19.13) Which of the following bases are strong enough to
deprotonate CH3COOH?
a) F-
pka of CH3COOH is 4.8.
b) (CH3)3CO-
b) has a pka of 18
c) has a pka of 50
c) CH3-
d) has a pka of 38
d) NH2e) Cl-
pka of conjugate acid must be greater
than that of the carboxylic acid being
deprotonated.
1
19.14) Rank the labeled protons in order of increasing
acidity.
Ha
H
OHb
OHc
O
Ha<Hb<Hc
The more stable the conjugate base the more acidic the
proton.
2
19.15) Match each pka value with each carboxylic
acid.(3.2, 4.9 and 0.2)
a) CH3CH2COOH
4.9
b) CF3COOH
0.2
c) ICH2COOH
3.2
Electron withdrawing groups make acids more
acidic.
3
19.16) Why is formic acid more acidic than
acetic acid?
H
OH
O
OH
O
The methyl group is electron donating and stabilizes
the acid while destabilizing the conjugate base thus
making it less acidic.
4
19.17) Rank the compounds within each group in order
of decreasing acidity.
a) CH3COOH, HSCH2COOH, HOCH2COOH
3
2
1
b) ICH2COOH, I2CHCOOH, ICH2CH2COOh
2
1
3
5
19.18) Rank each group of compounds in order of
decreasing acidity.
a)
CO2H
CO2H
CO2H
Cl
2
1
3
CO2H
b)
CO2H
CO2H
O
H3CO
2
1
3
6
19.19) Is the following compound more or less
acidic than phenol?
OH
HO
R
The more electron donating groups present, the less
acidic a compound is. This compound has an
additional hydroxy and alkyl group, both electron
donating. So it is less acidic.
7
19.22) Comparing CF3SO3H and CH3SO3H, which
has the weaker conjugate base? Which conjugate
base is the better leaving group? Which of these
acids has the higher pka?
CF3SO3H is the weaker conjugate base.
CF3SO3H is the better leaving group because it
is the weaker conjugate base.
CH3SO3H, with the electron donating methyl
group, has the higher pka and is thus a weaker
acid.
8
Introduction to Carbonyl Chemistry; Organometallic
Reagents; Oxidation and Reduction
Introduction
Two broad classes of compounds contain the carbonyl
group:
[1]
Compounds that have only carbon and hydrogen atoms
bonded to the carbonyl
[2]
Compounds that contain an electronegative atom bonded to
the carbonyl
9
• The presence or absence of a leaving group on the
carbonyl determines the type of reactions the carbonyl
compound will undergo.
• Carbonyl carbons are sp2 hybridized, trigonal planar, and
have bond angles that are ~1200. In these ways, the
carbonyl group resembles the trigonal planar sp2
hybridized carbons of a C=C.
10
• In one important way, the C=O and C=C are very
different.
• The electronegative oxygen atom in the carbonyl group
means that the bond is polarized, making the carbonyl
carbon electron deficient.
• Using a resonance description, the carbonyl group is
represented by two resonance structures.
11
General Reactions of Carbonyl Compounds
Carbonyls react with nucleophiles.
12
Aldehydes and ketones react with nucleophiles to form
addition products by a two-step process: nucleophilic
attack followed by protonation.
13
• The net result is that the  bond is broken, two new 
bonds are formed, and the elements of H and Nu are
added across the  bond.
• Aldehydes are more reactive than ketones towards
nucleophilic attack for both steric and electronic
reasons.
14
Carbonyl compounds with leaving groups react with nucleophiles
to form substitution products by a two-step process: nucleophilic
attack, followed by loss of the leaving group.
The net result is that Nu replaces Z, a nucleophilic substitution
reaction. This reaction is often called nucleophilic acyl
substitution.
15
16
• Nucleophilic addition and nucleophilic acyl substitution involve
the same first step—nucleophilic attack on the electrophilic
carbonyl carbon to form a tetrahedral intermediate.
• The difference between the two reactions is what then happens
to the intermediate.
• Aldehydes and ketones cannot undergo substitution because
they do not have a good leaving group bonded to the newly
formed sp3 hybridized carbon.
17
Preview of Oxidation and Reduction
• Carbonyl compounds are either reactants or products in
oxidation-reduction reactions.
18
The three most useful oxidation and reduction reactions of
carbonyl starting materials can be summarized as follows:
19
Reduction of Aldehydes and Ketones
• The most useful reagents for reducing aldehydes and ketones
are the metal hydride reagents.
• Treating an aldehyde or ketone with NaBH4 or LiAlH4, followed by
H2O or some other proton source affords an alcohol.
20
• The net result of adding H:¯ (from NaBH4 or LiAlH4) and H+ (from
H2O) is the addition of the elements of H2 to the carbonyl  bond.
21
• Catalytic hydrogenation also reduces aldehydes and ketones
to 1° and 2° alcohols respectively, using H2 and a catalyst.
• When a compound contains both a carbonyl group and a
carbon—carbon double bond, selective reduction of one
functional group can be achieved by proper choice of the
reagent.
A C=C is reduced faster than a C=O with H2 (Pd-C).
A C=O is readily reduced with NaBH4 and LiAlH4, but a
C=C is inert.
22
• Thus, 2-cyclohexenone, which contains both a C=C and a
C=O, can be reduced to three different compounds depending
upon the reagent used.
23
The Stereochemistry of Carbonyl Reduction
• Hydride converts a planar sp2 hybridized carbonyl carbon to a
tetrahedral sp3 hybridized carbon.
24
Enantioselective Carbonyl Reductions
• Selective formation of one enantiomer over another can occur if
a chiral reducing agent is used.
• A reduction that forms one enantiomer predominantly or
exclusively is an enantioselective or asymmetric reduction.
• An example of chiral reducing agents are the enantiomeric CBS
reagents.
25
• CBS refers to Corey, Bakshi and Shibata, the chemists who
developed these versatile reagents.
• One B—H bond serves as the source of hydride in this reduction.
• The (S)-CBS reagent delivers H:- from the front side of the C=O. This
generally affords the R alcohol as the major product.
• The (R)-CBS reagent delivers H:- from the back side of the C=O. This
generally affords the S alcohol as the major product.
26
• These reagents are highly enantioselective. For example,
treatment of propiophenone with the (S)-CBS reagent
forms the R alcohol in 97% ee.
27
Reduction of Carboxylic Acids and Their Derivatives
• LiAlH4 is a strong reducing agent that reacts with all
carboxylic acid derivatives.
• Diisobutylaluminum
hydride
([(CH3)2CHCH2]2AlH,
abbreviated DIBAL-H, has two bulky isobutyl groups
which makes this reagent less reactive than LiAlH4.
• Lithium tri-tert-butoxyaluminum hydride,
LiAlH[OC(CH3)3]3, has three electronegative O atoms
bonded to aluminum, which makes this reagent less
nucleophilic than LiAlH4.
28
• Acid chlorides and esters can be reduced to either aldehydes or 1°
alcohols depending on the reagent.
29
• In the reduction of an acid chloride, Cl¯ comes off as the
leaving group.
• In the reduction of the ester, CH3O¯ comes off as the
leaving group, which is then protonated by H2O to form
CH3OH.
30
• The mechanism illustrates why two different products are
possible.
31
• Carboxylic acids are reduced to 1° alcohols with LiAlH4.
• LiAlH4 is too strong a reducing agent to stop the reaction
at the aldehyde stage, but milder reagents are not strong
enough to initiate the reaction in the first place.
32
• Unlike the LiAlH4 reduction of all other carboxylic acid
derivatives, which affords 1° alcohols, the LiAlH4
reduction of amides forms amines.
• Since ¯NH2 is a very poor leaving group, it is never lost
during the reduction, and therefore an amine is formed.
33
34
35
Oxidation of Aldehydes
• A variety of oxidizing agents can be used, including
CrO3, Na2Cr2O7, K2Cr2O7, and KMnO4.
• Aldehydes can also be oxidized selectively in the
presence of other functional groups using silver(I) oxide
in aqueous ammonium hydroxide (Tollen’s reagent).
Since ketones have no H on the carbonyl carbon, they
do not undergo this oxidation reaction.
36
For Wednesday, 20.1-20.16.
37
20.1) What type of orbitals make up the indicated
bonds? And in what orbitals do the lone pairs on the
oxygen lie?
O
a
b
c
a. sp3-sp2
b. sp2-sp2, p-p
c. sp3-sp2
The lone pairs lie in sp2 hybridized
orbitals.
38
20.2) Which compounds undergo nucleophilic
addition and which substitution?
O
a)
O
b)
H3CH2CH2C
addition
Cl
substitution
O
c)
O
H3C
d)
H
OCH3
substitution
addition
39
20.3) Which compound in each pair is more reactive
toward nucleuphilic attack?
a)
H3CH2CH2C
H
H3C(H3C)HC
CH2CH3
H3CH2C
O
O
O
c)
H3CH2C
H
O
O
H3CH2C
d)
H3CH2CH2C
H3CH2CH2C
O
b)
O
O
O
Cl
O
OCH3
Cl
O
O
OCH3
H3CH2C
NHCH3
OCH3
40
20.4) What alcohol is formed when each
compound is treated with NaBH4 in MeOH?
a)
O
OH
NaBH4
H3CH2CH2C
MeOH
H
H3CH2CH2C
H
H
b)
OH
O
NaBH4
MeOH
c)
NaBH4
MeOH
O
OH
41
20.5) What aldehyde or ketone is needed to synthesize
each alcohol by metal hydride reduction?
a)
b)
OH
OH
O
O
c)
OH
O
42
20.6) Why can’t 1-methylcyclohexanol be prepared from
a carbonyl by reduction?
OH
Tertiary alcohols can not be made by reduction of a
carbonyl because there are no hydrogens on the
carbon with the -OH.
43
20.7) Draw the products of the following reactions?
a)
OH
O
LiAl4
H2O
O
b)
OH
NaBH4
MeOH
c)
O
O
H2 (1 equiv.)
Pd-C
d)
O
OH
H2 (excess)
Pd-C
44
e)
O
OH
NaBH4 (excess)
MeOH
f)
O
NaBD4
D
OH
MeOH
45
20.8) Draw the products when the following compounds
are treated with NaBH4 in MeOH.
a)
O
HO
NaBH4
H
H
MeOH
b)
+
NaBH4
O
OH
OH
MeOH
c)
(H3C)3C
O
NaBH4
MeOH
(H3C)3C
OH
(H3C)3C
+
OH
46
20.9) What reagent is needed to carry out the reaction
below?
O
HO
Cl
H
Cl
Two reagents are needed to carry out this
reaction. First, the (S)-CBS reagent to
produce the R-enantiomer. Followed by H2O
to protonate the alcohol.
47
20.10) Draw a stepwise mechanism for the following
reaction.
O
LiAlH4
OH
H 2O
Cl
O
O
O
Cl
Cl
H
H
H3Al
H
+ Cl
+ AlH3
O
O
H
OH
OH
H
H
H
H
H3Al
H
H
+ OH
+ AlH3
48
20.11) Draw an acid chloride and an ester that can be
used to produce each product.
a)
O
O
Cl
OCH3
CH2OH
Cl
OCH3
b)
O
OH
O
O
c)
OH
H3CO
H3CO
Cl
O
H3CO
OCH3
49
20.12) Draw the products of LiAlH4 reduction of each
compound.
O
a)
OH
OH
O
b)
NH2
NH2
c)
O
N(CH3)2
N(CH3)2
50
d)
O
NH
NH
51
20.13) What amide will form each of the following
amines when treated with LiAlH4?
O
a)
NH2
NH2
O
b)
N
c)
N
O
N
H
N
H
52
20.14) Predict the products of these compounds
when treated with the following reagents.
O
O
OH
a)
LiAlH4
OCH3
H2O
OH
OH
O
NaBH4
MeOH
b)
O
OCH3
O
LiAlH4
H3CO
OH
H2O
NaBH4
MeOH
HO
OH
No reaction
53
c)
H3CO
H3CO
O
OH
LiAlH4
H2O
NaBH4
H3CO
OH
MeOH
54
20.15) Predict the products in the following
reactions.
a)
OH
Ag2O
No Reaction
NH4OH
O
Na2Cr2O7
H2SO4, H2O
b)
OH
O
OH
OH
O
Ag2O
NH4OH
OH
Na2Cr2O7
O
O
H2SO4, H2O
OH
55
20.16) Predict the products of the compound below when
OH
reacted with each reagent.
O
HO
a)
OH
NaBH4
OH
MeOH
HO
b)
LiAlH4
OH
H2O
OH
HO
56
OH
O
HO
c)
O
PCC
O
O
OH
d)
O
Ag2O
OH
NH4OH
HO
O
e)
O
CrO3
OH
H2SO4, H2O
HO
O
57
Organometallic Reagents
• Other metals in organometallic reagents are Sn, Si, Tl, Al, Ti, and
Hg. General structures of the three common organometallic
reagents are shown:
58
• Since both Li and Mg are very electropositive metals,
organolithium (RLi) and organomagnesium (RMgX) reagents
contain very polar carbon—metal bonds and are therefore very
reactive reagents.
• Organomagnesium reagents are called Grignard reagents.
• Organocopper reagents (R2CuLi), also called organocuprates,
have a less polar carbon—metal bond and are therefore less
reactive. Although they contain two R groups bonded to Cu, only
one R group is utilized in the reaction.
• In organometallic reagents, carbon bears a - charge.
59
• Organolithium and Grignard reagents are typically
prepared by reaction of an alkyl halide with the
corresponding metal.
• With lithium, the halogen and metal exchange to form
the organolithium reagent. With Mg, the metal inserts in
the carbon—halogen bond, forming the Grignard
reagent.
60
• Grignard reagents are usually prepared in diethyl ether
(CH3CH2OCH2CH3) as solvent.
• It is thought that two ether O atoms complex with the Mg
atom, stabilizing the reagent.
61
• Organocuprates are prepared from organolithium
reagents by reaction with a Cu+ salt, often CuI.
62
• Acetylide ions are another example of organometallic
reagents.
• Acetylide ions can be thought of as “organosodium
reagents”.
• Since sodium is even more electropositive than lithium,
the C—Na bond of these organosodium compounds is
best described as ionic, rather than polar covalent.
63
• An acid-base reaction can also be used to prepare sp
hybridized organolithium compounds.
• Treatment of a terminal alkyne with CH3Li affords a
lithium acetylide.
• The equilibrium favors the products because the sp
hybridized C—H bond of the terminal alkyne is more
acidic than the sp3 hybridized conjugate acid, CH4, that
is formed.
64
• Organometallic reagents are strong bases that readily abstract a
proton from water to form hydrocarbons.
• Similar reactions occur with the O—H proton of alcohols and
carboxylic acids, and the N—H protons of amines.
65
• Since organolithium and Grignard reagents are
themselves prepared from alkyl halides, a two-step
method converts an alkyl halide into an alkane (or other
hydrocarbon).
• Organometallic reagents are also strong nucleophiles
that react with electrophilic carbon atoms to form new
carbon—carbon bonds.
• These reactions are very valuable in forming the carbon
skeletons of complex organic molecules.
66
Examples of functional
organometallic reagents:
group
transformations
involving
[1] Reaction of R—M with aldehydes and ketones to afford
alcohols
[2] Reaction of R—M with carboxylic acid derivatives
67
[3] Reaction of R—M with other electrophilic functional groups
68
Reaction of Organometallic Reagents with Aldehydes
and Ketones.
• Treatment of an aldehyde or ketone with either an
organolithium or Grignard reagent followed by water
forms an alcohol with a new carbon—carbon bond.
• This reaction is an addition because the elements of R’’
and H are added across the  bond.
69
• This reaction follows the general mechanism for
nucleophilic addition—that is, nucleophilic attack by a
carbanion followed by protonation.
• Mechanism 20.6 is shown using R’’MgX, but the same
steps occur with RLi reagents and acetylide anions.
70
Note that these reactions must be carried out under anhydrous
conditions to prevent traces of water from reacting with the
organometallic reagent.
71
• This reaction is used to prepare 1°, 2°, and 3° alcohols.
72
Retrosynthetic Analysis of Grignard Products
• To determine what carbonyl and Grignard components
are needed to prepare a given compound, follow these
two steps:
73
• Let us conduct a retrosynthetic analysis of 3-pentanol.
74
• Writing the reaction in the synthetic direction—that is,
from starting material to product—shows whether the
synthesis is feasible and the analysis is correct.
• Note that there is often more than one way to synthesize
a 20 alcohol by Grignard addition.
75
Protecting Groups
• Addition of organometallic reagents cannot be used with molecules
that contain both a carbonyl group and N—H or O—H bonds.
• Carbonyl compounds that also contain N—H or O—H bonds
undergo an acid-base reaction with organometallic reagents, not
nucleophilic addition.
76
Solving this problem requires a three-step strategy:
[1] Convert the OH group into another functional group that does
not interfere with the desired reaction. This new blocking
group is called a protecting group, and the reaction that
creates it is called “protection.”
[2] Carry out the desired reaction.
[3] Remove the protecting group. This reaction is called
“deprotection.”
A common OH protecting group is a silyl ether.
77
tert-Butyldimethylsilyl ethers are prepared from alcohols by
reaction with tert-butyldimethylsilyl chloride and an amine
base, usually imidazole.
The silyl ether is typically removed with a fluoride salt such as
tetrabutylammonium fluoride (CH3CH2CH2CH2)4N+F¯.
78
The use of tert-butyldimethylsilyl ether as a protecting
group makes possible the synthesis of 4-methyl-1,4pentanediol by a three-step sequence.
79
Figure 20.7
General strategy for using a
protecting group
80
For Friday, 20.17-20.25
81
20.17) Write out the rea tions needed to
convert CH3CH2Br to each of the following
reagents.
a) H3CH2C
H3CH2C
b)
Li
H3CH2C
H3CH2C
c)
+ 2 Li
Br
Li
H3CH2C
MgBr
+ Li
Br
MgBr
+ Mg
Br
H3CH2C
CuLi
H3CH2C
2 H3CH2C
H3CH2C
Br
Li
+ 2 Li
H3CH2C
Li
Li
Cu
+ CuI
H3CH2C
+ Li
Br
+ Li
CH2CH3
I
82
20.18) 1-octyne reacts readily with NaH, forming a gas
that bubbles out of the reaction mixture. 1-octyne also
reacts with CH3MgBr and a different gas is produced.
Write out balanced equations for each reaction.
HC
CCH2CH2CH2CH2CH2CH3
+ NaH
NaC
HC
CCH2CH2CH2CH2CH2CH3
CCH2CH2CH2CH2CH2CH3
+ H2
+ CH3MgBr
BrMgC
CCH2CH2CH2CH2CH2CH3
+ CH4
83
20.19) Draw the product of the following reactions.
a)
Li
+ H2O
+ LiOH
b)
MgBr
c)
+ H2O
MgBr
+ H2O
+ HOMgBr
+ HOMgBr
84
d)
LiC
CCH2CH3
+ H2O
HC
CCH2CH3
+ LiOH
85
20.20) Draw the product formed when each compound
is treated with C6H5MgBr followed by H2O.
a)
H
O
OH
H
b)
H
H
O
CH2CH3
OH
H3CH2C
CH2CH3
CH2CH3
O
c)
CH2CH3
H3CH2C
OH
H
H
86
d)
OH
O
87
20.21)Draw the products of each reaction.
a)
H3CH2CH2C
Li
+ LiOH
H2O
O
HO
b)
CH2CH2CH3
Li
H
O
+ LiOH
HO
H
H2O
H
H
OH
c)
C6H5Li
+ LiOH
O
H2O
88
d)
CNa
H2C
H2
C
O
C
H2O
OH
+ NaOH
89
20.22) Draw the products (including stereochemistry) of
the following reactions.
a)
O
H
H3CH2C
H3C
b)
H
OH
H
MgBr
OH
+
H2O
CH2CH3
H3CH2C
O
Li
OH
H2O
+
CH2CH3
OH
90
20.23) What Grignard and carbonyl are needed to
prepare each alcohol?
a)
O
OH
+
H3C
MgBr
H
MgBr
O
OH
b)
+
H
H
O
OH
c)
+
H3CH2C
MgBr
or
O
MgBr
|+
91
d)
OH
O
H3C
+
Br
or
O
MgBr
+
92
20.24) Tertiary alcohols with three different R groups on
the carbon attached to the OH can be prepared in three
different ways using the Grignard reagent. Show them.
a)
OH
O
H3C
CH2CH3
CH2CH3
CH2CH2CH3
+ H3C
MgBr
+ H CH C
3
2
MgBr
H3CH2CH2C
O
H3C
CH2CH2CH3
O
+ H CH CH C
3
2
2
H3C
MgBr
CH2CH3
93
b)
OH
O
+
H3C
MgBr
MgBr
O
+
MgBr
O
+
94
c)
OH
O
MgBr
+
MgBr
O
+
+ H3CH2C
MgBr
O
95
20.25) Show the steps for the following reaction.
CH2CH2CH2CH3
HO
O
HO
OH
TBDMS-Cl
HO
O
TBDMSO
H
N
O
N
BrMg
CH2CH2CH2CH3
H2O
CH2CH2CH2CH3
FN(CH2CH2CH2CH3)4
HO
OH
CH2CH2CH2CH3
TBDMSO
H2O
OH
96
Reaction of Organometallic Reagents with Carboxylic
Acid Derivatives.
• Both esters and acid chlorides form 3° alcohols when
treated with two equivalents of either Grignard or
organolithium reagents.
97
98
• To form a ketone from a carboxylic acid derivative, a less
reactive organometallic reagent—namely an organocuprate—is
needed.
• Acid chlorides, which have the best leaving group (Cl¯) of the
carboxylic acid derivatives, react with R’2CuLi to give a ketone
as the product.
• Esters, which contain a poorer leaving group (¯OR), do not react
with R’2CuLi.
99
Reaction of
Compounds
Organometallic
Reagents
with
Other
• Grignards react with CO2 to give carboxylic acids after
protonation with aqueous acid.
• This reaction is called carboxylation.
• The carboxylic acid formed has one more carbon atom than the
Grignard reagent from which it was prepared.
100
• The mechanism resembles earlier reactions of
nucleophilic Grignard reagents with carbonyl groups.
101
• Like other strong nucleophiles, organometallic
reagents—RLi, RMgX, and R2CuLi—open epoxide rings
to form alcohols.
102
• The reaction follows the same two-step process as opening
of epoxide rings with other negatively charged
nucleophiles—that is, nucleophilic attack from the back side
of the epoxide, followed by protonation of the resulting
alkoxide.
• In unsymmetrical epoxides, nucleophilic attack occurs at the
less substituted carbon atom.
103
,-Unsaturated Carbonyl Compounds
• ,-Unsaturated carbonyl compounds are conjugated molecules
containing a carbonyl group and a C=C separated by a single 
bond.
• Resonance shows that the carbonyl carbon and the  carbon
bear a partial positive charge.
104
• This means that ,-unsaturated carbonyl compounds can react
with nucleophiles at two different sites.
105
• The steps for the mechanism of 1,2-addition are exactly
the same as those for the nucleophilic addition of an
aldehyde or a ketone—that is, nucleophilic attack,
followed by protonation.
106
107
• Consider the conversion of a general enol A to the carbonyl
compound B. A and B are tautomers: A is the enol form and B is the
keto form of the tautomer.
• Equilibrium favors the keto form largely because the C=O is much
stronger than a C=C. Tautomerization, the process of converting
one tautomer into another, is catalyzed by both acid and base. 108
109
110
111
Summary
Reagents
of
the
Reactions
of
Organometallic
[1] Organometallic reagents (R—M) attack electrophilic
atoms, especially the carbonyl carbon.
112
[2]
After an organometallic reagent adds to the carbonyl group,
the fate of the intermediate depends on the presence or
absence of a leaving group.
[3]
The polarity of the R—M bond determines the reactivity of the
reagents:
—RLi and RMgX are very reactive reagents.
—R2CuLi is much less reactive.
113
Synthesis
Figure 20.8
Conversion of 2–hexanol into
other compounds
114
Reactions of Alcohols—Dehydration
• Dehydration, like dehydrohalogenation, is a  elimination
reaction in which the elements of OH and H are removed from
the  and  carbon atoms respectively.
• Dehydration is typically carried out using H2SO4 and other
strong acids, or phosphorus oxychloride (POCl3) in the
presence of an amine base.
115
• Typical acids used for alcohol dehydration are H2SO4 or ptoluenesulfonic acid (TsOH).
• More substituted alcohols dehydrate more easily, giving rise
to the following order of reactivity.
116
• When an alcohol has two or three  carbons, dehydration is
regioselective and follows the Zaitsev rule.
• The more substituted alkene is the major product when a
mixture of constitutional isomers is possible.
117
• Secondary and 3° alcohols react by an E1 mechanism,
whereas 1° alcohols react by an E2 mechanism.
118
• Since 1° carbocations are highly unstable, their dehydration
cannot occur by an E1 mechanism involving a carbocation
intermediate. Therefore, 1° alcohols undergo dehydration
following an E2 mechanism.
119
Dehydration of Alcohols Using POCl3 and Pyridine
• Some organic compounds decompose in the presence of
strong acid, so other methods have been developed to
convert alcohols to alkenes.
• A common method uses phosphorus oxychloride (POCl3) and
pyridine (an amine base) in place of H2SO4 or TsOH.
• POCl3 serves much the same role as a strong acid does in
acid-catalyzed dehydration. It converts a poor leaving group
(¯OH) into a good leaving group.
• Dehydration then proceeds by an E2 mechanism.
120
121
Conversion of Alcohols to Alkyl Halides with HX
• Substitution reactions do not occur with alcohols unless ¯OH
is converted into a good leaving group.
• The reaction of alcohols with HX (X = Cl, Br, I) is a general
method to prepare 1°, 2°, and 3° alkyl halides.
122
• More substituted alcohols usually react more rapidly with HX:
• This order of reactivity can be rationalized by considering the
reaction mechanisms involved. The mechanism depends on the
structure of the R group.
123
124
125
• The reactivity of hydrogen halides increases with increasing
acidity.
• Because Cl¯ is a poorer nucleophile than Br¯ or I¯, the
reaction of 10 alcohols with HCl occurs only when an
additional Lewis acid catalyst, usually ZnCl2, is added.
Complexation of ZnCl2 with the O atom of the alcohol makes a
very good leaving group that facilitates the SN2 reaction.
126
Conversion of Alcohols to Alkyl Halides with SOCl2
and PBr3
• Primary and 2° alcohols can be converted to alkyl
halides using SOCl2 and PBr3.
• SOCl2 (thionyl chloride) converts alcohols into alkyl
chlorides.
• PBr3 (phosphorus tribromide) converts alcohols into
alkyl bromides.
• Both reagents convert ¯OH into a good leaving group in
situ—that is, directly in the reaction mixture—as well as
provide the nucleophile, either Cl¯ or Br¯, to displace the
leaving group.
127
• When a 1° or 2° alcohol is treated with SOCl2 and
pyridine, an alkyl chloride is formed, with HCl and SO2 as
byproducts.
• The mechanism of this reaction consists of two parts:
conversion of the OH group into a better leaving group,
and nucleophilic cleavage by Cl¯ via an SN2 reaction.
128
129
• Treatment of a 10 or 20 alcohol with PBr3 forms an alkyl
halide.
• The mechanism of this reaction also consists of two
parts: conversion of the OH group into a better leaving
group, and nucleophilic cleavage by Br¯ via an SN2
reaction.
130
131
132
Tosylate—Another Good Leaving Group
• Alcohols can be converted into alkyl tosylates.
• An alkyl tosylate is composed of two parts: the alkyl group R,
derived from an alcohol; and the tosylate (short for ptoluenesulfonate), which is a good leaving group.
• A tosyl group, CH3C6H4SO2¯, is abbreviated Ts, so an alkyl
tosylate becomes ROTs.
133
• Alcohols are converted to tosylates by treatment with ptoluenesulfonyl chloride (TsCl) in the presence of
pyridine.
• This process converts a poor leaving group (¯OH) into a
good one (¯OTs).
• Tosylate is a good leaving group because its conjugate
acid, p-toluenesulfonic acid (CH3C6H4SO3H, TsOH) is a
strong acid (pKa = -7).
134
• (S)-2-Butanol is converted to its tosylate with retention
of configuration at the stereogenic center. Thus, the
C—O bond of the alcohol is not broken when tosylate is
formed.
135
• Because alkyl tosylates have good leaving groups, they
undergo both nucleophilic substitution and  elimination,
exactly as alkyl halides do.
• Generally, alkyl tosylates are treated with strong nucleophiles
and bases, so the mechanism of substitution is SN2, and the
mechanism of elimination is E2.
136
• Because substitution occurs via an SN2 mechanism, inversion
of configuration results when the leaving group is bonded to
a stereogenic center.
• We now have another two-step method to convert an alcohol
to a substitution product: reaction of an alcohol with TsCl and
pyridine to form a tosylate (step 1), followed by nucleophilic
attack on the tosylate (step 2).
137
• Step 1, formation of the tosylate, proceeds with retention of
configuration at a stereogenic center.
• Step 2 is an SN2 reaction, so it proceeds with inversion of
configuration because the nucleophile attacks from the
backside.
• Overall there is a net inversion of configuration at a
stereogenic center.
Example:
138
Figure 9.8
Summary: Nucleophilic
substitution and β elimination
reactions of alcohols
139
For Monday, 20.26-20.36
140