h

2.  Thiols
Thiols (R–S–H
R–S–H) and sulfides
sulfides (R–S–R’
R–S–R’) are sulfur
analogs of alcohols and ethers, respectively
 Sulfur replaces oxygen

3.  Thiols
Thiols (RSH
RSH), also known as mercaptans, are sulfur
analogs of alcohols
 They are named with the suffix –
–thiol
thiol
 SH
SH group is called “mercapto group
mercapto group” (“capturer of
mercury”)

4.  Sulfides
Sulfides (RSR’
RSR’) are sulfur analogs of ethers
◦ They are named by rules used for ethers, with sulfide
sulfide in place of
ether
ether for simple compounds and alkylthio
alkylthio in place of alkoxy
alkoxy

5. Practice Problem
Practice Problem: Name the following compounds:
: Name the following compounds:

6.  Alkanes have only strong, nonpolar σ bonds
 No reaction with nucleophiles or
electrophiles
 Not much reactivity - paraffins (little
affinity)
© Prentice Hall 2001
Chapter 8 6

7. © Prentice Hall 2001
Chapter 8 7

8.  Initiation: Homolytic cleavage
© Prentice Hall 2001
Chapter 8 8
Cl Cl
400
o
C
or hν
2 Cl
Br Br
400
o
C
or hν
2 Br
radicals
Note that when an
arrowhead with a single
barb is used, it denotes
movement of a single
electron

9. © Prentice Hall 2001
Chapter 8 9

10. © Prentice Hall 2001
Chapter 8 10

11. © Prentice Hall 2001
Chapter 8 11

12.  The very reactive chlorine atom will have lower
selectivity and attack pretty much any hydrogen
available on an alkane
 The less reactive bromine atom will be more
selective and tends to react preferentially with the
easy targets, i.e. tertiary hydrogens
© Prentice Hall 2001
Chapter 8 12

13. © Prentice Hall 2001
Chapter 8 13

14.  Benzylic and allylic radicals are even more
stable than tertiary alkyl radicals
 It should be easy for a halogen radical to
abstract a benzylic or allylic hydrogen
© Prentice Hall 2001
Chapter 8 14

15.  Problem is that for the allyl radical there is a
greater likelihood that the halogen will add
electrophilically to the adjacent double bond
© Prentice Hall 2001
Chapter 8 15

16.  Electrophilic addition can be minimized by
maintaining the halogen at a very low
concentration
 Under these conditions, halogens can
substitute for allylic and benzylic hydrogens
© Prentice Hall 2001
Chapter 8 16

17.  N-Bromosuccinimide (NBS) is a good
reagent for supplying low concentrations of
bromine radical
© Prentice Hall 2001
Chapter 8 17

18.  Bromine radical comes from
the homolytic cleavage of the
N–Br Bond
 Low concentration of Br2 is
generated by the reaction of
NBS with HBr
 Neither HBr nor Br2
accumulate, so electrophilic
addition is slow
© Prentice Hall 2001 Chapter 8 18
+ Br + HBr
N
O
O
Br + HBr N
O
O
H + Br2
+ Br2
Br
+ Br

19.  When a radical abstracts an allylic or benzylic
hydrogen, a radical that is stabilized by resonance
is obtained
© Prentice Hall 2001
Chapter 8 19

20.  If the resonance hybrid is not symmetrical, more
than one product is obtained
© Prentice Hall 2001
Chapter 8 20
CH3CHCH CH3CH CHCH2
CH3CH2CH CH2
Br CH2
Br2
CH3CHCH
Br
CH3CH CHCH2Br
CH2
+
3-bromo-1-butene 1-bromo-2-butene

21. © Prentice Hall 2001
Chapter 8 21

22.  If a chirality
center already
exists, it may
affect the
distribution of
products
 A pair of
diastereomers
will be formed,
but in unequal
proportions
© Prentice Hall 2001 Chapter 8 22

23. © Prentice Hall 2001
Chapter 8 23
 Cyclic alkanes react with halogens in much the
same way as acyclic compounds
+ Cl2
Cl
+ HCl

24. © Prentice Hall 2001
Chapter 8 24
 Cyclopropane undergoes electrophilic addition
much like an alkene

25. © Prentice Hall 2001
Chapter 8 25

26.  Ozone (O3) is a major constituent of smog
 In the stratosphere, a layer of ozone shields the
Earth from harmful solar radiation
 The ozone layer is thinnest at the equator and
thickest in polar regions
© Prentice Hall 2001
Chapter 8 26

27.  Ozone is formed in the stratosphere by interaction
of short-wavelength ultraviolet light with oxygen
© Prentice Hall 2001
Chapter 8 27
O2
O3
hν
O + O
O + O2

28.  The stratospheric ozone layer acts as a filter for
biologically harmful ultraviolet radiation
 Scientists have noted a precipitous drop in the
ozone concentrations over Antarctica since
1985
 Circumstantial evidence links the depletion in
ozone to synthetic chlorofluorocarbons (CFCs)
- used as refrigerants
© Prentice Hall 2001
Chapter 8 28

29.  Chlorofluorocarbons (CFCs) are exceptionally
stable, but under the intense ultraviolet
radiation present in the stratosphere, they
undergo a radical dissociation
© Prentice Hall 2001
Chapter 8 29
C Cl
F
Cl
F
hν
C
F
Cl
F
+ Cl

30.  The chlorine radicals are ozone-removing
reagents
 It has been estimated that each chlorine radical
destroys 100,000 ozone molecules in a radical
chain reaction
© Prentice Hall 2001
Chapter 8 30
Cl + O3 ClO + O2
ClO + O3 Cl + 2 O2
2 O3 3 O2
Overall

31.  Production and use of CFCs has been slowed, but
because these materials have a half-life of 70 -
120 years they will be around in the stratosphere
for a long time
 The ozone hole over Antarctica was observed in
October 1999 to be a little smaller than in October
1998
© Prentice Hall 2001
Chapter 8 31

32. Chapter 8 32
 Electrons in pi bond are loosely held.
 Electrophiles are attracted to the pi electrons.
 Carbocation intermediate forms.
 Nucleophile adds to the carbocation.
 Net result is addition to the double bond.
=>

33. Chapter 8 33
 Step 1: Pi electrons attack the electrophile.
C C + E
+
C
E
C +
C
E
C + + Nuc:
_
C
E
C
Nuc
=>
• Step 2: Nucleophile attacks the carbocation.

34. Chapter 8 34
=>

35. Chapter 8 35
Protonation of double bond yields the most stable
carbocation. Positive charge goes to the carbon
that was not protonated.
X
=>
+ Br
_
+
+
CH3 C
CH3
CH CH3
H
CH3 C
CH3
CH CH3
H
H Br
CH3 C
CH3
CH CH3

36. Chapter 8 36
CH3 C
CH3
CH CH3
H Br
CH3 C
CH3
CH CH3
H
+
+ Br
_
CH3 C
CH3
CH CH3
H
+
Br
_
CH3 C
CH3
CH CH3
H
Br
=>

37. Chapter 8 37
 Markovnikov’s Rule: The proton of an acid adds
to the carbon in the double bond that already has
the most H’s. “Rich get richer.”
 More general Markovnikov’s Rule: In an
electrophilic addition to an alkene, the
electrophile adds in such a way as to form the
most stable intermediate.
 HCl, HBr, and HI add to alkenes to form
Markovnikov products. =>

38. Chapter 8 38
 In the presence of peroxides, HBr adds to an
alkene to form the “anti-Markovnikov” product.
 Only HBr has the right bond energy.
 HCl bond is too strong.
 HI bond tends to break heterolytically to form
ions.
=>

39. Chapter 8 39
 Peroxide O-O bond breaks easily to form free
radicals.
+
R O H Br R O H + Br
O O
R R +
R O O R
heat
• Hydrogen is abstracted from HBr.
Electrophile
=>

40. Chapter 8 40
 Bromine adds to the double bond.
+
C
Br
C H Br
+ C
Br
C
H
Br
Electrophile =>
C
Br
C
C C
Br +
• Hydrogen is abstracted from HBr.

41. Chapter 8 41
 Tertiary radical is more stable, so that
intermediate forms faster. =>
CH3 C
CH3
CH CH3 Br
+
CH3 C
CH3
CH CH3
Br
CH3 C
CH3
CH CH3
Br
X

42. Chapter 8 42
 Reverse of dehydration of alcohol
 Use very dilute solutions of H2SO4 or H3PO4 to
drive equilibrium toward hydration.
=>
C C + H2O
H
+
C
H
C
OH
alkene
alcohol

43. Chapter 8 43
+
C
H
C
+
H2O C
H
C
O H
H
+
+ H2O
C
H
C
O H
H
+
C
H
C
O
H
H3O
+
+ =>
C C O
H H
H
+
+ + H2O
C
H
C
+

44. Chapter 8 44
 Markovnikov product is formed.
+
CH3 C
CH3
CH CH3 O
H H
H
+
+ H2O
+
H
CH3
CH
CH3
C
CH3
H2O
CH3 C
CH3
CH CH3
H
O
H H
+
H2O
CH3 C
CH3
CH CH3
H
O
H
=>

45. Chapter 8 45
 Oxymercuration-Demercuration
◦ Markovnikov product formed
◦ Anti addition of H-OH
◦ No rearrangements
 Hydroboration
◦ Anti-Markovnikov product formed
◦ Syn addition of H-OH
=>

46. Chapter 8 46
 Reagent is mercury(II) acetate which
dissociates slightly to form +
Hg(OAc).
 +
Hg(OAc) is the electrophile that attacks the pi
bond.
CH3 C
O
O Hg O C
O
CH3 CH3 C
O
O
_
Hg O C
O
CH3
+
=>

47. Chapter 8 47
The intermediate is a cyclic mercurinium ion, a
three-membered ring with a positive charge.
C C
+
Hg(OAc) C C
Hg
+
OAc
=>

48. Chapter 8 48
 Water approaches the mercurinium ion from
the side opposite the ring (anti addition).
 Water adds to the more substituted carbon to
form the Markovnikov product.
C C
Hg
+
OAc
H2O
C
O
+
C
Hg
H
H
OAc
H2O
C
O
C
Hg
H
OAc
=>

49. Chapter 8 49
Sodium borohydride, a reducing agent, replaces
the mercury with hydrogen.
C
O
C
Hg
H
OAc
4 4 C
O
C
H
H
+ NaBH4 + 4 OH
_
+ NaB(OH)4
+ 4 Hg + 4 OAc
_
=>

50. Chapter 8 50
Predict the product when the given alkene reacts
with aqueous mercuric acetate, followed by
reduction with sodium borohydride.
CH3
D
(1) Hg(OAc)2, H2O
(2) NaBH4
=>
OH
CH3
D
H
anti addition

51. Chapter 8 51
If the nucleophile is an alcohol, ROH, instead of
water, HOH, the product is an ether.
C C
(1) Hg(OAc)2,
CH3OH
C
O
C
Hg(OAc)
H3C
(2) NaBH4
C
O
C
H3C
H
=>

52. Chapter 8 52
 Borane, BH3, adds a hydrogen to the most
substituted carbon in the double bond.
 The alkylborane is then oxidized to the alcohol
which is the anti-Mark product.
C C
(1) BH3
C
H
C
BH2
(2) H2O2, OH
-
C
H
C
OH
=>

53. Chapter 8 53
 Borane exists as a dimer, B2H6, in equilibrium with
its monomer.
 Borane is a toxic, flammable, explosive gas.
 Safe when complexed with tetrahydrofuran.
THF THF .
BH3
O B2H6 O
+
B
-
H
H
H
+
2 2 =>

54. Chapter 8 54
 The electron-deficient borane adds to
the least-substituted carbon.
 The other carbon acquires a positive charge.
 H adds to adjacent C on same side (syn).
=>

55. Chapter 8 55
Borane prefers least-substituted carbon due to steric hindrance as
well as charge distribution.
=>
C C
H3C
H3C
H
H
+ BH3
B
C
C H
CH3
H3C
H
H
C
C
H
H
H
CH3
CH3
C
C
H
H
H3C
CH3
H
3

56. Chapter 8 56
 Oxidation of the alkyl borane with basic
hydrogen peroxide produces the alcohol.
 Orientation is anti-Markovnikov.
CH3 C
CH3
H
C
H
H
B
H2O2, NaOH
H2O
CH3 C
CH3
H
C
H
H
OH
=>

57. Chapter 8 57
Predict the product when the given alkene reacts with borane
in THF, followed by oxidation with basic hydrogen
peroxide.
CH3
D
(1)
(2)
BH3, THF
H2O2, OH-
=>
syn addition
H
CH3
D
OH

58. Chapter 8 58
 Alkene + H2 → Alkane
 Catalyst required, usually Pt, Pd, or Ni.
 Finely divided metal, heterogeneous
 Syn addition
=>

59. Chapter 8 59
 Insertion of -CH2 group into a double bond
produces a cyclopropane ring.
 Three methods:
◦ Diazomethane
◦ Simmons-Smith: methylene iodide and Zn(Cu)
◦ Alpha elimination, haloform
=>

60. Chapter 8 60
Extremely toxic and explosive. =>
N N CH2 N N CH2
diazomethane
N N CH2
heat or uv light
N2 +
carbene
C
H
H
C
H
H
C
C
C
C
C
H
H

61. Chapter 8 61
Best method for preparing cyclopropanes.
CH2I2 + Zn(Cu) ICH2ZnI
a carbenoid
CH2I2
Zn, CuCl
=>

62. Chapter 8 62
 Haloform reacts with base.
 H and X taken from same carbon
CHCl3 + KOH K
+ -
CCl3 + H2O
C
Cl
Cl
Cl Cl
-
+
C
Cl
Cl
Cl
Cl
CHCl3
KOH, H2O
=>

63. Chapter 8 63
Cis-trans isomerism maintained around carbons
that were in the double bond.
C C
H
CH3
H
H3C NaOH, H2O
CHBr3
C C
H
CH3
H
H3C
Br
Br
=>

64. Chapter 8 64
 Cl2, Br2, and sometimes I2 add to a double bond
to form a vicinal dibromide.
 Anti addition, so reaction is stereospecific.
C
C + Br2 C C
Br
Br
=>

65. Chapter 8 65
 Pi electrons attack the bromine molecule.
 A bromide ion splits off.
 Intermediate is a cyclic bromonium ion.
C
C + Br Br C
C
Br
+ Br =>

66. Chapter 8 66
Halide ion approaches from side opposite the
three-membered ring.
C
C
Br
Br
C
C
Br
Br
=>

67. Chapter 8 67
=>

68. Chapter 8 68
 Add Br2 in CCl4 (dark, red-brown color) to an
alkene in the presence of light.
 The color quickly disappears as the bromine adds
to the double bond.
 “Decolorizing bromine” is the chemical test for the
presence of a double bond.
=>

69. Chapter 8 69
 If a halogen is added in the presence of water,
a halohydrin is formed.
 Water is the nucleophile, instead of halide.
 Product is Markovnikov and anti.
C
C
Br
H2O
C
C
Br
O
H H
H2O
C
C
Br
O
H
+ H3O
+
=>

70. Chapter 8 70
The most highly substituted carbon has the most
positive charge, so nucleophile attacks there.
=>

71. Chapter 8 71
Predict the product when the given alkene reacts with
chlorine in water.
CH3
D
Cl2, H2O
=>
OH
CH3
D
Cl

72. Chapter 8 72
 Alkene reacts with a peroxyacid to form an
epoxide (also called oxirane).
 Usual reagent is peroxybenzoic acid.
C
C + R C
O
O O H C
C
O
R C
O
O H
+
=>

73. Chapter 8 73
One-step concerted reaction. Several bonds
break and form simultaneously.
O
C
O
R
H
C
C
O
O
H
O
C
O
R
C
C
+
=>

74. Chapter 8 74
Since there is no opportunity for rotation around
the double-bonded carbons, cis or trans
stereochemistry is maintained.
C
C
CH3 CH3
H H Ph C
O
O O H
C
C
CH3 CH3
H H
O
=>

75. Chapter 8 75
 Acid catalyzed.
 Water attacks the protonated epoxide.
 Trans diol is formed.
C
C
O
H3O
+
C
C
O
H
H2O
C
C
O
OH
H H H2O
C
C
O
OH
H
=>

76. Chapter 8 76
 To synthesize the glycol without isolating the
epoxide, use aqueous peroxyacetic acid or
peroxyformic acid.
 The reaction is stereospecific.
CH3COOH
O
OH
H
OH
H
=>

77. Chapter 8 77
 Alkene is converted to a cis-1,2-diol,
 Two reagents:
◦ Osmium tetroxide (expensive!), followed by hydrogen
peroxide or
◦ Cold, dilute aqueous potassium permanganate,
followed by hydrolysis with base
=>

78. Chapter 8 78
Concerted syn addition of two oxygens to form a
cyclic ester.
C
C
Os
O O
O
O
C
C
O O
O
O
Os
C
C
OH
OH
+ OsO4
H2O2
=>

79. Chapter 8 79
If a chiral carbon is formed, only one
stereoisomer will be produced (or a pair of
enantiomers).
C
C
CH2CH3
H CH2CH3
C
C
CH2CH3
CH2CH3
OH
OH
H
H
H2O2
H
(2)
(1) OsO4
cis-3-hexene meso-3,4-hexanediol
=>

80. Chapter 8 80
 Both the pi and sigma bonds break.
 C=C becomes C=O.
 Two methods:
◦ Warm or concentrated or acidic KMnO4.
◦ Ozonolysis
 Used to determine the position of a double bond in
an unknown.
=>

81. Chapter 8 81
 Permanganate is a strong oxidizing agent.
 Glycol initially formed is further oxidized.
 Disubstituted carbons become ketones.
 Monosubstituted carbons become carboxylic
acids.
 Terminal =CH2 becomes CO2.
=>

82. Chapter 8 82
C
C
CH3 CH3
H CH3 KMnO4
(warm, conc.)
C C
CH3
CH3
OH
OH
H3C
H
C
O
H3C
H
C
CH3
CH3
O
C
O
H3C
OH
+
=>

83. Chapter 8 83
 Reaction with ozone forms an ozonide.
 Ozonides are not isolated, but are treated with
a mild reducing agent like Zn or dimethyl
sulfide.
 Milder oxidation than permanganate.
 Products formed are ketones or aldehydes.
=>

84. Chapter 8 84
C
C
CH3 CH3
H CH3 O3
C
H3C
H
O O
C
CH3
CH3
O
Ozonide
+
(CH3)2S
C
H3C
H
O C
CH3
CH3
O CH3 S
O
CH3
DMSO
=>