Chap 8 thiols and sulfides (1)

 ThiolsThiols (R–S–HR–S–H) and sulfidessulfides (R–S–R’R–S–R’) are sulfur
analogs of alcohols and ethers, respectively
 Sulfur replaces oxygen

 ThiolsThiols (RSHRSH), also known as mercaptans, are sulfur
analogs of alcohols
 They are named with the suffix ––thiolthiol
 SHSH group is called “mercapto groupmercapto group” (“capturer of
mercury”)

 SulfidesSulfides (RSR’RSR’) are sulfur analogs of ethers
◦ They are named by rules used for ethers, with sulfidesulfide in place of
etherether for simple compounds and alkylthioalkylthio in place of alkoxyalkoxy

Practice ProblemPractice Problem: Name the following compounds:: Name the following compounds:

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

 Initiation: Homolytic cleavage
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

 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

 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

 Problem is that for the allyl radical there is a
greater likelihood that the halogen will add
electrophilically to the adjacent double bond

 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

 N-Bromosuccinimide (NBS) is a good
reagent for supplying low concentrations of
bromine radical

 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

 When a radical abstracts an allylic or benzylic
hydrogen, a radical that is stabilized by resonance
is obtained

 If the resonance hybrid is not symmetrical, more
than one product is obtained
CH3CHCH CH3CH CHCH2CH3CH2CH CH2Br CH2
Br2
CH3CHCH
Br
CH3CH CHCH2BrCH2
+
3-bromo-1-butene 1-bromo-2-butene

 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

 Cyclic alkanes react with halogens in much the
same way as acyclic compounds
+ Cl2
Cl
+ HCl

 Cyclopropane undergoes electrophilic addition
much like an alkene

 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

 Ozone is formed in the stratosphere by interaction
of short-wavelength ultraviolet light with oxygen
O2
O3
hν
O + O
O + O2

 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

 Chlorofluorocarbons (CFCs) are exceptionally
stable, but under the intense ultraviolet
radiation present in the stratosphere, they
undergo a radical dissociation
C ClF
Cl
F
hν
CF
Cl
F
+ Cl

 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
Cl + O3 ClO + O2
ClO + O3 Cl + 2 O2
2 O3 3 O2Overall

 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

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.
=>

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.

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

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
HBr
=>

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. =>

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.
=>

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

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

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

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

Chapter 8 43
+C
H
C
+
H2O C
H
C
O H
H
+
+ H2OC
H
C
O H
H
+
C
H
C
O
H
H3O+
+ =>
C C OH H
H
+
+ + H2OC
H
C
+

Chapter 8 44
 Markovnikov product is formed.
+
CH3 C
CH3
CH CH3 OH H
H
+
+ H2O+
H
CH3CH
CH3
CCH3
H2O
CH3 C
CH3
CH CH3
HO
H H
+
H2O
CH3 C
CH3
CH CH3
HO
H
=>

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
=>

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
+
=>

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
=>

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
=>

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
_
=>

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

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
=>

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
=>

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 =>

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).
=>

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
CC H
CH3
H3C
H
H
C
CH
H
H
CH3
CH3
C
C
H
H
H3C
CH3
H
3

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
=>

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

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

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
=>

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

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

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

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
BrBr
=>

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

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

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

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.
=>

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.
CC
Br
H2O
CC
Br
O
H H
H2O
CC
Br
O
H
+ H3O
+
=>

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

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

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

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

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

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

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
=>

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
=>

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

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
HH2O2
H
(2)
(1) OsO4
cis-3-hexene meso-3,4-hexanediol
=>

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.
=>

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.
=>

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

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.
=>

Chapter 8 84
CC
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
=>

Chap 8 thiols and sulfides (1)

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