)
I O
O
H⊖
Intramolecular SN2
SN2 with inversion

86. Ch. 6 - 86
Cl
t
Bu
CH3
CH3
t
Bu
Cl
+
( 50 : 50)
OH
t
Bu
HCl
CH3
(3)
CH3
t
Bu
O
t
Bu
CH3
H
H
sp2 hybridized
carbocation
Cl⊖ attacks
from top face
Cl⊖ attacks
from bottom
face
SN1 with racemization

87. Ch. 7 - 87
5. Synthesis of Alkenes via
Elimination Reactions
❖ Dehydrohalogenation of Alkyl Halides
C C
H
X
H
H
H
H
H
H
H
H
base
-HX

88. Ch. 7 - 88
6. Dehydrohalogenation of Alkyl
Halides
❖ The best reaction conditions to use
when synthesizing an alkene by
dehydrohalogenation are those that
promote an E2 mechanism
C C
H
X
B: C C
E2
B:H + X
+

89. Ch. 7 - 89
6A. How to Favor an E2 Mechanism
❖ Use a secondary or tertiary alkyl halide
if possible. (Because steric hinderance
in the substrate will inhibit substitution)
❖ When a synthesis must begin with a
primary alkyl halide, use a bulky base.
(Because the steric bulk of the base
will inhibit substitution)

90. Ch. 7 - 90
❖ Use a high concentration of a strong
and nonpolarizable base, such as an
alkoxide. (Because a weak and
polarizable base would not drive the
reaction toward a bimolecular reaction,
thereby allowing unimolecular
processes (such as SN1 or E1 reactions)
to compete.

91. Ch. 7 - 91
❖ Sodium ethoxide in ethanol
(EtONa/EtOH) and potassium tert-
butoxide in tertbutyl alcohol (t-BuOK/t-
BuOH) are bases typically used to
promote E2 reactions
❖ Use elevated temperature because
heat generally favors elimination over
substitution. (Because elimination
reactions are entropically favored over
substitution reactions)

92. Ch. 7 - 92
❖ Examples of dehydrohalogenations
where only a single elimination product
is possible
6B. Zaitsev’s Rule
Br
(79%)
(1)
EtONa
EtOH, 55
o
C
(2)
Br
EtONa
EtOH, 55
o
C
(91%)
(3) Br
t -BuOK
t -BuOH, 40
o
C
(85%)
( )
n
( )
n

93. Ch. 7 - 93
❖ Rate = H3C
H
C
Br
CH3 EtO
k
Br
Hb
Ha
B
2-methyl-2-butene
2-methyl-1-butene
̶̶̶̶̶̶̶̶̶ H
a
̶̶̶̶̶̶̶̶̶ H
b
(2nd order overall)
 bimolecular

94. Ch. 7 - 94
❖ When a small base is used (e.g. EtO⊖
or HO
⊖
) the major product will be the
more highly substituted alkene (the
more stable alkene)
❖ Examples:
Br
Hb
Ha
NaOEt
EtOH
70o
C
+
69% 31%
(eliminate Ha
) (eliminate Hb
)
(1)
(2)
Br KOEt
EtOH
51% 18% 31%
+ +
69%

95. Ch. 7 - 95
❖ Zaitsev’s Rule
● In elimination reactions, the more
highly substituted alkene product
predominates
❖ Stability of alkenes
C C
Me
Me
Me
Me
> C C
Me
Me
H
Me
> C C
H
Me
Me
H
> C C
H
Me
H
Me
> C C
H
Me
H
H

96. Ch. 7 - 96
❖ Hofmann’s Rule
● Most elimination reactions follow
Zaitsev’s rule in which the most
stable alkenes are the major
products. However, under some
circumstances, the major
elimination product is the less
substituted, less stable alkene
6C. Formation of the Less Substituted
Alkene Using a Bulky Base

97. Ch. 7 - 97
● Case 1: using a bulky base
CH3CH2CHCH3
EtO
(small)
CH3CH CHCH3
CH3CH2CH CH2
CH3CH CHCH3
CH3CH2CH CH2
+
(80%)
(20%)
+
(30%)
(70%)
t
BuO
(bulky)
Br
H C C C C
H
H
H
H
H
Br
H
H
H
EtO
⊖
(small base)
tBuO
⊖
(bulky base)

98. Ch. 7 - 98
● Case 2: with a bulky group next
to the leaving halide
C C C C
Me
H3C
Me H
H
Me
Br H
H
H C C C CH2
Me
H3C
Me H
H
Me
(mainly)
EtO
more crowded β-H
less crowded β-H

99. Ch. 7 - 99
❖ Zaitsev Rule vs. Hofmann Rule
● Examples
Br
H
b
H
a
NaOEt, EtOH, 70
o
C
+
69% 31%
(eliminate H
a
) (eliminate H
b
)
(1)
KO
t
Bu,
t
BuOH, 75
o
C 28% 72%

100. Ch. 7 - 100
● Examples
NaOEt, EtOH, 70
o
C
+
91% 9%
(eliminate H
a
) (eliminate H
b
)
(2)
KO
t
Bu,
t
BuOH, 75
o
C 7% 93%
Br
H
b
H
a

101. Ch. 7 - 101
10. Synthesis of Alkynes by
Elimination Reactions
❖ Synthesis of Alkynes by
Dehydrohalogenation of Vicinal
Dihalides
Br
C
H
C
Br
H
C C
NaNH2
heat

102. Ch. 7 - 102
❖ Mechanism
Br
R C
H
C R
Br
H
NH2
R R
NH2
H
R
R
Br
E2

103. Ch. 7 - 103
❖ Examples
Br
Br
H
H (78%)
NaNH2
heat
(1)
NaNH2
heat
Ph
Ph
Br2
CCl4 Ph
Br H
Br
H
Ph
Ph
Ph
(2)

104. Ch. 7 - 104
❖ Synthesis of Alkynes by
Dehydrohalogenation of Geminal
Dihalides
O
R CH3
PCl5
0o
C R CH3
Cl Cl
gem-dichloride
1. NaNH2 (3 equiv.), heat
2. HA
Ph H

105. Ch. 7 - 105
11. Replacement of the Acetylenic
Hydrogen Atom of Terminal
Alkynes
❖ The acetylide anion can be prepared by
R H
NaNH2
liq. NH3
R Na + NH3

106. Ch. 7 - 106
❖ Acetylide anions are useful
intermediates for the synthesis of other
alkynes
R R' X R R' X
+
❖ ∵ 2nd step is an SN2 reaction, usually
only good for 1
o
R’
❖ 2
o
and 3
o
R’ usually undergo E2
elimination

107. Ch. 7 - 107
❖ Examples Ph H
Ph Na
NaNH2
liq. NH3
CH3 I
I
H
SN2 E2
Ph CH3
+
NaI
Ph H
+
+
I

108. Addition Reactions of
Alkyl Halides
Ch. 6 - 108

109. Ch. 8 - 109
1A. How To Understand Additions
to Alkenes
❖ This is an addition reaction: E–Nu
added across the double bond
C C E Nu
E
C C
Nu
+
Bonds broken Bonds formed
p-bond s-bond 2 s-bonds

110. Ch. 8 - 110
❖ Since pi bonds are formed from the
overlapping of p orbitals, p electron
clouds are above and below the plane
of the double bond
C C
p electron
clouds

111. Ch. 8 - 111
❖ Electrophilic
● electron seeking
● C=C and C≡C p bonds are
particularly susceptible to
electrophilic reagents (electrophiles)
❖ Common electrophile
● H+, X+ (X = Cl, Br, I), Hg2+, etc.

112. Ch. 8 - 112
❖ In an electrophilic addition, the p
electrons seek an electrophile,
breaking the p bond, forming a s bond
and leaving a positive charge on the
vacant p orbital on the adjacent
carbon. Addition of B–
to form a s
bond provides an addition product

113. Ch. 8 - 113
C C
E Nu
+
−
C C
E
+ Nu
Nu
C C
E

114. Ch. 8 - 114
C C
A
C
C
E
C
C
Nu
Nu
E Nu
+
−
2. Electrophilic Addition of
Hydrogen Halides to Alkenes:
Mechanism and Markovnikov’s
Rule
❖ Mechanism

115. Ch. 8 - 115
C C
E
❖ Mechanism
● Sometimes do not go through a
“free carbocation”, may go via

116. Ch. 8 - 116
C C
H
H
H
H
E Nu
E
C
C H
H
H
H
E
C C
H
H
H
H
E
C C
H
H
H
H
Nu E
C
C H
H
H
H
Nu
Nu
same
as
same
as
❖ Markovnikov’s Rule
● For symmetrical substrates, no
problem for regiochemistry

117. Ch. 8 - 117
E Nu
C C
H
H3C
H
H E
C
C H
H
CH3
H
E
C C
CH3
H
H
H
E
C C
CH3
H
H
H
Nu E
C
C H
H
CH3
H
Nu
Nu Nu
or
different
from
❖ Markovnikov’s Rule
● But for unsymmetrical substrates,
two regioisomers are possible

118. Ch. 8 - 118
❖ Markovnikov’s Rule
● In the electrophilic addition of an
unsymmetrical electrophile across a
double bond of an alkene, the more
highly substituted and more
stabilized carbocation is formed as
the intermediate in preference to
the less highly substituted and less
stable one

119. Ch. 8 - 119
E Nu
E Nu E
+
−
❖ Markovnikov’s Rule
● Thus
E
NOT
Note: carbocation stability  3
o
> 2
o
> 1
o

120. Ch. 8 - 120
Br
Br
fast
❖ Addition of Hydrogen Halides
● Addition of HCl, HBr and HI across
a C=C bond
● H+
is the electrophile
slow
r.d.s
+
−
H Br
+
Br
NO

121. Ch. 8 - 121
2A. Theoretical Explanation of
Markovnikov’s Rule
H X
C C
H
H3C
H
H
+
−
H
C
C H
H
CH3
H
H
C C
CH3
H
H
H
or
2o
carbocation
(more stable)
1o
carbocation
(more stable)
step 1
(slow
r.d.s.)
❖ One way to state Markovnikov’s rule is to
say that in the addition of HX to an alkene,
the hydrogen atom adds to the carbon atom
of the double bond that already has the
greater number of hydrogen atoms

122. Ch. 8 - 122
H Br
H H
Br
Br
fast
(1o
cation)
(minor)
slow
(r.d.s.)
Br
fast
H
Br
H
(2o
cation)
(major)
☓
Step 1 Step 2

123. Ch. 8 - 123
❖ Examples
+
(1)
H Cl
Cl
H
H
Cl
(95 : 5)
+
(2)
H Br
(98 : 2)
H
Br
Br
H

124. Ch. 8 - 124
2C. Regioselective Reactions
❖ When a reaction that can potentially yield
two or more constitutional isomers actually
produces only one (or a predominance of
one), the reaction is said to be
regioselective
+
H Cl
Cl
H
H
Cl
95 : 5
(major) (minor)
Regioselectivity:
regioisomers

125. Ch. 8 - 125
2D. An Exception to Markovnikov’s Rule
❖ Via a radical mechanism
❖ This anti-Markovnikov addition does not
take place with HI, HCl, and HF, even when
peroxides are present
H Br
Br
(anti-Markovnikov's
product)
RO OR
heat H

126. Ch. 8 - 126
C C
H
H
Bu
H
H X
3. Stereochemistry of the Ionic
Addition to an Alkene
C CH2 H
H
Bu
X
X
achiral
trigonal planar
carbocation
C CH3
H
X
Bu
(S)-2-Halohexane
(50%)
C CH3
H
X
Bu
(R)-2-Halohexane
(50%)
attack from top
attack from bottom
racemate

127. Ch. 8 - 127
12. Electrophilic Addition of Bromine
and Chlorine to Alkenes
❖ Addition of X–X (X = Cl, Br) across a
C=C bond
Br2
C C
Br
C C
Br
CCl4
(vicinal
dibromide)

128. Ch. 8 - 128
❖ Examples
(anti addition of Br2)
Br
Br
Br
Br
+
Br2
−5o
C
(racemate)
(1)
(anti addition of Cl2)
Cl2
−10o
C
(2) Ph
Ph
Ph
Ph
Cl
Cl
Cl
Ph
Ph
Cl
same as
(rotation of C1-C2 bond)
1
2

129. Ch. 8 - 129
Br
+ Br
+
−
C C
Br
Br
12A. Mechanism of Halogen Addition
C C + Br Br
Br
Br
Br–Br bond becomes
polarized when close
to alkene
(vincinal
Dibromide) (bromonium)

130. Ch. 8 - 130
Br
Br
H H CCl4
Br Br
❖ Stereochemistry
● Anti addition
H
Br
Br
H
SN2 reaction
(anti)
enantiomer +

131. Ch. 8 - 131
18. Electrophilic Addition of
Bromine & Chlorine to Alkynes
C C
R H
X
C C
X
R
X
H
X
CH2Cl2
(X = Cl, Br, I)
X2 (excess)
C C
R H
X
C C
X
R
X
H
X
C C
H
X
X
H
X2
X2
(anti-addition)

132. Ch. 8 - 132
19. Addition of Hydrogen Halides
to Alkynes
C C
R H
X
C C
H
R
X
H
H
(X = Cl, Br, I)
X (excess)
H
❖ Regioselectivity
● Follow Markovnikov’s rule
C C
Br
CH3
Br
H
H
H
gem-dibromide
C C
H3C H
HBr
C C
CH3
Br
H
H HBr

133. Ch. 8 - 133
❖ Mechanism
C C
CH3 H
H Br
C C
H
H
CH3
Br
C C
CH3
Br
H
H
C C
H
H
H
CH3
Br
Br
C
H
H
H
C
Br
CH3
Br
H Br

134. Ch. 8 - 134
❖ Anti-Markovnikov addition of hydrogen
bromide to alkynes occurs when
peroxides are present in the reaction
mixture
H Br
peroxides
Br
H
(E) and (Z)
(74%)

135. Ch. 6 - 135
Radical Reactions

136. Ch. 10 - 136
1. Introduction: How Radicals Form
and How They React
❖ Heterolysis
A B
ions
heterolytic
bond
cleavage
+
A B
❖ Homolysis
A B
radicals
homolytic
bond
cleavage
+
A B

137. Ch. 10 - 137
3. The Reactions of Alkanes with
Halogens
❖ Alkanes have no functional group and
are inert to many reagents and do not
undergo many reactions
❖ Halogenation of alkanes is one of the
most typical free radical reactions

138. Ch. 10 - 138
❖ Alkanes react with molecular halogens
to produce alkyl halides by a
substitution reaction called radical
halogenation
R H X2
+ R X H X
+
heat
or
light (h)

139. Ch. 10 - 139
3A. Multiple Halogen Substitution
+ Cl2
H
C
H H
H
+
H
C
H Cl
H
+
Cl
C
H Cl
H
+
Cl
C
Cl Cl
H
+
Cl
C
Cl Cl
Cl
H Cl
heat
or
light

140. Ch. 10 - 140
3B. Lack of Chlorine Selectivity
❖ Chlorination of most higher alkanes
gives a mixture of isomeric monochloro
products as well as more highly
halogenated compounds
● Chlorine is relatively unselective;
it does not discriminate greatly
among the different types of
hydrogen atoms (primary,
secondary, and tertiary) in an
alkane

141. Ch. 10 - 141
● Because alkane chlorinations
usually yield a complex mixture of
products, they are not useful as
synthetic methods when the goal is
preparation of a specific alkyl
chloride
Cl2
+ + + HCl
Cl
light
Cl
Polichlorinated
products
(23%)
Isobutane Isobutyl
chloride
(48%)
tert-Butyl
chloride
(29%)

142. Ch. 10 - 142
● An exception is the halogenation of
an alkane (or cycloalkane) whose
hydrogen atoms are all equivalent.
[Equivalent hydrogen atoms are
defined as those which on
replacement by some other group
(e.g., chlorine) yield the same
compound.]

143. Ch. 10 - 143
● Bromine is generally less reactive
toward alkanes than chlorine, and
bromine is more selective in the
site of attack when it does react
+ Cl2 + H Cl
heat
or
light
Cl
Neopentane
(excess)
Neopentyl
chloride

144. Ch. 10 - 144
4. Chlorination of Methane:
Mechanism of Reaction
❖ Most radical reactions include 3 stages
(steps)
(1) chain initiation
(2) chain propagation
(3) chain termination

145. Ch. 10 - 145
❖ Mechanism of Free Radical Chlorination
of CH4
(1) Chain initiation
Cl Cl
h
(homolytic
cleavage)
2 Cl
● Radicals are created in this step

146. Ch. 10 - 146
(2) Chain propagation
H
C
H H
H
+
H Cl CH3
CH3 + Cl Cl CH3Cl +
(i)
(ii)
+
Cl
Cl
● Repeating (i) and (ii) in a chain
reaction provides the product CH3Cl
● In chain propagation, one radical
generates another and the process
goes on

147. Ch. 10 - 147
(2) Chain propagation
Cl
C
H H
H
+
H Cl CH2Cl
CH2Cl + Cl Cl CH2Cl2 +
(ia)
(iia)
+
Cl
Cl
● Other than CH3Cl, other
chlorination products can be formed
in the chain propagation step

148. Ch. 10 - 148
(2) Chain propagation
Cl
C
H Cl
H
+
H Cl CHCl2
CHCl2 + Cl Cl CHCl3 +
(ib)
(iib)
+
Cl
Cl
Cl
C
H Cl
Cl
+
H Cl CCl3
CCl3 + Cl Cl CCl4 +
(ic)
(iic)
+
Cl
Cl

149. Ch. 10 - 149
(3) Chain termination
Cl CH3
CH3
CH3 +
+
Cl
CH3 H3C CH3
CH2Cl2
CH2Cl
+
Cl
CHCl2 + CCl3 Cl2HC CCl3

150. Ch. 10 - 150
(3) Chain termination
● Free radical reactions cannot be
completed without chain
termination
● All radicals are quenched in this
step
● Radical reactions usually provide
mixture of many different products
● Synthesis of CH3Cl or CCl4 is
possible using different amounts of
reactants (CH4 and Cl2)

151. Ch. 10 - 151
e.g.:
CH4 (large excess) + Cl2
CH4 + Cl2 (large excess)
CH3Cl (mainly)
h
CCl4 (mainly)
h

152. Ch. 10 - 152
6. Halogenation of Higher Alkanes
❖ Mechanism for radical halogenation of
ethane
Cl2 2 Cl
Step 1
Chain initiation
+ Cl
Step 2
Chain propagation
Cl Cl
H
CH3CH2 +
CH3CH2 H Cl
+
CH3CH2 CH3CH2 Cl + Cl
Step 3
light
or heat

153. Ch. 10 - 153
Chain termination
Cl Cl
+
CH3CH2 CH3CH2 Cl
+ Cl
+
CH3CH2 CH3CH2 CH2CH3
CH3CH2
Cl
Cl

154. Ch. 10 - 154
Cl
Cl
Cl2
light
25o
C
+
Cl
Cl
Cl2
light
25o
C
+
Cl
Cl2
300o
C
+
Cl
Cl
Cl
+ +

155. Ch. 10 - 155
6A. Selectivity of Bromine
❖ Bromination is slower than chlorination
because the 1st propagation step is
more endothermic (overall still
exothermic). As a result, bromination
is more selective than chlorination
Br2
H Br
H
Br
+ +
h
(99%) (< 1%)

156. Ch. 10 - 156
❖ Mechanism
Br Br
H
Br H Br
H
Br H Br
H
H
h
2 Br
(major; 3o
radical more stable)
+
(minor; 1o
radical less stable)
+

157. Ch. 10 - 157
❖ Mechanism
Br
Br
Br
H H
Br
(major)
(minor)

158. Ch. 10 - 158
Br
Br
h
127o
C
+
Br2
(trace) (> 99%)
Cl
Cl
h
25o
C
+
(63%) (37%)
Cl2

159. Ch. 10 - 159
7. The Geometry of Alkyl Radicals
❖ Planar, similar to carbocation
R C
R
R
p-orbital
sp2 hybridized

160. Ch. 10 - 160
8. Reactions That Generate
Tetrahedral Chirality Centers
Cl *
achiral
+
Cl
Pentane
(achiral)
Cl2
1-Chloropentane
(achiral)
( )-2-Chloropentane
(a racemic form)
+
3-Chloropentane
(achiral)
Cl

161. Ch. 10 - 161
Cl
CH2CH2CH3
CH3
H Cl
H3CH2CH2C
H3C
H
(R)-2-Chloropentane
(50%)
+ Cl
+ Cl
(S)-2-Chloropentane
(50%)
❖ The Stereochemistry of chlorination at
C2 of pentane
CH3CH2CH2CH2CH3
C2
enantiomers
Cl
CH3
C
H CH2CH2CH3
trigonal planar
radical (achiral)
(b)
Cl2
(a)
Cl2

162. Ch. 10 - 162
9. Radical Addition to Alkenes: The
Anti-Markovnikov Addition of
Hydrogen Bromide
❖ Anti-Markovnikov addition of HBr to
alkenes – peroxide effect
● Addition of HBr to alkenes usually
follows Markovnikov’s rule
Br Br
HBr
H H
not

163. Ch. 10 - 163
● In the presence of peroxides (RO–
OR), anti-Markovnikov addition is
observed
H H
HBr
Br Br
not
RO OR
heat

164. Ch. 10 - 164
❖ Mechanism
● Via a radical mechanism
RO OR
H Br ROH Br
RO
2 RO
heat
(homolytic cleavage)
(chain initiation)
+ +

165. Ch. 10 - 165
Br
Br +
Br
+
not Br
Br
H Br
Br
H
Br
+
+
(3o
radical, more stable)
(1o
radical, less stable)

166. Ch. 10 - 166
❖ Synthetic application
HBr
RO-OR
heat
HBr
Br
Br
(via more stable
2o
carbocation)
(via more stable
2o
radical)

167. Ch. 10 - 167
❖ Hydrogen bromide is the only hydrogen
halide that gives anti-Markovnikov
addition when peroxides are present
❖ Hydrogen fluoride, hydrogen chloride,
and hydrogen iodide do not give anti-
Markovnikov addition even when
peroxides are present