Chapter 7 Study Guide.pdf

7/25/2019 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


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


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


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


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


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


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


SH-

EA

!H

EA

OTf

NaSH/H2O

SH1

2

3

1 2

3

OTf

NaOMe/MeOH

HOMe

MeO H

OTf

EA

!H

46

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