Chapter 6 Ionic Reactions---Nucleophilic substitution and elimination reactions of alkylhalides ( 卤代烃的亲核取代反应和消除反应）

Because halogen atoms are more electronegative than carbon, the carbon-halogen bond of alkyl halides is polarized; the carbon atom bears a

partial positive charge, the halogen atom a partial negative charge.

X X = F , Cl, Br, I+

_

Table 6.1 Carbon-halogen bond lengths

Bond Bond Length (A)

CH3F 1.39CH3Cl 1.78CH3Br 1.93CH3I 2.14

Vinyl halides ( 卤代乙烯） or phenyl halides （卤代苯基）

X

X

A vinylic halide A phenyl halide or ary halide

X = F, Cl, Br, ICompounds in which a halogen atoms is bonded to an sp2-hybridized carbon

Carbon-halogen bond is difficult to break

6.2 Physical properties of organic halides• Very low solubilities in water

• They are miscible with each other and with other relatively nonpolar solvents

• CH2Cl, CHCl3 and CCl4 are often used as solvents for nonpolar

• CHCl3 and CCl4 have a cumulative toxicity and are carcinogenic.

• Polyfluoroalkanes have low boiling points,• (Hexafluoroethane boils at –79o)

( 致癌物 ( 质 ) 的 )

6.3 Reaction Mechanisms

Mechanism of the reaction—The events that are postulated to take place at the molecular

level as reactants become products

If the reaction takes place in more than one step, then what are these steps, and what kinds of intermediates intervene between

reactants and products?

6.3A Homolysis and heterolysis of covalent bonds ( 共价键的均裂和异

裂）

A : B

A + B

A: + B

A+ + :B-

. .

- +

Homolysis£¨¾ùÁÑ£©

Heterolysis£¨ÒìÁÑ£©

Covent bond may break in three possible ways:

6.3B Reactive intermediates in organic chemistry

Organic reactions that take place in more than one step involve the formation of an intermediate----one that results from either homolysis or heterolysis of a bond. Homolysis of a bond to carbon leads to an intermediate known

as a carbon radical (free radical)

C : Zhomolysis

C . + Z.¾ùÁÑ

Carbon radical or free radical

Heterolysis of a bond can lead either to a trivalent carbon cation

or carbon anion

C : Zheterolysis

C + Z

ÒìÁÑ

Carbocation(or carbenium ion)£¨Ì¼ÑôÀë×Ó£©

C + Z

Carbanion£¨Ì¼ÒõÀë×Ó£©

+ : -

:- +

Carbanions are usually strong bases and strong nucleophiles. Nucleophiles are lewis basesthey are electron-pair donors. Carbon radicals, carbocations, and carbonions are usually highly reactive species.

6.3C Ionic reactions ( 离子反应） and radical reactions （自由基反

应 or 游离基反应）• In ionic reactions the bonds of the reacti

ng molecules undergo heterolysis;

• In radical reactions, they undergo homolysis (in detail in chapter 7)

• In this chapter we concern ourselves only with ionic reactions.

6.4 Nucleophilic substitution reactions ( 亲核取代反应）

:Nu - + R X R Nu + : -X

NucleophileÇ׺ËÊÔ¼Á

Alkyl halide(Substrate µ×Îï )

Product Halide ion

HO- + CH3 Cl CH3OH + Cl-

CH3O- + CH3 Br CH3OCH3 + Br-

I- + CH3CH2CH2 Cl CH3CH2CH2I + Cl-

Nucleophilic substitution reactions （亲核取代反应）

• The carbon-halogen bond of the substrate undergoes heterolysis, and the unshared pair of the nucleophile is used to form a new bond to the carbon atom

:Nu - + R X R Nu + : -X

NucleophileÇ׺ËÊÔ¼Á

Alkyl halide(Substrate µ×Îï )

Product Halide ion

Leaving group

What is it reaction mechanism?

Nu:- + RX Nu R X Nu R + X -1. SN2 reaction (Ë«·Ö×ÓÇ׺ËÈ¡´ú·´Ó¦)(one step)

- -transition state¹ý¶É̬

2. SN1 reaction (µ¥·Ö×ÓÇ׺ËÈ¡´ú·´Ó¦)(two steps)

RX R+ + X-Limiting step

carbocation

R++Nu:- Nu Rfast

6.5 Nucleophiles [nju:kliefail]n.[ 化 ] 亲核试剂

A nucleophile is a reagent that seeks a positive center

X + -

This is the positive center that the nucleophile seeks

The electronegative halogen polarize the C---X bond

Specific Example

(CH3)3C Cl + H2O (CH3)3C +

(CH3)3C + H2O+ O(CH3)3C H

H

+

(CH3)3C OH + H+

(SN1)

Nucleophile+ Cl-

tert-butyloxonium

tert-butyl alcohol

:

Slow

fast

HO- + CH3CH2 Br CH3CH2OH + Br- (SN2)

Nucleophile ethanol Leaving group

6.5 A Leaving groups ( 离去基团）

To be a good leaving group the substituent must be able to leave as a relatively stable, wea

kly basic molecule or ion

Nu:- + R L R Nu + :L-

:NH3 + BrCH3 CH3 NH3 + Br -+

CH3 OH2

+CH3OH +

Leaving group

CH3OCH3 + H2OH

+

6.7 Kinetics of a nucleophilic substitution reaction: An SN2 reaction

Table 6.3 Rate study of reaction of CH3Cl with OH at 60o-No initial [CH3Cl] initial [OH- ] initial rate (mol L-1 s-1)

1 0.0010 1.0 4.9 X 10-7

2 0.0020 1.0 9.8 X 10-7

3 0.0010 2.0 9.8 X 10-7

4 0.0020 2.0 19.6 X 10-7

:OH+ClCH3 CH3 OH + Cl -- 60o

H2O(SN2 reaction)

Rate [ CH3Cl] [OH-] Rate [ CH3Cl] [OH-]k

Second order reaction(¶þ¼¶·´Ó¦£©---- (SN2 reaction)

6.8 A mechanism for the SN2 reaction

（ SN2 的反应机理）

N + L CNu

Transition state £¨¹ý¶É̬£©

+ L:

Nu--C formsL--- C breaks

at the same time1.

2. Inversion of configuration (¹¹Ðη ת£©

SN2 Reaction

ClH

H

H

HO-H H

H

HO Cl-

-HO

H

H

H

transition state

+ Cl-

Generally speaking; Primary alkyl halides is a SN2 reaction

X R CH2X

伯卤代烃一般按 SN2 历程进行 .

6.9 Transition state theory: Free-energy diagrams (SN2)

ClH

H

H

HO- HO Cl-

-HO

H

H

H

transition state

+ Cl-

HO- + CH3Cl

CH3OH + Cl-Go

G

HO CH3 ClE

Reaction coordinate£¨·´Ó¦½ø³Ì£©

- -

G = H - ST

H H

H

Fig 6.6 A potential energy diagram for the reaction of methyl chloride

with hydroxide ion at 60 oC

HO- + CH3Cl

CH3OH + Cl-Go

G

HO CH3 ClE

Reaction coordinate£¨·´Ó¦½ø³Ì£©

- -

24.5

-24 kcal mol-1

#

G#less than 20 kcal mol-1, it will take place readily at room temp, if it is greater than 20 kcal mol-1, heating will be required

Fig 6.3 A free-energy diagram for a hypothetical reaction with a positive free- energy

change

X - + Y

GoG

X Y ZE

Reaction coordinate£¨·´Ó¦½ø³Ì£©

- -

Z

X Y + Z-

6.10 The stereochemistry of SN2 reactionsIn SN2 reaction the nucleophile attacks from the backside, that is, from the side directly opposite the leaving group. This mode of attack causes a change in the configuration of the carbon atom that is the object of nucleophilic attack

ClH

H

H

HO- HO Cl-

-HO

H

H

H

transition state

+ Cl-H

HH

An inversion of configuration £¨¹¹Ðη ת£©

+

SN2 reaction------a configuration inversion

An inversion of configuration £¨¹¹Ðη ת£©

Cl

HH

H3C + OH-H3C

H

H

OH

+ Cl-

cis-1-Chloro-3-methylcyclopentane trans-3-Methylcyclopentanol

SN2

H

H3C

H

Cl

OH

-

-

Leaving group departsfrom this side

Nucleophile attacks from this side

SN2 reactions always lead to inversion of configuration

An inversion of configuration £¨¹¹Ðη ת£©

+ NaOH- + NaCl

(R)-(-)-2-Bromooctane (S)-(+)-2-Octanol

SN2BrH

C6H13

CH3

HHO

C6H13

CH3

R S

[ = - 34.25o

Enantiomeric purity = 100%

E, P = 100 %

(CH3)3C Cl + OH-acetone (±ûͪ£©

H2O(CH3)3C OH + Cl-

Rate [(CH3)3CCl]

Rate [(CH3)3CCl]k first oder reaction (Ò»¼¶·´Ó¦£©

6.11 The reaction of tert-butyl chloride with hydroxide ion: An S

N1 reaction

6.12 A mechanism for the SN

1 reaction (multistep reactions)

Step 1 H3C C

CH3

CH3

ClSlow

(rate-determining step)H3C C

CH3

CH3

+ + Cl-

carbocation

Step 2H3C C

CH3

CH3

+

carbocation

+ :OH2H3C C

CH3

CH3

OH2+

tert-butyloxonium ion

fast

H3C C

CH3

CH3

O+

Step 3 + H2Ofast

H3C C

CH3

CH3

OH + H3O

tert-butyl alcohol

H

H

+

Fig 6.8 A free-energy diagram for the SN1 reaction of tert-butyl chlo

ride with water

(CH3)3CCl + H2O

(CH3)3C Cl-+

(CH3)3C++ H2O

(CH3)3C OH2

+

(CH3)3C OH2

+

(CH3)3COH

(CH3)3CO H

H

+G

G

Step1 Step2 Step 3

Reaction coordinate

E

The important transition state for the SN1 reaction is the transition state of the rate-determining step. In it the carbon-chlorine bond of tert-butyl chloride is largely broken and ions are beginning to develop. The solvent (water) stabilizes these developing ions

by solvation

Step 1 H3C C

CH3

CH3

ClSlow

(rate-determining step)H3C C

CH3

CH3

+ + Cl-

carbocation

-+

6.13 Carbocations ( 碳阳离子 )

CH3+

(CH3)3C+

H

H

H planar

sp2S

Vacant p orbital

methyl cation tert-butyl cation

H3C

H3C

CH3 planar

sp2sp3

Vacant p orbital

+ +

6.13A The structure of carbocations

The central carbon atom in a carbocation is electron deficient; it has only six electrons in its outside energy level.

6.13B The relative stabilities of carbocations

CH3+RCH2

+R2CH+R3C+Stability

3o 2o 1o

Tertiary C+ Secondary C+ Primary C+ methyl cation

(most stable) (least stable)

(ÊåÕý̼Àë×Ó ÖÙÕý̼Àë×Ó ²®Õý̼Àë×Ó ¼×»ùÕý̼Àë×Ó£©

Tertiary carbocations are the most stable, secondary carbocations are the stable, and the primary carbocations are not stable.

Fig 6.10 How a methyl group helps stabilize the positive charg

e of a carbocation.(CH3)3C+

H

H planarsp2 s

Vacant p orbital

ethyl cationtert-butyl cation

planar

sp2sp3

Vacant p orbital++

(most stable)

H

H

H

CH3CH2+

H

H

H

H

H

H

H

H

H

( less stable)

9 C H

( p) conjugate

3 C H

( p) conjugate

Because of sigama-p conjugating. As a result, the delocalization of charge and the order of stability of the carbocations as

follows

CH3+

2o 1o

Tertiary C+ Secondary C+ Primary C+ methyl cation

(least stable)

(ÊåÕý̼Àë×Ó ÖÙÕý̼Àë×Ó ²®Õý̼Àë×Ó ¼×»ùÕý̼Àë×Ó£©

C+

CH3

CH3

CH3

C+

CH3

CH3

H

C+CH3

H

H

3o

The relative stability of carbocations(Most stable)

6.14 The stereochemistry of SN1 reactions

CH3

C

H3C CH3

+

CH3

C

H3C CH3

Cl

H2O

H2O: :OH2

Frontside attackbackside attack

CH3

OH2H3C

H3CCH3

H2OCH3

CH3+

+

6.14A Reactions that involve racemization ( 涉及外消旋化反

应）

CH2CH3CH3

C

H3CH2C CH3

+

CH2CH2CH3

C

H3C CH2CH3

Br

H2O

H2O: :OH2

frontside attackbackside attackCH2CH2CH3

OHH3CH2C

H3C

CH2CH2CH3

HOCH2CH3

CH3

slow

fast fast

(3R)-3-Methyl-hexanol (3S)-3-Methyl-hexanol

Enantiomers ( a recemic form)

( SN1)

(3R)-3-Bromo-3-methylhexane

6.14B Solvolysis ( 溶剂化作用）

CH3

C

H3C CH3

Br H2O

CH3

H2OCH3

CH3+

slow

SN1+

CH3

HOCH3

CH3

-HBr

fast

+ CH3OH

tert-butyl alcohol

(CH3)3CCl (CH3)3COCH3

tert-butyl methyl etherMethanolysis ¼×´¼ÈܼÁ»¯

+ HCOOH(CH3)3CCl (CH3)3COOCH

tert-butyl formate (¼×ËáÊ嶡´¼õ¥£©

The SN1 reaction of an alkyl halide with water is an example of solvolysis

Examples of Solvolysis

In the last example the solvent is formic acid ( 甲酸，蚁酸 HCOOH)

and the following take place

slow

SN1

+

(CH3)3CClstep 1 (CH3)3C+ + Cl-

(CH3)3C+

O

CHO..

H

fast(CH3)3C O

H

O

H+

(CH3)3C O

H

O

H+ (CH3)3COCH

Otert-butyl formate

fast

step 2

step 3 -H+

6.15 Factors affecting the rates of SN1 and SN2 reactions( 影响 SN

1 and SN2 的反应速率因素）

• 1. The structure of the substrate

• 2. The concentration and reactivity of the nucleophile (for bimolecular reactions only)

• The effect of the solvent.

• The nature of the leaving group.

6.15A The effect of the structure of the substarate ( 底物结构的

影响）

Table 6.4 Relative rates of reactions of alkyl halides in SN2 reactions

Methyl

Substituent Compound relative rate

CH3X 30

1o CH3CH2X 1

2o

3o

(CH3)2CHX 0.02

Neopentyl ÐÂÎì»ù (CH3)3CCH2X 0.00001

(CH3)3X 0

General order of reactivity in SN2 reaction

Alkyl halides; methyl primary secondary tertiary

The important factor behind this order of reactivity is a steric effect ( 立体效应对 SN2 反应影响大）

H

XH X

H

H

XH

H

XH

H

H

H

H H

H

H

H

H

H

H

H

H

H

H

H

H

HH

H

H

H

H

HH

X

HH

H

Nu:-

Steric effects in the SN2 reaction

Steric hindrance (Á¢ÌåÕÏ° £©

SN1 reaction. The primary factor that determines the reactivity of organic substrates in an SN1 reaction is the relative stability of the c

arbocation that is formed

Tertiary C+

(ÊåÕý̼Àë×Ó£©

C+

CH3

CH3

CH3

3o

CH3

CH3C

CH3

Cl

Reactant

H2O+ Cl-

CH3

CH3C

CH3

Cl

Transition stateGo is positive

Stabilized cation

SN1 reaction

CH3+

2o 1o

Tertiary C+ Secondary C+ Primary C+ methyl cation

(least stable)

(ÊåÕý̼Àë×Ó ÖÙÕý̼Àë×Ó ²®Õý̼Àë×Ó ¼×»ùÕý̼Àë×Ó£©

C+

CH3

CH3

CH3

C+

CH3

CH3

H

C+CH3

H

H

3o

The relative stability of carbocations(Most stable)

An SN1 MechanismFor a methyl, primary, or secondary halide to react by an SN1 mechanism it would have to ionize to form a methyl, primary, or secondary carbocation. These carbocations, however, are much higher in energy than a tertiary carbocation, and the transition states leading to these carbocations are higher in

energy. The activation energy for an SN1 reaction of a simple methyl, primary or secondary halide, consequently, is so large (the re

action is so slow) .

6.15B The effect of the concentration and strength of the nucleoph

ile( 亲核试剂浓度和强度效应）Since the nucleophile does not participate in the

rate-determining step of an SN1 reaction, the rates of SN1 reactions are unaffected by either the concentration or the identity of the nucleophile. The rates of SN2 reactions, however, depend on both the concentration and the identity of the attacking nucleophile. We saw that how increasing the concentration of the nucleophile increases th

e rate of an SN2 reaction.

We describe nucleophiles as being strong or weak. When we do this we are really describing their relative reactives in SN2 reactions. A strong nucleophile is one that reacts rapidly with a given substrate. A weak nucleophile is one that reacts slowly with the same substrate under the same reaction conditions.

CH3O- + CH3Irapid

CH3OCH3 + I-

methoxide ion (¼×Ñõ»ùÀë×Ó£©

CH3OH + CH3Ivery slow

CH3OCH3 + I-

H

+

CH3O- CH3OHNucleophile(Ç׺ËÐÔ£©;

The relative strengths of nucleophiles can be correlated with two struc

tural features;• 1. A negatively charged nucleophile is always

a stronger nucleophile than its conjugate acid in a SN2 reaction.

CH3O- CH3OHNucleophile(Ç׺ËÐÔ£©;

HO- HOH

CH3COO- CH3COOH

RO- ROH

RS- RSH

(¸ºÀë×Ó) (ÖÐÐÔ·Ö×Ó)

2. In a group of nucleophiles in which the nucleophilic atom is the same, nucleophilicities parallel basic

ities

This is also their order of basicity. An alkoxide ion (RO-) is a slightly stronger base than a hydroxide ion (HO-), a hydroxide ion is a much stronger base than a carboxylate ion (RCOO-), and so on

Nucleophile(SN2Ç׺ËÐÔ´ÎÐò£©;

HO- HOHRCOO-RO- ROH

6.15C Solvent effects on SN2 reactions: Polar protic and aprotic solvents ( 极性质子溶剂和非质子溶剂）Protic solvent-----has a hydrogen atom attached to an atom of a strongly electronegative el

ement (oxygen). ( water, alcohol etc.)

Aprotic solvent------Aprotic solvents are those solvents whose molecules do not have a hydrogen atom that is attached to an atom of a strongly electronegative element.(Benzene, the

alkanes, etc.)

Relative Nucleophilicity in protic Solvents

I- Br- Cl- F-

SH- CN- I- OH- N3- Br- CH3CO2

- Cl- F- H2O

Why?

Because molecules of protic solvents can form hydrogen bonds to nucleophiles in the following way:

O

H

H

O

H H

H

O

HH

O

H

X-

Molecules of the protic solvent,water, solvate a halide ion by forminghydrogen bonds to it (a small nucleophile ½ÏСµÄÇ׺ËÄÜÁ¦£©

Polar Aprotic Solvents （极性非质子溶剂）

Aprotic solvents are those solvents whose molecules do not have a hydrogen atom that is attached to an atom of a strongly electronegative element.

O

S

O

C

O

H N

CH3

CH3 H3C CH3 N CH3

O

P

N(CH3)2(H3C)2N

N(CH3)2

N,N-Dimethylformamicde (DMF)

Dimethyl sulfoxide (DMSO)

CH3H3C

Dimethylacetamide (DMA)

Hexamethylphosphorictriamide (HMPT)

Na+ X- (F- Cl- Br- I-)Nucleophiles (Ç׺ËÐÔ£©

N,N-¶þ¼×»ù¼×õ£°· ¶þ¼×ÑÆí¿ N,N-¶þ¼×»ùÒÒõ£°· ÈýÁù¼×°·»ùì¢õ£°·

Problem 6.6 Classify the following solvents as being protic or apr

otic:• Formic acid HCOOH• Acetone CH3COCH3

• Acetonitrile CH3CN• Formamide HCONH2

• Sulfur dioxide SO2

• Ammonia NH3

• Trimethylamine N(CH3)2

• Ethylene glycol HOCH2CH2OH

Problem 6.7 Would you expect the reaction of propyl bromide

with sodium cyanide (NaCN), that is, to occur faster in DMF or in ethanol? Explain your answer.

CH3CH2CH2Br + NaCN CH3CH2CH2CN + NaBr

Explain; the reaction is an SN2 reaction. the nucleophile (CN-) will be relatively unencumbered ( 不受妨碍的 ) by solvent molecules, therefore, i

t will be more reactive than in ethanol.

S

O

H3C CH3

Dimethyl sulfoxide (DMSO)¶þ¼×ÑÆí¿

S

O

H3C CH3

Dimethyl sulfoxide (DMSO)¶þ¼×ÑÆí¿

Na+

O

S

O

S

CH3 H3C CH3H3C

CN-

naked anion

6.15D Solvent effects on SN1 reactions. The ionizing ability of

the solvent

Because of its ability to solvate cations and anions so effectively, the use of a polar protic solvent will greatly increase the rate of ionization of an alkyl halide in any SN1

reaction.

SN1 reaction in polar protic solvent

(CH3)3C Cl [(CH3)3C Cl] (CH3)3C+ + Cl-

reactant Transition stateseparated charge aredeveloping

(CH3)3C+ + Nu- (CH3)3CNufast

slow

ions solvated by molecules of the protic solvent

Water is the most effective solvent for promoting ionization, but most organic compounds do not dissolve appreciably in water. They usually dissolve, however, in

alcohols, and quite often mixed solvents (methanol-water) are used.

Table 6.5 Dielectric constants of common solvents

( 普通溶剂的介电常数）Solvent formula dielectric constant

water H2O 80

Formic acid HCOOH 59

Dimethyl sulfoxide (DMSO) HCONH2 49N,N-Dimethylformamide (DMF) HCON(CH3)2 37

acetonitrile CH3CN 36Ethanol CH3CH2OH 24Acetone CH3COCH3 21Acetic acid CH3COOH 6Polarity

6.15E The nature of the leaving group ( 离去基团的本性）

The Best leaving groups are those that become the most stable ions after they depart. Since most leaving groups leave as a negative ion, the best leaving groups are those ions that

stabilize a negative charge most effectively. Because weak bases do this best, the best

leaving groups are weak bases.

In either an SN1 or SN2 reaction the leaving group begins to

acquire a negative charge as the transition state is reached

SN1 Reaction (rate-limiting step)

X X + + X-

Transition state

SN2 Reaction

H

H

H

XNu-

H H

H

H

H

Nu

H

Nu X

Transition state

An iodide ion is the best leaving group and a fluoride ion is the

poorest:

I- Br- Cl- F-

Other weak bases that are good leaving groups

-O S

O

O

R -O S

O

O

OR -O S

O

O

CH3

An alkanesulfonate ion an alkyl sulfate ion p-toluenesulfonate ion

Íé»ù»ÇËáÑÎ Íé»ùÁòËáÑÎ ¶Ô¼×±½»ÇËáÑÎ

The trifluoromethanesulfonate ion (CF3SO3

-, commonyl called the triflate ion ( 三氟甲磺酸离

子）CF3SO3

-

Trifate ion (a ‘ super’ leaving group)

ROH (OH- rarely act as leaving groups).

X- R OH

H

+R X + H2O

Leaving group is aweak base (H2O)

Very powerful bases such as hydride ions (H-) and alkanide ions

(R-) never act as leaving groups.

Nu- CH3CH2 H CH3CH2 Nu + H-

Nu- CH3 CH3 CH3 Nu + CH3-

These are not leaving group

6.15F Summary: SN1 versus SN2

CH3X CH3CH2X CH3CHXCH3 (CH3)3CX

SN2 SN1rate

²®Â±´úÍé(CH3X)Ö÷Òª°´SN2Àú³Ì½øÐУ¬Êå±´úÍé(R3CX)Ö÷Òª°´ SN1Àú³Ì½øÐУ¬ÖÙ±´úÍé(CH3)2CHX¼È¿É°´SN2 ÓÖ¿ÉÒÔ°´ SN1Àú³Ì½øÐÐ)

6.16 Organic synthesis 有机合成 : Functional group transformations using SN2 reactions

The process of making one compound from another is called synthesis. SN2 reactions are highly useful in organic synthesis because they enable us to convert one functional group into another--- a process that is called a functionalgroup transformation or a functional group interconversion （官能

团转换）

SN2 reactions in organic synthesis

R X

(R = Me, Io, 2o)

(X = Cl, Br, or I)

OH- R OH

R OR'

R SH

R SR'

R CN

R OOCR

R NR3X-

R N3

R'O-

SH-

R'S-

CN-

R'COO-

R'3N

N3-

+

Alcohol ´¼

ether ÃÑ

Thiol Áò́¼

thioether ÁòÃÑ

nitrile ëæ

ester õ¥

quanternary ammonium halide±́ú ¼¾°· ÑÎalkyl azide µþµª»¯ÄÆ

Alkyl chlorides and bromides are also easily converted to alkyl iodides by nucleophilic substituti

on reactions.

R Cl

R Br

I-

R I (+Cl- or Br-)

If we have available ®-2-bromobutane, we can carry out

the following synthesis;

N C- +

H3C

H3CH2C

H Br

( R )-2-Bromobutane

SN2

( Inversion )

CH3

CH2CH3

HNC

( S )-2-methylbutanenitrile( S )-2-¼×»ù¶¡ëæ

+ Br-

( R ) ( S )Inversion · ת

Problem 6.11 Starting with ( S)-2-brombutane, outline syntheses

of each of the following compounds:

(a) (R)-CH3CHCH2CH3

OCH2CH3

( C ) (R)- CH3CHCH2CH3

SH

( b ) (R )-CH3CHCH2CH3

OOCCH3

( d ) ( R )-CH3CHCH2CH3

SCH3

Answer (a)

+

H3CH2C

H3C

H Br

( S )-2-Bromobutane

SN2

( Inversion )

CH2CH3

CH3

HH3CH2CO

( R)-2-ethyloxylbutane( R)-2-ÒÒÑõ»ù¶¡Íé

+ Br-

( S ) ( R )Inversion · ת

CH3CH2O-

Na+

6.16A The unreactivity of vinylic ( 乙烯式） and phenyl halides

（卤代苯）A halogen atom attached to one carbon atom of a double bond are called vinylic halides; those that have a halogen atom attached to a benzene ring are called phenyl

halides

X (F, Cl, Br, I)

X

a vinylic halide phenyl halide

Both compounds are unreactive in SN1 or SN2Â±Ô ×ÓÁ¬ÔÚË«¼üÉϷdz£²»»îÆÃ

6.17 Elimination reactions of alkylhalides ( 卤代烃的消除反应））Another characteristic of alkyl halides is that they undergo elimination reactions. In an elimination reaction the fragments of some molecule

(YZ) are removed (eliminated) from adjacent atoms of the reactant. This elimination leads to t

he introduction of a multiple bond ( 重键） :

C C

Y Z

Elimination

+ Y-Z

6.17A Dehydrohalogenation ( 脱卤化氢）

A widely used method for synthesizing alkenes is the elimination of HX from

adjacent atoms of an alkyl halide. Heating the alkyl halide with a strong base causes

the reaction to take place.H

CH3C C

H

H

Br H

H3C

H

H

HElimination

+ H-BrCH3CH2ONa

CH3CH2OH

CH3

CH3C C

H

H

Br H

H3C

H3C

H

HElimination

+ H-BrCH3CH2ONa

CH3CH2OH

79%

91%

Dehydrohalogenation ( 脱卤化氢）

H

C C

H

C

X H

C

(Beta Elimination)+ H-X

Base (¼î£©

X = Cl, Br, I

(1,2-Ïû³ý£©

CH3CH2ClKOH

CH3CH2OHCH2=CH2 + KCl + H2O

CH3CH2CHCH3KOH

CH3CH2OHCH3CH=CHCH3 + CH3CH2CH=CH2

Cl 70% 30%

6.17B Bases used in dehydrohalogenation

Various strong bases have been used for dehydrohalogenations.

1. 2R OH + Na 2 R ONa + H2

Alcohol Sodium alkoxide £¨´¼ÄÆ£©

3. 2 HOH + 2 Na 2 HONa + H2

Sodium hydroxide

4.

Ethyl Alcohol Sodium ethoxide £¨ÒÒ´¼ÄÆ£©

2. 2 CH3CH2OH + Na 2 CH3CH2ONa + H2

H3C

CH3

CH3

OH + K H3C

CH3

CH3

OK + H2

Potassium tert-butoxide£¨Êå-¶¡´¼¼Ø£©

tert-Butyl alcohol £¨Êå-¶¡´¼£©(excess)

6.17C Mechanisms of Dehydrohalogenations ( 脱卤化氢机理）

One mechanism is a bimolecular mechanism called the E2 reaction; the other is a unimolecular mechanism calle

d the E1 reaction

6.18 The E2 reaction

CH3CHCH3

Br

CH3CH2ONa

CH3CH2OHCH3CH=CH2

isopropyl bromide Propene

Rate [CH3CHBrCH3] [CH3CH2O-]

Rate = k[CH3CHBrCH3][CH3CH2O-]

Second order reaction¶þ¼¶·´Ó¦

The E2 reaction

CH3CHCH3

Br

CH3CH2ONa

CH3CH2OHCH3CH=CH2

isopropyl bromide Propene

H

Br

H

H

H

CH3

CH3CH2O-

H

Br

H

H

H

CH3

CH3CH2O-

transition state for an E2 reaction

H

H

H

CH3

+ C2H5OH + Br-

6.19 The E1 reaction

CH3

H3C

CH3

Cl

tert-Butyl chloride

80% C2H5OH

20% H2O, 25oC

SN1CH3

H3C

CH3

OH

CH3

H3C

CH3

OC2H5+

tert-butyl alcohol tert-Butyl ethyl ether

E1H2C C

CH3

CH3

2-Methylpropene 17%

83%

The E1 reaction mechanismCH3

H3C

CH3

Cl

tert-Butyl chloride

80% C2H5OH

20% H2O, 25oC

H2C CCH3

CH3

2-Methylpropene 17%

Step 1

CH3

C +H3C

CH3

+ Cl-

Step 2

CH3

C +H3C

CH3

H2O SN1

CH3

CH3C

CH3

OH

C2H5OH

CH3

CH3C

CH3

OC2H5

CH3

C +H3C

CH2 H

E1

SN1

Slow

fast

6.20 Substitution (SN1 and SN2) versus Elimination (E2 and E1)

Since eliminations occur best by an E2 path when carried out with a high concentration of a strong base ( and thus a high concentration of a strong nucleophile), substitution reactions by an SN2 path often compete with the elimination reaction. When the nucleophile attacks the carbon atom bearing the leaving g

roup, substitution results.

SN2 compete with E2, SN1 compete with E1.

The SN2 compete with E2

H C

C X

+ Nu H + X-

H C

CNu

b

SN2

a

E2

Nu-

b

a

+ Cl-

CH3CH2ONa + CH3CH2BrCH3CH2OH

55oCCH3CH2OCH2CH3 + CH2=CH2

SN2 (90%) E2 (10%)

CH3CH2ONa + CH3CHCH3

CH3CH2OH

55oC SN2 (21%) E2 (79%)

Br

(CH3)CHOC2H5 + CH2=CHCH3

CH3CH2ONa + CH3CCH3

CH3CH2OH

25oC(CH3)3COCH2CH3 +CH2=C(CH3)2

SN1 (9%) E2 (91%)

CH3CH2ONa + CH3CCH3

CH3CH2OH

55oC E1+E2 (100%)

Br

CH2=C(CH3)2 + CH3CH2OH

Br

CH3

CH3

majorminor

CH3OH

65oC

E2 (1%)

minorCH3O- + CH3(CH2)15CH2CH2 Br

CH3(CH2)15CH=CH2

CH3(CH2)15CH2CH2OCH3

SN2 (99%)Major

(CH3)3COH

55oC

E2 (85%)

majorCO- + CH3(CH2)15CH2CH2 Br

CH3(CH2)15CH=CH2

CH3(CH2)15CH2CH2OC(CH3)3

SN2 (15%)Minor

CH3

CH3

H3C

Sample problem; In each case give the mechanism (SN1, SN2,

E1, or E2)

(a ) CH3CH2CH2Br + CH3O-CH3OH

50oCCH3CH2CH2OCH3 + CH3CH=CH2

SN2 major (Ö÷£© E2 minor(¸¶£©

( b) CH3CH2CH2Br + (CH3)3CO-(CH3)3COH

50oCCH3CH2CH2OC(CH3)3 + CH3CH=CH2

SN2 minor £¨ ¸¶ £© E2 major £¨ Ö÷£©