8-1 8 Organic Chemistry William H. Brown & Christopher S. Foote.

Organic Organic ChemistryChemistry

William H. Brown & William H. Brown & Christopher S. FooteChristopher S. FooteWilliam H. Brown & William H. Brown & Christopher S. FooteChristopher S. Foote

Nucleophilic Nucleophilic Substitution Substitution

and and -Elimination-Elimination

Chapter 8

Chapter 8Chapter 8

88 Nucleophilic SubstitutionNucleophilic Substitution

Nucleophilic substitution:Nucleophilic substitution: any reaction in which one nucleophile is substituted for another at a tetravalent carbon

Nucleophile:Nucleophile: a molecule or ion that donates a pair of electrons to another molecule or ion to form a new covalent bond; a Lewis base

nucleophilicsubstitution

Nucleophile

++ C NuC XNu:- :X-

leavinggroup

88 Nucleophilic SubstitutionNucleophilic Substitution An important reaction of alkyl halides

+an alcohol (after proton transfer)

an alkylammonium ion

an alkyl iodide I -

CH3NH3+

CH3O-H

a nitrileCH3C N- C N

an alkyneCH3C CHHC C -

a thiol (a mercaptan)HS - CH3SH

an ether

an alcohol

Reaction: + +CH3NuCH3Br Br-

::::::

88 SolventsSolvents Protic solvent:Protic solvent: a solvent that is a hydrogen bond

donor • the most common protic solvents contain -OH groups

Aprotic solvent:Aprotic solvent: a solvent that cannot serve as a hydrogen bond donor• nowhere in the molecule is there a hydrogen bonded

to an atom of high electronegativity

88 Dielectric ConstantDielectric Constant Solvents are classified as polar and nonpolar• the most common measure of solvent polarity is

dielectric constant

Dielectric constant:Dielectric constant: a measure of a solvent’s ability to insulate opposite charges from one another• the greater the value of the dielectric constant of a

solvent, the smaller the interaction between ions of opposite charge dissolved in that solvent

• polar solvent: dielectric constant > 15• nonpolar solvent: dielectric constant < 15

88 Protic SolventsProtic Solvents

Solvent StructureDielectric Constant (25°C)

Formic acid

Methanol

Ethanol

H2O 79

CH3CH2OH

Acetic acid CH3COOH 6

88 Aprotic SolventsAprotic Solvents

2.3Toluene

4.3Diethyl ether

9.1Dichloromethane

SolventDielectricConstantStructure

Dimethyl sulfoxide (DMSO)

Acetonitrile

Acetone

N,N-Dimethylformamide (DMF)

Nonpolar

CH3C N

CH2Cl2CH3CH2 OCH2CH3

Hexane CH3(CH2)4CH3 1.9

(CH3)2S=O

(CH3)2NCHO

(CH3)2C=O

C6H5CH3

88 MechanismsMechanisms Chemists propose two limiting mechanisms for

nucleophilic displacement• a fundamental difference between them is the timing

of bond breaking and bond forming steps

At one extreme, the two processes take place simultaneously; designated SN2• S = substitution• N = nucleophilic• 2 = bimolecular (two species are involved in the rate-

determining step)

88 Mechanism - SMechanism - SNN22• both reactants are involved in the transition state of

the rate-determining step

HO:- + C

HO Brδ- δ-

T ransition state with simultaneous bond breaking and bond forming

HO + :Br -

88 Mechanism - SMechanism - SNN22

QuickTime™ and aPhoto - JPEG decompressor

are needed to see this picture.

88 Mechanism - SMechanism - SNN11 Bond breaking between carbon and the leaving

group is entirely completed before bond forming with the nucleophile begins

This mechanism is designated SN1 where• S = substitution• N = nucleophilic• 1 = unimolecular (only one species is involved in the

rate-determining step)

88 Mechanism - SMechanism - SNN11• Step 1: ionization of the C-X bond gives a carbocation

intermediate

CH3H3C

slow, ratedetermining

A carbocation intermediate; its shape is trigonal planar

+ :Br-

88 Mechanism - SMechanism - SNN11• Step 2: reaction of the carbocation with methanol

gives an oxonium ion. Attack occurs with equal probability from either face of the planar carbocation

• Step 3: proton transfer completes the reaction

+++ fastC

H3CH3C

CH3CH3

CH3CH3C

Lewis acid Nucleophile(Lewis base)

Oxonium ions

CH3H3C

88 Mechanism - SMechanism - SNN11

88 Evidence of SEvidence of SNN reactions reactions1. What is the rate of an SN reaction affected by: • the structure of Nu?• the structure of RX?• the structure of the leaving group?• the solvent?

2. What is the stereochemistry of the product if the Nu attacks at a stereocenter?

3. When and how does rearrangement occur?

88 KineticsKinetics For an SN1 reaction, the rate of reaction is first

order in haloalkane and zero order in nucleophile

d[(CH3)3CBr]dt

Rate = - = k[(CH3)3CBr]

CH3CBr + CH3OH CH3COCH3 + HBr

2-Bromo-2-methyl-propane

(tert-Butyl bromide)

Methanol 2-Methoxy-2-methyl-propane

(tert-Butyl methyl ether)

88 KineticsKinetics For an SN2 reaction, the rate is first order in

haloalkane and first order in nucleophile

d[CH3Br]

dtrate = = k[CH3Br][OH-]

CH3Br + Na+OH- CH3OH + Na+Br-

Bromomethane Methanol

88 NucleophilicityNucleophilicity Nucleophilicity:Nucleophilicity: a kinetic property measured by

the rate at which a Nu causes a nucleophilic substitution under a standardized set of experimental conditions

Basicity:Basicity: a equilibrium property measured by the position of equilibrium in an acid-base reaction

Because all nucleophiles are also bases, we study correlations between nucleophilicity and basicity

88 NucleophilicityNucleophilicity

Br-, I-

HO-, CH3O-, RO-CH3S-, RS-

CH3COO-, RCOO-

CH3OH, ROH

CH3COOH, RCOOH

NH3, RNH2, R2NH, R3NCH3SH, RSH, R2S

Nucleophile

Moderate

CN-, N3-

Effectiveness

88 NucleophilicityNucleophilicity Relative nucleophilicities of halide ions in polar

aprotic solvents are quite different from those in polar protic solvents

How do we account for these differences?

Increasing NucleophilicitySolvent

Polar aprotic

Polar protic F- < Cl- < Br- < I-

I- < Br- < Cl- < F-

88 NucleophilicityNucleophilicity A guiding principle is the freer the nucleophile,

the greater its nucleophilicity Polar aprotic solvents (e.g., DMSO, acetone,

acetonitrile, DMF) • are very effective in solvating cations, but not nearly

so effective in solvating anions. • because anions are only poorly solvated, they

participate readily in SN reactions, and

• nucleophilicity parallels basicity: F- > Cl- > Br- > I-

88 NucleophilicityNucleophilicity Polar protic solvents (e.g., water, methanol)• anions are highly solvated by hydrogen bonding with

the solvent• the more concentrated the negative charge of the

anion, the more tightly it is held in a solvent shell• the nucleophile must be at least partially removed

from its solvent shell to participate in SN reactions

• because F- is most tightly solvated and I- the least, nucleophilicity is I- > Br- > Cl- > F-

88 NucleophilicityNucleophilicity Generalizations• within a period, nucleophilicity increases from left to

right; that is, it increases with basicity

Increasing NucleophilicityPeriod

Period 2

Period 3

F- < OH- < NH2- < CH3

Cl- < SH- < PH2-

88 NucleophilicityNucleophilicity Generalizations• in a series of reagents with the same nucleophilic

atom, anionic reagents are stronger nucleophiles than neutral reagents

Increasing Nucleophilicity

ROH < RO-

H2 O < OH-

NH3 < NH2-

RSH < RS-

88 NucleophilicityNucleophilicity• when comparing groups of reagents in which the

nucleophilic atom is the same, the stronger the base, the greater the nucleophilicity

Hydroxide ion

Alkoxide ion

Carboxylate ion

Decreasing Acidity

Increasing Nucleophilicity

Conjugate acid

Nucleophile

RCOOH4-5

88 StereochemistryStereochemistry For an SN1 reaction at a stereocenter, the

product is almost completely racemized

-H+ CH3O C

C OCH3

R EnantiomerS Enantiomer

R EnantiomerA racemic mixture

Planar carbocation (achiral)

88 StereochemistryStereochemistry For SN1 reactions at a stereocenter• examples of complete racemization have been

observed, but• partial racemization with a slight excess of inversion

is more common

Approach of the nucleophile from this side is less hindered

Approach of the nucleophile fromthis side is partially blocked by chloride ion, which remains associated with the carbocation as anion pair

88 StereochemistryStereochemistry For SN2 reactions at a stereocenter, there is

inversion of configuration at the stereocenter Experiment of Hughes and Ingold

Iacetone

SN2-131+

I I131

2-Iodooctane

88 Hughes-Ingold ExptHughes-Ingold Expt• the reaction is 2nd order, therefore, SN2

• the rate of racemization of enantiomerically pure 2-iodooctane is twice the rate of incorporation of I-131

I:- CH

I :I-SN2

acetone+131 131

(S)-2-Iodooctane (R)-2-Iodooctane

88 Structure of RXStructure of RX SN1 reactions governed by electronic factors; • the relative stabilities of carbocation intermediates

SN2 reactions governed by steric factors; • the relative ease of approach of the nucleophile to the

site of reactionGoverned byelectronic factors

Governed bysteric factors

R3CX R2CHX RCH2X CH3X

Access to the site of reaction

(3°) (methyl)(2°) (1°)

Carbocation stability

88 Effect of Effect of -Branching-Branching

1.2 x 10-51.2 x 10-3Relative Rate

Alkyl Bromide

-Branches 0 1 2 3

1.0 4.1 10x -1

Br Br Br Brβ β β β

88 Effect of Effect of -Branching-Branching

CH3CH2Br

freeaccess

Bromoethane(Ethyl bromide)

CH3CCH2Br

1-Bromo-2,2-dimethylpropane(Neopentyl bromide)

blockedaccess

88 Allylic HalidesAllylic Halides Allylic cations are stabilized by resonance

delocalization of the positive charge• a 1° allylic cation is about as stable as a 2° alkyl cation

Allyl cation(a hybrid of two equivalent contributing

structures)

CH2=CH-CH2 CH2-CH=CH2

88 Allylic CationsAllylic Cations 2° & 3° allylic cations are even more stable

As also are benzylic cations

A 2° allylic carbocationA 3° allylic carbocation

CH2=CH-CH-CH3 CH2=CH-C-CH3

C6H5-CH2+CH2

Benzyl cation(a benzylic carbocation)

The benzyl cation is also writtenin this abbreviated form

88 The Leaving GroupThe Leaving Group The more stable the anion, the better the leaving

ability• the most stable anions are the conjugate bases of

strong acids

I- > Br- > Cl- >> F- > CH3CO- > HO- > CH3O- > NH2-

OReactivity as a leaving group

Stability of anion; strength of conjugate acid

88 The Solvent - SThe Solvent - SNN22 The most common type of SN2 reaction involves

a negative Nu and a negative leaving group

• the weaker the solvation of Nu, the less the energy required to remove it from its solvation shell and the greater the rate of SN2

Transition state

negatively chargednucleophile

negatively chargedleaving group

negative charge dispersed in the transition state

XC XCNu CNuNu:-δ−δ−

88 The Solvent - SThe Solvent - SNN22Br N3

CH3C N

CH3OHH2 O

(CH3 )2S=O

(CH3 )2NCHO

N3 Br-

SolventType

polar aprotic

polar protic

k(methanol)

k(solvent)

Solvent

+solvent

88 The Solvent - SThe Solvent - SNN11 SN1 reactions involve creation and separation of

unlike charge in the transition state of the rate-determining step

Rate depends on the ability of the solvent to keep these charges separated and to solvate both the anion and the cation

Polar protic solvents (formic acid, water, methanol) are the most effective solvents for SN1 reactions

88 The Solvent - SThe Solvent - SNN11

80% water: 20% ethanol

40% water: 60% ethanolethanol

Solvent

solvolysis++

k(solvent)

k(ethanol)

14,000

100,000

CH3 CH3

CH3CCl ROH CH3COR HCl

88 Rearrangements in SRearrangements in SNN11 Rearrangements are common in SN1 reactions if

the initial carbocation can rearrange to a more stable one

2-Methoxy-2-phenylbutane2-Chloro-3-phenylbutane

++ CH3OH + Cl -

CH3OHCH3OH

88 Rearrangements in SRearrangements in SNN11 Mechanism of a carbocation rearrangement

Cl ++ Cl

A 2° carbocation

A 3° benzylic carbocation

An oxonium ion

88 Summary of SSummary of SNN1 & S1 & SNN22Type of Alkyl Halide

CH3XMethyl

RCH2XPrimary

R2CHXSecondary

R3CXTertiary

SN2 SN1

Substitution at a stereocenter

SN2 is favored. SN1 does not occur. The methyl cationis so unstable, it is never observedin solution.

SN1 rarely occurs. Primary cations are so unstable, that they are rarely observed in solution.

SN1 is favored in protic solvents withpoor nucleophiles. Carbocation rearrangements may occur.

SN2 is favored in aproticsolvents with goodnucleophiles.

SN2 does not occur becauseof steric hindrance aroundthe reaction center.

SN1 is favored because of the ease of formation of tertiary carbocations.

Inversion of configuration.The nucleophile attacksthe stereocenter from theside opposite the leavinggroup.

Racemization is favored. The carbocationintermediate is planar, and attack of thenucleophile occurs with equalprobability from either side. There is often some net inversion of configuration.

SN2 is favored.

88 Neighboring GroupsNeighboring Groups In an SN1 reaction, departure of the leaving

group is not assisted by Nu In an SN2 reaction, departure of the leaving

group is assisted by Nu These two types are distinguished by their order

of reaction; SN2 reactions are 2nd order, and SN1 reactions are 1st order

But some reactions are 1st order and yet involve two successive SN2 reactions

88 Mustard GasesMustard Gases Mustard gases • contain either S-C-C-X or N-C-C-X

• what is unusual about the mustard gases is that they undergo hydrolysis so rapidly in water, a very poor nucleophile

Cl 2H2O HOS

OH 2HCl+ +

Bis(2-chloroethyl)sulfide(a sulfur mustard gas)

Bis(2-chloroethyl)methylamine(a nitrogen mustard gas)

Cl ClN

88 Mustard GasesMustard Gases• the reason is neighboring group participation by the

adjacent heteroatom

• proton transfer to solvent completes the reaction

ClA cyclic

sulfonium ion

an internal SN2 reaction

a secondSN2 reaction

fast +

88 SSNN1/S1/SNN2 Problems2 Problems Problem 1:Problem 1: predict the mechanism for this

reaction, and the stereochemistry of each product

Problem 2:Problem 2: predict the mechanism of this reaction

+ CH3OH/ H2O HClR enantiomer

+ +Na+CN- Na+Br -DMSOBr CN

88 SSNN1/S1/SNN2 Problems2 Problems Problem 3:Problem 3: predict the mechanism of this

reaction and the configuration of product

Problem 4:Problem 4: predict the mechanism of this reaction

+ +acetoneCH3S-Na+ Na+Br-

R enantiomer

Br SCH3

+acetic acidBr OCCH3CH3COH + HBr

88 SSNN1/S1/SNN2 Problems2 Problems Problem 5:Problem 5: predict the mechanism of this

reaction

toluene(CH3 )3P P(CH3)3 Br-Br

88 Phase-Transfer CatalysisPhase-Transfer Catalysis A substance that transfers ions from an aqueous

phase to an organic phase An effective phase-transfer catalyst must have

sufficient• hydrophilic character to dissolve in water and form an

ion pair with the ion to be transported• hydrophobic character to dissolve in the organic

phase and transport the ion into it

The following salt is an effective phase-transfer catalysts for the transport of anions

(CH3CH2CH2CH2)4N+Cl-

Tetrabutylammonium chloride(Bu4N

+Cl- )

88 Phase-Transfer CatalysisPhase-Transfer Catalysis

88 -Elimination-Elimination -Elimination:-Elimination: a reaction in which a small

molecule, such as HCl, HI, or HOH, is split out or eliminated from a larger molecule

CH3CH2O-Na+

CH3CH2OH

CH3CH2OH Na+X -

A haloalkane Base

An alkene

88 -Elimination-Elimination Zaitsev rule:Zaitsev rule: the major product of a -elimination

is the more stable (the more highly substituted) alkene

2-Methyl-2-butene (major product)

CH3CH2O-Na+

CH3CH2OH 2-Bromo-2-methylbutane

2-Methyl-1-butene

1-Methyl-cyclopentene

(major product)

CH3O-Na+

1-Bromo-1-methyl-cyclopentane

Methylene-cyclopentane

88 -Elimination-Elimination There are two limiting mechanisms for -

elimination reactions E1 mechanism:E1 mechanism: at one extreme, breaking of the R-X

bond is complete before reaction with base to break the C-H bond• only R-X is involved in the rate-determining step

E2 mechanism:E2 mechanism: at the other extreme, breaking of the R-X and C-H bonds is concerted• both R-X and base are involved in the rate-determining step

88 E1 MechanismE1 Mechanism• ionization of C-X gives a carbocation intermediate

• proton transfer from the carbocation intermediate to the base (in this case, the solvent) gives the alkene

CH3-C-CH3

+(A carbocation intermediate)

H3CH-CH2-C-CH3

H3CH CH2=C-CH3

CH3fast

88 E1 MechanismE1 Mechanism

88 E2 MechanismE2 Mechanism

88 Kinetics of E1 and E2Kinetics of E1 and E2 E1 is a 1st order reaction; 1st order in RX and

zero order is base

E2 is a 2nd order reaction; 1st order in base and 1st order in RX

Rate == k[RX]

d[RX]dt

Rate = = k[RX][Base]

88 Regioselectivity of E1/E2Regioselectivity of E1/E2 E1: major product is the more stable alkene E2: with strong base, the major product is the

more stable alkene• double bond character is highly developed in the

transition state• thus, the transition state of lowest energy is that

leading to the most stable (the most highly substituted) alkene

88 Stereoselectivity of E2Stereoselectivity of E2 E2 is most favorable (lowest activation energy)

when H and X are oriented anti and coplanar

CH3O:-

-H and -X are anti and coplanar(dihedral angle 180°)

88 Stereochemistry of E2Stereochemistry of E2 Consider E2 of these stereoisomers

3-Isopropyl-cyclohexene

(major product)

cis-1-Chloro-2-isopropyl-

cyclohexane

CH3O-Na+

trans-1-Chloro-2-isopropyl-

cyclohexane

Cl3-Isopropyl-cyclohexene

88 Stereochemistry of E2Stereochemistry of E2• in the more stable chair of the cis isomer, the larger

isopropyl is equatorial and chlorine is axial

CH3O:-

CH3OH :Cl-

6 + +E2

88 Stereochemistry of E2Stereochemistry of E2• in the more stable chair of the trans isomer, there is

no H anti and coplanar with X, but there is one in the less stable chair

More stable chair (no H is anti and coplanar to Cl)

Less stable chair(H on carbon 6 is

anti and coplanar to Cl)

88 Stereochemistry of E2Stereochemistry of E2• it is only the less stable chair conformation of this

isomer that can undergo an E2 reaction

HCH3O:

CH3OH :Cl-

88 Stereochemistry of E2Stereochemistry of E2Problem:Problem: account for the fact that E2 reaction of the meso-dibromide gives only the E-alkene

meso-1,2-Dibromo-1,2-diphenylethane

(E)-1-Bromo-1,2-diphenylethylene

C6H5C6H5

CH3O- Na+

CH3OHC6H5CH-CHC6H5

88 Summary of E2 vs E1Summary of E2 vs E1

Alkyl halide E1 E2

Primary

Secondary

Tertiary

E1 does not occur.Primary carbocations areso unstable, they are neverobserved in solution.

E2 is favored.

Main reaction with strong bases such as OH- and OR-.

Main reaction with weak bases such as H2O, ROH.

Main reaction with strong bases such as OH- and OR-.

Main reaction with weak bases such as H2O, ROH.

88 SSNN vs E vs E Many nucleophiles are also strong bases (OH-

and RO-) and SN and E reactions often compete

The ratio of SN/E products depends on the relative rates of the two reactions

nucleophilicsubstitution

-eliminationC CH X + Nu-

C CH Nu X-+

C C H-Nu+ X-+

88 SSNN vs E vs E

RCH2 X

SN1 and E1 reactions of primary halides are never observed.

bases such as I- and CH3COO-.

SN2 SN1 reactions of methyl halides are never observed.The methyl cation is so unstable that it is never formed in solution.

E2 The main reaction with strong, bulky bases such as potassium tert-butoxide.

Primary cations are never formed in solution, and, therefore,

Halide Reaction Comments

Methyl

Primary The main reaction with good nucleophiles/weak

88 SSNN vs E (cont’d) vs E (cont’d)

The main reaction with bases/nucleophiles where the

pKa of the conjugate acid is 11 or less, as for exampleI- and CH3COO-.

R2 CHX

Main reaction with strong bases such as HO- and RO-.

Main reactions with poor nucleophiles/weak bases.

The main reaction with bases/nucleophiles where

SN2 reactions of tertiary halides are never observed

SN1/E1

Secondary

Tertiary

because of the extreme crowding around the 3° carbon.

SN1/E1 Common in reactions with weak nucleophiles in polarprotic solvents, such as water, methanol, and ethanol.

pKa of the conjugate acid is 11 or greater, as for exampleOH- and CH3CH2O

88 Prob 8.9Prob 8.9 Draw a structural formula for the most stable

carbocation of each molecular formula.

(a) (b)

(c) (d)

C4H9+ C3H7

C3H7O+C8H15+

88 Prob 8.11Prob 8.11 From each pair, select the stronger nucleophile.

(a) (b)

orH2 O OH-

CH3SH CH3S-

CH3COO-

oror(e)

I- in DMSO

CH3OCH3 CH3SCH3Cl-

I- in methanol

88 Prob 8.12Prob 8.12 Draw a structural formula for the product of each SN2

reaction.

acetone+

ethanol+

(a) CH3CH2 ONaCH3CH2 CH2Cl

(CH3 )3N CH3I

ethanol+

+ acetone

(c) CH2Br

H3C Cl

CH3SNa

88 Prob 8.12 (cont’d)Prob 8.12 (cont’d)

ethanol

CH3CH2 CH2Cl

CH3Cdiethylether

acetone+(h)

(g) +ethanol

NHO CH3(CH2)6 CH2Cl

CH3CH2 CH2Br NaCN

88 Prob 8.14Prob 8.14 Account for the fact that the rate of this reaction is 1000

times faster in DMSO than it is in ethanol.

2KCN 2KCl1,3-Dichloropropane

+ +Propanedinitrile

Cl Cl NC CN

88 Prob 8.15Prob 8.15 The following reaction involves two successive SN2

reactions. Propose a structural formula for the product.

1-Aminoadamantane Methyl 2,4-dibromobutanoate

NH2R3 N

C15H23NO2Br

88 Prob 8.16Prob 8.16 Which member of each pair shows the greater rate of SN2

reaction with KI in acetone?

(a) or

(b) or Br

(c) or

88 Prob 8.17Prob 8.17 Which member of each pair gives the greater rate of SN2

reaction with KN3 in acetone?

88 Prob 8.19Prob 8.19 Limiting yourself to a single 1,2-shift, suggest a

structural formula for a more stable carbocation.

(f)(e)

88 Prob 8.21Prob 8.21 Draw a structural formula for the product of each SN1

reaction.

CH3CH2 OH

ClCH3OH(b)

+ methanol

ethanol

S enantiomer

(d) methanol+

acetic acid(c) CH3COH

CH3OHBr

88 Prob 8.24Prob 8.24 From each pair, select the compound that undergoes SN1

solvolysis in ethanol more rapidly.

(a) or or(b)Cl Cl Cl Br

(d)or(c) orClCl ClCl

(f) oror(e)Br Br

88 Prob 8.25Prob 8.25 Account for the following relative rates on solvolysis

under SN1 conditions.

Relative rate ofsolvolysis (SN1) 10.2 109

ClOCl O Cl

88 Prob 8.26Prob 8.26 Explain why the following compound is very unreactive

under SN1 conditions.

1-Iodobicyclo[2.2.2]octane

88 Prob 8.27Prob 8.27 Propose a synthesis for each compound from a

haloalkane and a nucleophile.

(b) (c)(a) CNO

(d) SH

O (g)SH

88 Prob 8.29Prob 8.29 Propose a mechanism for the formation of each product.

CH3CH=CHCH2ClH2O

CH3CH=CHCH2OH + CH3CHCH=CH2

OH1-Chloro-2-butene

2-Buten-1-ol 3-Buten-2-ol

88 Prob 8.30Prob 8.30 Propose a mechanism for the formation of this product.

If the configuration of the starting material is S, what is the configuration of the product?

88 Prob 8.31Prob 8.31 Propose a mechanism for the formation of the products

of this solvolysis reaction.

BrCH3CH2OH

+OCH2CH3 + HBr

OSO2Ar

CH3COHH

OCCH3O

88 Prob 8.33Prob 8.33 Which compound in each set undergoes more rapid

solvolysis when refluxed in ethanol?Br Br

or(b) BrCl

ClClor(d)

88 Prob 8.34Prob 8.34 Account for these relative rates of solvolysis in acetic

(CH3)3CBr

Br Br Br

1 10-2 10-7 10-12

88 Prob 8.35Prob 8.35 On SN1 solvolysis in acetic acid, (1) reacts 1011 times

faster than (2). Furthermore, solvolysis of (1) occurs with complete retention of configuration. Draw structural formulas for the products of each solvolysis and account for the difference in rates.

OSO2Ar OSO2Ar

(1) (2)

88 Prob 8.36Prob 8.36 Draw structural formulas for the alkene(s) formed on

treatment of each compound with sodium ethoxide in ethanol. Assume reaction by an E2 mechanism.

(c)(b)

(a) ClCl

(f)(e) ClBr

88 Prob 8.37Prob 8.37 Draw structural for all chloroalkanes that undergo

dehydrohalogenation when treated with KOH to give each alkene as the major product.

(c)(b)(a)

(e)(d)

88 Prob 8.38Prob 8.38 On treatment with sodium ethoxide in ethanol, each

compound gives 3,4-dimethyl-3-hexene. One compound gives an E alkene, the other gives a Z alkene. Which compound gives which alkene?

C CEt Et

88 Prob 8.39Prob 8.39 On treatment with sodium ethoxide in ethanol, this

compound gives a single stereoisomer. Predict whether the alkene has the E or Z configuration.

BrH C6H5

H C6H5 CH3CH2O-Na+

CH3CH2OHC6H5CH=CC6H5

1-Bromo-1,2-diphenylpropane

1,2-Diphenylpropene

88 Prob 8.40Prob 8.40 Elimination of HBr from 2-bromonorbornane gives only

2-norbornene. Account for the regiospecificity of this elimination reaction.

2-Bromonorbornane 2-Norbornene 1-Norbornene

88 Prob 8.43Prob 8.43 Arrange these haloalkanes in order of increasing ratio of

E2 to SN2 products on reaction of each with sodium ethoxide in ethanol.

CH3CCH2 CH3

CH3CH2Br

CH3CHCH2CH2Br(d)

CH3CHCH2Br(b)

CH3 CH3

88 Prob 8.44Prob 8.44 Draw a structural formula for the major product of each

reaction and specify the most likely mechanism for its formation.

80°(b) +

+ methanol(a)

CH3CCH2 CH3 NaOHH2 O

(c) +DMSO

(R)-CH3CHCH2CH3 CH3CO-Na

+ methanol

R enantiomer

+ acetone

CH3O-Na

(g) + ethanol

R enantiomer

CH3CHCH2CH3

CH3CH2 ONa

CH2=CHCH2Cl

formic acid

(3)(2)

CH3CH2 OH

the transisomer

the cisisomer

88 Prob 8.46Prob 8.46 Show how to bring about each conversion.

(a) (b)Cl Br

Cl OH(c) (d) Br

(e) (f)Br Br

88 Prob 8.46 (cont’d)Prob 8.46 (cont’d) Show how to bring about each conversion.

(g) (h)Br

88 Prob 8.47Prob 8.47 Which reaction gives the tert-butyl ether in good yield?

What is the product of the other reaction?CH3

CH3CO- K

CH2O-K+

CH3CCl

CH3COCH2

CH3COCH2 + KCl(a) +

DMSO + KCl

88 Prob 8.48Prob 8.48 Each ether can, in principle, by synthesized by a

Williamson ether synthesis forming bond (1) or bond (2). Which combination gives the better yield?

(2)(1)

(a) (b)

(1) (2)O-CH2 CH3

CH3 O-CCH3

CH2=CHCH2-O-CH2CCH3

88 Prob 8.49Prob 8.49 Propose a mechanism for this reaction.

ClCH2CH2OHNa2CO3, H2O

H2C CH2

88 Prob 8.50Prob 8.50 Each compound can be synthesized by an SN2 reaction.

Propose a combination of haloalkane and nucleophile that will give each product.

CH3OCH3(a) (b)CH3SH

CH3CH2CH2PH2(c) (d)CH3CH2CN

CH3SCH2C(CH3)3(e) (f)(CH3)3NH+ Cl -

C6H5COCH2C6H5(g) (h)(R)-CH3CHCH2CH2CH3

CH2=CHCH2OCH(CH3)2(i) (j)CH2=CHCH2OCH2CH=CH2

(k) (l)N O O

Nucleophilic Nucleophilic Substitution Substitution

and and -Elimination-Elimination

End Chapter 8End Chapter 8

8-1 8 Organic Chemistry William H. Brown & Christopher S. Foote.

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