Download Chlorination of Propane

Alkyl Halides
Nomenclature
Preparation (Quick Review)
Reactions
Name the following alkyl halides
Cl
Br
Cl
Cl
(3S)-3-chloro-2-methylpentane
Cl
(R)
(S)
(R)
(R)
Br
Cl
Cl
(2R)-2-bromobutane
(1R,3R)-1,3-dichlorocyclohexane
Study nomenclature. It will be part of the next quiz.
How are alkyl halides prepared?
- From alkanes
- Free radical halogenation
- From alkenes
- Hydrohalogenation (alkene + HX)
- Halogenation (alkene + X2)
- From alcohols
Free Radical Chlorination
Chlorination of Propane
30%
70%
Chlorination of Methylpropane
.
CH3
CH3
C
H + Cl .
CH3
CH3
CH3
CH2Cl
H
+ Cl2
CH3
C
CH3
CH3
CH3
CH3
C.
CH3
C.
CH3
CH3
CH2
C
H + CH3
C
.
CH3
CH3
CH2
+ Cl2
CH3
C
CH3
H 65% + Cl.
Cl 35% + Cl .
Consider the free radical monochlorination of
2,2,5-trimethylhexane. Draw all of the
unique products. Which are chiral?
Consider the free radical monochlorination of 1,4dimethylcyclohexane. Draw all of the unique
products. Which are chiral?
Conversion of Alcohols into
Alkyl Halides
Reactions with HX, SOCl2, PBr3
Alcohols to Alkyl Halides
OH
HX (HCl or HBr)
X
rapid S N1
+ HOH
o
3 alcohol
OH
o
2 alcohol
HX
moderate S N1
X
+ HOH
Lucas Test
CH3
ZnCl 2
12M HCl
CH3COH
CH3
CH3CCl forms is seconds
CH3 + HOZnCl 2
CH3
CH3
CH3C
CH3
OZnCl2
CH3 H
CH3C
CH3
Cl
Qualitative test for Alcohol
Characterization
primary
OH
>10 minutes
(if at all)
OH
ZnCl 2, HCl
secondary
OH
tertiary
Cl
<5 minutes
Cl
1-2 seconds
Cl
1o and 2o Alcohols: best to use
SOCl2, PBr3, or P/I2
All are SN2 Reactions
SOCl2
pyridine
OH
PBr 3
P, I2
(in situ prep.
of PI3)
Cl
Br
I
Thionyl chloride mechanism
O
Cl
S
Cl
SOCl2
OH
O
O
H
S
Cl + SO2 + HCl
pyridine
O
Cl
O
Cl
H
N
S
+
Cl
-H
O
Cl
O
S
Cl
Reactions of Alkyl Halides
Formation of Grignard Reagents
What makes Grignard reagents interesting?
Haven’t we seen this before….
A (partial) negatively charged
carbon is always interesting!
Mechanisms: Fill in the missing arrows to show the flow of electrons.
Victor Grignard
The Nobel Prize in Chemistry 1912
Main Event:
Substitution vs Elimination
Substitution, Nucleophilic,
Bimolecular – SN2
Nuc :

C

X

Nuc
C

X
Nuc
transition state
Rate = k[Nuc: ][R-X]
Second Order Rate Kinetics
C
+ X
Reaction Profile for SN2 Reaction
Stereochemistry of SN2 Reaction
Inversion of Configuration
CN
Br
+ KCN
(S)
+ KBr
(R)
Proof of Inversion of Configuration
at a Chiral Center
CH2
benzyl (Bz)
O
OCCH3
-OAc, acetate
OH
H
Bz
OTs
TsCl
H
Bz
CH3
(S)(-)
[]D = -33o
CH3
(S)
KOAc
SO2Cl
p-toluenesulfonyl chloride
(Ts-Cl)
O
CH3
CH3
RO-H
S O R
O
a tosylate (ROTs)
H
Bz
CH3
OH
(R)(+)
[]D = +33o
H2O
H
Bz
CH3
OAc
(R)
Acetate Approaches from 180o
Behind Leaving Group
Bz
AcO
OTs
H
CH3
(S)

AcO
Bz
CH3 H

OTs
Bz
AcO
(R)
H
CH3
OTs
Inversion on a Ring is often more
Obvious: Cis -> Trans
Substrate Reactivity
A primary substrate will react more rapidly than secondary (which
is much more rapid than tertiary).
R
Rate: ~0
(CH3)3CBr
tertiary
Br + Cl
R
Cl + Br
6
1
500
40,000
(CH3)3CCH2Br
(CH3)2CHBr
CH3CH2Br
CH3Br
secondary
primary
methyl
neopentyl
2 x 10
1o > 2o >> 3o
Bulkiness of Substrate
Nucleophilicity
Nucleophile strength roughly parallels basicity
-
-
-
CH3 > NH2 > OH > F
-
Nucleophile strength increases going down a group
OH < SH
-
-
-
-
F < Cl < Br < I
NH3 < PH3
A base is always a stronger nucleophile than its conjugate acid
-
NH2 > NH3
-
OCH3 > CH3OH
Nucleophiles (preferably non-basic)
basic
-
-
non-basic
-
-
-
-
-
-
HS > :P(CH 3)3 > CN > I > OCH3 > OH > Br > Cl > NH3 > OAc
Strong bases favor elimination over substitution.
Good Leaving Groups are Weak
Bases
C
LG bond is broken during RDS
Quality of leaving groups is crucial
Sulfonates are excellent leaving groups
O
SO
CH3
O
O
CH3SO
tosylate
O
mesylate
TsO-
MsO-
Common Leaving Groups
TsO- = MsO- > NH3- > I- > H 2O- = Br- > Cl- >> F-
Sulfonates are easily prepared from alcohols
O
CH3OH + ClSR
in pyridine
O
CH3OSR + HCl
O
O
tosylate R =
mesylate R = CH
CH3
3
Polar, Aprotic Solvents favor SN2
Solvents should be able to "cage" the metal cation
O
CH3SCH3
DMSO
O
O
CH3CN
HCN(CH3)2 CH3CCH3
acetonitrile
acetone
DMF
Polar, protic solvents lower energy of nucleophile
by solvation
HOCH3
CH3OH
Br
CH3OH
HOCH3
SN2 VS E2
SN2
H
R1 C
R2
Nuc:
C
H
Nuc
R1 C
R2
Br
C
+ Br
E2
H
R1 C
R2
C
B:
Br
rate = k[R-Br][B -]
R1
C
R2
C
+ B-H + Br
Bimolecular Elimination - E2
Nucleophile acts as Bronsted Base
Base:
H
C

C

C
Br
+ base-H
+ Br
-Elimination

Base
C
H
C

C


Br
SN2 Competes with E2
Depends on the Nature of the Nucleophile
CH3CO2
wk. base
Br
CH3CHCH3
CH3CH2O
str. base
Substitution
OAc
CH3CHCH3
100%
OEt
CH3CHCH3
20%
Elimination
CH2=CHCH3
0%
CH2=CHCH3
80%
Stereochemistry of E2
rate = k[R-X][base]
second order rate kinetics
CH3O
H
C

C
C
Br
H on  carbon is anti to leaving group
C
+ CH3OH
+ Br
Anti-Coplanar Conformation
3(R),4(R) 3-Bromo-3,4-dimethylhexane
CH2CH3
Br
CH3
NaOCH3
H
CH3
in CH 3OH
heat
CH2CH3
H and Br must be in anti-coplanar
orientation
CH3O
H
Me Et
C
C C
Et

Me (R) (R) Br
Me
C
Et
OCH3
H
Me
Et
Et
Me
Br
Me
Et
Me
Et
Et
Me
In a Cyclohexane,
Leaving Group must be Axial
KOC(CH3)3
OTs
in t-BuOH / 
+ KOTs
OTs
OTs
has no anti-coplanar H
H
OtBu
H
Zaitsev’s Rule
NaOCH3
in CH 3OH
Br
+
85%
15%
Zaitsev's Rule: In an elimination reaction, the
more highly substituted alkene (usually) predominates
More Stable Alkene Predominates
Hyperconjugation
p bond associates with adjacent C-H s bond
1-butene
trans 2-butene
C
C
C
C
mono-substituted
disubstituted
Which will react more rapidly?
CH3
Cl
NaOEt in EtOH
heat
CH(CH3)2
Menthyl chloride
CH3
Cl
CH(CH3)2
Neomenthyl chloride
NaOEt in EtOH
heat
Find the Reactive Conformations
Menthyl chloride
(CH3)2CH
Neomenthyl chloride
Cl
CH3
CH3
(CH3)2CH
Cl
stable
H
H
stable and reactive
flip
NaOEt
CH(CH3)2
CH3
CH3
CH(CH3)2
CH(CH3)2
H
NaOEt
Cl
reactive
CH3
E2 Reaction of
(R,R) 2-iodo-3-methylpentane
I
CH3CHCHCH2CH3
CH3
H
NaOCH2CH3
C
in ethanol 
C
CH3
CH3
(R,R)
CH2CH3
OR
CH3
CH2CH3
H
CH2=CHCHCH2CH3
C
OR
CH3
C
CH3
Stereochemistry is Important
reactive conformation
I
H
CH3
C
C
CH2CH3
CH3
OEt
(R,R)
I
H
H
CH3
CH3
CH3CH2
C=C
CH3
H
CH3CH2
H
CH3
Unimolecular Substitution and
Elimination – SN1 and E1
CH3
CH3
C
Br
in warm CH 3OH
CH3
CH3
CH3
C
CH3
SN1
Rate = k[R-Br]
1st order rate kinetics
CH3
OCH3 +
C=CH2
CH3
+ HBr
E1
SN1 mechanism
1st step is rate determining
Reaction Profiles
SN1
S N2
SN1 Transition State
SN1 Solvent Effects
CH3
CH3
C
Cl
ROH
react.:
1
CH3
C
OR + HCl
CH3
CH3
EtOH
CH3
40% H 2O / 60% EtOH
100
80% H 2O / 20% EtOH
14,000
H 2O
100,000
Transition state energy is lowered by polar protic solvents
Partial Racemization in SN1
Carbocation Stability
more highly substituted, lower energy
Carbocation Stability
CH3
CH3
C
H
> CH3
CH3
tertiary
>
C
= CH2=CH CH2 =
CH3
secondary = primary allylic
=
CH2 > CH3CH2
primary benzylic > primary
resonance stabilized
Carbocations can Rearrange
1,2-Hydride Shift
Br
CH3 C
H
H
C
CH3
CH3
H2O
H
CH3 C
H
OH
C
CH3 + HBr
CH3
Hydride shift
H
2
o
Hydride
shift
H
o
3
E1 Mechanism
E1 and SN1 Compete
b)
a)
OTs
CH3OH / 
CH3
+
Zaitsev
a) CH3OH
H
H
CH3
CH3
b) CH3OH
CH3
OCH3
CH3
Dehydration of Alcohols – E1
OH
H
H2SO4 (aq) cat.
+ H2O
H
regenerated
H
O
HSO 4
or H2O
H
-H2O
H
E1cB Reaction
- Proceeds via a carbanion intermediate
- Leaving group (say for example, halide or
alcohol) is two carbons away from a carbonyl
group (aldehyde, ketone)
Give the Major Product & Predict
the Mechanism
OH
CH3
6M H2SO4
120 oC, distill
Good leaving group favors unimolecular reactions
High temperature favors elimination
OH
CH3
6M H 2SO 4
120 oC, distill
E1
CH3
H
CH3
CH2CH3
OTs
KBr
in acetone, 20 oC
Non–basic nucleophile favors substitution
Low temperature favors substitution
Polar aprotic solvent
H
CH3
CH2CH3
OTs
KBr
in acetone, 20 oC
SN2
Br
CH3
CH2CH3
H
CH3CH2CH2OH
Br
warm
Formation of stable carbocation favors unimolecular reactions.
Non–basic nucleophile favors substitution
High temperature favors elimination
Br
CH3CH2CH2OH
warm
SN1/E1
OCH2CH2CH3
+
CH3
Br
NaSCH2CH3
in CH 3CN
Non–basic nucleophile favors substitution
Polar aprotic solvent
CH3
Br
NaSCH 2CH3
CH3
in CH 3CN
SN2
SCH2CH3
I
CH3
CH3
NaOCH2CH3
in refluxing ethanol
I
CH3
CH3
NaOCH2CH3
in refluxing ethanol
E2
CH3
CH3
Which Reacts More Rapidly in E2
Reaction?
(CH3)2CH
I
A
(CH3)2CH
I
B
Cis Reacts more Rapidly. Why?