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1
Lecture 26
Extending the R-X Spectrum: Making Alcohols (ROH) into Reactive SN and E Reactants (Two Ways)
So far we have mostly limited our “good” leaving groups to the very similar, stable halide anions,
Cl , Br and I-.
-
-
H
H (solvated)
Base
+
Base
...very stable conjugate bases,
which make very good leaving groups = I , Br , Cl .
strong acids (HI, HBr, HCl) have....
A select number of acid Ka/pKa values are provided below. The top five acids are all strong acids
having excellent leaving group conjugate bases. The last example is the weak acid water, which means
its conjugate base (hydroxide) is a relatively poor leaving group.
Conjugated Acid
Ka
pKa
Conjugated Base
(Leaving Group)
H
I
1010
-10
I
very good
H
Br
109
-9
Br
very good
H
Cl
107
-7
Cl
very good
O
O
H
O
S
R
103
-3
O
S
R very good
O
O
H
OH2
102
H
OH
10-16
-2
+16
H2O
O-H
very good
poor
Our general R-X pattern can be expanded with two new versions having an attached oxygen atom.
This is accomplished by changing the poor leaving group (OH) of an alcohol (ROH) into a good leaving
group using two different strategies. There are many commercially available alcohols as starting
structures and there are a large number of reactions known to produce alcohol structures from other
functional groups, and alcohols can react in many different ways. This makes alcohols a central group in
synthetic methodology, as the following examples demonstrate.
1. Sulfonated Alcohols as Good Leaving Groups
A common strategy to make the poor “OH” leaving group of an alcohol into a good leaving group
is to convert it into a sulfonate ester group (-OSO2R). The new leaving group is the conjugate base of a
sulfonic acid (example four in the table, just above). The conjugate base and/or leaving group is referred
to as a sulfonate anion. As an anion, its stability and appearance are very similar to the bisulfate
conjugate base of strongly acidic sulfuric acid (see below), so it is not surprising that it is such a good
leaving group and stable conjugate base.
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2
Lecture 26
Sulfuric acid ionization to form its very stable conjugate base, bisulfate
O
O
H O
S
O H
O H
H O
S
H
O
pKa
1010
-10
H
O
reference
base
sulfuric
acid
H O H
O
Ka
bisulfate
(very stable anion)
Sulfonic acid ionization and its conjugate base, alkyl sulfonate (same as leaving group from carbon)
O
O
R
S
O H
R
O H
O
H O H
O
pKa
103
-3
H
O
H
Sulfonates are stable anions,
which make excellent leaving
groups when bonded to carbon.
reference
base
sulfonic
acids
S
Ka
Alcohols are converted into sulfonates by a different type of substitution reaction than presented
in this topic. A possible mechanism is presented below for their formation. There are many variations of
the sulfonate leaving group, but the most common is the toluenesulfonyl ester (tosylate ester). All of the
forms of sulfur, below, have close analogies with carbon functional groups that we will study in the
coming chapters. The sulfur functional groups necessary to prepare tosylates are presented along with the
analogous carbon functional groups.
Formation of an inorganic sulfonate ester
O
S
Cl
O
O
O
R
S
H
toluenesulfonyl chloride
(tosyl chloride)
pyridine
(base)
N
R
H
pyridinium
hydrochloride
(precipitates
from solution)
alkyl toluene sulfonate
(tosylate ester)
Formation of an analogous organic alkanoate ester
O
R
C
O
Cl
carboxylic
acid chloride
O
R'
H
alcohol
R
R
R
N
R
tertiary amine
(base)
Cl
O
N
alcohol with a
poor "HO"
leaving group
O
C
O
R'
carboxylic
ester
R
R
N
H
Cl
R
trialkylammonium
chloride
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3
Lecture 26
Possible Mechanism for Tosylate Formation from an Alcohol and Tosyl Chloride
(making a poor leaving group, -OH, into a good leaving group, -O3SR)
The sulfur atom is positively polarized in its highly oxidized state as an acid chloride. This is
easily seen in the charged resonance structures below. Sulfur has a +2 formal charge in the only structure
showing an octet. We draw the other resonance structures because sulfur’s valence electrons are in the
n=3 subshell, where 3d orbitals are available.
Resonance structures of tosyl chloride before attack of alcohol
O
S
O
O
Cl
S
O
S
Cl
+2
Cl
O
O
The sulfur still has an octet,
but is highly positive and is
strongly electrophilic. Also
sulfur has available 3d orbitals
and can exceed the octet rule.
Because sulfur has these 3d orbitals available, it can expand its octet by taking on a fifth group.
The negatively polarized alcohol oxygen atom’s lone pairs are strongly attracted to the positively
polarized sulfur forming a new bond. We already know that chloride, Cl--, is a very good leaving group
and it can leave allowing sulfur to go back to its normal tetravalent state. The alcohol oxygen atom now
has three bonds and it has acquired a positive formal charge. The positive formal charge can be easily
neutralized with any available base (pyridine is added for just this purpose). Overall, the chlorine atom is
replaced by the “O-R” of the alcohol group, forming an inorganic sulfur ester (analogous to an organic
carbon ester). The HCl that is also formed in the reaction is neutralized by the pyridine base.
A Different Mechanism for Substitution Reaction at Sulfur
(addition of the nucleophile before the leaving group leaves)
nucleophilic oxygen
attacks electrophilic
sulfur
O
S
Cl
O
O
S
R
R
E1
O
O
R
H
N
Pyridine is added to
neutralize any acidic
protons. It is often
called a "proton sponge".
SN2
SN1
S
Cl
N
The second resonance
structure emphasizes the
electrophilic nature of
the sulfur.
E2
O
O
H
O
R
O H
chloride is
eliminated
by a negative
push from
oxygen
O
Ts
O
R
=
Tosylate leaving group,
similar to halides = "TsO-".
This becomes one of our
"X" leaving groups.
S
O
R
O
alkyl sulfonate,
the alcohol oxygen is
now a good leaving
group
Z:\classes\How to Cure the Benzene Blues\90 lectures to organic\314 Lectures\lect_26, 19pp SN & E #4 (ROH,summary) for web.doc
N
Cl
salt
(ppt)
H
4
Lecture 26
This mechanism is not possible at tetravalent carbon (sp3) because carbon cannot bond to five
groups or have more than eight electrons in its valence. However, there is a similar mechanism at carbon
where trivalent carbon (sp2) of an acid chloride becomes tetravalent (sp3) in an intermediate and then goes
back to being trivalent (sp2). That mechanism is presented in just a bit as a problem for you to solve. The
important result of this transformation is that a poor leaving group, the HO- of an alcohol, is made into a
very good leaving group, a sulfonate group RSO3-.
The reaction presented above is a general reaction of acid chlorides, which can produce esters (of
carbon = organic, nitrogen = inorganic, sulfur = inorganic, phosphorous = inorganic and others). The
oxygen atoms and the chlorine atom inductively pull electron density from the central atom (C, N, S or P)
and make it very partial positive. This inductive effect is reinforced by a resonance effect where we
explicitly write a positive charge on the central atom as electrons are shifted to oxygen. In the case of
carbon, theoretical calculations estimate this contribution at 10-15 percent. It is definitely reasonable to
write it and is very helpful in understanding the electrophilic nature of carbonyl carbons (C=O).
Once the tosylate is made, and the alcohol oxygen is attached to the sulfonyl portion, the oxygen
has become a very good leaving group and can leave as a very stable anion. (Think again of a stable base
vs. its strong acid Ka/pKa.)
R
O
O
O
S
R
O
O
O
O
S
O
O
O
S
S
O
O
The stable anion leaving group makes ionization process much easier in
SN1/E1 reactions and it is also a very good leaving group in SN2/E2
reactions. Its conjugate acid has a pKa = -3.
Compare this to the “HO” leaving group of an alcohol.
R
O
H
"HO" is a poor
leaving group.
O
R
H
conjugate acid (H2O)
pKa = 16
This is a difficult reaction
because both products are
high energy intermediates.
Problem 1 - Show analogous resonance structures for the carbon, nitrogen and phosphorous acid chlorides
to those given above for tosyl chloride, which emphasizes their electrophilic character.
a
b
O
H3C
C
Cl
c
O
N
Cl
O
R
P
R
Cl
Problem 2 - Show a similar mechanism of substitution (to that of tosyl chloride, above) for each of the
above acid chlorides with methanol as the attacking alcohol (add the nucleophile and then eliminate the
leaving group). Use a generic base, B:, to remove the acidic proton. What is the nucleophile? What is
the electrophilic atom in each example? What is the leaving group in each example?
Problem 3 - Show how each of the following compounds can react in an SN2 reaction (mechanism).
Identify the nucleophile, substrate, leaving group and product in each equation (-OSO2C6H4CH3 = -OTs)
Z:\classes\How to Cure the Benzene Blues\90 lectures to organic\314 Lectures\lect_26, 19pp SN & E #4 (ROH,summary) for web.doc
5
Lecture 26
a
O
O
CH3
C
H 3C O
CH2
c
b
S
N C
CH3CH2
OTs
N3
OTs
O
Problem 4 - What alcohol and reactions would generate each of the above tosylate esters? Write out the
reaction equation, including the necessary reagents to produce the desired transformation. (Don’t forget
the proton sponge.)
Problem 5 - Alcohols can be converted into inorganic esters and organic esters. An example of each
possibility is shown below.
an organic ester
an inorganic ester
O
R O H
py
Cl S
O
S
R O H
Cl C CH
3
alcohol
ethanoyl chloride
(acetyl chloride)
O
tosyl chloride
alcohol
R O
O
O
O
alkyl tosylate
R3N
C
CH3
R O
alkyl ethanoate
(alkyl acetate)
Use these two reactions and the fact that -OTs is a good leaving group and CH3CO2-- is a good
nucleophile to propose a synthesis of both enantiomers shown below from the chiral alcohol shown.
Classify all chiral centers as R or S.
O
O
H
CH3
O
CH3
C
H
C
H
C CH
3
CH2CH3
CH2CH3
H
H3C C CH2CH3
O
C CH3
O
Problem 6 - (R)-2-butanol is subjected to the following reaction sequence. The reaction with hydroxide
in the second step occurs at carbon. What is the absolute configuration of the final 2-butanol product?
Draw out the reaction in 3D.
O
O
18
OH
H3C CH2 CH CH3
Cl S
O
CH3
18
S
O
CH3
O
H3C CH2 CH CH3
HO
OH
H3C CH2 CH CH3
(R)-2-butanol
Z:\classes\How to Cure the Benzene Blues\90 lectures to organic\314 Lectures\lect_26, 19pp SN & E #4 (ROH,summary) for web.doc
O
18
S
O
CH3
O
6
Lecture 26
Problem 7 - Suggest a mechanism for each of the following transformations.
RO
/ ROH
(R)
O
Cl
OR
S
O
OH
OTs
N
(S)
ROH
OR
OR
(R / S)
racemic
achiral
Problem 8 - Predict reasonable products (major and/or minor) for each of the following reaction
conditions. If certain choices are excluded, state why.
a
b
OTs
RO / ROH
c
OTs
RO / ROH
d
OTs
ROH
OTs
ROH
2. Protonated Alcohols - Water as a Good Leaving Group
A second method for making an alcohol, OH, into a good leaving group involves a very different
set of conditions. Strong protic acids are used to extensively protonate the alcohol OH. When the alcohol
OH is protonated, the leaving group is water, not hydroxide. Water’s conjugate acid is H3O+,
(pKa = -2). If substitution is the desired goal, then the strong halide acids are normally used, HCl, HBr or
HI. If elimination is the desired goal, then concentrated sulfuric acid (H2SO4) and/or concentrated
phosphoric acid (H3PO4) are used at an elevated temperature (∆).
Using the hydrohalic acids (HCl, HBr or HI), very polar, strongly acidic conditions encourage SN1
reactions, and these are assumed to be operating at all tertiary and secondary alcohol (ROH) centers.
Rearrangements are frequently observed under these conditions. When methyl or primary alcohols are
used, the large energy expense of carbocation formation prevents the escape of water on its own. The
H2O at a primary ROH2+ is assumed to be pushed off by the conjugate halide base partner of the strong
acid used to form a methyl or primary alkyl halide without rearrangement (SN2).
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7
Lecture 26
a. 1o, 2o and 3o ROH reacted with HX acids (HCl, HBr, HI) - usually SN chemistry
i. primary alcohols (SN2 emphasized, no rearrangement, similar at methanol)
H
H
O
Br
O
H
conc.
H Br
Br
H O
SN 2
H 3O
H
ii. secondary alcohols (SN1 emphasized, rearrangements possible)
O
H
H
I
O
I
I
H
H
H
top and bottom
attack
I
H3O
H2O
iii. tertiary alcohols (SN1 emphasized, rearrangements possible)
H
O
H
H Cl
O
CH3
H
H2O
H3O
Cl
Cl
top and bottom
attack but can't tell
in this example
b. 1o, 2o and 3o ROH reacted with H2SO4 and/or H3PO4 and high temperature - E1 chemistry
Using strongly acidic sulfuric acid, H2SO4, at elevated temperatures favors E1 reactions and these
are assumed to be operating in all of the reactions below. Rearrangements are possible and observed.
i. primary alcohols (with high temperature, E1 is proposed, rearrangements possible)
O
conc. H2SO4
H
very high Temp
(difficult)
distills out
ii. secondary alcohols (E1 emphasized, rearrangements possible)
H
O
∆
H2SO4
distills out
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8
Lecture 26
iii. tertiary alcohols (E1 emphasized, rearrangements possible)
O
H
∆
H2SO4
major
(>90%)
distills out
We have some, limited control to direct the alcohol functionality toward SN or E choices. The
conditions to effect these different pathways are important, so you must be aware of the details mentioned
above (halide acids = SN reactions and H2SO4/∆ = E1 reactions). Heat is a crucial aspect of the E1
reaction, since it allows the lower boiling alkene to escape from the reaction mixture by distillation, while
the higher boiling alcohol or inorganic ester remains, in the reaction pot to reestablish equilibrium by
forming more alkene, which distills......etc. The alkene boils at a much lower temperature than the
alcohol it came from because it does not have an “OH” to form hydrogen bonds with.
Examples of Boiling Point Differences Between Alcohols and Possible Alkene Products
boiling points
of alcohols (oC)
boiling point
of alkenes (oC)
o
OH (79 C)
(82 oC)
OH
OH
(97 oC)
H2C
CH2
∆Tbp
(-104 oC)
(183 oC)
(-47 oC)
(129 oC)
(-47 oC)
(144 oC)
(-6 oC)
(106 oC)
(1 oC)
(96 oC)
(4 oC)
(99 oC)
OH
(100 oC)
OH
(161 oC)
(83 oC)
(77 oC)
Problem 9 (R)-2-butanol retains its optical activity indefinitely in aqueous base (-OH), but is rapidly
converted to optically inactive 2-butanol (racemic) when in contact with dilute sulfuric acid
(H2SO4 + H2O Æ H3O+ + HSO4 --). Explain with detailed mechanisms.
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9
Lecture 26
Problem 10 - Summary problem, Suggested Strategy
1.
Look at the reaction conditions first. A strong base/nucleophile favors the bimolecular processes
(SN2,E2). A weak base/nucleophile favors the unimolecular processes (SN1,E1)
O
oxygen examples of strong base/nucleophiles in our course:
H
O
R
O
O
O
oxygen examples of weak base/nucleophiles in our course:
H
O
H
R
O
H
H
O
2. Next, look to the reactant structures (CH3X, 1oRX, 2oRX, 3oRX, hindered 1oRX, allylic, benzylic,
etc.)
a. Methyl RX
i. bimolecular conditions - only SN2
ii. unimolecular conditions – no reaction, we don’t propose methyl carbocations
b. 1oRX
i. bimolecular conditions – mainly SN2, some E2, The amount of E2 increases with substitution
at Cβ or increasing bulk in the base/nucleophile (i.e. much more E2 with t-butoxide than with less
sterically bulky alkoxides, such as ethoxide). Allylic and benzylic RX are especially reactive by
SN2. If Cβ is fully substituted then neither SN2 or E2 is possible at 1o RX.
ii. Unimolecular conditions - no reaction due to the high energy of 1oR+.
c. 2oRX =
i. bimolecular conditions – with hydroxide and alkoxides there is generally more E2 than SN2, but
significant amounts of both observed. If the nucleophile is less basic, like carboxylates, more SN2
is observed (i.e. more SN2 with acetate than ethoxide). In our course we will assume mainly SN2
with acetate and mainly E2 with hydroxide and alkoxides. Allylic and benzylic RX are especially
reactive by SN2. If Cβ is fully substituted then SN2 is not possible. E2 may be possible at the
other Cβ position.
ii. unimolecular conditions – generally more SN1 than E1. If a highly substituted alkene can
form, it can become a significant product. Greater amounts of more substituted alkenes form than
less substituted alkenes. Rearrangements are common. In our course we will assume that an
alcohol reacts by E1 in strong acid/high temperature conditions (H2SO4,/H3PO4/∆) and the alkene
is distilled out (E1).
d. 3oRX
i. bimolecular condition - only E2 products are obtained (no SN2). The most stable alkene is
usually the major product, but other alkene products obtained too.
ii. unimolecular conditions - generally SN1 > E1. See 2oRX comments above. Because R+s are
flat (sp2), Cβ-Hs can be lost from either side, leading to a greater variety of elimination products.
Rearrangements are common.
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10
Lecture 26
In each of the following substrates consider two sets of conditions: (1) RO--/ROH (strong) and
(2) ROH (weak). Also remember that cyclohexanes have two possible chair conformations and each
conformation should be considered for the E2 reactions (anti elimination required). In all unimolecular
reactions only consider up to one rearrangement and only if a more stable carbocation forms.
a
b
X
d
c
X
X
f
e
X
X
X
h
g
X
X
Challenging Carbocation Rearrangements and Reactions
Problem 11 - Propose an arrow pushing mechanism for the following transformations. What is the
purpose of the strong acid (H3O+ or BF3)? What is the “good” leaving group in a, b, c and d? Part e is
less obvious, but the purpose of the BF3 is similar to the purpose of H3O+. Both Lewis acids make
oxygen a better leaving group when bonded to an oxygen lone pair. You need to make a good leaving
group, have it leave to form the best possible carbocation. In parts d and e, there are multiple
rearrangements and ring strain is part of the problem. In part d form the most stable carbocation first. In
part e you don’t have a choice.
a.
b.
HO
O
OH
O
HO
OH
+
H3O
H2O
H3O+
H2O
Why does methyl migrate instead of phenyl?
c.
d.
O
OH
OH
OH
H3O+
H2O
d.
H3O+
H2O
O
O
C
H3C
H
C
H
H
BF3 (strong Lewis acid)
no H2O
H3C
C
C
H2
H
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11
Lecture 26
Additional carbocation rearrangements to consider. (Some of these are tough.)
O
H3O+
R O H
O
H-X (acid)
O
OH
7
H2O
3
1
OH
6
TsO
5
7
3
O
H
BF3 is a Lewis acid
similar role to H3O+
6
4
O
O
BF3
Et2O
H
2
1
2
4
HO
O
H3O
+
5
H
HCl
CH3OH
O
O
O
Problem 12 – In the body cholesterol is constructed from eighteen 2 carbon acetate units (mevalonate is
an intermediate) that are converted into a 30 carbon hexaene (squalene, with 6 double bonds), which is
then oxidized to an epoxide (squalene oxide) and protonated in an acid catalyzed reaction (enzyme)
involving four carbocation rearrangements to form the cholesterol precursor, “lanosterol”. Lanosterol
becomes chloesterol after 19 additional steps! Cholesterol is the lead steroid for all of the body’s other
steroids.
O
H 3C
C
SCoA
via mevalonate,
isoprenoids, geranyl
farnesyl structures
oxidation
8 x acetyl CoA
O
squalene
squalene oxide
H-Enzyme
other steroids
estrogen
testosterone
cortisol
bile acids
etc.
H
additional
steps
H
H
HO
H
cholesterol
1. protonated epoxide
ring opening,
2. pi bond nucleophilic
attack,
3. rearrangement and
4. E1 reaction
H
many
additional
steps
HO
H
lanosterol
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12
Lecture 26
i. The cyclization step to form lanosterol involves enzyme acid catalyzed protonation to make a good
leaving group, which undergoes subsequent reactions you have studied. The complex system has been
simplified below. Show how this key cyclization process begins by writing an arrow pushing mechanism.
Hint – In acid what is your first expectation? Look for the bonds that have changed (broken and formed).
Consider acid/base, carbocations, pi bond nucleophiles…etc.
R
R
H-Enzyme
O
HO
H
squalene
oxide analog
ii. Show how the tetracyclic carbocation rearranges to lanosterol. (Four rearrangements and an
elimination reaction.)
R
H
R H
CH3
H
CH3
H
H B
H
HO
HO
H
lanosterol
precursor
H
lanosterol
Problem 13 – A spectacular example of rearrangement in an acid catalyzed alcohol dehydration is shown
below. When 3ß-friedelanol is treated with acid, a carbocation forms, and then a total of seven methyl
and hydrogen migrations occur before the loss of a proton produces 13(18)-oleanene. Each migration is a
1,2-shift of a methanide ion (CH3: --) or hydride ion (H: --). Show how each of these migrations occur.
H A
HO
3ß-friedelanol
13(18)-oleanene.
Z:\classes\How to Cure the Benzene Blues\90 lectures to organic\314 Lectures\lect_26, 19pp SN & E #4 (ROH,summary) for web.doc