Chapter 6 Study Guide.pdf

7/25/2019 Chapter 6 Study Guide.pdf

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Chapter 6

Organohalides/Vinyl halides/Phenyl halides/Aryl halides/Polyfluoroalkanes

Nucleophilic Substitution Reactions

Nucleophile/ Electrophile

Substrate

Heterolysis

Alkyloxonium ion

Leaving group

Bimolecular/Unimolecular

Back side

Inverted(inversion of configuration)

Concerted Mechanism

Energy of Activation/Energy Barrier

Transition State

Exergonic/Endergonic

Bond breaking-bond making

Downhill/Uphill

Second-order/First-order/Kinetics

Electron-releasing

Carbocation

Hyperconjugation

Racemization

Steric effects/steric hindrance

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Types of Halides

Physical Properties of Organohalides

Low solubility in water/Miscible in nonpolar solvents

Very toxic and carcinogenic/High density

Structuresdichloromethane(methylene chloride), chloroform, carbon tetrachloride

Methylene refers to CH2

CH3Xif X = I, then liquid; if X = F, Cl, Br, then gas.

In the ethyl series, F and Cl are gases; I and Br are liquids.

In the propyl series, F is a gas; I, Br and Cl are liquids.

In the butyl series, all are liquids.

Polyfluoroalkanes have unusually low bp.

Nucleophilic Substitution Reactions(SN)

Substitution reactions are characterized by having one group leave a molecule(leaving group)while another group(entering group) replaces it on the molecule. Since this series of reactions

involves a nucleophile they are called nucleophilic substitution reactions.

The series of reactions we study involve the breaking of a polarized carbon bond(heterolysis).There are two main types of reactions involved here. The main difference is when the polarized

carbon bond break. Does it break because of an incoming nucleophile(SN2) or does it breakbefore the nucleophile attacks the carbon(SN1).

Nucleophiles

Nucleophiles are negative or neutral atoms with an unshared pair of electrons. The less stablethe negative charge or lone pair is the more powerful the molecule will act as a nucleophile.

X

X

Vinyl halide Phenyl Halide Aryl Halide

Ar-X

Nu

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Leaving Group(L)

The leaving group is a functional group attached to carbon that has polarized the carbon-FGbond. The more stable the leaving group is when it is detached from the carbon the better a

leaving group it will become. The best leaving groups produces stable, weakly basic ions ormolecules.

Substitution, Nucleophilic, bimolecular(SN2)

In performing Kinetics on different substitution reactions, Scientists were able to identify two

trends. The first trend lead to the labeling of some substitution reactions as Substitution,Nucleophilic, bimolecular. This means a substitution reaction(one group comes in another group

leaves) is initialized by nucleophilic(nucleophile) attack on a polarized carbon bond. The ratedetermining step is bimolecular(involves two species).

In short, by kinetic terms this reaction is second order. Experiments carried out determined thatthe rate is affected by the nucleophile and the substrate.

Mechanism for SN2 reactions

The SN2 reaction has a concerted mechanism. In this reaction, the nucleophile enters on the backside forcing a new covalent bond to form. In this process as the new bond forms the old covalent

bond to the leaving group(polarized bond) is broken. This is referred to as bond making/bondbreaking. As the new nucleophile pushes the old leaving group out the configuration of the other

substituents attached to carbon is reversed(inversion of configuration). A transition state brieflyoccurs when the bond making/bond breaking process is going on. In energy terms, these

reactions are exergonic and proceed downhill after overcoming a small energy barrier. There is atransition state but no intermediate.

Know the energy diagram and the mechanism. Know the 10 degree rule and the !G = 84 rule.

Nu

L

R'

R''

Nu

R'

R''

Nu L

R' R''

transition state

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SN2

OCH3

OH

Br

NaCl/Acetone

MECHANISM

OCH3

OH

Br

Cl-

OCH3

OH

Cl

OCH3

OH

Cl

ENERGY DIAGRAM/TRANSITION STATE

1 3

Cl

Br

OH

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2

3!H

EA

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Stereochemistry of SN2 reactions

Since the reactions proceed through an inversion of configuration it usually leads to a change instereochemistry(if stereocenter is present at substitution carbon) but not always.

(R)-2-bromobutane + Iodide produces (S)-2-iodobutane

However, (R)-2-bromo-2-chlorobutane + Fluoride produces (R)-2-chloro-2-fluorobutane

Draw the molecules and the mechanisms and prove this to yourself.

The vast majority of the time there is an inversion of configuration.

Substitution, Nucleophilic, unimolecular(SN1)

In performing Kinetics on different substitution reactions, Scientists were able to identify two trends.The second trend lead to the labeling of some substitution reactions as Substitution, Nucleophilic,unimolecular. This means a substitution reaction(one group comes in another group leaves) is initialized

by nucleophilic(nucleophile) attack on a polarized carbon bond. The rate determining step isunimolecular(involves only one species).

In short, by kinetic terms this reaction is first order. Experiments carried out determined that the

rate is affected ONLYby the substrate. The nucleophile has NObearing on the outcome of thereaction. This means if you double the amount of substrate the reactions rate increases 2 times.

If you double, triple, quadruple, 100Xs the concentration of the nucleophile, nothing happens tothe rate. The nucleophile is not involved in the reactions kinetics.

This is actually a multi-step reaction mechanism(SN2 was concertedone step). One of the

steps is much slower than the others and is the rate-determining step. Unlike SN2, SN1 has adistinct intermediate that forms. This means two energy hills(activation energy) must be

overcome for the reaction to proceed.

Know the energy diagrams(including deprotonation, if necessary).

Mechanism for SN2 reactions

The SN1 reaction is a multi-step mechanism. In this reaction, the highly polarized covalent bond

is broken(due to polarizability and solvent effects). A highly energetic, unstable carbocationintermediate is formed that will react with any nucleophile present(including the solvent and theleaving group). Then the new covalent bond is formed(and any deprotonation that is necessary

will occur). Often a nucleophile is not even present as the solvent acts as anucleophile(solvolysis).

Know the mechanism.

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OCH3

OH

NaCl/Water

MECHANISM

OCH3

OH


1 3

BrCl

OCH3

OHBr

OCH3

OH

2

Cl-

OCH3

OHCl

OCH3

OH

2

Intermediate

SN1

12

3!H

EAEA

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Stereochemistry of SN1 reactions/Carbocations

Carbocations are extremely unstable. However, there are some factors that increase stability.

Alkyl substitutions are electron releasing, which means the electron density flows towards thecarbocation through the covalent bond in essence stabilizing the carbocation(hyperconjugation).

See page 248 for a good description.

This fact leads to the following:

Stability of Carbocations 3o> 2o> 1o> Methyl

Would CF3Cl make a good carbocation if the Cl is a leaving group?

Since the reactions proceed through a carbocation(trigonal planar) the molecule will lose allstereochemistry if the carbocation is a stereocenter. Carbocations are achiral. Nucleophiles canattack from either face of the carbocation equally. This leads to a mixture of

products(racemization).

See page 249.

Solvolysis

As stated previously, the nucleophile can act as the solvent. If the solvent is water this is calledhydrolysis. If solvent is methanol(methanolysis).

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OCH3

OH

MECHANISM

OCH3

OH


1

3

BrOCH2CH3

OCH3

OHBr

OCH3

OH

2

OCH2CH3

OCH3

OH

OCH2CH3

OCH3

OH

2

Intermediate

SN1-Solvolysis

H

H

OCH3

OH

OCH2CH3

Br-

4

CH3CH2OH

1

2

3!H

EAE

A

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SN1 SN2

Substrate Structure neopentyl or 3o>2

o>1

o>methyl Methyl>1

o>2

o>3

oor neopentyl

Formation of Carbocation most important Mainly this is due to Steric Effects

Tertiary and neopentyl exclusively Steric Hindrance

Allylic and benzylic by this mechanism Methyl's and primary exclusively

Secondary possible Secondary possible

Nucleophile Reactivity No effect Increased [Nu] = increased rate

Full negative better than neutral

RO->OH

->>RCO2

->ROH>H2O

Solvent Polar Protic Polar Aprotic(DMSO, DMF, DMA, HMPA)

Increase the rate of carbocation solvation Dissolve ionics well without solvation or

higher dielectric constant = higher rate hydrogen bonding

Reactivity order in protics Solvates cations well but not anions

SH->CN

->I

->OH

->N3

->Br

->CH3CO2

-> Creates "Naked" Anion--very reactive

>Cl->F->H2O

Greatly increased rate in polar aprotics

I>Br>Cl>F--reactivity order in polar protics F>Cl>Br>I--reactivity order in polar aprotic

smaller atoms easier to solvate No solvation so reverts to natural order

Leaving Group Ability to Leave Ability to Leave

I>Br>Cl>F I>Br>Cl>F

alkane sulfonate, alkyl sulfite, triflate and alkane sulfonate, alkyl sulfite, triflate and

p-toluenesulfonate are all good leaving p-toluenesulfonate are all good leavinggroups. Triflate is excellent. groups. Triflate is excellent.

If leaving group produces a strongly basic If leaving group produces a strongly basic

ion, it will be a weak leaving group(i.e. OH) ion, it will be a weak leaving group(i.e. OH)

OH can be dissolved in acid to turn it into OH can be dissolved in acid to turn it into

water molecule which is a good leaving water molecule which is a good leaving

group. group.

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SN1 SN2

Substrate Tertiary Methyl>Primary>Secondary

Secondary ok Secondary ok

Nucleophile weak lewis base, neutral molecule or strong lewis base in high concentration

solvent

Solvent Polar Protic Polar Aprotic

Stereochemistry Racemization Inversion of Configuration

Note: Vinylic and phenylic halides do not undergo either type of substitution. The

carbocation is too unstable for SN1 and the double bonds is too nucleophilic to withstand

a nucleophilic substitution such as SN2(it will repel the nucleophile, 2 negatives repeleach other).

Note: See page 261 for examples of SN2 substitutions.

Elimination reactions

Organohalides(and other good leaving groups) can undergo elimination as easily as substitution.

In elimination, the leaving group and a hydrogen on the "#carbon are removed and replaced with

a double bond.

These are often called beta-eliminations or 1,2-eliminations.

There have to be hydrogen on the beta carbon for eliminations to occur("-hydrogens).

In these reactions you need to have a good base to accomplish the elimination.

Alkoxide bases in general are good enough.

Methoxide(CH3O-) and Ethoxide(CH3CH2O

-) are used mostly.

Methoxide/methanol and ethoxide/ethanol are the usual solvent pairs.

Just like with substitution, there are two possibilities. The unimolecular(E1) and the bimolecularmechanism(E2).

E1 proceeds through carbocation and looks similar to SN1

E2 is a concerted reaction similar to SN2.

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Potassium tert butoxide is a fantastic base with very low nucleophilicity. If base is also a goodnucleophile, then you can get competition between substitution and elimination.

E1 E2

Substrate Tertiary Methyl, primary, secondary, tertiary are

all ok

Secondary ok

Nucleophile needs more basic nucleophile with low needs more basic nucleophile with low

nucleophilicity nucleophilicity

Solvent Polar Protic Polar Aprotic

Hydrogens needs "-hydrogen needs antiperiplanar "-hydrogen

L

R'

R''

R'

R''

CH3O-/CH3OH

CH3O-

Nu

L

R'

R''

HNu

transition state

H

R'

R''

H

R'

R''

H2C C

L

R' R''

H

H

E2

E1

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OCH3

OCH3

MECHANISM

OCH3


1

E2

3

OCH3

Br

OCH3

OCH3

Br

H

O-K

O-

OCH3

OCH3

H

Br

OCH3

OCH3

O-

1

2

3!H

EA

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OCH3

OCH3

NaOH/H2O/!

MECHANISM

OCH3


1

Br

OCH3

OCH3Br

OCH3

OCH3

2

OCH3

OCH3

2

Intermediate

E1

3

H

-OH

OCH3

OCH3

OCH3

12

3!H

EAEA

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SN2 E2

Substrate Primary halides preferred Primary ok

Secondary ok Secondary prefers eliminationTertirary never Tertiarly only elimination

Nucleophile nucleophilic with low basicity needs more basic nucleophile with low

nucleophilicity

Solvent Polar aprotic Polar Aprotic

Hydrogens no need for hydrogens needs antiperiplanar "-hydrogen

Temperature Increasing temperature favors eliminations

Base weak polarizable base favors substitution: strong, sterically hindered base favorsI-, Cl

-, Br

-, RS

-, CH3CO2

-eliminations

strong slightly polarizable bases favor

elimination: NH2-, RO

-

Leaving group no effect no effect

Note: Primary halide with alkoxide base favors substitution. But with potassium tert

butoxide will undergo elimination.

SN1 E1

Substrate stable carbocation favors substitution stable carbocation favors elimination

Nucleophile poor nucleophile favors substitution poor nucleophile favors elimination

Solvent polar protic favors substitution polar protic favors elimination

Hydrogens no need for hydrogens needs antiperiplanar "-hydrogen

Temperature low temperature favors substituions Increasing temperature favors eliminations

Leaving group no effect no effect

In General substitution favored over elimination best to use E2 if you want elimination

strong, hindered base

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SN1 SN2 E1 E2

Methyl NEVER YES NEVER NEVER

primary NEVER YES NEVER POSSIBLE

PREFERRED STRONG HINDERED

BASE NEEDED

secondary USUALLY NEVER YES USUALLY NEVER YES

WITH WEAK BASE STRONG HINDERED

BASE NEEDED

tertiary YES NEVER YES YES

WITH LOW WITH HIGH STRONG HINDERED

TEMPERATURE TEMPERATURE BASE NEEDED

Note: SN1 is best with tertiary halide, low temperature, weak base and polar proticsolvent.

SN2 is best with methyl or primary halide, low temperature, weak base and polar

aprotic solvent.

E1 is not a good method for elimination usually. However with tertiary halideand high temperature eliminations can occur.

E2 is the favored elimination method. All substrates other than methyl can

undergo E2 elimination. Strong, sterically hindered bases favor E2(potassium tertbutoxide).

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HOMEWORK #3 Name:

Fill in the products of these reactions:

NaSH/DMSO

1 equivalent

NaOMe/MeOH

Use the same above starting

material for all 3 reactions.

1 equivalent of reagent will

react only once.

I

OCH3

Br

2 equivalents

NaOMe/MeOH

OTf

NaI/DMSO

NaBr/H2O

OK

Use the same above startingmaterial for all 3 reactions.

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Draw the product, mechanism and energy diagram for the following.

NaOMe/MeOH

H

TfO

OCH3

Br

NaOMe/MeOH

BrH3C

NaSH/ethanol

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Br

H3C

CH

CH3

OH

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For the following, fill in the reagent(above the arrow) needed.

OH

SH

ICl

OH

OH

OH

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HOMEWORK #3 Name:

Fill in the products of these reactions:

OTf

NaI/DMSO

NaBr/H2O

OK

Use the same above starting

material for all 3 reactions.

Br

I

NaSH/DMSO

1 equivalent

NaOMe/MeOH

Use the same above startingmaterial for all 3 reactions.

1 equivalent of reagent willreact only once.

I

OCH3

Br

2 equivalents

NaOMe/MeOH

OCH3

Br

OCH3

Br

OCH3

SH

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NaOMe/MeOH

H

TfO

OCH3

Br

NaOMe/MeOH

OCH3

rotate to get H and OTf in plane

OTf

H

H3CO the two wedged groups go on same side

and the two dashed groups go on the

same side.

OMe

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Draw the product, mechanism and energy diagram for the following.

BrH3C

NaSH/ethanol

SHH3C

SH-

1

2

3


1

2

3

!H

EA

EA

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Br

H3C

CH

CH3

OH

H3C

CH

CH3

OH

O H

B-

1

2

3

4 O


1

2

3

!H

EA

EA

4

EA

2

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OH

SH

I

Cl

OH

OH

OH

H2SO4, heat

85% H3PO4, heat

dilute H2SO4, heat

****All of the above could also be done by:

1) TfCl, pyr 2) NaOMe, MeOH

1) TsCl, pyr

2) NaSH, DMF

1) NaSH, DMSO

2) NaCl, DMSO

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