Nucleophilic Substitution and b

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Transcript Nucleophilic Substitution and b

Stereochemistry of E2
• E2 is most favorable (lowest activation energy)
when H and Lv are anti and coplanar
CH3 O:
-
D
H
C
E
C
A
Lv
B
-H an d -Lv are anti and cop lanar
(dih edral an gle 180°)
CH3 O
H
A
B
C
D
E
C
Lv
Examples of E2 Stereochemistry
Explain both regioselectivity and relative rates of reaction.
CH3O -
Cl
cis
Faster
reaction
+
Major product.
Zaitsev product
But
CH3O -
Cl
trans
Slower
reaction
+
Only product
Anti-Zaitsev
Principles to be used in analysis
Stereochemical requirement: anti conformation for departing groups. This
means that both must be axial.
Dominant conformation: ring flipping between two chair conformations,
dominant conformation will be with iso propyl equatorial.
First the cis isomer.
Reactive Conformation; H
and Cl are anti to each other
-
CH3 O:
H
2
H
E2
6
H
+ CH3 OH + :Cl
H
1
Cl
In order for the H and the
Cl to be anti, both must be
in axial positions
1-Isopropylcycloh exene
Iso-propyl groups is in more stable
equatorial position. Dominant
conformation is reactive conformation.
Now the trans
In the more stable chair of the trans isomer, there is no H anti and
coplanar with Lv, but there is one in the less stable chair
HH
HH
66
HH
Cl
Cl
11
HH
Cl
Cl
22
HH Unreactive
conformation
More
s
table
chair
More s table chair
(no
(noHHisisanti
antian
andd
coplanar
coplanartotoCl)
Cl)
66
11
22
HH
Reactive but only with
the H on C 6
HH
HH
Less
Lessstable
stablech
chair
air
(H
(Hon
oncarb
carbon
on66isis
an
antitiand
andcop
coplan
lanar
artotoCl)
Cl)
Most of the compound exists in the
unreactive conformation. Slow reaction.
Cl
H
-
CH3 O:
6
H
2
1
H
E2
+ CH3 OH + Cl
H
(R)-3-Is op ropylcyclohexene
Anti Zaitsev
Example, Predict Product
Ph
base
H3C
Ph
H3C
Ph
CH3
H
H3C
Br
Ph
Problem!: Fischer projection diagram represents an eclipsed structure.
Task: convert to a staggered structure wherein H and Br are anti and predict
product. We will convert to a Newman and see what we get…
Ph
H
rotate upper
chiral C by 180 Ph
Ph
CH3
H
CH3
H3C
H
Br
H3C
Br
=
H3C
Br
Ph
H3C
rotate 120
further
Ph
CH3
H
Ph
H3C
Br
Ph
H & Br not anti
yet!
CH3
Br
H3C
H
Ph
H3C
Ph
H3C
Ph
Ph
Ph
H3C
base
H3C
Br
H
Ph
Ph
Now anti and
we can see
where the pi
bond will be.
Alternative Approach: CAR
Ph
base
H3C
Ph
H3C
Ph
CH3
H
H3C
Br
Ph
The H and Br
will be
leaving: just
indicate by
disks.
Anti
Geometry
A
CR
Meso or
Racemic??
This may be recognized
as one of the
enantiomers of the
racemic mixture.
A
C < -- > R
Relationship
works in both
directions.
Should get cis
isomer.
Note: As we have said
before it may take some
work to characterize a
compound as “racemic” or
“meso”.
E1 or E2
Alkyl halide
Primary
E1 (Carbocation)
E2
E2 is favored.
RCH2 X
E1 does not occur.
Primary carbocations are
so unstable, they are never
observed in solution.
Secondary
R2 CHX
Main reaction with weak
bases such as H 2O, ROH.
Main reaction with strong
bases such as OH - and OR -.
Tertiary
R3 CX
Main reaction with weak
bases such as H 2O, ROH.
Main reaction with strong
bases such as OH - and OR -.
1o good nucleophiles, aprotic solvents
1o
2o good nucleophiles but also poor bases, 2o lower hinderance, better nucleophile than
aprotic solvents
base
30
3o lower hinderance, better nucleophile than
R-Nuc
base
SN1
R-Nuc SN2
good
nucleophile
R-X
weak nucleophile
ionization
Rearrange ?
R+ + X-
1o,
strong
base
2o, 3o polar solvents,
weak nucleophiles,
weak bases
alkene
E2
weak base
alkene
1o strong, bulky bases
E1
2o strong bases
1o
3o strong bases
2o heat, more hindered
3o heat, more hindered
Recall Halohydrins and Epoxides
Cl2, H2O
Cl
base
OH
Cl
H2O
Cl
Creation of Nucleophile
Internal SN2 reaction
with inversion
O
RO
ROH
H
O
Creation of
good
leaving
group.
O
H
Attack by poor
nucleophile
OH
Neighboring Group Effect
• Mustard gases
– contain either S-C-C-X or N-C-C-X
N
S
Cl
Cl
Bis(2-chloroethyl)sulfide
(a sulfur mustard gas)
Cl
Cl
Bis(2-chloroethyl)methylamine
(a nitrogen mustard gas)
– what is unusual about the mustard gases is that they undergo
hydrolysis rapidly in water, a very poor nucleophile
Cl
S
Cl
+
2 H2 O
HO
S
OH + 2 HCl
– the reason is neighboring group participation
by the adjacent heteroatom
:
slow , rate
d etermining
S
Cl
Cl
an internal
SN 2 reaction
Good nucleophile.
S
Cl
Cl
S
+
Cl
A cyclic
sulfonium ion
:
+
+
+ : O-H
H
fas t
a second
S N 2 reaction
S
Cl
H
+O
H
– proton transfer to “solvent” completes the reaction
From an old quiz
5. Provide a clear, unambiguous mechanism to explain
the following stereochemical results.
Complete structures of intermediates, if any, should be shown.
Use curved arrow notation consistently.
SCH3
H
SCH3
H2O
CH3
H
SCH3
CH3
H3C
H
+
H3C
H3C
H
H
Cl
H
OH
CH3
OH
Here is the crux of the matter: how can the non-reacting carbon change its
configuration??? Further it does not always change but only if configuration of the
reacting carbon changes!! We got a mixture of enantiomers, a racemic mixture.
Something strange is happening!!
Expect sulfur to attack the C-Cl, displacing the Cl and forming a three membered ring.
Like this…
S
H
CH3
CH3
SCH3
S
CH3
H3C
H
H2O
Cl
But we have to be careful with stereochemistry
OH
We have to put the molecule in the correct conformation.
SCH3
SCH3
H
H
H
CH3
=
H3C
CH3
H
=
H3C
H
H3C
CH3
H
Cl
Cl
Cl
SCH3
S and Cl are
eclipsed, not
anti.
Reactive conformation reached by 180 rotation around C-C bond
H
Cl
CH3
H
H
H3C
SCH3
H
H3C
CH3
S
CH3
And then the ring is opened by
attack of water
But let’s pause for a moment. Our reactant was optically active with two chiral carbons.
Recall the problem: If reaction occurs only at the C bearing the Cl the other should
remain chiral! Hmmmm?
But now notice that the intermediate sulfonium ion is achiral. It has a mirror plane of
symmetry. Only optically inactive products will result.
Two modes of attack by water.
OH2
SCH3
H
HO
H
H3C
H
H3C
H
CH3
CH3
H3C
CH3
=
H
S
CH3
H
H
CH3
=
H3C
SCH3
HO
H
SCH3
180 rotation
OH
Enantiomers,
racemic
mixture
And…
OH2
SCH3
H
H
H
H3C
CH3
OH
H
H3C
H
H3CS
S
CH3
=
H3C
CH3
H
H3C
H
=
CH3
H
H3CS
CH3
OH
180 rotation
Again note the ring structure is achiral and that we must,
of course, produce optically inactive product.
OH
Alcohols
Hydrogen Bonding
Three ethanol molecules.
Hydrogen Bonding & boiling point
Increases boiling point, higher temperature needed to separate the
molecules.
Hexane
69 deg.
1-pentanol
138
1,4-butanediol
230
Ethanol
78 deg
Dimethyl ether
24
Earlier Discussion of Acidity
RO-H

RO – (solvated) +
H + (solvated)
Alkoxide ion, base
Increasing Hinderance of Solvation
Methanol
Ethanol
2-Propanol
2-Methyl-2-propanol
Increasing Basicity of Alkoxide Anion, the conjugate base
Increasing Acidity of the alcohol
Alkoxides can be produced in several ways…
Recall: H2O + Na Na+ + OH- + ½ H2(g)
Alcohols behave similarly
ROH + Na Na+ + OR- + ½ H2(g)
Also: ROH + NaH Na+ + OR- + ½ H2(g)
Alkoxide, strong base, strong
nucleophile (unless sterically
hindered)
-OH as a Leaving Group
Poor leaving group, hydroxide
ion.
R-OH
+ H+

R-OH2+
Protonation of the alcohol sets-up a good leaving group,
water.
Another way to turn the –OH into a leaving group…
Conversion to Alkyl Halide,
HX + ROH  RX + H2O
When a carbocation can be formed (Tertiary, Secondary alcohols)
beware of rearangement. SN1
H+
R3COH
XR3COH2
R3C + + H2O
R3CX
Expect both
configurations.
When a carbocation cannot be formed. Methanol, primary. SN2
H
RCH2OH
X-
+
RCH2OH2
RCH2X
But sometimes experiment does
not agree with our ideas…
X
HX
CH2OH
Observed reaction
CH2X
The problem:
•Rearrangement of carbon skeleton which usually indicates carbocations.
•Reacting alcohol is primary; do not expect carbocation.
•Time to adjust our thinking a bit….
H3C
H
CH2OH
+
X
H3
C
H3C
CH2OH2+
CH3
CH2.......OH2
H2O
Not a primary carbocation
X-
Other ways to convert: ROH  RX
We have used acid to convert OH into a good leaving group
H+
XR3C + + H2O
R3COH2
R3COH
R3CX
There are other ways to accomplish the conversion to the halide.
primary, secondary
Br -
PBr3
RCH2-O(H)PBr2
RCH2-OH
RCH2Br + HOPBr2
Leaving group.
primary, secondary, tertiary
Cl -
SOCl2
RCH2-OS(O)Cl
RCH2-OH
RCH2Cl + SO2
amine
Leaving group.
Next, a very useful alternative to halide…
An alternative to making the halide:
ROH  ROTs
CH3
CH3
Preparation from
alcohols.
ROH +
O
S
O
O
S
O
Cl
p-toluenesulfonyl
chloride
Tosyl chloride
TsCl
The configuration of
the R group is
unchanged.
O
R
Tosylate
group, -OTs,
good leaving
group,
including the
oxygen.
Example
CH3
CH3
TsCl
H
H
OTs
C3H7
CH3
OH
C3H7
CH3
C2H5
Preparation of tosylate.
Retention of configuration
C2H5
Substitution on a tosylate
The –OTs group is an excellent leaving group
Acid Catalyzed Dehydration of an Alcohol,
discussed earlier as reverse of hydration
Secondary and tertiary alcohols, carbocations
Protonation,
establishing of good
leaving group.
Elimination of water to
yield carbocation in
rate determining step.
Expect tertiary faster
than secondary.
Rearrangements can
occur.
Elimination of H+
from carbocation to
yield alkene.
Zaitsev Rule
followed.
Primary alcohols
Problem: primary carbocations are not observed. Need a modified, non-carbocation
mechanism.
Recall these concepts:
1. Nucleophilic substitution on tertiary halides invokes the carbocation but nucleophilic
substitution on primary RX avoids the carbocation by requiring the nucleophile to
become involved immediately.
2. The E2 reaction requires the strong base to become involved immediately.
Note that secondary and tertiary protonated alcohols eliminate the water to yield a
carbocation because the carbocation is relatively stable. The carbocation then
undergoes a second step: removal of the H+.
The primary carbocation is too unstable for our liking so we combine the departure of
the water with the removal of the H+.
What would the mechanism be???
Here is the mechanism for acid catalyzed
dehydration of Primary alcohols
1. protonation
2. The
carbocation is
avoided by
removing the
H at the same
time as H2O
departs (like
E2).
As before,
rearrangements can
be done while
avoiding the primary
carbocation.
Principle of Microscopic Reversibility
Same mechanism in either
direction.
Pinacol Rearrangement: an example of
stabilization of a carbocation by an adjacent lone
pair.
Overall:
Mechanism
Reversible
protonation.
Elimination of
water to yield
tertiary
carbocation.
1,2
rearrangement
to yield
resonance
stabilized cation.
Deprotonation.
This is a
protonated
ketone!
Oxidation
Primary alcohol
RCH2OH
Na2Cr2O7
Na2Cr2O7
RCH=O
RCO2H
Na2Cr2O7 (orange)  Cr3+
(green) Actual reagent is
H2CrO4, chromic acid.
Secondary
Na2Cr2O7
R2CHOH
R2C=O
KMnO4 (basic) can also be
used. MnO2 is produced.
Tertiary
R3COH
NR
The failure of an attempted
oxidation (no color change) is
evidence for a tertiary alcohol.
Example…
OH
OH
Na2Cr2O7
acid
HO
CH2OH
O
CO2H
Oxidation using PCC
Primary alcohol
PCC
RCH2OH
RCH=O
Secondary
PCC
R2CHOH
R2C=O
Stops here, is not oxidized to
carboxylic acid
Periodic Acid Oxidation
OH
O
OH
HIO4
glycol
O
+
HIO3
two aldehydes
OH
O
O
HIO4
HO
O
+
aldehydes
carboxylic acid
O
O
O
HO
HIO4
O
+
OH
carboxylic acid
HIO3
carboxylic acid
OH
O
O
2 HIO4
+ 2 HIO3
HO
O
OH
OH
O
HIO3
Mechanistic Notes
Cyclic structure is formed during
the reaction.
Evidence of cyclic intermediate.
Sulfur Analogs, Thiols
Preparation
RI +
HS-
 RSH
SN2 reaction. Best for primary, ok secondary, not tertiary (E2 instead)
Oxidation
Acidity
H2S pKa = 7.0
RSH
pKa = 8.5