Vollhardt  Capítulo 8 (Álcoois)
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Vollhardt Capítulo 8 (Álcoois)


DisciplinaQuímica Orgânica II1.705 materiais36.340 seguidores
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Callout
A Ácidez média de um OH ligado a uma cadeia carbonica se mostra entre 12 e 18, sendo que, onde o carbono está mais substituído, o pKa mostra-se maior pela estabilidade do composto após a desprotonação.
 C h a p t e r 8 293
It is sometimes suffi cient to generate alkoxides in less than stoichiometric equilibrium 
concentrations. For this purpose, we may add an alkali metal hydroxide to the alcohol.
pKa \ufffd 15.7
H2O
pKa \ufffd 15.9 
K
H ðO\u161\u161 Na\ufffd \ufffd OHCH3CH2O \u161\u161\ufffd \u161\u161 \u161\u161ðCH3CH2O \ufffd Na\ufffd \ufffd
With this base present, approximately half of the alcohol exists as the alkoxide, if we assume 
equimolar concentrations of starting materials. If the alcohol is the solvent (i.e., present in 
large excess), however, essentially all of the base exists in the form of the alkoxide.
Steric disruption and inductive effects control the acidity of alcohols
Table 8-2 shows an almost million-fold variation in the acidity of the alcohols. A closer 
look at the fi rst column reveals that the acidity decreases (pKa increases) from methanol to 
primary, secondary, and fi nally tertiary systems.
Relative pKa Values of Alcohols (in Solution)
 CH3OH , primary , secondary , tertiary
 Strongest acid Weakest acid
 Increasing acidity
This ordering has been ascribed to steric disruption of solvation and to hydrogen bonding 
in the alkoxide (Figure 8-4). Because solvation and hydrogen bonding stabilize the negative 
charge on oxygen, interference with these processes leads to an increase in pKa.
The second column in Table 8-2 reveals another contribution to the pKa of alcohols: The 
presence of halogens increases acidity. Recall that the carbon of the C \u2013 X bond is positively 
Exercise 8-3
Working with the Concepts: Estimating Acid-Base Equilibria
You want to prepare potassium methoxide by treatment of methanol with KCN. Will this procedure 
work?
Strategy
We need to visualize the desired reaction by writing it down. We then add to the equation the 
pKa values of the acids on each side (consult Table 2-2 or 6-4, and Table 8-2). If the pKa of 
the (conjugate) acid on the right is more than 2 units larger than that of methanol on the left, the 
equilibrium will lie .99% to the right (K . 100).
Solution
\u2022 The equilibrium reaction and the associated pKa values are
pKa \ufffd 9.2pKa \ufffd 15.5 
K\ufffd CN\ufffdCH3OH \ufffd HCNCH3O\ufffd K\ufffd \ufffd
\u2022 The pKa of HCN is 6.3 units smaller than that of methanol; it is a much stronger acid.
\u2022 The equilibrium will lie to the left; K 5 1026.3. This approach to preparing potassium methoxide 
will not work.
Exercise 8-4
Try It Yourself
Which of the following bases are strong enough to cause essentially complete deprotonation of 
methanol? The pKa of the conjugate acid is given in parentheses.
(a) CH3CH2CH2CH2Li (50); (b) CH3CO2Na (4.7); (c) LiN[CH(CH3)2]2 (LDA, 40);
(d) KH (38); (e) CH3SNa (10).
8 - 3 A l c o h o l s a s A c i d s a n d B a s e s
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294 C h a p t e r 8 H y d r o x y F u n c t i o n a l G r o u p : A l c o h o l s
polarized as a result of the high electronegativity of X (Sections 1-3 and 6-1). Electron 
withdrawal by the halogen also causes atoms farther away to be slightly positively charged. 
This phenomenon of transmission of charge, both negative and positive, through the s bonds 
in a chain of atoms is called an inductive effect. Here it stabilizes the negative charge on 
the alkoxide oxygen by electrostatic attraction. The inductive effect in alcohols increases 
with the number of electronegative groups but decreases with distance from the oxygen.
Figure 8-4 The smaller methoxide 
ion is better solvated than is the 
larger tertiary butoxide ion. 
(S 5 solvent molecules).
\u2212
\u2212
S
S
S
S
S
S
S
S S
SO
O
Exercise 8-5
Rank the following alcohols in order of increasing acidity.
OH OH OH
Cl
Cl
OH
Cl
Exercise 8-6
Which side of the following equilibrium reaction is favored (assume equimolar concentrations of 
starting materials)?
(CH3)3CO
2 1 CH3OH \u394 (CH3)3COH 1 CH3O2
The lone electron pairs on oxygen make alcohols weakly basic
Alcohols may also be basic, although weakly so. Very strong acids are required to protonate 
the OH group, as indicated by the low pKa values (strong acidity) of their conjugate acids, 
the alkyloxonium ions (Table 8-3). Molecules that may be both acids and bases are called 
amphoteric (ampho, Greek, both).
The amphoteric nature of the hydroxy functional group characterizes the chemical reac-
tivity of alcohols. In strong acids, they exist as alkyloxonium ions, in neutral media as 
alcohols, and in strong bases as alkoxides.
RO \ufffdROH
Alcohols Are Amphoteric
Strong acid
Mild base
Alkyloxonium
ion
Alcohol Alkoxide
ion
Strong base
Mild acid
\u161\ufffd \u161\ufffdðR
H
H
O
\ufffd
O
i
f
ð
Table 8-3
pKa Values of 
Four Protonated 
Alcohols
Compound pKa
CH3O
1
H2 22.2
CH3CH2O
1
H2 22.4
(CH3)2CHO
1
H2 23.2
(CH3)3CO
1
H2 23.8
Cl
Inductive Effect
of the Chlorine
in 2-Chloroethoxide
OCH2O CH2 \ufffdO O ð\u161\ufffd
Increasing 
inductive effect
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 C h a p t e r 8 295
In Summary Alcohols are amphoteric. They are acidic by virtue of the electronegativity 
of the oxygen and are converted into alkoxides by strong bases. In solution, the steric bulk 
of branching inhibits solvation of the alkoxide, thereby raising the pKa of the corresponding 
alcohol. Electron-withdrawing substituents close to the functional group lower the pKa by 
the inductive effect. Alcohols are also weakly basic and can be protonated by strong acids 
to furnish alkyloxonium ions.
8-4 Industrial Sources of Alcohols: Carbon Monoxide and Ethene
Let us now turn to the preparation of alcohols. We start in this section with methods of 
special importance in industry. Subsequent sections address procedures that are used more 
generally in the synthetic laboratory to introduce the hydroxy functional group into a wide 
range of organic molecules.
Methanol is made on a multibillion-pound scale from a pressurized mixture of CO and 
H2 called synthesis gas. The reaction involves a catalyst consisting of copper, zinc oxide, 
and chromium(III) oxide.
Cu\u2013ZnO\u2013Cr2O3, 250\u2daC, 50\u2013100 atmCO 2 H2 CH3OH\ufffd
Changing the catalyst to rhodium or ruthenium leads to 1,2-ethanediol (ethylene glycol), 
an important industrial chemical that is the principal component of automobile antifreeze.
2 CO 3 H2 CH2 CH2
OH
1,2-Ethanediol
(Ethylene glycol)
Rh or Ru, pressure, heat
\ufffd
A
OH
A
O
Other reactions that would permit the selective formation of a given alcohol from syn-
thesis gas are the focus of much current research, because synthesis gas is readily available 
by the gasifi cation of coal or other biomass in the presence of water, or by the partial oxi-
dation of methane.
Coal x CO y H2 CH4\ufffd
Air, H2O, \u394 Air
The methanol produced from synthesis gas costs only about $1 per gallon and, because its 
heat content is high [182.5 kcal mol21 (763.6 kJ mol21); compare values in Table 3-7], it 
has become the cornerstone of a methanol-based fuel strategy. A key development has been 
the methanol fuel cell, in which the electrons released upon the conversion of methanol to 
CO2 are used to power cars and electronic devices, such as cell phones, laptop computers, 
and small power-generating units for domestic use.
Ethanol is prepared in large quantities by fermentation of sugars (see Chapter Opening) 
or by the phosphoric acid \u2013 catalyzed hydration of ethene (ethylene). The hydration (and 
other addition reactions) of alkenes are considered in detail in Chapter 12.
CH2 CH2 HOH CH2 CH2
H
H3PO4, 300ºC
\ufffd
A
OH
A
OP
In Summary The industrial preparation of methanol