Nucleophilic Substitution & Elimination Chemistry
Beauchamp
1
Problem 1 - How can you tell whether the SN2 reaction occurs with front side attack, backside attack or front and
backside attack? Use the two molecules to explain you answer. Use curved arrow formalism to show electron
movement for how the reaction actually works.
Problem 2 - Why might C3 and C4 rings react so slowly in SN2 reactions? (Hint-think about bond angles in the
transition state versus bond angles in the starting ring structure.)
Problem 3 - Why might C6 rings react slower in SN2 reactions? What are the possible conformations from which a
reaction is expected? Trace the path of approach for backside attack across the cyclohexane ring to see what positions
block this approach. Which chair conformation would have the leaving group in a more reactive position (axial or
equatorial)? Is this part of the difficulty (which conformation is preferred and present in higher concentration)?
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Nucleophilic Substitution & Elimination Chemistry
Beauchamp
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Problem 4 - In each of the following pairs of nucleophiles one is a much better nucleophile than its closely related
partner. Propose a possible explanation.
Problem 5 – Write out the expected SN2 product for each possible combination (4x8=32 possibilities).
1
Nu
=
2
3
O
4
O
S
O
H
H3C
O
N
S
H
N
C
H3C
8
D
D
Al
D
D
Li
7
6
5
N
N
O
A
N
C
O
OH
O
S
SH
N
N
D
N
B
O
O
OH
S
SH
C
N
O
OH
C
N
N
O
SH
O
C
S
D
N
N
D
N
O
N
N
D
OH
O
N
SH
O
S
N
C
N
N
D
O
Problem 6 – Using R-Br compounds from page 2 and reagents from page 3 to propose starting materials to make each
of the following compounds. One example is provided. TM-1 is an E2 product (see page 15), all the others are SN2
products.
2-azidopropane
N
N
solution
problem
N
N3
Br
NaN3
(buy)
?
SN2
TM = target molecule
TM-1
(make)
K
Br
TM-2
O
(make)
TM-3
Br
E2
(make)
Na
SN2
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O
O
R
Na
R
SN2
Br
Nucleophilic Substitution & Elimination Chemistry
TM-4
O
(make)
Beauchamp
O
Br
O
TM-6
TM-5
N
N
3
O
(no structure)
O
Br
O
(make)
O
O
SN 2
SN 2
O
TM-7
TM-9
Br
H3C
SN2
OH Br
H3C
O
H3C
Ph
TM-8
Ph
Ph
P
Br
Ph
Ph
P
Ph
SN1
(buy)
(buy)
N
N
C
C
Br
SN2
Na
Br
SN2
TM-11
TM-10
Ph
S
Ph
Br
Ph
(buy)
Br
Br
(make)
SN2
SN2
Ph
TM-13
TM-12
HO
TM-14
Na
Na
Na
OH
SH
R
(buy)
Br
SN2
TM-15
O
Li
O
S
(buy)
SN 2
S
Br
(make)
S N2
R
TM-16
O
O
S
Br
S Li
Li
S
Br
O
O
(make)
SN2
H
S
Br
H
S
(make)
SN2
TM-17
Li
AlD4
(buy)
D
Br
SN2
Problem 7 – Show the acyl substitution mechanism for each functional group with hydroxide as the nucleophile.
O
C
R
O
Na
O
Cl
R
H
C
H
O
O
H
acyl
substitution
R
H
O
H
R
R
C
acid/base
O
C
O
C
O
Cl
O
O
C
R
Cl
R
acyl
substitution
O
O
O
O
C
O
C
R
R
O
C
O
H
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H
O
C
O
R
Cl
Nucleophilic Substitution & Elimination Chemistry
O
O
Na
C
R
R
O
O
R
H
O
R
R
N
O
C
R
O
O
O
R
H
O
R
O
acid/base
O
R
C
acyl
substitution
R
H
O
4
C
R
O
H
O
H
N
R
H
Na
C
O
R
C
acyl
substitution
Beauchamp
N
C
R
R
O
H
O
O
R
N
C
R
R
O
R
acid/base
H
R
Problem 8 – Write an arrow pushing mechanism for each of the following reactions.
H
O
-17
Keq = 10-37 = 10+20
10
H
Na
O
acid/base
H
H
pKa
(gas)
pKa = 17
H
O
pKa = 19
-19
Keq = 10-37 = 10+18
10
H
K
H
O
H
(gas)
ROH
16-19
H-H
37
acid/base
Problem 9 – Write an arrow pushing mechanism for the following reaction.
a. D
H
Al
D
D
b.
D
+
R
Li
Cl
H
CH3
D
C
H
B
H
H
C
Na
+
S
D
CH3
SN2
S
C
Cl
H
CH3
H
SN2
Cl
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C
S
D
Cl
CH3
The center inverted, but the
priorities changed, so still S.
Nucleophilic Substitution & Elimination Chemistry
Beauchamp
5
Problem 10 – It is hard to tell where the hydride was introduced since there are usually so many other hydrogens in
organic molecules. Where could have X have been in the reactant molecule? There are no obvious clues. Which
position(s) for X would likely be more reactive with the hydride reagent? Could we tell where X was if we used
LiAlD4?
X
SN2
X
Where was "X"?
H
H2
C
Al
H
H
H2
C
C
H2
+
CH3
X
Li
H
X could be on any sp3 carbon atom.
If we used LiAlD4 there would be
a 'D' where 'X' was.
X
X
Problem 11 - How many total hydrogen atoms are on C carbons in the given RX compound? How many different
types of hydrogen atoms are on C carbons (a little tricky)? How many different products are possible? Hint - Be
careful of the simple CH2. The two hydrogen atoms appear equivalent, but E/Z (cis/trans) possibilities are often
present. (See below, problem 7, for relative expected amounts of the E2 products.)
CH3
H3C
X
=
R
H
H
H
C
C1 C C
CH2
H3C
H
C1
H3C
H
C C
CH2
H3C
H
H
H
H
H3C
H
C
C1
H
H
CH2
C
H3C
H
C C
H
E and Z possible here depending
on stereochemistry of C and C1.
CH3
CH3
CH3
H
C
C1 has only 1 H, could be R or S stereochemistry
C2 has two H, they are different, lead to E/Z stereoisomers
C3 has three H and they are all equivalent, no E/Z possible
H
X
CH3
H3C
H
H3C
C1
CH2
H
H3C
H
C C
H
H
H
H3C
E and Z possible here depending
on which proton is lost.
H
C
C1
C
CH2
H
H
C
H
No E and Z is possible here.
stability or alkenes = tetrasub > trisub > trans-disub > cis-disub = gem disub > monosub.
Problem 12 - Reconsider the elimination products expected in problem 11 and identify the ones that you now
expect to be the major and minor products. How would an absolute configuration of Cα as R, compare to Cα as S?
What about C1 as R versus S?
H
H
C
H3C
H
O
H
H
C3
C
Br
H3C
H
C1
R
etc.
CH2CH3
(3S,4R)-3-bromo-3,4-dimethylheptane
H
H
H
C
H3C
C3
H
H
C1
C
H
H3C
C
CH2CH3
E alkene
tetrasub = most stable, #1
H3C
H
C3
H
C
H
H3C
H
H
C
H3C
H
C3
C
H3C
CH2CH3
No Z or E alkene
gem disub = less stable, #4
H
H
H
C1
H
C1
H
C
CH2CH3
E alkene
trisub = middle stable, #2
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H
H3C
H
C3
C
H3C
H
C1
CH2CH3
Z alkene
trisub = middle stable, #3
Nucleophilic Substitution & Elimination Chemistry
Beauchamp
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Problem 13 -Provide an explanation for any unexpected deviations from our general rule for alkene stabilities
above. A more negative potential energy is more stable.
7 C(gr) + 7 H2(g) = zero reference energy
most
stable
Hf
least
stable
Hf
Hf
Steric crowding
by t-butyl group
changes order of
stability.
Expected this
alkene to be
the least stable,
but it's not.
Hof = -18.7 kcal/mole
Hof = -19.97 kcal/mole
Hof = -22.7 kcal/mole
Our rules predict that stability is trans-di > cis-di > monosubstituted alkene. However, the data show that the
cis alkene is less stable than the monosubstituted alkene. This is due to the large size of the t-butyl group crowding
the medium methyl group. This would probably cause twisting of the pi bond and weaken pi overlap.
Problem 14 – Order the stabilities of the alkynes below (1 = most stable). Provide a possible explanation.
H
C
C
H
R
C
C
H
R
C
C
R
"R" = a simple alkyl group
Our rules predict that more substitution at pi bonds should make them more stable by extra inductive donating
effect (compared to H) at more electronegative sp hybrid orbitals, so disubstituted > monosubstituted > unsubstituted.
Problem 15 - Propose an explanation for the following table of data. Write out the expected products and state by
which mechanism they formed. Nu: -- /B: -- = CH3CO2 -- (a weak base, but good nucleophile).
O
percent
substitution
percent
elimination
our rules
H3C
H2
C
Br
100 %
0%
SN2 > E2
H3C
H
C
Br
100 %
0%
SN2 > E2
Br
11 %
89 %
SN2 > E2
(wrong
prediction)
Br
0%
100 %
O
O
O
CH3
H3C
O
CH
O
H3C
H
C
CH3
CH3
O
H3C
O
C
CH3
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only E2
Acetate is a pretty stable anion
due to resonance stabilization on
two oxygen atoms. We predict
that it will favor SN2 reaction at
methyl, primary and secondary
RX electrophiles and E2 at
tertiary RX. Everything looks as
expected here except example 3,
which is secondary, but C is
tertiary so there is extra steric
hindrance at both C and C which
appears to push it to E2 > SN2.
We will still follow our simplistic
rules in this course, but here is an
example that shows our rules are
a bit too simplistic.
Nucleophilic Substitution & Elimination Chemistry
Beauchamp
7
Problem 16 – One of the following reactions produces over 90% SN2 product and one of them produces about 85%
E2 product in contrast to our general rules (ambiguity is organic chemistry’s middle name). Match these results
with the correct reaction and explain why they are different.
Cl
O
pKa = 16
O
H
SN2 > E2
Cl
H
 90% SN2
O
pKa = 19
Cl
Cl
E2 > SN2
H
H
O
 85% E2
Sterically larger t-butoxide forces E2 reactions even at primary RX.
Problem 17 - A stronger base (as measured by a higher pKa of its conjugate acid) tends to produce more relative
amounts of E2 compared to SN2, relative to a second (weaker) base/nucleophile. Greater substitution at C and C
also increases the proportion of E2 product, because the greater steric hindrance slows down the competing SN2
reaction. Use this information to make predictions about which set of conditions in each part would produce
relatively more elimination product. Briefly, explain your reasoning. Write out all expected elimination products.
Are there any examples below where one reaction (SN2 or E2) would completely dominate?
a.
Cl
Cl
I
I
more E2 because of extra steric
hindrance at C position.
...versus...
b.
O
Cl
Cl
...versus...
O
pKa(RCO2H) = 5
more E2 because base is less stable
(higher pKa of conjugate acid, more
basic)
O
pKa(ROH) = 16
c.
Cl
Cl
more E2 because of extra steric
hindrance at C position.
...versus...
d.
Cl
Cl
N
C
N
C
...versus...
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more E2 because of extra steric
hindrance at C position, only
E2 at tertiary RX center.
Nucleophilic Substitution & Elimination Chemistry
Beauchamp
8
e.
Br
R
Br
O
R
more E2 because of extra steric
hindrance at C position, which
is quaternary, so no SN2, only E2.
O
...versus...
f.
more E2 because base is less stable
(higher pKa of conjugate acid, more
basic)
...versus...
Cl
pKa(RCCH) = 25
Cl
N
C
pKa(NCH) = 9
Problem 18 - (2R,3S)-2-bromo-3-deuteriobutane when reacted with potassium ethoxide produces
cis-2-butene having deuterium and trans-2-butene not having deuterium. The diastereomer
(2R,3R)-2-bromo-3-deuteriobutane under the same conditions produces cis-2-butene not having deuterium and
trans-2-butene having deuterium present. Explain these observations by drawing the correct 3D structures, rotating
to the proper conformation for elimination and showing an arrow pushing mechanism leading to the observed
products. (Protium = H and deuterium = D; H and D are isotopes. Their chemistries are similar, but we can tell
them apart.)
D
CH3
Br
H
D
CH3
cis-2-butene
(D present)
O
Br
Na
H
H
H
CH3
cis-2-butene
(no D present)
O
Na
D
CH3
trans-2-butene
(no D present)
(2R,3S)
(2R,3R)
(2R,3R)
(2R,3S)
H
H
CH3
O
One 3D
example.
D
D
H
C
Br
H
C
H3C
H3C
CH3
D
H3C
H3 C
D
Br
(2R,3S)
D
H
Z alkene, with D
H
C
C
D
CH3
E alkene (with D)
Br
O
D
H
H3CC
H
3
H3C
Et
O
H
One 3D
example.
H3 C
Et
O
H
O
CH3
Z alkene (with D)
Et
H3 C
trans-2-butene
(D present)
D
HD
C
H
H3CH
H
CH3
H
CH3
Br
(2R,3S)
H
H
DHCH
H3 C
H
E alkene, without D
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H
Br
(2R,3S)
H3 C
H
H
1-alkene, no E/Z
Nucleophilic Substitution & Elimination Chemistry
Beauchamp
9
Problem 19 - Draw a Newman projection of the two possible conformations leading to E2 reaction. Show how the
orientation of the substituents about the newly formed alkene compares.
B
H
staggered
anti E2
reaction
B
R3
C
R1
R2
R4
R4
C
H
R3
R1
R2
X
reactant conformation 1
rotate about
center bond < 1%
eclipsed
syn E2
reaction
R1
Newman
projections
R3
C
H
B
alkene stereochemistry
is different
X
> 99%
R2
H
R2
R4
R1
R4
C
R1
R3
R2
R4
diasteromers
R3
X
reactant conformation 2
alkene stereochemistry
is different
B
H
X
R2
R3
R1
R4
Problem 20 – What are the possible products of the following reactions? What is the major product(s) and what is the
minor product(s)? There are 55 possible combinations.
1
=
Nu
2
O
H
hydroxide
7
D
N
B
D
D
Na
N
N
azide
H3C
R
alkoxide
D
8
t-butoxide
D
OR
1
Nu
D
Al
OH
8
O
O
9
N3
5
O
carboxylate
CH3
9
O
H
S
OR
2
R
Ph
S
dithiane anion
O
3
H
12
acetylide
O
N
CH2
sulfur
ylids
imidate
5
4
C
N
10
Ph
S
D
11
S
12
O
CH2
N
Ph
S
O
Nu
X
O
O
10
9
8
7
(E2 > SN2)
3 possible
(E2 > SN2)
3 possible
D
S
(E2 > SN2)
3 possible
C
N
11
Ph
S
6
SH
S
N3
5
4
3
2
1
B secondary RX
12
O
CH2
Ph
O
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O
6
C
SH
O
C
C
N
cyanide
S
enolates
6
C
R
alkylthiolate
hydrogenthiolate
11 Ph
10
S
D
1
X
7
4
S
Li
D
aluminumdeuteride
borodeuteride
A primary RX
O
3
O
C
O
OR
H3C
N
O
CH
Nucleophilic Substitution & Elimination Chemistry
C
Beauchamp
primary RX
secondary RX
D
X
10
R
Nu
Nu
S
Nu
X
SN2 > E2 products
(all except t-butoxide)
no R/S
Nu
SN2 > E2 products
(all except hydroxide, alkoxide,
t-butoxide, acetylide)
E2 products
no E/Z
E
E2 products
(enantiomers)
tertiary RX
Nu
X
only E2 products (3 ways, tri > gem di)
Problem 21 - Explain the differences in stability among the following carbocations (hydride affinities = a larger
number means a less stable carbocation = more energy released when combined with hydride).
methyl = 315
H
primary = 270
H
R
secondary = 249
H
H
C
C
H
R
C
H
allyl = 256
tertiary = 232
H
C
R
R
R
C
CH2
H2C
H2C
CH2
R
Primary carbocations are 270 kcal/mole, but
this primary allylic carbocation is only 256,
so it is 14 kcal/mole more stable than expected
because of resonance sharing by adjacent pi bond.
R groups inductively donate electrons to electron deficient centers like
carbocations (and free radicals). The more R groups the more stable.
The carbocation that releases the most energy when combined with
hydride was the least stable starting energy.
benzyl = 239
CH2
H
C
CH2
CH2
CH2
Primary carbocations are 270 kcal/mole, but this primary benzylic carbocation is only 239, so it
is 31 kcal/mole more stable than expectedbecause of resonance sharing by adjacent pi bonds.
= 248
H
O
CH2
CH2
H
O
H
N
= 218
H
CH2
N
= 230
CH2
O
C
CH3
O
C
CH3
H
H
Primary carbocations are 270 kcal/mole, but the primary carbocation next to oxygen is only 248, so it is 22 kcal/mole more
stable than expected even though oxygen's inductive effect works in the opposite direction. When next to nitrogen it is more
stable than expected by 52 kcal/mole because less electronegative nitrogen shares better than more electronegative oxygen.
The last example, lone pair next to oxygen is 40 kcal/mole more stable than expected. For all of them it is because of resonance
sharing by adjacent lone pair of electrons. In each case this helps fill the octet of the carbocation and adds an extra bond.
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Nucleophilic Substitution & Elimination Chemistry
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11
Problem 22 - Why are vinyl carbocations so difficult to form? (Hint – What is their hybridization?) How does an
empty sp2 orbital (phenyl carbocation) or empty sp orbital (ethynyl carbocation) compare to a typical sp2
carbocation carbon with an empty 2p orbital? (An empty 2p orbital is also present on the vinyl carbocation, but the
carbon hybridization is sp.)
H
C
C
H
H
2p orbital, but
on sp carbon
H 287
C
C
386
sp2 orbital
very electronegative
sp orbital
very electronegative
300
sp hybridized carbon is the most electronegative carbon because it is 50% s character and makes it more difficult (higher
energy) to form. The empty orbital is 2p (the least electronegative). The second example is also sp carbon, but the
orbital is sp, which makes it much more difficult to take electrons from. The phenyl example cannot change its shape
because of the ring and that makes the empty orbital sp2, which is also more electronegative than 2p and harder to form.
The difficulty to form can be seen from the energy released when each carbocation reacts with the hydride donor. The
more energy released, the less stable the starting point.
electronegativity
of orbitals
hold on
electrons
least
least
most
most
2p
3
sp
sp2
sp
2s
Problem 23 – The bond energy depends on charge effects in the anions too. Can you explain the differences in
bond energies below? (Hint: Where is the charge more delocalized?) We won’t emphasize these differences.
CH3
H3C
C
CH3
H
X
H3C
X = gas phase bond energies
X
C
CH3
Cl
Br
I
CH3
+157
+149
+140
The difference in energy cost is due to the different anions since the carbocation is the same in every instance.
The larger and more delocalized the electrons the more stable is the anion and easier to form, just like we saw
in our acid/base topic.
Problem 24 – Draw in all of the mechanistic steps in an SN1 reaction of 2R-bromobutane with a. water, b. methanol
and c. ethanoic acid. Add in necessary details (3D stereochemistry, curved arrows, lone pairs, formal charge).
What are the final products?
H
b
Br
a
O
H
H
H
O
H
H
H
H
c
H
H
H
O
H
a
H
H
H
H
H
O
H
H
H
H
O
H
H
a
H
Adds from both faces,
makes R and S chiral
center. (SN1)
H
H
H
(SN1)
R/S
b and c
2E alkene
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from c
2E alkene
1 alkene
from b
Nucleophilic Substitution & Elimination Chemistry
Beauchamp
H
b
Br
a
O
H3C
H3C
H
H
H
H
c
H
H
H
H
O
H
H
2E alkene
a
O
c
H
(SN1)
R/S
H
H
from c
C
H
H
H
H
1 alkene
2E alkene
from b
O
H
C
H3 C
H
b and c
b
O
H
H
a
H
H
O
Adds from both faces,
makes R and S chiral
center. (SN1)
similar steps and
similar products
(ether instead of
an alcohol)
Br
H3C
H
H
H3C
O
H
a
H
O
12
H
O
H
H3 C
H
H
H
H
H
(SN1)
R/S
H
similar steps and
similar products
(ester instead of
an alcohol or ether),
plus some minor
alkene products.
Problem 25 – What are the likely SN1 and E1 products of the initial carbocation and the rearranged carbocations
from “a”, “b” and “c”?
a
H3C
H3C
H:
hydride
migrates
SN1 and E1 are
possible here
CH2 H
C
C
O
CH3
H
3o carbocation
looks very good
b
H3C
CH2 H
C
H
CH3
2o carbocation
looks OK
E1
E and Z possible
E1
no E and Z possible
trisubst. > gem disubst.
H3C:
methyl
migrates
C
R
SN1 (achiral)
SN1 and E1 are
possible here
CH3
H
O
R
E1
E and Z possible
SN1
R and S possible
y:\files\classes\315\315 Handouts\315 Fall 2013\2b 315 SN and E & chem catalog, answers.doc
E1
no E and Z possible
trans disubst. < trisubst.
Nucleophilic Substitution & Elimination Chemistry
c
H3CH2C:
ethyl
H3C migrates
CH2
H3C
C
C
SN1 and E1 are
possible here
H
Beauchamp
13
R
O
CH3
H
H
2 carbocation
looks OK
E1
E and Z possible
SN 1
R and S possible
o
E1
no E and Z possible
trisubst. > monosubst.
Problem 26 – Write out your own mechanism for all reasonable products from the given R-X compound in water
(2-halo-3-methylbutane).
CH3 H
H3C
C
C
H
X
CH3 H
O
H
CH3
H
H3C
C
C
CH3
CH3 H
C
H3C
C
C
H
SN1
E1
CH3 H
C
CH2
H
H
H
c
H
H3C
O
a
C
C
CH2
H
O
H
E1
H3C
C
CH3 H
C
CH2
H3C
C
C
CH3
H
H
H
b
O
SN1
Also, from second carbocation.
H
CH3 H
CH3 H
C
C
CH2
H
O
H
H
E1
CH3 H
H
H 3C
CH3
H
SN 1
H3C
CH3 H
rearrangement
2o to 3o R+
H3C
R and S
C
C
O
H
CH3 H
CH3
H2C
C
C
CH3
H
H
Problem 27 - Consider all possible rearrangements from ionization of the following RX reactants. Which
are reasonable? What are the possible SN1 and E1 products from the reasonable carbocation possibilities?
a.
H3C
CH3
X
C
CH3
c.
b. H
2
C
CH3
C
CH
CH3
initial R+
subsequent R+
SN1 = 2
SN 1 = 1
E1 = 1
E1 = 2
total = 6 products
CH2
X
CH
CH3
subsequent R+
initial R+
SN1 = 6
SN1 = 2
E1 = 8
E1 = 1
total = 17 products
y:\files\classes\315\315 Handouts\315 Fall 2013\2b 315 SN and E & chem catalog, answers.doc
H3C
CH3
X
C
H 2C
CH
CH2
initial R+ subsequent R+
SN1 = 6
SN 1 = 2
E1 = 1
E1 = 8
total = 17 products
Nucleophilic Substitution & Elimination Chemistry
Beauchamp
14
H2
C
a
CH3
C
poor = 2o to 1o
CH2
b. H
2
C
H2
CH3
C c
H
b C
C
b
CH2
C a
H
a a
H
H
CH3
C
X
CH
ionization
CH2
CH3
This rearrangement
is not going to happen.
CH2
C
H
H
CH3
rearrange
CH3
b'
b
good = 2o to 3o
H
H
these two
are equivalent
CH3
CH3
OK, 3o to 3o
add
nucleophile
H2
C
c
lose
beta H
o
CH3
C
o
good = 2 to 3
CH3
H2
C
c'
C
CH2
add
nucleophile
H
H
C
these two
are different
CH3
C
CH2
CH3
OK, 3o to 3o
lose
beta H
add
nucleophile
lose
beta H
Assume water is the solvent. Possible SN1/E1 products are:
b
good = 2o to 3o
b. H
2
C
CH3
C
X
CH
CH2
ionization
CH3
H2
CH3
C c
H
b C
C
b
CH2
C
H
Possible products
from initial R+
E1
add
nucleophile
SN1
R and S
c
good = 2o to 3o
CH3
rearrange
H2
C
CH3
C
H
CH3
around 17 possible products
H2
C
H
H
C
C
CH2
CH2
CH3
H
CH3
C
CH3
H
lose
beta H
OH
add
add
lose
nucleophile beta H
nucleophile
from subsequent R+
*
*
E1
R and S
E1
R and S
lose
beta H
*
*
add
lose
nucleophile beta H
E1
E1
E1
OH
* = chiral center
c'
OK, 3o to 3o
OH
SN1
(RR, SS, RS SR)
OH
y:\files\classes\315\315 Handouts\315 Fall 2013\2b 315 SN and E & chem catalog, answers.doc
SN1
SN1
E1
Nucleophilic Substitution & Elimination Chemistry
Beauchamp
CH3
CH3
d
3o carbocation
e
3o carbocation CH
c.
CH3
15
3
H2
C
CH3
d
e
H
H
CH3
CH3
CH3
add
nucleophile
CH3
CH3
same as c' above
same as b above
add
lose
nucleophile beta H
lose
beta H
H2
C
CH3
C
C
CH2
H
2o carbocation
CH3
C
CH3
*
X
H
3o carbocation
CH2
OH
add
lose
nucleophile beta H
SN1
R and S
Similar to above.
* = chiral center
H
E1
around 17 possible products
d. What would happen to the complexity of the above problems with a small change of an ethyl for a
methyl? Use the key of “b” and “c” as a guide. This problem is a lot more messy than those above,
(which is the point of asking it). There are too many possibilities to consider listing every answer.
CH3
H2C
CH3
H2C
initial R+ subsequent R+
SN1 = 2 SN1 = 12
E1 = 1
E1 = 16
X
CH
CH3
similar
X
CH3
d. Possible products
rearrange
add
lose
nucleophile beta H
add
lose
nucleophile beta H
add
lose
nucleophile beta H
add
lose
nucleophile beta H
add
lose
nucleophile beta H
OH
*
E1
SN1
R and S
* = chiral center
around 31 possible products
E1
R and S
*
E1
R and S
*
E1
R and S
E and Z
E1
E1
E1
E and Z
*
E1
E and Z
E1
R and S
OH
*
*
*
*
*
OH
SN1
(RR, SS, RS SR)
OH
SN1
(RR, SS, RS SR)
y:\files\classes\315\315 Handouts\315 Fall 2013\2b 315 SN and E & chem catalog, answers.doc
*
SN1
R and S
CH3
same as c above
H
*
H
C
SN1
R and S
OH
Nucleophilic Substitution & Elimination Chemistry
Beauchamp
16
Problem 28 – Lanosterol is the first steroid skeletal structure in your body on its way to cholesterol and other
steroids in your body. It is formed in a spectacular enantiospecific cyclization of protonated squalene oxide. The
initially formed 3o carbocation rearranges 4 times before it undergoes an E1 reaction to form lanosterol. Add in the
arrows and formal charge to show the rearrangements and the final E1 reaction.
R
R
squalene
acid
catalysis
H
H
CH3
H
H
CH3
Requires 5
arrows to
show the
reaction.
H
R
CH3
H
rearrangement
1
CH3
H
R
H
CH3
CH3
rearrangement
4
CH3
CH3
CH3
rearrangement
3
H
H
B
R
E1 reaction
CH3
CH3
H
CH3
HO
H
H
H
H
H3C
CH3
CH3
H
CH3
CH3
19 more
steps
CH3
HO
H
H
H
H
H
other
body
steroids
H
HO
cholesterol
lanosterol
Problem 29 - Propose a synthesis of monodeuterated cyclohexane from cyclohexanol.
H
O
HBr
H
SN 1
H
Br
y:\files\classes\315\315 Handouts\315 Fall 2013\2b 315 SN and E & chem catalog, answers.doc
1. Mg
2. D2O
H
CH3
H
H
HO
HO
R
CH3
CH3
CH3
rearrangement
2
H
H
B
CH3
HO
lanosterol precursor
CH3
CH3
H
H
R
H
CH3
HO
protonated
squalene
O
CH3
H
D
Nucleophilic Substitution & Elimination Chemistry
Beauchamp
17
Problem 8 (29b) – Write a detailed arrow-pushing mechanism for each of the following transformations.
a.
H
H
O
S
O
Cl
S
H
O
H
N
N
O
N
H
N
Cl
O
S
Cl
O
toluenesulfonyl chloride
H
Cl
H
N
O
S
N
O
N
H
N-ethyltoluenesulfonamide
b.
H
O
H
O
N
Cl
N
NR3
H
O
Cl
H
O
H
N
Cl
ethanoyl chloride
(acetyl chloride)
Cl
N
H
H
NR3
N-ethylethanamide
(N-ethylacetamide)
Problem 30 – We can now make the following molecules. Propose a synthesis for each. (Tosylates formed from
alcohols and tosyl chloride/pyridine via acyl substitution reaction, convert “OH” from poor leaving group into a very
good leaving group, similar to iodide).
H3C
OTs
OTs
OTs
OTs
2
1
OTs
OTs
3
5
4
OTs
OTs
6
OTs
Br
NaBr
(SN2)
no rearrangement
7
8
9
All of these tosylates can be made from alcohols + tosyl chloride/pyridine.
TsCl / pyridine
O
R
H
O
R
OH
Br
Ts
Tosylate is a very good leaving group (SN/E chemistry is possible).
Tosylates can be used when a usual 'OH' to 'Br' transformation would
lead to rearrangement.
Br
OH
HBr
rearrangement
major product
y:\files\classes\315\315 Handouts\315 Fall 2013\2b 315 SN and E & chem catalog, answers.doc
1. TsCl/py
2. NaBr
no
rearrangement
major product
Nucleophilic Substitution & Elimination Chemistry
Beauchamp
18
Problem 31– Look back at the table of R-Br structures on page 2. Include stereoisomers together. Be able to list
any relevant structures under each criteria below.
1. Isomers that can react fastest in SN2 reactions
This would include all primary RBr, except when a neopentyl center is at a C position.
2. Isomers that give E2 reaction but not SN2 with sodium methoxide
This would include all tertiary RBr and secondary neopentyl, when there is a C-H.
3. Isomers that react fastest in SN1 reactions
This would include all tertiary RBr.
4. Isomers that can react by all four mechanisms, SN2, E2, SN1 and E1 (What are the necessary
conditions?)
This would include secondary RBr. SN1/E1 conditions would use weak nucleophile/bases (neutral) and
SN2/E2 conditions would use strong nucleophile/bases (usually anions). When a neopentyl center is at
a C position the SN2 reaction will not work.
5. Isomers that might rearrange to more stable carbocation in reactions with methanol.
This would include secondary RBr structures, especially if a tertiary carbocation can form.
6. Isomers that are completely unreactive with methoxide/methanol
This would include primary neopentyl centers (unreactive by all four mechanisms).
7. Isomers that are completely unreactive with methanol, alone.
This would include all primary RBr compounds.
y:\files\classes\315\315 Handouts\315 Fall 2013\2b 315 SN and E & chem catalog, answers.doc
Nucleophilic Substitution & Elimination Chemistry
Beauchamp
19
Problem 32 – Predict possible products of a. water and b. ethanoate (acetate) with structures 2, 10 and 13. Only
consider rearrangements to more stable carbocations where appropriate.
Cyclohexane Examples
2
1
X
4
X
X
6
X
8
10
X
X
7
5
3
X = Cl
Br
I
OTs
9
11
X
X
X
X
1. Which conformation is reactive?
2. Is SN2 possible? Requires an open approach
at C and C .
3. Is E2 possible? Requires anti C -H.
4. How many possible products are there?
5. What is the relationship among the starting
structures?
6. What is the relationship among the products?
7. Are any of the starting structures chiral?
8. Are any of the product structures chiral?
C
H3C
chair 1
Cyclohexane structures
have two chair conformations
possible.
H
H3C
X
chair 2
H
H
H3C
t-butyl substituent locks
in chair conformation
with equatorial t-butyl
Keq
H3C
1
9,999
CH3
CH3
H
severe steric
repulsion
14
13
12
C
15
X
X
X
Mechanism predictions with:
H
O
weak
nucleophiles
O
Some examples
Na
R
O
Na
O K
R
X
Na
H
O
H
O
R
O
O
H
strong nucleophile/bases + other anions shown above
R
y:\files\classes\315\315 Handouts\315 Fall 2013\2b 315 SN and E & chem catalog, answers.doc
OH
Nucleophilic Substitution & Elimination Chemistry
Beauchamp
20
Only axial 'X' conformations are shown, equatorial 'X' is too slow reacting in SN2/E2 reactions. Axial or equatorial RX can react via SN1/E1.
1 = 3o RX
2 = 2o RX
H
4 = 2o RX
3 = 2o RX
CH3
H
H
H
H
H
X
Q1. SN1/E1 both conformations
Q2. No SN2 because anti CH3
Q3. E2 is OK at at rear C-H
Q4. SN1 = 2, E1 = 2, rearrangement
likely to 3o, SN2 = 0, E2 = 1
Q5. enantiomer with 3,
diastereomer with 4 and 5
Q6. diastereomer products are possible
Q7. chiral
Q8. chiral
6 = 2o RX
H
H
H
H
Q1. SN1/E1 both conformations
Q2. No SN2 because 3o RX
Q3. E2 is OK at both C-H
Q4. SN1 = 1, E1 = 1, SN2 = 0, E2 = 1
Q5. NA
Q6. NA
Q7. achiral
Q8. achiral
5 = 2o RX
H
H
CH3
X
H
H
H
H
CH3
X
H
H
H
X
X
Q1. SN1/E1 both conformations
Q2. SN2 is possible
Q3. E2 is OK at at both C-H
Q4. SN1 = 2, E1 = 2, rearrangement
likely to 3o, SN2 = 1, E2 = 2
Q5. enantiomer with 5,
diastereomer with 2 and 3
Q6. diastereomer products are possible
Q7. chiral
Q8. chiral
7 = 2o RX
8 = 2o RX
H
H3C
H3C
CH3
H
H
Q1. SN1/E1 both conformations
Q2. No SN2 because anti CH3
Q3. E2 is OK at at front C-H
Q4. SN1 = 2, E1 = 2, rearrangement
likely to 3o, SN2 = 0, E2 = 1
Q5. enantiomer with 2,
diastereomer with 4 and 5
Q6. diastereomer products are possible
Q7. chiral
Q8. chiral
H
H
H
H
CH3
H
H
H
X
H
H
H
H
H
H
CH3
H
X
H
H
X
Q1. SN1/E1 both conformations
Q2. SN2 is possible
Q3. E2 is OK at at both C-H
Q4. SN1 = 2, E1 = 2, rearrangement
likely to 3o, SN2 = 1, E2 = 2
Q5. enantiomer with 4,
diastereomer with 2 and 3
Q6. diastereomer products are possible
Q7. chiral
Q8. chiral
Q1. SN1/E1 both conformations
Q2. SN2 is possible
Q3. E2 is OK at at both C-H
Q4. SN1 = 2, E1 = 2, rearrangement
possible, SN2 = 1, E2 = 2
Q5. enantiomer with 7,
diastereomer with 8 and 9
Q6. diastereomer products are possible
Q7. chiral
Q8. chiral
Q1. SN1/E1 both conformations
Q2. SN2 is possible
Q3. E2 is OK at at both C-H
Q4. SN1 = 2, E1 = 2, rearrangement
possible, SN2 = 1, E2 = 2
Q5. enantiomer with 6,
diastereomer with 8 and 9
Q6. diastereomer products are possible
Q7. chiral
Q8. chiral
Q1. SN1/E1 both conformations
Q2. SN2 is possible
Q3. E2 is OK at at both C-H
Q4. SN1 = 2, E1 = 2, rearrangement
possible, SN2 = 1, E2 = 2
Q5. enantiomer with 9,
diastereomer with 6 and 7
Q6. diastereomer products are possible
Q7. chiral
Q8. chiral
9 = 2o RX
10 = 2o RX
11 = 2o RX
12 = 2o RX
H
CH3
H
H
H
CH3
H
X
H
H
H
H
H
H
H3C
X
H
H
H
H 3C
X
H
H
C
CH3
Q1. SN1/E1 both conformations
Q1. SN1/E1 both conformations
Q1. SN1/E1 both conformations
Q2. SN2 is possible
Q2. SN2 is possible
Q2. SN2 is possible
Q3. E2 is OK at at both C-H
Q3. E2 is OK at at both C-H
Q3. E2 is OK at at both C-H
Q4. SN1 = 2, E1 = 2, rearrangement
Q4. SN1 = 2, E1 = 2, rearrangement
Q4. SN1 = 2, E1 = 2, rearrangement
possible, SN2 = 1, E2 = 2
possible, SN2 = 1, E2 = 2
possible, SN2 = 1, E2 = 2
Q5. achiral, diastereomer with 10
Q5. enantiomer with 8,
Q5. achiral, diastereomer with 11
diastereomer with 6 and 7
Q6. diastereomer products are possible Q6. diastereomer products are possible
Q7. achiral
Q6. diastereomer products are possible Q7. achiral
Q8. achiral
Q7. chiral
Q8. achiral
Q8. chiral
y:\files\classes\315\315 Handouts\315 Fall 2013\2b 315 SN and E & chem catalog, answers.doc
H
H3C
H
H
H
H
H
H
X
H
Q1. SN1/E1 above conformation only
Q2. SN2 is possible
Q3. E2 is OK at at both C-H
Q4. SN1 = 2, E1 = 2, rearrangement
possible, SN2 = 1, E2 = 2
Q5. achiral, diastereomer with 13
Q6. diastereomer products are possible
Q7. achiral
Q8. achiral
Nucleophilic Substitution & Elimination Chemistry
13 = 2o RX
14 = 2o RX
H
H3C
H3C
Beauchamp
H
C
H
CH3
H
15 = 2o RX
H
H3C
H
H3C
H
H
H
CH3
H
X
Q1. SN1/E1 above conformation only
Q2. SN2 is possible
Q3. E2 is OK at at both C-H
Q4. SN1 = 2, E1 = 2, rearrangement
possible, SN2 = 1, E2 = 2
Q5. achiral, diastereomer with 12
Q6. diastereomer products are possible
Q7. achiral
Q8. achiral
H
H3C
X
C
21
H
H3C
H
C
H
H
CH3
H
X
H
Q1. SN1/E1 above conformation only
Q2. SN2 is possible
Q3. E2 is OK at at both C-H
Q4. SN1 = 2, E1 = 2, rearrangement
possible, SN2 = 1, E2 = 2
Q5. achiral, diastereomer with 15
Q6. diastereomer products are possible
Q7. achiral
Q8. achiral
H
Q1. SN1/E1 above conformation only
Q2. SN2 is possible
Q3. E2 is OK at at both C-H
Q4. SN1 = 2, E1 = 2, rearrangement
possible, SN2 = 1, E2 = 2
Q5. achiral, diastereomer with 14
Q6. diastereomer products are possible
Q7. achiral
Q8. achiral
Problem 33 – What are the oxidation states of each carbon atom below? All of the atoms in these examples have a
formal charge of zero.
H
H
H
C
H
H
C
H
OH
C
oxidation state
oxidation state
-4
-2
of carbon = _____?
of carbon = _____?
H3C
H
C
H
C
H3 C
H
oxidation state
-3
of carbon = _____?
OH
OH
C
C
O
oxidation state
+1
of carbon = _____?
C
O
oxidation state
+4
of carbon = _____?
oxidation state
+2
of carbon = _____?
H
H3C
oxidation state
-1
of carbon = _____?
O
O
HO
oxidation state
0
of carbon = _____?
H
C
O
H
H
H
H
H
 H2O
HO
C
H3C
O
O
HO
oxidation state
oxidation state
+3
+4
of carbon = _____?
of carbon = _____?
Problem 34 – What are the oxidation states below on the carbon atom and the chromium atom as the reaction
proceeds? Which step does the oxidation/reduction occur? (PCC, B: = pyridine and Jones, B: = water)
H
R
C
Cr
O
H
R
partial negative of oxygen
bonds to partial positive of
chromium
O
O
oxidation state of C = 0
R
C
C
H
R
O
Cr
O
R.D.S. = slow step
oxidation state of C = +2
E2 reaction
(step 3)
oxidation state of Cr = +4
Carbon oxidation
occurs in this step.
y:\files\classes\315\315 Handouts\315 Fall 2013\2b 315 SN and E & chem catalog, answers.doc
oxidation state of C = 0
B
oxidation state of Cr = +6
acid/base
O
O
H
(step 2)
O
R
B
C
R
B
H
O
H
R
O
Cr
O
R
(step 1)
oxidation state of Cr = +6
B
O
H
O
O
O
Cr
O
oxidation state of C = 0
oxidation state of Cr = +6
Nucleophilic Substitution & Elimination Chemistry
Beauchamp
22
Problem 35 – Supply all of the mechanistic details in the sequences below showing 1. the oxidation of a primary
alcohol, 2. hydration of the carbonyl group and 3. oxidation of the carbonyl hydrate (Jones conditions).
Under aqueous conditions, a hydrate of a carbonyl group has two OH groups which allow a second oxidation, if another C-H bond is
present. This is only possible if the starting carbonyl group was an aldehyde (true when starting with methyl and primary alcohols).
O
H
H3C
Cr
C
C
H2
H
O
H
H
O
nucleophile
addition at Cr
H3C
H
H3C
resonance
C
H2
H
nucleophile
addition at C
H3C
O
H
O
H3C
C
C
H2
H
H
O
H
H
acid/base
H
H3C
H
C
C
H2
H
H3C
Cr
O
H
H3C
H
H
O
O
H
C
C
H2
O
y:\files\classes\315\315 Handouts\315 Fall 2013\2b 315 SN and E & chem catalog, answers.doc
O
H
H
acid/base
O
O
H
O
H
H
carboxylic acid from
an aldehyde or a
primary alcohol
O
O
Cr
O
O
C
C
H2
H3C
O
O
O
O
C
C
H
H2
aldehydes
nucleophile
addition at Cr
Cr
O
second oxidation of the carbonyl hydrate
O
3
H
H
H
Cr
O
H
O
O
O
H
H
Cr
O
1. oxidation of 'OH' to C=O
2. hydration of C=O group
3. oxidation of 'OH' to C=O
H
O
O
H
O
acid/base
H
O
H
H
O
H
H
H
C
C
H2
H
O
H
H
O
inorganic
ester
E2 reaction
(oxidation here)
2
C
C
H2
hydration of the aldehyde
O
O
O
C
H3C
H
H
O
Cr
H
acid/base
H
Jones = CrO3 / H2O / acid
primary alcohols oxidize to carboxylic acids
(water hydrates the carbonyl group,
which oxidizes a second time )
primary alcohols
H
O
C
C
H2
O
O
Cr
H
O
chromic
anhydride
H
O
O
O
O
H
O
1
H3C
E2 reaction
(oxidation here)
C
C
H2
O
H
H
O
H
H
inorganic
ester
Nucleophilic Substitution & Elimination Chemistry
Beauchamp
23
Problem 36 – We can now make the following molecules. Propose a synthesis for each from our starting materials.
aldehydes
O
O
O
C
H
H3C
C
H
H3C
H3 C
C
C
H2
H
C
C
H3C
H3C
C
O
O
C
C
CH
H3 C
H3C
CH
Cl
N
H
CH3
esters
CH3
N
H
H3C
C
C
H2
C
O
CH3
C
H3C
CH
O
CH3
H3C
N
H
O
CH3
C
C
H2
CH3
O
C
C
H2
C
CH3
N
H
O
O
O
C
CH3
N
H
O
CH3
H3C
Cl
O
O
C
H3 C
C
C
H2
CH3
H3C
C
H
Cl
O
O
O
C
Cl
amides
OH
O
C
C
C
H2
Cl
C
CH3
O
C
H
C
H2
O
H3C
OH
OH
OH
O
O
C
C
H2
acid chlorides
H
H
H
O
H3 C
C
H2
OH
(not stable)
O
O
C
OH
C
CH3
O
H
O
CH
H
carboxylic acids
O
O
O
C
CH
C
CH3
O
O
O
CH3
C
C
H2
CH3
conjugated aldehydes, carboxylic acids and esters
O
H2C
H2C
C
C
H
O
O
H
H2C
C
C
H
O
C
C
H
OH
O
CH3
H2C
O
C
C
H2 C
H
O
C
C
CH3
OH
H2 C
C
C
CH3
CH3
O
CH3
O
ketones
O
O
O
C
H3C
OH
CH3
C
O
1. LDA
2. workup
PCC
CH3
anhydrides
O
O
O
O
O
O
C
C
C
C
C
C
H3C
O
H
H3C
O
CH3
H3C
O
nitriles
C
H2
CH3
O
O
C
C
H3C
O
C
N
C
N
C
We haven't learned how
to make this epoxide yet.
We will when we cover
alkenes.
O
CH3
CH
O
C
H3C
C
O
CH3
N
N
CH3
N
H
C
N
C
H3C
y:\files\classes\315\315 Handouts\315 Fall 2013\2b 315 SN and E & chem catalog, answers.doc
C
N
N
C
O
CH3
Nucleophilic Substitution & Elimination Chemistry
H
H3C
alkane
Br2
h
NaOH
SN2
Br
H3C
bromoalkane
N
S
H
1. NaOH
2. CH3Br
C
carboxylic OH
acid
O
epoxide
H
C
N
o
1 amide
o
1 amine
H
C
O
CH3
We can't make the 1C acid chloride,
but we can make the 1C anhydride,
which does similar reactions
O
O
NH2
H3C
O
ester
C
H3C
Cl
acid chloride
O
Ph
NH2
H3C
O
CrO3
H2O
O
1. LiAlH4
2. workup
H3C
nitrile
H2C
aldehyde
24
Ph n-BuLi
H3C
C
O
Ph-S-Ph
N3
H3C
CrO3
py.
OH
H3C
alcohol
NaN3
SN2
NaCN
SN2
Beauchamp
C
H3C
CH3
C
O
H
more reactive C=O
anhydride
H
excess
1. NaNR2
2. workup
(zipper)
1. NaNR2
2. CH3Br
1. NaNR2
2.
Br
H2SO4/
2 Br2
h
t-butoxide (E2)
Br2
h
H
Br
NaOH
SN2
NaN3
SN2
NaCN
SN2
1. LiAlH4
2. workup
C
O
OH
H
O
CH3
O
n-BuLi
Ph
S
O
CrO3
H2O
O
Ph-S-Ph
N3
N
CrO3
py.
OH
SOCl2
Cl
Ph
O
O
NH2
N
CH3
H3C
NH2
O
O
O
H
Many of the syntheses of other target molecules are similar. A few that go farther are shown. Not all are shown.
Br2
h
Br
HBr
ROOR
h
t-butoxide
(E2)
O
NaOH
Br SN2
O
CrO3
OH
py.
H
Br2
h
O
Br t-butoxide
(E2)
OH
1. LDA
2. workup
Br
O
CrO3
H2O
NaOH
SN2
OH
1. LDA
2. workup
OH
CrO3
py.
CrO3
H2O
y:\files\classes\315\315 Handouts\315 Fall 2013\2b 315 SN and E & chem catalog, answers.doc
H
1. NaOH
2. CH3Br
O
H
O
CrO3
H2O
OH
O
OH
O
CrO3
py.
O
H
O
OH
O
HBr
ROOR
h
CrO3
py.
CrO3
H2O
O
O
OH
1. NaOH
2. CH3Br
etc.
O