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1
SUBSTITUTION AND ELIMINATION
REACTIONS OF ALKYL HALIDES
SN1, SN2, E1 & E2 REACTIONS
2
Reactions of Alkyl Halides (R-X): [SN1, SN2, E1, & E2 reactions]
The -carbon in an alkyl halide is electrophilic (electron accepting) for either or both of two reasons…
a) the C to X (F, Cl, Br) bond is polar making carbon +
(4.0 – 2.5) = 1.5
(3.0 – 2.5) = 0.5
(2.8 – 2.5) = 0.3
FH3C
EN (F-C) =
EN (Cl-C) =
EN (Br-C) =
EN (I-C) = (2.5 – 2.5) = 0.0
b) X (Cl, Br, I) is a leaving group
pKb = 23 pKb = 22 pKb = 21 pKb = 11 pKb = -1.7
I -
Br -
Cl -
F -
HO -
30,000 10,000 200 1 0
decreasing basicity, increasing stability
increasing leaving ability
BrH3C
ClH3C
IH3C
The best leaving groups are the weakest bases.
The poorest leaving groups are the strongest bases.
3
Reactions of Alkyl Halides (R-X): [SN1, SN2, E1, & E2 reactions]
When a nucleophile (electron donor, e.g., OH-) reacts with an alkyl halide, the halogen leaves as a halide
There are two competing reactions of alkyl halides with nucleophiles….
1) substitution
2) elimination
C C
H
X
Nu:-+ C C
H
Nu
+ X-
+ C C
H
X
Nu:- C C + X- + Nu H
The Nu:- replaces the halogen on the -carbon.
The Nu:- removes an H+ from a -carbon & the halogen leaves forming an alkene.
BrR.... :
..
.. ::BrNu:
4
There are two kinds of substitution reactions, called SN1 and SN2. As well as two kinds of elimination reactions, called E1 and E2.
Let’s study SN2 reactions first. SN2 stands for Substitution, Nucleophilic, bimolecular. Another word for bimolecular is ‘2nd order’.
Bimolecular (or 2nd order) means that the rate of an SN2 reaction is directly proportional to the molar concentration of two reacting molecules, the alkyl halide ‘substrate’ and the nucleophile:
Rate = k [RX] [Nu:-] (This is a rate equation and k is a constant). The mechanism of an SN2 reaction is the one shown on slide #2:
2nd Order Nucleophilic Substitution Reactions, i.e., SN2 reactions
C C
H
X
Nu:-+ C C
H
Nu
+ X-
Note that the nucleophile must hit the back side of the -carbon. The nucleophile to C bond forms as the C to X bond breaks. No C+ intermediate forms. An example is shown on the next slide.
5
2nd Order Nucleophilic Substitution Reactions, i.e., SN2 reactions
The rate of an SN2 reaction depends upon 4 factors: 1. The nature of the substrate (the alkyl halide)2. The power of the nucleophile3. The ability of the leaving group to leave4. The nature of the solvent
1. Consider the nature of the substrate:
Unhindered alkyl halides, those in which the back side of the -carbon is not blocked, will react fastest in SN2 reactions, that is:
Me° >> 1° >> 2° >> 3°
While a methyl halides reacts quickly in SN2 reactions, a 3° does not react. The back side of an -carbon in a 3° alkyl halide is completely blocked.
O H....: C Br
..
.. :
H
H
H+
transition state
C Br.... :
H H
H
OH.... +
..
.. :Br:C
H
HH
OH....
6
Me°>> 1° >> 2° >> 3°
Effect of nature of substrate on rate of SN2 reactions:
CH3 Br CH3 CH2 BrCH Br
CH3
CH3
C Br
CH3
CH3
CH3
t-butyl bromidemethyl bromide ethyl bromide isopropyl bromide
Back side of -C of a methyl halide
is unhindered.
Back side of -C of a 1° alkyl halide is slightly hindered.
Back side of -C of a 2° alkyl halide is mostly hindered.
Back side of -C of a 3° alkyl halide is
completely blocked.
decreasing rate of SN2 reactions
SPACE FILLING MODELS SHOW ACTUAL SHAPES AND RELATIVE SIZES
7
The -carbon in vinyl and aryl halides, as in 3° carbocations, is completely hindered and these alkyl halides do not undergo SN2 reactions.
Effect of the nucleophile on rate of SN2 reactions:
CH2 CH Br
H2C CH Br
Br
Br
vinyl bromide bromobenzene
The overlapping p-orbitals that form the -bonds in vinyl and aryl halides completely block the access of a nucleophile to the back side of the -carbon.
Nu:-Nu:-
8
2. Consider the power of the nucleophile: The better the nucleophile, the faster the rate of SN2 reactions.
The table below show the relative power or various nucleophiles. The best nucleophiles are the best electron donors.
Effect of nature substrate on rate of SN2 reactions:
Reactivity Nu:- Relative Reactivity
very weak HSO4-, H2PO4
-, RCOOH < 0.01
weak ROH 1
HOH, NO3- 100
fair F- 500
Cl-, RCOO- 20 103
NH3, CH3SCH3 300 103
good N3-, Br- 600 103
OH-, CH3O- 2 106
very good CN-, HS-, RS-, (CH3)3P:, NH2- ,RMgX, I-, H- > 100 106
incr
easi
ng
9
3. Consider the nature of the leaving group:
The leaving group usually has a negative charge Groups which best stabilize a negative charge are the best leaving groups,
i.e., the weakest bases are stable as anions and are the best leaving groups.
Weak bases are readily identified. They have high pKb values.
Iodine (-I) is a good leaving group because iodide (I-) is non basic.
The hydroxyl group (-OH) is a poor leaving group because hydroxide (OH-) is a strong base.
Effect of nature of the leaving group on rate of SN2 reactions:
pKb = 23 pKb = 22 pKb = 21 pKb = 11 pKb = -1.7 pKb = -2 pKb = -21
I- Br - Cl- F- HO- RO- H2N-
30,000 10,000 200 1 0 0 0
Increasing leaving ability
10
4. Consider the nature of the solvent:
There are 3 classes of organic solvents: Protic solvents, which contain –OH or –NH2 groups. Protic solvents
slow down SN2 reactions.
Polar aprotic solvents like acetone, which contain strong dipoles but no –OH or –NH2 groups. Polar aprotic solvents speed up SN2 reactions.
Non polar solvents, e.g., hydrocarbons. SN2 reactions are relatively slow
in non polar solvents.
Effect of the solvent on rate of SN2 reactions:
Protic solvents (e.g., H2O, MeOH, EtOH, CH3COOH, etc.) cluster around the
Nu:- (solvate it) and lower its energy (stabilize it) and reduce its reactivity via H-bonding.
X:-
H
H
HH OR
OR
OR
RO
+
+
+
+
-
-
-
-
A solvated anion (Nu:-) has reduced nucleophilicity,reduced reactivity and increased stability
A solvated nucleophile has difficulty hitting the -carbon.
11
Polar Aprotic Solvents solvate the cation counterion of the nucleophile but not the nucleophile.
Examples include acetonitrile (CH3CN), acetone (CH3COCH3),
dimethylformamide (DMF) [(CH3)2NC=OH], dimethyl sulfoxide, DMSO
[(CH3)2SO], hexamethylphosphoramide, HMPA {[(CH3)2N]3PO} and
dimethylacetamide (DMA).
Effect of the solvent on rate of SN2 reactions:
DMF
C
O
H N
CH3
CH3
C
O
N
CH3
CH3
DMSO
S
O
CH3H3C
HMPA
[(CH3)2N]3P O H3C
DMA
: : : : : :.. .. ..
....
..
CH3 C N :
acetonitrileC
O
CH3H3C
: :
acetone
Polar aprotic solvents solvate metal cationsleaving the anion counterion (Nu: -) bare andthus more reactive
CH3C O
O: :....
:_
Na+
Na+
N C CH3
N C CH3
N C CH3NCH3C
-
-
-
-
+
+
+
+
+ CH3C O
O: :....
:_
CH3CN::
..
..
:
12
Non polar solvents (benzene, carbon tetrachloride, hexane, etc.) do not solvate or stabilize nucleophiles.
SN2 reactions are relatively slow in non polar solvents similar to that
in protic solvents.
Effect of the solvent on rate of SN2 reactions:
benzene
C
Cl
ClCl Cl
carbontetrachloride
CH3CH2CH2CH2CH2CH3
n-hexane
13
1st Order Nucleophilic Substitution Reactions, i.e., SN1 reactions
C
CH3
H3C
CH3
Br + Na+ I- C
CH3
H3C
CH3
I + Na+ Br-3°rapid
3 alkyl halides are essentially inert to substitution by the SN2 mechanism
because of steric hindrance at the back side of the -carbon.
Despite this, 3 alkyl halides do undergo nucleophilic substitution reactions quite rapidly , but by a different mechanism, i.e., the SN1 mechanism.
SN1 = Substitution, Nucleophilic, 1st order (unimolecular).
SN1 reactions obey 1st order kinetics, i.e., Rate = k[RX].
The rate depends upon the concentration of only 1 reactant, the alkyl halide-not the nucleophile
The order of reactivity of substrates for SN1 reactions is the reverse of SN2
3 > 2 > 1 > vinyl > phenyl > Me°
R3C-Br R2HC-Br RH2C-Br CH2=CH-Br -Br H3C-Br
increasing rate of SN1 reactions
14
The mechanism of an SN1 reaction occurs in 2 steps:
Reaction Steps …
1. the slower, rate-limiting dissociation of the alkyl halide forming a C+ intermediate
2. a rapid nucleophilic attack on the C+
Mechanism of SN1 reactions
C
CH3
H3C
CH3
3°Br....
: + Na+ Br-C
CH3
H3C
CH3
I....
:1.
Br --C
CH3
H3C
CH3
+
3° C+
rapid
Na+ I -....: :
2.
Note that the nucleophile is not involved in the slower, rate-limiting step.
15
The rate of an SN1 reaction depends upon 3 factors:
1. The nature of the substrate (the alkyl halide)
2. The ability of the leaving group to leave
3. The nature of the solvent
The rate is independent of the power of the nucleophile.
1. Consider the nature of the substrate:
Highly substituted alkyl halides (substrates) form a more stable C+.
The Rate of SN1 reactions
C
H
H
H +C
CH3
H
H +C
CH3
H
H3C +C
CH3
CH3
H3C +
tertiary 3º
secondary 2º
primary 1º
methyl
morestable
lessstable
> > >
increasing rate of SN1 reactions
16
Alkyl groups are weak electron donors. They stabilize carbocations by donating electron density by induction
(through bonds)
They stabilize carbocations by hyperconjugation (by partial overlap of the alkyl C-to-H bonds with the empty p-orbital of the carbocation).
Stability of Carbocations
C
CH3
CH3
H3C +
Inductive effects: Alkyl groups donate (shift) electrondensity through sigma bonds toelectron deficient atoms.This stabilizes the carbocation.
vacant p orbitalof a carbocation
sp2
hybridizedcarbocation
Csp3-Hssigma bondorbital
overlap (hyperconjugation)
HYPERCONJUGATION
+C C
..H
H
H
17
Allyl and benzyl halides also react quickly by SN1 reactions because their carbocations are unusually stable due to their resonance forms which delocalize charge over an extended system
Stability of Carbocations
H2C CH +CH2 CH2HCH2C+
1º allyl carbocation
H2C CH +CHR CHRHCH2C+
2º allyl carbocation
2º benzylic1º benzylic
C
H
R
+C
H
H
+C
H
HC
H
H
C
H
H
+ +
+
18
Relative Stability of All Types of Carbocations
2º allylic
> >
3º allylic
> > >
3º C+
CCH3
CH3
CH3
+
CH2 CH CHR+CH2 CH CR2
+
C R2
+
3º benzylic
C HR+
2º benzylic
1º allylic
CCH3
CH3
H
+
2º C+
CH2 CH CH2
+
C H2
+
1º benzylic
1º C+
CCH3
H
H
++
+
CH
H
H
methyl C
+
phenyl
> CH2 CH+
+vinyl C
Increasing C+ stability and rate of SN1 reaction
Note that 1° allylic and 1° benzylic C+’s are about as stable as 2°alkyl C+’s.
Note that 2° allylic and 2° benzylic C+’s are about as stable as 3° alkyl C+’s.
Note that 3° allylic and 3° benzlic C+’s are more stable than 3° alkyl C+’s
Note that phenyl and vinyl C+’s are unstable. Phenyl and vinyl halides do not usually react by SN1 or SN2 reactions
19
2. Consider the nature of the leaving group:The nature of the leaving group has the same effect on both SN1 and SN2
reactions.
The better the leaving group, the faster a C+ can form and hence the faster will be the SN1 reaction.
The leaving group usually has a negative charge Groups which best stabilize a negative charge are the best leaving groups,
i.e., the weakest bases are stable as anions and are the best leaving groups.
Weak bases are readily identified. They have high pKb values.
Effect of nature of the leaving group on rate of SN1 reactions:
pKb = 23 pKb = 22 pKb = 21 pKb = 11 pKb = -1.7 pKb = -2 pKb = -21
I- Br - Cl- F- HO- RO- H2N-
30,000 10,000 200 1 0 0 0
Increasing leaving ability
Iodine (-I) is a good leaving group because iodide (I-) is non basic.
The hydroxyl group (-OH) is a poor leaving group because hydroxide (OH-) is a strong base.
20
3. Consider the nature of the solvent: For SN1 reactions, the solvent affects the rate only if it influences the
stability of the charged transition state, i.e., the C+. The Nu:- is not involved in the rate determining step so solvent effects on the Nu:- do not affect the rate of SN1 reactions.
Polar solvents, both protic and aprotic, will solvate and stabilize the charged transition state (C+ intermediate), lowering the activation energy and accelerating SN1 reactions.
Nonpolar solvents do not lower the activation energy and thus make SN1
reactions relatively slower
Effect of the solvent on rate of SN1 reactions:
reaction rate increases with polarity of solvent
The relative rates of an SN1 reaction due to solvent effects are given
(CH3)3C-Cl + ROH (CH3)3C-OR + HCl
H2O 20% EtOH (aq) 40% EtOH (aq) EtOH
100,000 14,000 100 1
21
Solvent polarity is usually expressed by the “dielectric constant”, , which is a measure of the ability of a solvent to act as an electric insulator.
Polar solvents are good electric insulators because their dipoles surround and associate with charged species.
Dielectric constants of some common solvents are given in the following table
Effect of the solvent on rate of SN1 reactions:
name dielectric constant name dielectric constant
aprotic solvents protic solvents
hexane 1.9 acetic acid 6.2
benzene 2.3 acetone 20.7
diethyl ether 4.3 ethanol 24.3
chloroform 4.8 methanol 33.6
HMPA 30 formic acid 58.0
DMF 38 water 80.4
DMSO 48
22
Consider the nature of the Nucleophile: Recall again that the nature of the nucleophile has no effect on the rate of
SN1 reactions because the slowest (rate-determining) step of an SN1 reaction is the dissociation of the leaving group and formation of the carbocation.
All carbocations are very good electrophiles (electron acceptors) and even weak nucleophiles, like H2O and methanol, will react quickly with them.
The two SN1 reactions will proceed at essentially the same rate since the only difference is the nucleophile.
Effect of the nucleophile on rate of SN1 reactions:
C
CH3
H3C
CH3
Br + Na+ I- C
CH3
H3C
CH3
I + Na+ Br-3°
C
CH3
H3C
CH3
Br + C
CH3
H3C
CH3
F + K+ Br-3° K+ F-
23
We have seen that alkyl halides may react with basic nucleophiles such as NaOH via substitution reactions.
Also recall our study of the preparation of alkenes. When a 2° or 3° alkyl halide is treated with a strong base such as NaOH, dehydrohalogenation occurs producing an alkene – an elimination (E2) reaction.
bromocyclohexane + KOH cyclohexene (80 % yield)
Substitution and elimination reactions are often in competition. We shall consider the determining factors after studying the mechanisms of elimination.
Elimination Reactions, E1 and E2:
BrKOH in ethanol + KBr + H2O
-HBr
O H....: C Br
..
.. :
H
H
H+
transition state
C Br.... :
H H
H
OH.... +
..
.. :Br:C
H
HH
OH....
24
+ C C
H
Br
OH- C C + Br- + HO H
There are 2 kinds of elimination reactions, E1 and E2.
E2 = Elimination, Bimolecular (2nd order). Rate = k [RX] [Nu:-] E2 reactions occur when a 2° or 3° alkyl halide is treated with a strong
base such as OH-, OR-, NH2-, H-, etc.
E2 Reaction Mechanism
The Nu:- removes an H+ from a -carbon & the halogen leaves forming an alkene.
All strong bases, like OH-, are good nucleophiles. In 2° and 3° alkyl halides the -carbon in the alkyl halide is hindered. In such cases, a strong base will ‘abstract’ (remove) a hydrogen ion (H+) from a -carbon, before it hits the -carbon. Thus strong bases cause elimination (E2) in 2° and 3° alkyl halides and cause substitution (SN2) in unhindered methyl° and 1° alkyl halides.
25
In E2 reactions, the Base to H bond formation, the C to H bond breaking, the C to C bond formation, and the C to Br bond breaking all occur simultaneously. No carbocation intermediate forms.
Reactions in which several steps occur simultaneously are called ‘concerted’ reactions.
Zaitsev’s Rule:
Recall that in elimination of HX from alkenes, the more highly substituted (more stable) alkene product predominates.
E2 Reaction Mechanism
B:-
C
H
CR
R
X
B
R
R
H
C CRR
R
R X
+
-
C C
R R
RR
+ B H + X-
CH3CH2CHCH3
Br CH3CH2O-Na+
EtOH
CH3CH CHCH3 + CH3CH2CH CH2
2-butene 1-butene
major product( > 80%)
minor product( < 20%)
26
E2 reactions, do not always follow Zaitsev’s rule.
E2 eliminations occur with anti-periplanar geometry, i.e., periplanar means that all 4 reacting atoms - H, C, C, & X - all lie in the same plane. Anti means that H and X (the eliminated atoms) are on opposite sides of the molecules.
Look at the mechanism again and note the opposite side & same plane orientation of the mechanism:
E2 Reactions are ‘antiperiplanar’
B:-
C
H
CR
R
X
B
R
R
H
C CRR
R
R X
+
-
C C
R R
RR
+ B H + X-
27
Antiperiplanar E2 Reactions in Cyclic Alkyl Halides
When E2 reactions occur in open chain alkyl halides, the Zaitsev product is usually the major product. Single bonds can rotate to the proper alignment to allow the antiperiplanar elimination.
In cyclic structures, however, single bonds cannot rotate. We need to be mindful of the stereochemistry in cyclic alkyl halides undergoing E2 reactions.
See the following example.
Trans –1-chloro-2-methylcyclopentane undergoes E2 elimination with NaOH. Draw and name the major product.
H
Cl
H3C
H
H
H
Na+ OH-
E2
H3C
HH
H
Non Zaitsev productis major product.
H
HH3C
H
Little or no Zaitsev (more stable) product is formed.
HOH
NaCl+
+
3-methylcyclopentene 1-methylcyclopentene
28
Just as SN2 reactions are analogous to E2 reactions, so SN1 reactions have
an analog, E1 reaction. E1 = Elimination, unimolecular (1st order); Rate = k [RX]
E1 eliminations, like SN1 substitutions, begin with unimolecular
dissociation, but the dissociation is followed by loss of a proton from the -carbon (attached to the C+) rather than by substitution.
E1 & SN1 normally occur in competition, whenever an alkyl halide is treated
in a protic solvent with a nonbasic, poor nucleophile.
Note: The best E1 substrates are also the best SN1 substrates, and
mixtures of products are usually obtained.
E1 Reactions
CCH3
CH3
CH3
Br
slow
B:-
CCH3
CH3
C H
H
H
+ rapidC C
CH3 H
CH3H
+ B H + Br-
Br--
29
As with E2 reactions, E1 reactions also produce the more highly substituted alkene (Zaitsev’s rule). However, unlike E2 reactions where no C+ is produced, C+ rearrangements can occur in E1 reactions.
e.g., t-butyl chloride + H2O (in EtOH) at 65 C t-butanol + 2-methylpropene
In most unimolecular reactions, SN1 is favored over E1, especially at low temperature. Such reactions with mixed products are not often used in synthetic chemistry.
If the E1 product is desired, it is better to use a strong base and force the E2 reaction.
Note that increasing the strength of the nucleophile favors SN1 over E1. Can you postulate an explanation?
E1 Reactions
CCH3
CH3
CH3
Cl +
H2O, EtOH
65ºCCCH3
CH3
CH3
OH C C
CH3 H
CH3H
64%36%
SN1product
E1product
30
1. Non basic, good nucleophiles, like Br- and I- will cause substitution not elimination. In 3° substrates, only SN1 is possible. In Me° and 1° substrates, SN2 is faster. For 2° substrates, the mechanism of substitution depends upon the solvent.
2. Strong bases, like OH- and OR-, are also good nucleophiles. Substitution and elimination compete. In 3° and 2° alkyl halides, E2 is faster. In 1° and Me° alkyl halides, SN2 occurs.
3. Weakly basic, weak nucleophiles, like H2O, EtOH, CH3COOH, etc., cannot react unless a C+ forms. This only occurs with 2° or 3° substrates. Once the C+ forms, both SN1 and E1 occur in competition. The substitution product is usually predominant.
4. High temperatures increase the yield of elimination product over substitution product. (G = H –TS) Elimination produces more products than substitution, hence creates greater entropy (disorder).
5. Polar solvents, both protic and aprotic, like H2O and CH3CN, respectively, favor unimolecular reactions (SN1 and E1) by stabilizing the C+ intermediate. Polar aprotic solvents enhance bimolecular reactions (SN2 and E2) by activating the nucleophile.
Predicting Reaction Mechanisms
31
Predicting Reaction Mechanisms
alkyl halide
(substrate)
good Nu-
nonbasic e.g., bromide
Br-
good Nu-
strong base e.g., ethoxide
C2H5O
-
good Nu-
strong bulky base e.g., t-butoxide
(CH3)3CO
-
very poor Nu-
nonbasic e.g., acetic acid
CH3COOH
Me
1°
2
3
SN1, E1
SN2
E2
SN2
SN2
SN1
SN2
SN2
E2
E2
SN2
E2 (SN2)
E2
no reaction
no reaction
SN1, E1
Strong bulky bases like t-butoxide are hindered. They have difficulty hitting the -carbon in a 1° alkyl halide. As a result, they favor E2 over SN2 products.
32
The nucleophiles in the table on slide 30 are extremes. Some nucleophiles have basicity and nucleophilicity in between these extremes. The reaction mechanisms that they will predominate can be interpolated with good success.
Predict the predominant reaction mechanisms the following table.
Predicting Reaction Mechanisms
alkyl halide
(substrate)
……….. Nu- ………. .base
e.g., cyanide CN-
pkb = ………
……….. Nu- ……….. base
e.g., alkyl sulfide RS-, also HS-
pkb = ………
……….. Nu-
……….. base e.g., carboxylate
RCOO-
pkb = ………
alcohol (substrate)
HI
HBr
HCl
Me Me
1 1
2 2
3 3
4.7
v. gd.
moderate
SN2
SN2
SN2
E2
6.0 / 7.0
moderate
v. gd.
SN2
SN2
E2
SN2
9
weak
fair
SN2
SN2
E2
E2
SN2
SN2
SN1
SN1
HCl, HBr and HI are assumed to be in aqueous solution, a protic solvent.
33
Recall the preparation of long alkynes.
1. A terminal alkyne (pKa = 25) is deprotonated with a very strong base…
R-C C-H + NaNH2 R-C C:- Na+ + NH3
2. An alkynide anion is a good Nu:- which can substitute (replace) halogen atoms in methyl or 1 alkyl halides producing longer terminal alkynes
R-C C: - Na+ + CH3CH2-X R-C C-CH2CH3 + NaX
The reaction is straightforward with Me and 1 alkyl halides and proceeds via an SN2 mechanism
Alkynide anions are also strong bases (pKb = -11) as well as good Nu:-’s, so E2 competes with SN2 for 2 and 3 alkyl halides
CH3(CH2)3CC:- Na+ + CH3-CH(Br)-CH3 CH3(CH2)3CCCH(CH3)2 (7% SN2)
+ CH3(CH2)3CCH + CH3CH=CH2 (93% E2)
Alkylation of Alkynides
34
Alkyl halides can be prepared from alcohols by reaction with HX, i.e., the substitution of a halide on a protonated alcohol.
Preparation of Alkyl Halides from Alcohols
+ H Cl(CH3)3C OH..
..(CH3)3C OH2..
+H2O-
(CH3)3C + (CH3)3C..
: H2O+
:-
..
..:Cl
Cl..SN1
(Lucas Test)3º
..1º
CH3CH2 OH..+ H Cl CH3CH2 OH2
+
:-
..
..:Cl
CH3CH2Cl + H2OSN2
Very slow. Protic solvent inhibits the nucleophile.
Rapid. 3° C+ is stabilized by protic sovent (H2O)
Draw the mechanism of the reaction of isopropyl alcohol with HBr.What products form if concentrated H2SO4 is used in place of aq. HCl?
OH- is a poor leaving group, i.e., is not displaced directly by nucleophiles. Reaction in acid media protonates the OH group producing a better leaving group (H2O). 2 and 3 alcohols react by SN1 but Me° and 1 alcohols react by
SN2.
35
Alternative to using hydrohalic acids (HCl, HBr, HI), alcohols can be converted to alkyl halides by reaction with PBr3 which transforms OH- into a better leaving group allowing substitution (SN2) to occur without
rearrangement.
Preparation of Alkyl Halides from Alcohols
SN2 CH3(CH2)4CH2 OH
PBr
BrBr
:
etherCH3(CH2)4CH2 O
PBr2
..
..
..
+
H
Br-
CH3(CH2)4CH2Br1º
36
On Slide 22 we noted that 2° and 3° alkyl halides can be dehydrohalogenated with a strong base such as OH- producing an alkene.
bromocyclohexane + KOH cyclohexene (80 % yield)
Clearly, this is an E2 reaction.
Preparation of Alkenes from Alkyl Halides
BrKOH in ethanol + KBr + H2O
-HBr
Predict the mechanism that occurs with a Me° or 1° alkyl halide.
Predict the products and mechanism that occur with isopentyl chloride and KOH
37
Alkyl Halide Substrate Reactivity:
Summary of SN /Elimination Reactions
unhindered substrates favor SN2
do not form a stable C +do not react by S N1 or E1
hindered substrates. SN2 increasingly unfavorable, E2 is OK
form increasingly stable C+ favors SN1 and E1. E2 is OK
CH
H
H
Br CCH3
H
H
Br CCH3
CH3
H
Br CCH3
CH3
CH3
Br
E2 reactions possible with strong bases
E2 reactions possible with strong bulky bases (t-butoxide)
methyl 1º 2º 3º
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Reactivity of Nucleophiles:
Note that poor nucleophiles that are also weak bases (H2O, ROH, CH3COOH,
etc.) do not undergo any reaction unless a C+ is formed first. If a C+ can form (as with a 2º, 3º, any benzylic, or any allylic halides), then E1 and SN1
generally occur together.
Leaving Group Activity:
Summary of SN /Elimination Reactions
HS- CN- I- CH3O- HO- NH3 Cl- H2O
125,000 125,000 100,000 25,000 16,000 1000 700 1
good nucleophiles which are good nucleophiles which are weak bases favor SN reactions also strong bases favor elimination
pKb = 23 pKb = 22 pKb = 21 pKb = 11 pKb = -1.7 pKb = -2 pKb = -21
I- Br - Cl- F- HO- RO- H2N-
30,000 10,000 200 1 0 0 0
good leaving groups favor both poor leaving groups make both substitution and elimination reactions substitution and elimination
reactions unfavorable