Transcript Reactions of Alkyl Halides (SN1, SN2, E1, and E2 reactions)

FOUR
REACTIONS
OF
ALCOHOLS
1
4 REACTIONS OF ALCOHOLS
1.
2.
3.
4.
SUBSTITUTION WITH HALIDES TO FORM ALKYL HALIDES
ELIMINATION OF WATER (DEHYDRATION TO ALKENES)
ACID/BASE REACTIONS WITH STRONG BASES/ACIDS
OXIDATION TO ALDEHYDES, KETONES, OR CARBOXYLIC
ACIDS.
Alcohols have several polar atoms, several sites of reactivity.
Alcohols can undergo substitution, elimination and acid/base reactions.
An alcohol, like HOH, can act as a Nu:- or base as well as an E+ or acid.
+
d
H3C
E
+
d
-
..
O
..
-
Nu:
& base
+
d
H
E+
& acid
2
1. SUBSTITUTION WITH HALIDES
1. SUBSTITUTION WITH HALIDES TO FORM ALKYL HALIDES
Hydrohalic acids (HCl, HBr, HI) convert alcohols to alkyl halides
SN1 or SN2
R
OH
+
H
alcohol
X
R
X
+
H
OH
alkyl halide
 Even in concentrated form, these acids contain a lot of H2O
 H2O is a protic solvent. It stabilizes C+’s and weakens (solvates) Nu:’s
alcohol
(substrate)
HI
HBr
HCl
Me
SN2
1 unhindered
SN2
2
SN1
3
SN1
At room temperature, the SN2 reaction is slow.
 3° alcohols react rapidly
 2° alcohols react moderately
 1° and Me°alcohols are unreactive
3
1. SUBSTITUTION WITH HALIDES
Mechanism:
H
C
C
C
C
..
..OH
..
Cl
.. :
C
conc. (37%)
+ H
O
..
H
C
C
fast SN1 rxn.
C
H2O
C
C
C
..
Cl
.. :
conc. (37%)
isopropyl moderate
alcohol
SN1 rxn.
(2º alcohol)
C
C
C
H2O
+ H
O
..
H
C
C+
2º C+
C
..
Cl :
..
+ H2O
C
t-butyl chloride
C
C
C
C
3º C+
H
..
..OH
C+
C
t-butyl alcohol
(3º alcohol)
C
..
: Cl : ..
..
..
C
: Cl : C
C
..
Cl :
..
+
H2O
isopropyl
chloride
The hydroxyl (-OH) group is a poor leaving group unless protonated.
4
1. SUBSTITUTION WITH HALIDES
H
C
C
..
..OH
..
Cl
.. :
H2O
1º C+
conc. (37%)
C
C
needs heat
ethyl alcohol
slow SN2 rxn.
+ H
O
..
H
C
C+
not formed
too unstable
(1º alcohol)
C
..
..
: Cl : -
C
..
Cl :
..
+
H2O
ethyl
chloride
 In these mechanisms, even though H-Cl is shown as the acidic species,
it is understood that H2O is present (37% HCl contains 63% H2O).
 H-Cl is largely present as hydronium ion (H3O+ Cl-) and H3O+
protonates the alcohol in the first step.
 Name the mechanism & show the product of the following reaction
conc. aq. HBr
CH2
CH
CH2
allyl alcohol
2-propen-1-ol
CH2
OH
SN1
CH
CH2
Br
+
H2O
allyl bromide
3-bromo-1-propene
5
1. SUBSTITUTION WITH HALIDES
 1° and 2°alcohol reactivity can be improved by reacting with a polar
aprotic reagent/solvent, such as phosphorus tribromide (PBr3) or
thionyl chloride (SOCl2).
 These are fuming, corrosive liquids (nasty) but they do the job well.
 1° and 2° alcohols react quickly via SN2 without heating.
(1º alcohol)
R OH
+
E+
Cl
O
Cl
S
Fast
SN2
H
-
O
R
+
Cl
+
O
S
Cl
R
Cl + SO2 + HCl
E
(1º alcohol)
R OH
+
E+
Br
P
Br
Br
-
Br
R
+
H
+
O
Fast
SN2
PBr2
R
Br +
HOPBr2
E
Note that SOCl2 produces alkyl chlorides & PBr3 produces alkyl bromides.
6
1. SUBSTITUTION WITH HALIDES
 The difference in the rate of reaction of alcohols with conc. HCl is
used as a qualitative test (the Lucas Test) to determine the degree of
substitution of an alcohol.
 Lucas reagent is conc. HCl, saturated with ZnCl2 salt. The Zn+2 ion
coordinates (bonds) with the alcohol oxygen even better than H+ and
speeds up the rate at which the C+ can form.
 About ½ mL of alcohol and 3 mL Lucas reagent are mixed in a test
tube. 3°, allyl and benzyl alcohols react instantly (via SN1) to produce
an insoluble alkyl halide, which appears initially as a hazy emulsion
that separates into two liquid layers on standing.
 2° alcohols also react quickly (SN1) in Lucas reagent (R-Cl separates).
 1° alcohols do not react (no R-Cl layer forms).
 2° and 3° alcohols can be differentiated by repeating the test in conc.
HCl (no ZnCl2 present). 3° alcohols react quickly but the solution
remains clear in 2° alcohols.
7
2. DEHYDRATION OF ALCOHOLS TO ALKENES
2. DEHYDRATION TO ALKENES WITH H2SO4 OR H3PO4 AND HEAT
H
OH
C
C
alcohol
H+HSO4dehydration
E1
C
C
+
H2O
alkene
 H2SO4 and H3PO4 acids are dehydrating; they react strongly with
water and will remove H2O from other chemicals. H2SO4 is
sometimes used as a desiccant.
 When heated with alcohols, elimination (E1) occurs. The more stable,
more substituted alkene (Zaitsev product) is the major product.
 With only gently heating (50°C) 20% H2SO4 dehydrates 3° alcohols.
 2°alcohols require 60% H2SO4 @ 100°C.
 1°alcohols need 95% H2SO4 @ 170°C (!), however, much of the
product is lost. The 1° alcohol is badly charred.
8
2. DEHYDRATION OF ALCOHOLS TO ALKENES
Mechanism:
Fast E1 rxn.
20%
+
H HSO4-
CH3
..
O
..
H
1-methylcyclohexanol
(3º alcohol)
..
..OH
C
isopropyl
alcohol
(2º alcohol)
THF
50ºC
C
CH3
CH3
+
H
H
H2O
3º C+
C
60%
100ºC
C
C
+ H
O
H
moderate
..
OH
..
ethyl alcohol
(1º alcohol)
C H
E1
C
+
H2O
H
HSO4-
1-methylcyclohexene
HSO4-
C
C
C
H2O +
+
H2SO4
2º C+
propene
E1
95%
170ºC
slow
E1 rxn.
C+
H2O
E1 rxn.
H+ HSO4C
E1
+ H
O
.. H
H+ HSO4-
C
C
CH3
C
H
C
Unstable 1 C+
requires high temp.
+ H
O
..
H
C
H
H2O
C+
1º C+
C
C
+
H2O +
H2SO4
HSO4ethylene
9
2. DEHYDRATION OF ALCOHOLS TO ALKENES
 Note that even 1° alcohols dehydrate via the E1 mechanism. The
high temperature (170°C) allows the formation of the unstable 1°C+.
 With weak nucleophiles, no reaction will occur unless a C+ forms.
 HSO4- (or H2PO4-) are very weak, non basic Nu:-’s. They have no
propensity to remove H+ from a b-carbon via the E2 mechanism.
 Recall that E1 and SN1 occur in competition. High temperature
favors elimination over substitution.
 Draw and name the major products of the following reactions.
CH3
OH
H2SO4

CH 3
H
1-methylcyclopentene
Zaitsev product
(major product)
2-methylcyclopentanol
10
2. DEHYDRATION OF ALCOHOLS TO ALKENES
Recall that in unimolecular reactions, C+ rearrangements can occur.
With that in mind, draw the major (Zaitsev) product from the dehydration
of neohexyl alcohol.
CH3
H3C
C
CH2
CH2
OH
H3C

CH3
CH3
CH3
H2SO4
C
CH2
+
CH2
H3C
+
C
CH
CH3
CH
CH3
CH3
CH3
neohexyl alcohol
2,3-dimethyl-2-butene
H3C
H2O
+
Zaitsev product
(major product)
H3C
CH3
CH3
C
H3C
C
CH3
C
+
CH3
11
2. DEHYDRATION OF ALCOHOLS TO ALKENES
 1° and 2° alcohols are charred by the severe conditions of acid
dehydration and the E1 mechanism.
 There is a way to safely dehydrate these alcohols with good yields
even at room temperature. The reagents needed however are
unpleasant and corrosive.
 Phosphorus oxychloride (POCl3) in pyridine solvent (C5H5N:) will
dehydrate 1°, 2° and 3° alcohols to alkenes via an E2 mechanism.
pKb = 5.3
..
OH
..
Cl
+
O
P
E+
Cl
H
:N
H
..
P
O
E2
H
O
+
Cl
pyridine is a moderately
strong base & very bulky
Cl
Cl
good leaving
group
cyclopentene
Recall that strong, bulky bases like t-butoxide cause E2 even w. 1°substrates.
12
2. DEHYDRATION OF ALCOHOLS TO ALKENES
 Pyridine, like t-butoxide causes E2 reactions even w. 1°substrates.
 Pyridine is very bulky because of its large p-system of overlapping p-orbitals.
H2O
:N
-
-
HSO4-
alkyl
halide
(substrate)
good Nu
strong base
e.g., bromide
e.g., ethoxide
strong bulky base
e.g., t-butoxide
Br
C2H5O
(CH3)3CO
Me
SN2
SN2
SN2
no reaction
1°
SN2
SN2
E2 (SN2)
no reaction
2
SN2
E2
E2
SN1, E1
3
SN1
E2
E2
SN1, E1
-
-
good Nu
-
good Nu
nonbasic
-
very poor Nu
nonbasic
e.g., acetic acid
CH3COOH
 HSO4- and H2O are nonbasic, weak nucleophiles (like CH3COOH).
 They will not react unless a C+ forms. This occurs rapidly with 3° alcohols,
moderately with 2° alcohols and only at high temp. with 1° alcohols.
13
PROBLEMS
 Write equations showing how the following transformations can be carried
out. More than one step may be necessary. Mechanism are not required
but show all reagents used.
OH
CH2
CH2
OH
CH
H2O
E2
POCl3,
CH
CH2
CH3
Markovnikov
H2SO4 product
H2O
N
Br
1°
OH
CH2
CH3 Markovnikov
product
KOH
SN2
H2SO4
H2O
OH
POCl3,
CH2
1°
CH2
N
E2
14
3. ACID/BASE REACTIONS OF ALCOHOLS
3. ACID & BASE REACTIONS WITH STRONG BASES & ACIDS
Very strong bases abstract (remove) the weakly acidic H+ (pKa = 16).
C
..
O
..
+
H
alcohol
(as acid)
B:
.. O
..:
C
strong base
+
BH+
alkoxide
pKb = -2
pKa = 16
Very strong acids protonate the weakly basic –OH group (pKb = 16).
C
..
O
..
H
H
alcohol
(as base)
+
H
A
C
+
O
..
H
+
A
-
strong acid
protonated alcohol
pKb = 16
pKa = -2
The acid/base behavior of alcohols is much the same as that of H2O.
15
PROBLEMS
Draw products, show mechanisms and calculate extent of reaction (%).
CH3
CH2
..
O
..
Na+HCO3H
pKa = 16
ethanol
CH3
CH2
..
O
..
H
pKb = 7.6
sodium ethoxide
CH3
pKeq = pKa + pKb -14
CH3
CH3
CH2
+
CH2
Na :NH2-
CH3
pKb = -21
+
H2CO3
..
O:
..
Na+
+
H2O
pKeq = 16 -1.74 -14 = +.3
Extent of reaction < 50%
..
H
Na+
pKeq = 16 + 7.6 -14 = 9.4
Extent of reaction = 0%
Na+ OHpKb = -1.74
..
O
..
CH2
..
O:
..
CH2
..
O:
..
Na+
+
pKeq = 16 -21 -14 = -19
Extent of reaction = 100%
:NH3
16
3. REACTION OF ALCOHOLS WITH A BASE
 Active metals such as Na, Li, K, Ca, etc. are very strong bases.
 They deprotonate alcohols liberating H2 gas
 Draw products and show mechanisms for the following.
CH3CH2O
+
H
ethanol
Na
pKb = ca. -30
.
H3C
CH
H3C
O
H
+
CH3
/ 2 H2
Extent of reaction = 100%
H3C
CH
H3C
O- K+
+
1
/2 H2
potassium isopropoxide
CH3
H
+
CH3
CH3CHCH2O
1
+
sodium ethoxide pKeq = 16 -30 -14 = -28
K
isopropyl alcohol
CH3CHCH2O
CH3CH2O- Na+
H
.
. Ca
(CH3CHCH2O- )2 Ca+ +
isobutyl alcohol
H2
calcium diisobutoxide
17
3. REACTION OF ALCOHOLS WITH A BASE
Show the reaction of neopentyl alcohol with aluminum metal.
CH3
3
CH3CCH2OH
+
..
Al
.
CH3
neopentyl alcohol
CH3
(CH3CCH2O- )3 Al+3
+
3
/ 2 H2
CH3
aluminum trineopentoxide
 The evolution of H2 gas from alcohols in the presence Na metal is a
quick qualitative test for alcohols. The reaction is vigorous but not
explosive as it is in H2O.
 Aldehydes (pKa = 17) and ketones (pKa = 19) release only trace
amounts of H2 gas when mixed with Na metal. [The aldehydes and
ketones tested (and the test tube) must be free of water, i.e., dry! or
a false positive result will be seen.]
18
ACIDITY OF ALCOHOLS
Alcohols typically have a pKa of ca. 16, e.g., CH3CH2OH and CH3OH.
Alcohol
(CH3)3COH
pKa
CH3CH2OH
HOH
CH3OH
16
15.74
15.5
CF3CH2OH
(CF3)3COH
12
5.4
18
 Tert-butyl alcohol is less
acidic because its conjugate
base (alkoxide) is bulky and
not easily solvated by water.
 Electron withdrawing groups
such as –F make alcohols
more acidic. They stabilize
the alkoxide (by induction). A
weaker conjugate base
necessitates a stronger acid.
 Note the increased acidity of 2,2,2-trifluoroethanol and especially
nonafluoro – t-butyl alcohol.
 Unlike resonance effects, the inductive effect is only significant over
a short range.
19
PROBLEMS
 Write equations showing how the following transformations can be carried
out. More than one step may be necessary. Mechanism are not required
but show all reagents used.
OH
O
CH3
CH3
Na
CH2CH3
or
CH3CH2Cl
Na+NH2-
SN2
-
+
O Na
CH3
CH2
CH2 OH
Gilman Reagent
HBr
fast SN1
CH2 Br
1
2
LI +
Cu
2
H3O+
20
4. OXIDATION OF ALCOHOLS
4. OXIDATION OF ALCOHOLS ( TO CARBONYLS)
H
:O :
oxidation
:O :
C
C
reduction
alcohol
carbonyl
compound
 Alcohols are oxidized to
carbonyl compounds:
aldehydes, ketones and
carboxylic acids.
 In turn, these carbonyl
compounds can be
reduced to alcohols.
 The most common oxidants are oxides of nonmetals that are in a
high oxidation state, e.g.,HNO3, KMnO4, Na2Cr2O7 in H2SO4, and
NaOCl in HAc.
 Calculate the oxidation state of:
(1) + N + (-2)3 = 0
N = +5
N in HNO3
Mn + (-2)4 = -1
Mn = +7
Mn in MnO4(1)2 + Cr2 + (-2)7 = 0
Cr = 12/2 = +6
Cr in Na2Cr2O7
(-2) + Cl = -1
Cl = +1 21
Cl in OCl-
4. OXIDATION OF ALCOHOLS
 Oxidants are usually classified into one of 3 groups (strengths).
1. Mild oxidants include Collins reagent (CrO3 in pyridine), and
PCC (pyridinium chlorochromate, (C5H5NCrO3Cl in
dichloromethane solvent.) These oxidants are dissolved in cold
anhydrous solvents. (no water– this weakens the oxidant.)
2. Moderate-to-strong oxidants are cold to warm, aqueous
solutions of HNO3, acidic or basic KMnO4, NaOCl in aq. HAc,
Jones reagent (CrO3 in aq. H2SO4 + acetone).
3. Severe oxidants are hot, aqueous solutions of HNO3, acidic or
basic aq. KMnO4, or Na2Cr2O7 in H2SO4
 KMnO4 and Cr+6 reagents are often used for qualitative tests.
When reacting, these oxidizing agents are reduced and show
distinct color changes. List the color changes that occur.
neutral or OH+3
Cr+6
MnO2 brown
Cr
orange
green
or
blue
precipitate
MnO4-
purple
+
H
Mn+2
colorless
22
4. OXIDATION OF ALCOHOLS
 1 alcohols are easily oxidized. They have 2 -hydrogens.
 Mild oxidants remove 1 hydrogen yielding an aldehyde.
 Moderate-to-strong (or severe) oxidants remove both hydrogens
yielding a carboxylic acid.
mild oxidation
(oxidants in non aq. solvents)
OH
R
PCC (in CH2Cl2)
Collins reagent (CrO3 in pyridine)
CH2
O
R
C
H
aldehyde
O
1º alcohol
moderate or strong oxidation
(oxidants in aq. solvents)
Jones reagent (CrO3 in aq. H2SO4 & acetone)
KMnO4, HNO3, etc.
R
C OH
carboxylic
acid
23
4. OXIDATION OF ALCOHOLS
 2 alcohols have only 1 -hydrogen. Mild or moderate-to-strong
alcohols remove the only -hydrogen yielding a ketone.
 Under severe oxidation conditions, the ketone product is further
oxidized. In the study of alkynes, we learned that a ketone is
always in equilibrium with a minute amount of an enol, its ‘tautomer’.
 In our study of alkenes we learned that hot KMnO4, hot H2CrO4, or
hot conc. HNO3 oxidize enols, cleaving the C-to-C p-bond and
yielding carboxylic acids or ketones (which will again be cleaved
until only carboxylic acids remain).
O
mild to moderate oxidation
OH
R CH R
2º alcohol
(in non aq. solvents or any cold aq. solvent)
PCC (in CH2Cl2)
Collins reagent (CrO3 in pyridine)
Jones reagent (CrO3 in aq. H2SO4 & acetone)
cold acidic or basic KMnO4, HNO3, etc.
severe oxidation
(hot aq. KMnO4 or HNO3) R
R
O H
R
ketone
R
ketone
O
C
C
R
C
enol
R
O
2 R C
OH
carboxylic acid
24
4. OXIDATION OF ALCOHOLS
 3 alcohols are not oxidized by any of the above oxidants except hot
KMnO4 or hot conc. HNO3 which dehydrate them to alkenes and
subsequently cleave the alkenes to carboxylic acids.
 Recall that 3 alcohols are easiest to dehydrate to alkenes.
no rxn.
any anhydrous or cold aq. oxidant
OH
R
C
R
R
3º alcohol
O
severe oxidation
(hot aq. KMnO4 or HNO3)
R
dehydration
CH
R
alkene
2 R C
OH
carboxylic acid
 The mechanism of oxidations are either not fully understood or are
very complex. The student will need to memorize the oxidation
products of alcohols with the various classes of oxidants.
25
PROBLEMS
 Draw the products of the following oxidations.
CH2OH Jones
reagent
1-hexanol
OH
cold
Na2CrO4
2-hexanol
in H2SO4
CH2OH
1-hexanol
PCC or
Collins
reagent
hexanoic acid
COH
O
O
2-hexanone
hexanal
CH
O
26
PROBLEMS
 Draw the products of the following oxidations.
H3C
cold KMnO4
aqueous
H3C
OH
O
3-methylcyclohexanol
H3C
hot concentrated
HNO3
O
H3C
C
Cleavage occurs on either
side of the -carbon
H3C
O
OH
C
OH
OH
OH
C
+
C
OH
O
O
27
IODOFORM REACTION (OXIDATION)
The Iodoform reaction: is a qualitative oxidation test used to identify
the structure of a 1° or 2° alcohol. The alcohol is treated with a
solution of iodine and a solution of sodium hydroxide.
NaOH + I2  NaOI (sodium hypoiodite)
 NaOI is analogous to sodium hypochlorite (NaOCl, the so-called
‘liquid chlorine’ used as laundry bleach,e.g., Javex, and used as
swimming pool disinfectant).
 NaOI is a mild oxidant. It oxidizes 1° alcohols to aldehydes and
2° alcohols to ketones; but will not oxidize 3° alcohols.
 If the 1° or 2° alcohol had a methyl group (-CH3) attached to the
-carbon, the three hydrogens of the –CH3 group are substituted by
iodine (-CH3  -CI3) and the product cleaves producing iodoform
(CHI3), a pale yellow precipitate.

CH3CH2OH
1° alcohol
NaOI
O
3 NaOI
CI3-C-H
CH3-C-H
aldehyde
O
-CH3
-CI3
Na+ OHcleaves
O
+
-O-C-H
CHI3 + Na
iodoform
28
IODOFORM REACTION (OXIDATION)
 Ethanol is the only 1° alcohol that gives a positive iodoform test
because it is the only 1° that has a –CH3 group attached directly to
the -carbon.
 By contrast, many 2° alcohols have –CH3 group attached directly to
their -carbon and this can be identified by the iodoform test.
 Write equations for the iodoform reaction with 2-butanol.
CH3CH2CHCH3 +
OH
CH3CH2CCH3 +
OXIDATION
CH3CH2CCH3 + NaI + H2O
NaOI
IODINATION
3 NaOI
O
O
CH3CH2CCI3 +
O
O
CLEAVAGE
CH3CH2CCI3
+
NaOH
3 NaOH
CHI3 +
+
Na-O-C-CH2-CH3
O
29
PROBLEMS
 Using the Lucas test, oxidation with Jones reagent, and the
iodoform test, show how a student could distinguish between
2-pentanol and 3-pentanol.
2-pentanol
OH
Lucas test
ZnCl2, con. HCl
Jones reagent
CrO3 in H2SO4
Iodoform
NaOI
3-pentanol
2-chloropentane
Cl
Cl
3-chloropentane
2-pentanone
O
OH
O
O
3-pentanone
O- Na+
no iodoform
precipitate
+
CHI3
30
PROBLEMS
 Write equations showing how the following transformations can be carried
out. More than one step may be necessary. Mechanism are not required
but show all reagents used.
OH
OH
CH3
Cr+6, H+
1 CH3MgBr
O
CH2 CH2
POCl3,
OH
CH2OH
KMnO4
H2O
N
or
CH
+
2 H3O
CH2
1 LiAlH4
+
2 H3O
O
C
OH
hot KMnO4
31
SOLUBILITY AND BOILING POINTS OF ALCOHOLS
 Alcohols are the first organics we have studied that are polar.
 Alkanes, alkenes, alkynes and arenes are all hydrocarbons with non
polar bonds. Even alkyl halides are only weakly polar. These
compounds all exhibit characteristically low boiling points are are
insoluble in polar solvent like water.
HC and RX
n-pentane
carbon tetrachloride
n-butyl bromide
H2O Solubility
(g/100 mL, 25 C)
0.05
0.08
0.06
Describe the trend in solubility of
alcohols re: MW and branching.
Explain the unusually high solubility
of a polyol such as 1,4-butanediol.
Alcohols/phenol
H2O Solubility
(g/100 mL, 20 C)
n-butyl alcohol
sec-butyl alcohol
isobutyl alcohol
tert-butyl alcohol
n-pentyl alcohol
n-hexyl alcohol
1-heptanol
phenol
1-octanol
1,4-butanediol
9
12
10
miscible
2.7
0.6
0.2
6.7
0.05
miscible
Explain why cyclohexanol boils at 160 °C and cyclohexene at 83 °C.
32
PHENOLS
 Phenols have the hydroxyl (-OH) functional group of an alcohol
directly bonded to an aromatic ring; a fully conjugated ring; a ring
with a complete pattern of alternating single and double bonds.
alcohols
OH
cyclohexyl alcohol
CH3
phenols
OH
phenol
OH
CH3CHCH2OH
isobuty alcohol
b-naphthol
 Despite their similar appearance, the chemistry of phenols is
different than that of alcohols.
33
PHENOLS
 Phenol is manufactured in two steps from benzene.
 Benzene undergoes Electrophilic Aromatic Substitution (EAS) with
H2SO4 and SO3 (a mixture called ‘oleum’) producing
benzenesulfonic acid.
 The sulfonic acid group is replaced by the –OH group by reacting
with NaOH at high temperature (300°C).
 Write equations showing the preparation of phenol from benzene.
SO3, H2SO4
40 ºC
O
S
1 NaOH 300°C
OH
+
2 H3O
O
OH
phenol
benzenesulfonic acid
 The hydroxyl group donates electron density to the aromatic ring by
resonance, making phenols reactive toward EAS.
34
REACTIONS OF PHENOLS
 Phenols have the same hydroxyl (-OH) functional group as alcohols
but, while alcohols and H2O are essentially neutral (pKa = 16),
phenols are weak acids (pKa = 10). pKeq = 9.9 -1.74 -14 = -5.8
Extent -of reaction = 100%
..
O
..
..
O:
..
H
pKa = 9.9
phenol
..
O
..
Na+ HCO3-
pKb = 7.6
+
H2O
sodium- phenoxide
..
O:
..
H
+
pKb = 4.1
Na+ OH-
+
Na+
Na+
+
H2CO3
pKeq = 9.9 + 7.6 -14 = 3.5
Extent of reaction = 0%
 Phenols react 100% with aqueous NaOH but not with NaHCO3
(pKb = 7.6). This difference is often used differentiate phenols from
other stronger organic acids, such as carboxylic acids (pKa = 5).
 Carboxylic acids react 100% with both NaOH and NaHCO3.
35
REACTIONS OF PHENOLS
 Phenols are 106 times more acidic than alcohols (pKa 10 vs. pKa 16)
because the conjugate base, phenoxide ion (pKb = 4), is 106 times
more stable than the alkoxide (pKb = -2).
 In the alkoxide, the negative charge resides on only one atom, the
oxygen.
 In the phenoxide, the negative charge is ‘delocalized’ (spread out)
over 4 atoms by resonance throughout the fully conjugated p system
of the aromatic ring.
..
O
..:
..
..
O
..
:
..
O
..
..
..
O
..
Sodium phenoxide has 4 different resonance structures.
36
REACTIONS OF PHENOLS
 Phenols become stronger acids when electron withdrawing groups
are attached to the phenyl ring, e.g., -Br, -NO2, -CF3, etc.
 Phenols become weaker acids when electron donating groups are
attached to the phenyl ring, e.g., -CH3, -OCH3, -NH2, etc.
Draw these structures
p-aminophenol
phenol
p-nitrophenol
2,4,6-trinitrophenol
pKa
10.5
9.9
7.2
0.60
..
HO
..
..
HO
..
..
NH2
NO2
O2N
..
HO
..
O2N
NO2
37
REACTIONS OF PHENOLS
 Recall from our study of electrophilic aromatic substitution (EAS)
reactions that electron donor groups make the ring a better Nu:(electron donor); thus accelerating EAS reactions of aromatics.
 These same ring activators (electron donors) will reduce the acidity
of phenols by making conjugate base (phenoxide) less stable.
 Ring deactivators are electron withdrawing groups. They increase
the acidity of phenols by making phenoxides more stable.
o- and pdirecting
NH2
OCH3
o- and pdirecting
F
CH3
m-directing
Br
O
O
CH
C OH
O
NH C CH3
activators
NO2
Reactivity
Reactivity
OH
SO3H
H
Cl
deactivators
I
O
O
COCH3
C CH3
C N
+
N R3
deactivators
38
PROBLEM
Rank the following phenols in order of acidity, where 1 is most acidic.
Then name their conjugate bases.
m-cresol
m-nitrophenol
benzoic acid
p-chlorophenol
phenol
OH
HO
HO
OH
CH3
C
4
5
OH
3
1
O
NO2
Cl
2
..
..
: ..
O
CH3
..
O:
..
C
..
O
.. :
: O:
..
:O
..
Cl
NO2
O
m-cresoxide
phenoxide
benzoate
p-chlorophenoxide
m-nitrophenoxide
39
PROBLEM
 Write equations showing how the following transformation can be carried
out. More than one step may be necessary. Mechanism are not required
but show all reagents used.
an ether
phenyl n-propyl ether
O
CH2CH2CH3
CH3CH2CH2I
H2SO4
SN2
SO3
300° C
SO3H 1 NaOH
+
2 H3O
benzenesulfonic acid
OH
NaOH
O- Na+
acid/base
sodium phenoxide
40
Do the practice
problems in your
purchased notes.
41
-
-
-
alkyl
halide
(substrate)
good Nu
strong base
e.g., bromide
e.g., ethoxide
strong bulky base
e.g., t-butoxide
Br
C2H5O
(CH3)3CO
Me
SN2
SN2
SN2
no reaction
1°
SN2
SN2
E2 (SN2)
no reaction
2
SN2
E2
E2
SN1, E1
3
SN1
E2
E2
SN1, E1
-
-
good Nu
-
good Nu
nonbasic
-
very poor Nu
nonbasic
e.g., acetic acid
CH3COOH
42