chapter 6 ionic reactions--- nucleophilic substitution and elimination reactions of alkylhalides (...
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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÷£©