chapter 7 study guide.pdf
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Chapter 7
Definitions
Olefin
Alkene
Alkyne
Unsaturated fatty acids
vic-dibromides
gem-dibromides
Biochemistry
Increasing the unsaturation of a fatty acid decreases mp(makes it turn from solid to liquid)
Increasing the unsaturation decreases the order which decreases mp
Cis double bonds have less order than trans double bonds
Cis double bonds decrease mp
The fluidity of cellular membranes is determined by unsaturation
Reindeers/artic fish have adapted to have a higher degree of unsaturation; this leads to more fluidity
in membrane to counterbalance extreme temperatures
Fatty acids are a long chain of carbons with a carboxylic acid on the end.
Humans and other higher animals have fatty acids that contain 16 or 18 carbons(always even). If
the chain of carbons contains no double bonds, they are saturated. One double bond means
monounsaturated. More than one is polyunsaturated. 3 fatty acids will combine with glycerol to
make a fat/oil molecule. Saturated fats contain fatty acids with no double bonds. Monounsaturated
fats contain fatty acids with one double bond.
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Alkenes/alkynes
Ethylene/Acetylene/Olefin/Olefiant gas
Physical properties are similar to alkanes
Soluble in nonpolar liquids; slightly soluble in water
Density lower than water
E/Z nomenclature
Cahn/Ingold/Prelog systemR/S and E/Z. You check each half of the double bond. Identify the
largest(by atomic #) group on each half of double bond. If both of the largest groups are on the
same side it is Z. If the largest groups are on opposites sides then it is E. In first ex, Cl is bigger
than chain of 4 carbons and on the other half chain of three carbons is bigger than H. Since Cl and
three carbon chain are on same side it is Z.
E = entgegen(opposite sides)
Z = zusamment(same side)
Cl
H
OH
CH3Cl
Br
Z
Z
Z
E
Cl
E
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Stability of alkenes
H of alkene = -120 kj/mole on average
H of cis butene = -120 kj/mole
H of trans butene = -115 kj/mole
Cis are less stable than trans
This is due to steric interactions in cis vs trans
C C
H
HH
H
H
H)(C
C
H
HH
H
H
H
steric hindrance
free of sterichindrance
More substituted double bonds are more stable than less substituted
tetrasubstituted--most stable
trisubstituted--2nd most stable
disubstituted(same carbon)--3rd most stable
disubstituted(trans)--4th most stable
disubstituted(cis)--5th most stable
monosubstituted--6th most stable
unsubstituted--least stable
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Cyclic structures
In cyclopropene, cyclobutene and cyclopentene only cis double bonds are possible
In cyclohexene although the trans is hypothetically possible it is so unfavored that cyclohexenes
almost always adopt the cis configuration.
In cycloheptene the trans form has been observed spectroscopically for a very brief time. However
cycloheptene is almost exclusively cis.
In cyclooctene, cis predominates but the trans is isolatable as a pair of enantiomers.
all 3,4,5,6,7 carbon ringsform predominantly the cisform. Although for 6 and 7it is possible to observe transform fleetingly.
H
H
cis predominatesin cyclooctene but thetrans forms is alsoseen.
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Chemical Reactions
Synthesis of alkenes via elimination reaction
Dehydrohalogenation
H
Br
NaOMe
Dehydrohalogenation of alkyl halides is best carried out by E2 mechanism. In E1
there are too many side reactions possible. One is rearrangement due to migration
of carbocation.
To force reaction to go E2 you need to start with secondary or tertiary alkyl halide. If
you start with primary alkyl halide, use a bulky base(potassium tert-butoxide). High
temperature favors the elimination over the substitution.
Also to increase the chances of E2 mechanism use a high concentration of strong
base(alkoxide base). Also the use of a base pair is often used(ethoxide in ethanol,
methoxide in methanol, tert-butoxide in tert-butanol).
Zaitsevs Rule
When a mixture of double bonds is possible from an elimination reaction, the more
substituted double bond will form. Energetically, this is explained by a lower energy
transition state for the more substituted double bond than for a less substituted double
bond.
Exception to the ruleUsing potassium tert-butoxide often leads to the less
substituted double bond due to the size of the base and steric hindrances. This is
known as the Hoffmann Rule.
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Stereochemistry of double bond
In the transition state of the elimination reaction the leaving group, the carbons of the double
bond and the hydrogen being removed are all in the same plane. This is the anti periplanar
conformation talked about in last chapter.
If anti periplanar is not possible syn periplanar eliminations will occur.
Br H
H3C H
rotate to getanticoplanar,
antiperiplanar
H
H3C
CH2
H3C Br H
H
H3C
CH2
H3C Br
CH2
H
CH3OMe
NaOMe/MeOH
CH3
Ethyl
H
Ethyl
wedged groupsstay on same side.
In cyclic structures, axial Hydrogens are needed in elimination reactions. This often leads to
Hoffman products forming instead of Zaitsev products.
In cyclics, eliminations must be done with H and leaving group axial If the most stable form
of a molecule has the leaving group equatorial, it must ring flip to the axial form to undergo
elimination reactions. This leads to larger reaction rates.
If there is only one axial Hydrogen available for elimination the elimination will form that
product regardless if it is the Zaitsev or the Hofmann products. If there are two axial
hydrogens then the Zaitsev Rule will determine which product forms.
See example page 296-297.
In summary, the leaving group must be axial even if that is the unfavored ring flip. If
the axial position is in the unfavored ring flip the reaction will go much slower. With leaving
group in axial position, if there are axial hydrogens on either side, the reaction will follow
Zaitsevs rule. If there is only one axial hydrogen, the reaction will proceed with that
hydrogen even if it leads to a Hofmann product.
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Potassium tert-butoxide
NaOMe/MeOH
Br
H
Ethyl
HH3C
Propyl
H3C Br
CH3
EthylH
Propyl
H3C Br
H
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O Tf
CH 3H 3C O
N aO M e/M e OH
C H3H 3C O
N OT T HIS
CH 3H 3C OO Tf
C H 3
H
H
H 3C O
O nly ax ial hydrogen on the molecule
Dehydration of alcohol
H C l /
O H
H
C l
Experimental conditions
High temperature and Bronsted acids(phosphoric and sulfuric acid) favor dehydration ofalcohols.
Primary alcohols proceed through an E2 mechanism.
With primary alcohols(very difficult) concentrated sulfuric acid and high
temperatures(180 degrees) is needed.
Secondary and tertiary alcohols proceed through an E1 mechanism.
With secondary alcohols mild conditions such as 85% phosphoric acid with moderateheat(85 degrees) is needed.
With tertiary alcohols very mild conditions will easily dehydrate the alcohol.
20% sulfuric acid and 85 degrees is a common set of conditions.
Ease of dehydrationtertiary > secondary > primary
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OH
OH
OH
H2SO4/
180o-??
85% H3PO4/
80-180o
20% H2SO4
20-80o
Mechanism for primary alcohol dehydration
Mechanism for secondary and tertiary alcohol dehydration.
H 2 S O 4 /
1 8 0 o - ? ?
O S
O
O
O HH
O H
O
H
H
H
O S
O
O
O H
OH
20% H2SO 4
20-80o
O S
O
O
OHH
OS
O
O
H O
O
H
H
H
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Rearrangement of carbocations
Dehydration reactions of secondary and tertiary alcohols proceed through an E1
mechanism. Since E1 goes through a carbocation rearrangements are possible.
Tertiary carbocations are more stable than secondary and secondary are more stablethan primary. Therefore if a secondary carbocation forms during dehydration and a
rearrangement can lead to a tertiary carbocation then that rearrangement will occur.
Pages 303-306.
Rearrangement of carbocations can proceed through methyl migration(1,2 shift) or
hydride migration(1,2 shift).
Rings can expand due to carbocation formations(305).
Secondary to tertiary
OH
H 2SO 4/heat
Explain these three products.
Which is the major product?
OH
H2SO 4/heat
Explain these three products.
Which is the major product?
1
2
3
HB
straightforward elimination leads to #3
OR
HB
Methyl
shift leads to#1
H
ORMethyl shiftfollowed by
hydride shiftleads to #2
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From primary:
If there is a primary alcohol, it will proceed through the E2 mechanism. There is a
possibility for the double bond to reprotonate and form a carbocation. At this point the
original double bond may reform. A new double bond using a different-hydrogen
may form. A migration of methyl or hydrogen to form a new carbocation(then doublebonds could form). Finally a substitution may occur.
OH
HCl
Cl
Explain these products.
Which is the major?
Which is the first formed?
1
2
3
4
1 forms from straight E2 elimination.
A small fraction of 1 may reprotonateand then encounter shifts...
H+(from HCl or water)
Now elimination leads to #2
H B-
H
H B-
Now leadsto elimination
#3.
Substituion
leads to #4
Cl-
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Debromination of vi c-dibromides(not in text)
Br
BrBr Zn, Acetic Acid(or ethanol) Br
Br
BrBrNaI/acetone Br
Synthesis of alkynes via elimination reaction(double E2 elimination)
See page 307 for mechanism.
Vic-dibromides can eliminate twice to form an alkyne. Either the vic-dibromide will already
be present or it can be generated from any alkene.
Br
Br
1 eq. Br2
2 NaNH2
Br Br
NaNH2/NH3
The first product forms not the second. Why??
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Strong bases are needed to do the double eliminations(sodium amide is a good base).
How would you turn pentan-1-ol into 1-pentyne?
Br
Br
1 eq. Br2
2 NaNH2
OH
H2SO4(conc.)
Because of the acidity of terminal alkynes, three equilvalents of sodium amide is used to
deprotonate the alkyne as it forms and drive the reaction to completion.
The solvent is usually liquid ammonia or mineral oil.
Ketones can be converted to gem-dichlorides which can be eliminated to alkynes.
O
PCl5/Oo
Cl Cl
1) 3 eq. NaNH2/NH3
2) H+
Why is the H+added as a second step?
What does the product look like before the
H+is added?
pKa revisited
Terminal alkynes are extremely acidic when compared to alkenes and alkanes.
Looking at basicities on top of p. 309, will hydroxide deprotonate a terminal alkyne??
What base will deprotonate a terminal alkyne?
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SOLUTION VS. GAS PHASE
In solution alkynides are more basic than hydroxides. But in gases, hydroxides are more
basic than alkynides. This is due to solvation effects. Solvent solvate smaller ions better
thus making smaller anions less basic. In gas, larger ions are more polarized than smaller.
The polarization of larger ions make them more stable. Bottom line. Smaller ions are more
stabilized in solvents. Larger ions are more stabilized in the gas phase.
Substitution of terminal alkynes
If you recall, vinylic and phenylic halides do not undergo substitution due to the presence of
the double bond. We can take advantage of the acidity of the terminal alkyne to in essence
carry out a substitution reaction. Once the alkyne is deprotonated, it acts as a nucleophile in
a SN2 type reaction.
HNaNH2/NH3 Propylbromide
The reaction takes place with primary halides only. In secondary and tertiary halides, the
terminal alkyne acts as a base to form an E2 elimination product.
HNaNH2/NH3
2-propylbromide
H
+
Hydrogenation of Alkenes/Alkynes
The hydrogenation of alkenes/alkynes are examples of addition reactions. The double bond
is replaced by the two hydrogens as they add to the double bond.
Biochemically, liquid vegetable oils which contain multiple double bonds usually are partially
hydrogenated to semi-solid fats/margarines
Fats/oils are glycerol molecules with three fatty acids on them. The fatty acids can be
saturated, unsaturated or polyunsaturated. The more the unsaturation the more the subtance
will act as an oil(liquid) instead of a fat(solid). Hydrogenating fats often leads to
isomerization of some double bonds from the naturally occurring and safer cis form to the
nonnatural more dangerous trans form.
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Hydrogenation is accompanied by a catalyst to make it occur at a favorable rate. Although
this is an exothermic reaction it has a very high activation energy and requires a catalyst to
make it go under normal conditions. Page 312.
Hydrogen gas absorbed onto rhodium, ruthenium, platinum, palladium or nickel is a very
common reagent set. Wilkinsons catalyst(Rh[(C6H5)3P]Cl) is increasingly being used.With the presence of the catalyst the reaction can take place at room temperature and room
pressure. Harder to hydrogenate double bonds often take place at very high pressure in
special chambers called hydrogenation chambers. The stereochemistry of these reaction is
a syn addition of hydrogen.
Syn addition means both groups come from the same side; anti addition means the groups
come from opposite sides. See 313-314.
H2/Pd
With alkynes, platinum is usually used as the catalyst. The reaction usually proceeds from
alkyne to alkane(not stopping at the alkene level). This reaction also leads to
syn addition.
H2/Pt
If you want to stop at the alkene level, the use of a poisoned catalyst such as P-2 or
Lindlars Catalyst will yield to a Z-alkene.
H2/P-2
P-2 = Ni2B; Ni2B is generated from Nickel acetate reduction by NaBH4 see p. 315
Lindlar's catalyst
Lindlar's Catalyst = H2, Pd, CaCO3in quinoline, see p. 316.
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If you want to stop at the alkene level, the use of dissolving metal reactions will yield an E-alkene.
1) Li or Na/ethylamine
-78o
2) NH4Cl
Mechanism for E-alkene.
1) Li, C2H5NH2
2) NH4Cl
Li
C2H5N
H
H
H
H
H
Li H
C2H5N
H
H
Index of hydrogen deficiency(IHD)
Alkanes = CnH2n + 2 Alkenes = CnH2n Alkynes = CnH2n 2
Oxygen = 0 H Nitrogen = -1 H Halogens = +1 H
Each unit of unsaturation equals a double bond or a ring.
i.e. 1 unit of unsaturation could equal a double bond or a ring.
2 units of unsaturation could equal 2 double bonds, 2 rings, 1 ring 1 double
bond or a triple bond
STUDY PAGES 320-322
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Chapter 7--Chemical Reactions with Mechanisms
Synthesis of alkenes via elimination reaction
Dehydrohalogenation
H
Br
NaOMe
Mechanism for Dehydrohalogenation
H
Br
NaOMe
OMe
Dehydration of alcohol
Mechanism for primary alcohol dehydration
H 2 S O 4 /
1 8 0o
- ? ?
O S
O
O
O HH
O H
O
H
H
H
O S
O
O
O H
OH
OH
OH
H2SO4/
180o-??
85% H3PO4/
80-180o
20% H2SO 4
20-80o
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Mechanism for secondary and tertiary alcohol dehydration.
Synthesis of alkynes via elimination reaction(double E2 elimination)
Br Br
Br
Br
Br Br
NaNH2
NaNH2
NaNH2
Mechanism
OH
20% H2SO 4
20-80o
O S
O
O
OHH
OS
O
O
H O
O
H
H
H
Br
Br
N aN H2H
H
NH2-
B r
H
NH 2-
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Substitution of terminal alkynes
1 NaNH2
2) Pentyl Bromide
Mechanism
H NH2-
Cl
Hydrogenation of alkyne to E-alkene
Mechanism for E-alkene.
1) Li, C2H5NH2
2) NH4Cl
Li
C2H5N
H
H
H
H
H
Li H
C2H5N
H
H
1) Li or Na/ethylamine
-78o
2) NH4Cl
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Chapter 7--Chemical Reactions without Mechanisms
Formation of gem-dihalide
O
PCl5/0o
Hydrogenation reactions.
Debromination of vi c-dibromides(not in text)
H2/Pd
H2/Pt
H2/P-2
P-2 = Ni2B; Ni2B is generated from Nickel acetate reduction by NaBH4 see p. 315
Lindlar's catalyst
Lindlar's Catalyst = H2, Pd, CaCO3in quinoline, see p. 316.
Br
BrBrZn, Acetic Acid(or ethanol) Br
Br
BrBrNaI/acetone Br
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Homework Chapter 7 Name:
Br
Br
OH
O
1) NaNH2
2) octyl chloride
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H2/Pt
H2/P-2
Lindlar's Catalyst
1) Li, ethylamine
2) NH4Cl
H2/Pt
H2/P-2
Lindlar's Catalyst
1) Li, ethylamine
2) NH4Cl
H2/Ni
H2/Pd
H2/Wilkinson's catalyst
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Give the mechanism for the following:
H
Br
NaOMe
OH
H2SO4/!
180o-??
20% H2SO4
20-80o
OH
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Show the reagents for the following transformations
H
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Homework Chapter 7 Key Name:
Br
Br
Zn, AcOH
or
NaI/acetone
3 eq. NaNH2
OH
1) 85% H3PO4, heat
2) Br23) 3 eq. NaNH2
O
1) PCl5/0o
2) 3 eq. NaNH2
1) NaNH22) Pentyl chloride
1) NaNH2
2) octyl chloride
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H2/Pt
H2/P-2
Lindlar's Catalyst
1) Li, ethylamine
2) NH4Cl
H2/Pt
H2/P-2
Lindlar's Catalyst
1) Li, ethylamine
2) NH4Cl
H2/Ni
H2/Pd
H2/Wilkinson's catalyst
No Reaction
No Reaction
No Reaction
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Give the mechanism for the following:
H
Br
NaOMe
OH
H2SO4/!
180
o
-??
20% H2SO4
20-80o
OH
OH
180o-??
OH2
H
B
H2SO4
H
Br
NaOMe
OMe
20% H2SO4
20-80o
OH
OH2
H
B
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Show the reagents for the following transformations
H
H
H
1) NaNH22) Pentyl chloride
3) H2/Pt
1) NaNH22) Ethyl chloride
3) Li, ethyl amine
4) NH4Cl
1) NaNH22) Butyl chloride
3) H2/P-2
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Homework #3 (Chapter 6 and 7)
Name:
Part I: Fill in the products, show stereochemistry (if racemic mixture,
you need to show only one isomer).
NaCl, Acetone
NaSH, H2O
OTf
NaOMe/MeOH
OK
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OH
Br
1) NaH, TfCl
2) 1 eq. of NaOMe/MeOH
2 eq. NaCN/Acetone
H2/Pd
H2/P-2
1) NaNH22) Butyl chloride
3) H2/Pt
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Part II: Fill in the reagent(s) for the following reactions.
OH Cl
OH
OH
Br
Br
Br
Br
Br
O
CN
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Part III: Mechanisms.
A. For the following 2 reactions, draw the product, mechanism and energy diagram
(include EA, H and all important intermediates or transition states).
Br
NaSH/H2O
NaOMe/MeOH
TfO
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Homework #3 (Chapter 6 and 7) KEY
Part I: Fill in the products, show stereochemistry (if racemic mixture,
you need to show only one isomer).
NaCl, Acetone
NaSH, H2O
OTf
NaOMe/MeOH
OK
Cl
SH
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OH
Br
1) NaH, TfCl
2) 1 eq. of NaOMe/MeOH
2 eq. NaCN/Acetone
Br
OH
CN
H2/Pd
H2/P-2
1) NaNH22) Butyl chloride
3) H2/Pt
3
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Part II: Fill in the reagent(s) for the following reactions.
OH Cl
OH
OH
Br
Br
Br
Br
Br
O
CN
1) NaH, TfCl2) NaCl, Acetone
1) NaH, TfCl
2) KOtBu
1) NaH, TfCl
2) KOtBu 3) Br24) NaNH2 5) H2/P-2
Zn, AcOH
or
NaI/Acetone
3 eq. NaNH2
1) PCl5/OoC
2) 3 eq. NaNH2
3) Propyl Cl 4) H2/Pt
1) NaCl, Acetone
2) NaCN, Acetone
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EA EA
H
EA
H
TfO
H
OMe
Part III: Mechanisms.
A. For the following 2 reactions, draw the product, mechanism and energy diagram
(include EA, H and all important intermediates or transition states).
Br
NaSH/H2O
SH-
SH
NaOMe/MeOH
TfO
HOMe
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Homework #3 Name:
A. Show the product of the following reactions:
OTf
NaCN/DMSO
NaOMe/MeOH
Potassium tert-butoxide
NaBr/Acetone
NaSCH3/H2O
OTs OMe
1 eq. NaOMe/MeOH
2 eq. NaOMe/MeOH
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NaOMe/MeOH
1) NaH/TsCl2) 2 eq. NaCN/DMF
I
OH
OK
NaSH/H2O
NaOH/H2O
NaSH/acetone
OTf
MeO OH
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NaSCH3/DMSO
NaOMe/MeOH
H
MeO
Cl
H
H2/Pd
1) Br22) NaNH23) Pentyl chloride
4) H2/P-2
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B. Fill in the reagents for the following reactions.
OTs
Br CN
Br
Br SH
O
Br
Br
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C. For the following 2 reactions, draw the product, mechanism and energy diagram
(include EA, H and all important intermediates or transition states).
OTf
NaSH/H2O
OTf
NaOMe/MeOH
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Homework #3KEY
A. Show the product of the following reactions:
OTf
NaCN/DMSO
NaOMe/MeOH
Potassium tert-butoxide
CN
OMe
NaBr/Acetone
NaSCH3/H2O
OTs OMe
1 eq. NaOMe/MeOH
2 eq. NaOMe/MeOH
OMe
OMe
Br
SCH3
OMe
OMe
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NaOMe/MeOH
1) NaH/TsCl2) 2 eq. NaCN/DMF
I
OH OK
OH
I
H
H
H
OH
CN
CN
NaSH/H2O
NaOH/H2O
NaSH/acetone
OTf
MeO OH
MeO OH
SH
NO REACTION
MeO OH
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NaSCH3/DMSO
NaOMe/MeOH
H
MeO
Cl
H
t-Bu
ClH
Ethyl
MeO H
t-Bu
ClH
OMe
H Ethyl
OMe
H
H3CS
MeO
H
H
H2/Pd
1) Br22) NaNH23) Pentyl chloride
4) H2/P-2
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B. Fill in the reagents for the following reactions.
OTs
Br CN
Br
Br SH
O
Br
Br
NaOMe/MeOH
NaCN/Acetone
Zn/AcOH
1) KOtBu
2) Br2- 3) NaNH24) Ethyl Cl 5) H2/Pt
NaSH/H2O
1) Na, ethylamine
-78OC
2) NH4Cl
1) PCl5, OoC
2) 3 eq. NaNH23)
4) H2/Pd
Br
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C. For the following 2 reactions, draw the product, mechanism and energy diagram
(include EA, H and all important intermediates or transition states).
SH-
EA
!H
EA
OTf
NaSH/H2O
SH1
2
3
1 2
3
OTf
NaOMe/MeOH
HOMe
MeO H
OTf
EA
!H
46