Chapter 8 Study Guide.pdf

7/25/2019 Chapter 8 Study Guide.pdf

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Chapter 8

Addition reactions

Nucleophiles/Electrophiles

Regioselective/Stereoselective

Carbenes

Reactions of alkenes

For the reactions of HX with an alkene, the order of reactivity is HI> HBr> HCl> HF.HCl is so slow to react (except to highly substituted alkeneWHY???) that a new

methodology has been developed to carry out these reactions. The hydrogen halide isdissolved in alumina or silica along with the alkene in a solution of dichloromethane.

HF is so slow to react we can basically call it unreactive.

HBr

H2SO4

H2O/H+

Br2

Cl2

Br

OSO3H

OH

Br

B

Cl

Cl

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The pi bond of an alkene is a highly electron rich region, which can act as a nucleophile.The alkene will attack an electrophile attaching the electrophile to the least substituted

carbon. This will leave a carbocation on the most substituted carbon, which will thenreact with any available nucleophile. This is called Markovnikov addition after the

Russian chemist who came up with the addition. The most stable carbocation forms and

reacts with the nucleophile. Since it proceeds through a carbocation the product isracemic. The mechanism is as follows:

Know the energy diagram p. 333.

If you add HBr in the presence of peroxides you can get the anti-Markovnikov product toform.

The other addition reactions proceed through the same basic mechanism with the same

Markovnikov rules. In the hydration of the double bond, the reaction is similar to thedehydration of an alcohol. In the dehydration reaction, concentrated acid and heat wereadded to an alcohol to make an alkene. In the hydration reaction, dilute acid is added to

an alkene to make an alcohol. Because all of these reactions proceed throughcarbocation, rearrangements are possible. Two modern reactions were created to avoidthe carbocation problem. Oxymercuration-demercuration gives us Markovnikov addition

without rearrangements. Hydroboration-oxidation gives us the anti-Markovnikovaddition without rearrangements.

HBr

H

Br-

Br

HBr/Peroxides

Br

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OXYMERCURATION/DEMERCURATIONMARKOVNIKOV ADDITION

OH

1) HgAcetate/THF/H2O

2) NaOH/NaBH4

HgAcetate = Hg(OCCH3)2

O O

OCCH3 = Ac

MECHANISM

Hg(OAc)2 HgOAc + OAc

HgOAc

+

+

H

O

H

HgOAc

O

H

H

H

O

H

HgOAc

O H

other steps

Reaction is regioselective. The new group(OH) will always go to one carbon inpreference of another. Regioselective, reaction forms one regioisomer(by location) overanother.

The next reactions is regio stereoselective. Not only does the new group go to one

carbon in preference of another. It goes in a specific stereochemistry. Specifically synaddition(same side).

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HYDROBORATION/OXIDATIONANTI-MARKOVNIKOV/SYN ADDITION

1) BH3:THF

2) H2O2/OH-

OH

H

MECHANISM

BH

H

H

H B H

H

H B

H

B

3

+ enatiomer

H O O

R

B

R

O OH

unstable

intermediate

R

B

R

ORrepeat 2x's

RO

B

RO

OR

OH

BRO

OR

OH

OR

BRO

OR

O

O

R

H

ROH

OH

H=

R

R

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The addition of Br2and Cl2proceed through a slightly different mechanism from eachother and from the other addition mechanism. The Br2mechanism is the important one to

us. This mechanism proceeds through a Bromonium(cyclopropane like) ion and leads toanti addition. This reaction is stereospecific.

See mechanisms pages 350, 353 and 354

Br

Br-

Br-

Br

Br

BrBr

Br

Br

This leads to a pair of trans(anti) enantiomers.

Note: The addition of Br2to an unknown is a chemical test for the presence ofunsaturation. Br2is a reddish-brown color to begin with. As it reacts with thealkene/alkyne it becomes clear. This is a good test to see if your hydrogen

deficiency is due to rings or to unsaturation. Why must there be an excess of thealkene/alkyne and not the Br2.

Page 352. It may look like the reaction doesnt give anti product. But the bond has just

been rotated to show the meso structure more clearly.

Br

Br

H

CH3

H

H3C

Br H

Br H

CH3

CH3

These are the same.They are just rotationsaround the central

carbon-carbon bond.They are called confomers

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The bromonium ion is stereospecific as we see above. However it can also beregiospecific. If you perform a halohydrin (Halogen and alcohol) reaction using BrOH,

the bromonium ion will form as in the Br2 reaction. Here however when the OH groupattacks in the anti (stereospecifically) conformation it also attacks with a high degree of

regioselectivity (attacking the more substituted carbon of the alkene). The reason for this

is the formation of a carbocation- like transition state. In that Transition State thecarbocation would prefer to be on the most substituted carbon.

BrOH

BrOH

BrOH

OH

OH

Br

Br

OH

OH

BrHere since one side of the double bond

is disubstituted and the other side ismonosubstituted, the OH will attachonly on the more substituted side.

Here since both sides of the double bondare monosubstituted, the OH will attach

equally on both sides.

Here the OH attaches only to theore substituted side of the do ble bo d.

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The mechanism for the regiospecific Transition State is as follows:

Reactions of carbenes(diazomethane)

Carbenes are a group of compounds where the carbon has formed only 2 bonds. Theseare extremely reactive intermediates. The reactions are highly stereospecific and are very

useful in forming hard to form cyclopropanes. Carbenes can readily form from

diazomethane or from chloroform. When forming dihalocarbenes there must be an hydrogen but there can not be a hydrogen or elimination will occur instead.

BrOH

Br

+

OH-

OH

Br

+

H2C N Nor h

CH2 + N2

Highly reactive carbene.

OK + CHCl2CH2 OH

Cl

+

OK + CHCl3 OH + CCl3

K+

CCl2

DihalocarbeneHighly reactive

hydrogens lead to elimination

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A carbene derivative called a carbenoid can be synthesized using the Simmons-Smithcyclopropane synthesis. In this reaction diiodomethane and a Zinc-Copper couple are

stirred with an alkene to produce a reactive intermediate, the carbenoid, which forms acyclopropane ring with the alkene in situ.

Diol Addition to an Alkene

Syn hydroxylation(1,2-diols or glycols). There are two sets of reagents that lead to syn-hydroxylation. The first is potassium permanganate stirred cold in a basic water solution.The seconds is osmium tetroxide and pyridine followed by aqueous acidic solution.

Although the osmium reagent is better, the potassium reagent is cheaper and much safer.However, the potassium reagent often oxidizes the reaction too much and offers low

yields. The osmium reagent has been modified to allow catalytic amounts of osmium tobe introduced thus saving cost and offering less toxicity. Barry Sharpless of Stanford isone of the leading proponents of this method(Sharpless dihydroxylation) Who is Barry

Sharpless?

K M n O 4/OH-/H 2O/cold

1) OsO 4, pyridine

2) NaHSO 3/H 2O

O H

O H

W hat is the enantiomer of this?Is this a true stereospecific reaction?

Cl

Br

Diazomethane

OK

CHCl3CH2I2/Zn(Cu)

Diethyl ether

Cl

Br

Cl

Br

Cl

Br

Cl

Cl

This reaction produces a pair of enantiomer, what is the

other enantiomer for the diazomethane(for ex.)

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OH

OH

OH

OH

OH

OH

CIS TRANS

OH

OH

OH

OH

KMnO4/OH

-/H

2O/cold

SHARPLESS

ASYMMETRIC

DIHYDROXYLATION

4 POSSIBLE ONLY PAIR ACTUAL

GIVES ONLY 1 ENANTIOMER OUT OF 4 POSSIBLE STEREOISOMERS

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Oxidative Cleavage of Alkenes

Alkenes can be cleaved(cut in half) with reagents like hot potassium permanganate andozone. Hot potassium permanganate yields an oxidative workup that gives ketones, acids

and carbon dioxide.

Note: The addition of hot potassium permanganate is a chemical test to find thefunctionality of alkene groups in a molecule. The terminal carbon of a terminal

alkene is converted to carbon dioxide by hot potassium permanganate.Monosubstituted carbons are converted to acids and disubstituted carbons areconverted to ketones.

1) KMnO4, OH-, H2O

heat

2) H3O+

O

OH

+ CO2

HO

OH

O

O

HO

O

O

+

+

1) KMnO4, OH-, H2O

heat

2) H3O

+

1) KMnO4, OH-, H2O

heat

2) H3O+

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f

e

d

c

b

a

g

1) KMnO4/OH-/H2O/heat

2) H+

f

e

d

c

b

a

OO

g

OH

O

f

e

d

c

b

a

OH

O

+ CO2

terminal alkene leads to CO2

d

c

b

a

1


2) H+

d

c

b

a

OO

1

OH

internal alkene leads to acids and/or ketones

2

3

2

3

OH

d

c

b

a

OH

3

2

1

O

O

OH

d

c

b

a

1


2) H+

d

c

b

a

OO

1

internal alkene leads to acids and/or ketones

2

3

2

3

OH

d

c

b

a

3

2

1

O

O

OH

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Ozone will also cleave a double bond. However ozone is followed by a reductive workupthat leaves ketone and aldehydes only.

The reaction mechanism is fairly complicated. First an ozonide is formed. This collapses

and reforms in a new configuration. Finally Zinc as a Lewis Acid catalyzes a finalcollapse of the intermediate to form the reductive product.

O

O

O O O

O

O

O

O

O

O

O

i n i ti a l oz o n ide

o z o n i d e r e f o r m s

Z n / A c O H

O

O

H

H

1) O 3, CH 2Cl2, -78oC

2) Zn , AcOH

O

H

H

H

O

O

H

O

O

+

+

1) O 3, CH 2Cl2, -78oC

2) Zn , AcOH

1) O 3, CH 2Cl2, -78oC

2) Zn, AcOH

O

H H

+

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HBr

H2O/H+

H2/Pt

Br2

HBr/H2O2

1) I22) 3 eq. NaNH 23) Pentyl chloride

Br

Br

OH

Br

OH

Br

Br

Br

1

2

3

4

5

HBr

H2/Pt

HBr/H2O2

H2O/H+

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Chemical Reactions with MechanismsReactions of alkenes

Mechanism for addition of HX to alkene

Mechanism for addition of X2to alkene.

HBr

H2SO4

H2O/H+

Br2

Cl2

Br

OSO 3H

OH

Br

BrCl

Cl

HBr

H

Br-

Br

Br2Br

Br-

Br-

Br

Br

Br

Br

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OXYMERCURATION/DEMERCURATIONMARKOVNIKOV ADDITION

HYDROBORATION/OXIDATIONANTI-MARKOVNIKOV/SYN ADDITION

O H

1 ) H g A c e t ate / T H F /H 2O

2 ) N a O H /N a B H 4

H g A c e t a te = H g (O C C H 3 ) 2

O O

O C C H 3 = A c

M E C H A N I S M

H g ( O A c ) 2 H g O A c + O A c

H g O A c

+

+

H

O

H

H g O A c

O

H

H

H

O

H

H g O A c

O H

o t h e r s t e p s

1) B H 3: T H F

2 ) H 2 O 2 /O H-

OH

H

M E C H A N I S M

BH

H

H

H B H

H

H B

H

B

3

+ ena tiom er

H O O

R

R

B

R

O OH

u n s t a b l e

in t e rmedia t e

R

B

R

O Rrepeat 2x's

R O

B

R O

O R

O H

BR O

O R

O H

O R

BR O

O R

O

O

R

H

R O H

O H

H=

R

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Halohydrin reaction

BrOH

OH

Br

Mechanism for the reaction.

Ozonolysis of double bonds.

1) O3, CH2Cl2, -78oC

2) Zn, AcOH

O

H

O

H H+

Mechanism.

BrOH

Br

+

OH-

OH

Br

+

O

OO

O

OO

O

O

O

H

H

O

OO

many steps

O

H

O

H H+

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Chemical Reactions without Mechanisms

Addition reactions with alkynes

Formation of a ketone

Anti-Markovnikov addition of HBr.

Carbene addition to alkene.

HBr/Peroxides

Br

1 mole HI

1 mole Br2

2 moles of HCl

Acetyl Bromide

Alumina

2 moles of

I2

I

Br

Br

Cl Cl

I I

I I

Br

H

H2O/HgSO4

H2

SO4

O

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Syn-dihydroxylation of alkene.

Oxidation of alkene.

KM nO 4/O H-/ H2O/cold

1) OsO4 , pyridin e

2) NaHSO3/H

2O

OH

OH

W hat is the enantiom er of this?

Is this a true stereospec ific reaction?

Cl

Br

Diazomethane

OK

CHCl3CH2I2/Zn(Cu)

Diethyl ether

Cl

Br

Cl

Br

Cl

Br

Cl

Cl

This reaction produces a pair of enantiomer, what is the

other enantiomer for the diazomethane(for ex.)

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Oxidation of alkyne.

1) KMnO4, OH-, H2O

heat

2) H3O+

O

OH

+

CO2

HO

OH

O

O

HO

O

O

+

+

1) KMnO4, OH-, H2O

heat

2) H3O+

1) KMnO4, OH-, H2O

heat

2) H3O+

1) O3

2) AcOH

Or

1) KMnO4, OH-

2) H+

OH

OO

HHO

H

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Homework #4

Part I: Indicate if the following are (Markovnikov or anti-Markovnikov)and/or (syn or anti addition).

HC l

B r2

H 3O+

HBr/H2 O2

1) HgAcetate, THF, H2 O

2) NaBH 4, NaOH

HBr/CCl4

1) BH3:THF

2) H2 O2 , O H-

1) Hg(O2 CC F3) , CH3 CH 2C H2 OH

2) NaBH4 , NaOH

Br2/H 2O

C l

Br

Br

O H

Br

Br

Br

O H

OH OH

OCH 2CH 2CH3

1

2

3

4

5

6

7

8

9

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Diazomethane

OH

CH2I2/Zn(Cu)/diethyl ether

KMnO4, OH-, H2O, cold

1) OsO4, pyridine

2) NaHSO3, H2O

OH

OH

OH

10

11

12

13

Part II: Give the structures of the products

A BC

D

E

2mole

s ofB

r2

1mole

ofCl2

2 moles o f HB r

1 ) K M n O 4 /O H-

H 2O/hea t

2 ) H 3O+

2 moles of

HC l

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Part III: Give the structures of the products

A B

I

H

F

G

C

D

E

H 2 O /H+

B r2

1 ) H g A c e ta te

T H F /H 2 O

2 ) N a O H /N a B H 4

1 ) B H 3 : T H F

2 ) H 2 O 2 /O H-

H B r / C C l4

Br 2/H 2

O

D i a z o m e t h a n e

K M n O 4/ O H-

H 2 O /c o l d

1 ) K M n O 4/ O H-

H 2 O / h e a t

2 ) H 3 O+

Part IV: Show mechanisms for the following.

HBr/H2O2

Br

Br

HBr

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Part II:

A

B

C

D

E

2mole

s ofB

r2

1moleo

fCl2

2 moles of HBr

1) KMnO4/OH-

H2O/heat

2) H3O+

2 moles of

HCl

Cl Cl

B r B r

Br Br

OH

O

2

Cl

Cl

Br Br

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Part III:

5

4

3

2

a

b

c

d

1

A B

I H

F

G

C

D

E

H2O/H+

Br2

1) HgAcetate

THF/H2O

2) NaOH/NaBH4

1) BH3:THF

2) H2O2/OH-

HBr/CCl4

Br2/H2

O

Diazomethane

KMnO4/OH-

H2O/cold

1) KMnO4/OH-

H2O/heat

2) H3O+

HO

Br

Br

1

2

3

4

5

O

d

c

b

a

OH

O

HO

Br

HO

Br

H

OHOH

OH

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Part IV: Show mechanisms for the following.

HBr/H2O2

Br

H O O H

Heat or h

H Br

HOH + Br

Br

H Br

Br

30

radical better

than 2o

radical.

2 HO

HO

Br

+

Br

H Br

H

Br-

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Homework--Chapter 8 Name:

Draw the product for the following reactions.

HBr

HBr/H2O2

Br2/H2O

1) Hg(O2CCF3), butanol

2) NaBH4, NaOH

CH2I2/Zn(Cu), diethyl ether

KMnO4/OH-/H2O/cold


2) H3O+

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KEY

Draw the product for the following reactions.

HBr

HBr/H2O2

Br2/H2O

1) Hg(O2CCF3), butanol

2) NaBH4, NaOH

CH2I2/Zn(Cu), diethyl ether

KMnO4/OH-/H2O/cold


2) H3O+

Br

O

BrOH

Br

OH

OH

O

HO

O

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