WWU -- Chemistry REAGENTS WITH CARBON- METAL BONDS; ORGANOMETALLIC SYNTHESIS OF ALCOHOLS Chapter 15

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REAGENTS WITH CARBON-METAL BONDS;

ORGANOMETALLIC SYNTHESIS

OF ALCOHOLS

Chapter 15

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Assignment

• DO: Sections 15.0 through 15.7• READ: Sections 15.8 and 15.10• SKIP: Section 15.9• DO: Section 15.11• DO: Problems

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Problem Assignment

• In Text Problems– 15-1 through 15-13

• End-of-Chapter Problems– 1 through 3

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Reagents with Carbon-Metal Bonds

• How do we make large molecules when most of our available reagents are relatively simple in structure?

• How do we “dock” two large molecular fragments together?

• What we need are methods of forming carbon-carbon bonds.

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• Up to now, we really haven’t looked at methods of forming C-C bonds. We’ve formed C-O bonds, C-Cl bonds, and C-Br bonds in many examples, but what about C-C bonds?

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Can anyone suggest a C-C bond formation reaction that we have already encountered?

The Diels-Alder reaction!

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Let’s go back to a very familiar reaction,

nucleophilic substitution:

Now, if our nucleophilic atom were carbon, we would have a method that we could adapt and develop.

C Br

R

HH

Nu: Nu C

R

H

H+ Br-

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Consider:

Here is the theme of this chapter. It introduces a new class of reagents that are capable of acting as carbon nucleophiles, opening the door to our being able to combine small molecular fragments and build large molecules from them.

R

C

HH

+ C Br

R

HH

C C

RR

H HH H

+ Br-

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Generalized Method

R X + 2 M R M + MX

M = a metal

R X + M R M X

For a monovalent metal:

For a divalent metal:

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Formation of Organolithium Reagents

R X R Li+ +

R = 1°, 2°, 3°, aryl

X = I > Br > Cl

2 Li Li X

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Example:

2 Li + CH3 CH2 CH2 CH2 Brhexane

CH3 CH2 CH2 CH2 Li

+ LiBr

Typical solvents:

•diethyl ether

•tetrahydrofuran (THF)

•hydrocarbons (pentane, hexane, etc.)

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Some important points to consider:

• organosodium and organopotassium reagents are difficult to form -- this method is best for organolithium reagents.

• E2 dehydrohalogenation is an important side reaction, especially if the alkyl halide is secondary or tertiary. This problem is particularly serious with R-Na’s or R-K’s.

Who knows why E2 dehydrohalogenation happens in this reaction?

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Formation of Grignard Reagents

R X R Mg+ Mgether

X

R = 1°, 2°, 3°, aryl

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Example:

CH3 C CH3

CH3

Cl

+ Mgether

CH3 C CH3

CH3

MgCl

Typical solvents:

•Diethyl ether (b.p. 35 °C)

•Tetrahydrofuran -- THF (b.p. 65 °C)

•Dioxane (b.p. 101 °C)

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An ether is required to form a stable Grignard complex.

R

Mg

X

CH3 CH2

O

CH2

CH3CH2

O

CH2CH3 CH3

Formation of this complex is exothermic; the reaction is sufficiently exothermic to boil the solution without having to add external heat!

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Why might you need different solvents?

Br + Mgether

35 °CMgBr

Cl + Mg65 °C

MgClTHF

This reaction is too slow at 35 °C.

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The complete structure of the Grignard reagent is quite complex. It is probably an equilibrium mixture of the type:

2 R-MgX R2Mg + MgX2

While this may be more correct, it is easier to treat the Grignard reagent as if it were R-MgX, which is what we shall do in this course.

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Owing to the electronegativity difference between the metal and carbon, the carbon-metal bond has a great deal of partial ionic character. The bonds are polar covalent in nature.

This means that we can write:

R MgX

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In fact, we can treat the Grignard (or any organometallic) reagent according to:

Thus, the organometallic reagent acts as a source of “R:-”, which is the conjugate base of an alkane.

We therefore expect the organometallic reagents to be very basic and strongly nucleophilic.

R MgX R: MgX

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If the organometallic reagents are basic, then we should see them react readily with acids.

Any source of H+ will bring about this reaction:

acids, carboxylic acids, water, alcohols, amines, even atmospheric moisture

R MgX + R H + MgX+

R Li + R H + Li+

H+

H+

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We can use the reaction of organometallic reagents with sources of proton deliberately

CH3 CH2 CH CH3

Br

CH3 CH2 CH CH3

Mg

CH3 CH2 CH CH3

D

Mg

ether

Br

D2O

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Do You Remember This?

Why does the nucleophile go to the CH2 group and not the R-CH group?

Nu: CH CH2

O

R+ R CH CH2 Nu

OH

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Reaction with Epoxides

Notice that:

•whatever the length of the carbon chain in R, the product has added two carbons

•the product is a terminal alcohol

CH2CH2

O

R MgX R CH2 CH2 O

R CH2 CH2 OH

+ether ..

:.. MgX

+_

H2O

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Crude outline of a mechanism

CH2 CH2

O

"R:-" + R CH2 CH2 O

H+

R CH2 CH2 OH

ether

(from RMgX)

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Examples:

CH3 MgI + CH2 CH2

O

ether H2OCH3 CH2 CH2 OH

60% yield

Li

CH2CH

O+

CH2 CH

OHether H2O

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Reasoning by analogy, you could do...

CH2

CH2 O

CH2

+ R MgXether

H2OR CH2 CH2 CH2 OH

Oxetane(trimethylene oxide)

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Reaction with Carbonyl Compounds

R MgX R C

R

R

O R C OH

R

R

O

CR R

+..

:.. MgX

+_ H2Oether

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Crude outline of the mechanism of carbonyl addition

"R:-"

(from R-MgX)

C O

R'

R

R C O

R

R'

H+

R

C

R'

OHR

R, R' = H, alkyl, or aryl

+

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Outcome of the reaction of an organometallic with carbonyl compounds

R M + C O

H

H

R C OH

H

Hformaldehyde primary alcohol

R M + C O

H

R

R C OH

R

Hsecondary alcohol

R M + C O

R'

R

R C OH

R

R'ketones tertiary alcohol

other aldehydes

(M = Li or MgX)

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Example

CH3 CH2 CH2 MgBr + CH3 CH2 C H

O

ether

H2O, H+

CH3 CH2 CH2 C

OH

H

CH2 CH3

Propylmagnesium bromide Propanal (Propionaldehyde)

3-Hexanol

The product is a secondary alcohol

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Example #2O

+ CH3 MgIether

H2O, H+

OHCH3

Cyclohexanone

Methylmagnesium iodide

1-Methylcyclohexanol

The product is a tertiary alcohol.

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From “PLKE-Micro-3”...

MgBr + C

O

1) ether

2) H2O, H+

C

OH

Phenylmagnesium bromide

Benzophenone

Triphenylmethanol

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Preparation of Alkanes

Wurtz Reaction

R X + 2 Na2 R R + 2 NaX

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Example of a Wurtz Reaction

2 CH3 CH2 CH2 Br + 2 Na CH3 CH2 CH2 CH2 CH2 CH3

+ 2 NaBr

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The reaction occurs in two steps:

1)

2)

CH3 CH2 CH2 Br 2 Na CH3 CH2 CH2 Na

+

CH3 CH2 CH2 Na

+

CH3 CH2 CH2 Br

CH3 CH2 CH2 CH2 CH2 CH3

Br

+ Br

+

Na

Na

The second step is an SN2 reaction with the organosodium compound acting as the nucleophile.

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Characteristics of the Wurtz Reaction:

• Characteristically poor yields• “Worst Reaction”• Works only with primary alkyl halides• With secondary and tertiary alkyl

halides, all you get is alkene.• Why?• Only even-numbered alkanes can be

prepared -- both halves have to be the same.

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The Wurtz Reaction is an example of an Alkylation

Reaction• Alkylation: a reaction to attach an alkyl

group to some other atom.• Other alkylations we have encountered

include:– Williamson ether synthesis (alkylation of

oxygen)– Wurtz reaction (alkylation of carbon)– Alkylation of amines (nitrogen)– S-AdM (biological methylation)

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• Can we do an alkylation of carbon?

• Can we do it better?

• Can we make odd-numbered alkanes?

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• Obviously, the answer to the previous questions is “yes”!

• A new type of organometallic reagent, a lithium dialkylcuprate, affords us the possibility of alkylating carbon in good yield

• We also have a route to the synthesis of an odd numbered alkane -- the two halves being joined do not have to be the same.

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Lithium Dialkylcuprates

a lithium dialkylcuprate

R X R Li

R Li Cu

R

R

Li

2 + CuI0°C

etherLi

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Example

2 CH3 Li + CuI0 °C

etherCu

CH3

CH3

Li

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The dialkylcuprate is a very good alkylating agent.

R X + R2CuLi0 °C

ether

X = Cl, Br, I

R R + R Cu

+ Li X

•This reaction is known as the Corey-House synthesis.

•Note that the two alkyl groups do not have to be identical! -- (unlike the Wurtz reaction)

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Example

CH3

CH2CH2

CH2CH2

I+ (CH3)2CuLi

0 °Cether

CH3

CH2CH2

CH2CH2

CH3

+ CH3 Cu + Li I

98% yield

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This wouldn’t work by a Wurtz synthesis...

I

+ (CH3)2CuLi0 °C

ether

CH3

75% yield

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Also...

Br

+ (CH3)2CuLi0 °C

ether

CH3

•In general, allylic halides are unreactive in organometallic reactions.

•Not here!

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This would be impossible by other methods:

C C

H

C8H17

I

H

(E)-1-Iodo-1-decene

+ (C4H9)2CuLi0 °C

ether

C C

H

C8H17

C4H9

H

(E)-5-TetradeceneStereospecific!

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Also...

R C Cl

O

+ R2CuLi0 °C

etherR C R

O

+ R Cu

+ Li Cl

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Synthesis of Manicone

CH3

CH2CH

CHC

CCl

O

CH3 CH3

+ (C2H5)2CuLi

-78.5 °C ether

CH3

CH2CH

CHC

CCH2

CH3

O

CH3CH3

Manicone

Manicone is a pheromone secreted by certain male ants as they swarm. It causes female ants of the same species to swarm at the same time the males do. This facilitates mating!

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Alkynylorganometallic Compounds

R M C CH R C C R

R H+ + M+

:

_

Section 15.8 -- assigned as reading

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Other Organometallic Reagents

Organozinc reagents are used in synthesis owing to their greater selectivity (see J. Vyvyan)

We can also make R-Zn, R-Sb, R-As, R-Be, R-Ca, R-Hg, R-Sn, … reagents. We choose other metals for different degrees of reactivity and for greater selectivity.

CH2 ICH3 +ether

CH3 CH2 ZnZn

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If the reaction of alkyl halide with metal is too slow, one can make a metal alloy with sodium or potassium. For example, lead, by itself is too unreactive. But we can do...

4 CH3 CH2 Br + 4 Na-Pb

CH2 Pb CH2

CH2

CH2

CH3

CH3

CH3

CH3

sodium-lead alloy

Tetraethyllead

Tetraethyllead (TEL) used to be used in gasoline as an anti-knock agent.

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Reactions with Metal Salts

• We can transfer an R group from one metal to another.

• Generally this works when we transfer an alkyl group from a more active to a less active metal (from a negative E° to a positive E°)

• This reaction is energetically favorable -- exothermic

• We need to consider reduction potentials

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Example

2 RMgX + CdCl2 R Cd R + 2 MgXCl

We are transferring the R group from Mg to Cd.

Mg: E° = - 2.38 volts

Cd: E° = - 0.40 volts

Organocadmium reagents are very useful (see Chapter 17), but they cannot be made directly.

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Preparation of Tetraphenyltin

4 BrNa

toluene111 °C

4 Na

SnCl4

Snm.p. 229 °C

Na: E° = -2.71 volts

Sn: E° = +0.01 volts

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Preparation of an Organosilane

4 CH3 CH2 MgBrSiCl4

CH3 CH2 Si CH2 CH3

CH2

CH2

CH3

CH3

How would you make TMS?

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Speculate:

4 CH3 CH2 Br + 4 Na-Pb

CH2 Pb CH2

CH2

CH2

CH3

CH3

CH3

CH3

sodium-lead alloy

Tetraethyllead

Na: E° = - 2.71 volts

Pb: E° = - 0.13 volts

Perhaps an organosodium reagent is formed initially, and then the ethyl group is transferred from the sodium to the lead.