Transcript Chlorination of Propane

Alkyl Halides
Name the following alkyl halides
Cl
(R)
Br
(S)
(R)
Cl
(S)
(R)
(S)
Cl
Cl
Cl
Cl
(R)
Br
(S)
(R)
Cl
(S)
(R)
(S)
Cl
Cl
Cl
How are alkyl halides prepared?
- From alkanes
- Free radical halogenation
- From alkenes
- Hydrohalogenation
- Halogenation
- From alcohols
Free Radical Chlorination
Experimental Evidence Helps to
Determine Mechanism
• Chlorination does not occur at room
temperature in the dark.
• The most effective wavelength of light is
blue that is strongly absorbed by Cl2 gas.
• The light-initiated reaction has a high
quantum yield (many molecules of product
are formed from each photon of light).
Free Radical Species are Constantly
Generated Throughout the Reaction
Propagation
Termination: Reaction of any 2 Radicals
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 .
3o Radicals are Easiest to Form
Stability of Free Radicals
(1) Consider the free radical monochlorination of
2,2,5-trimethylhexane. Draw all of the unique
products (Ignore stereoisomers).
(2) Consider the free radical monochlorination
of 1,4-dimethylcyclohexane. The reaction
affords three different products (all C8H15Cl).
Draw the three products (Ignore stereoisomers).
Conversion of Alkenes into Alkyl
Halides
Br
Br2, CCl4
+
Br
HBr
Br
Br
Br
Br
+
Conversion of Alcohols into
Alkyl Halides
Reactions with HX, SOCl2, PBr3
Conversion of Alcohol into a
Leaving Group
• Form Tosylate (p-TsCl, pyridine)
• Use strong acid (H3O+)
• Convert to Alkyl Halide (HX, SOCl2, PBr3)
Alcohols to Alkyl Halides
OH
HX (HCl or HBr)
X
rapid SN1
+ HOH
o
3 alcoh ol
OH
o
2 alcoh ol
HX
moderat e SN1
X
+ HOH
What the **** is SN1?
Unimolecular Nucleophilic Substitution
+ X
X
Nu
Nu
Lucas Test
CH3
ZnCl2
12M HCl
CH3COH
CH3
CH3CCl forms is seconds
CH3 + HOZnCl2
CH3
CH3
CH3C
CH3
OZnCl2
CH3 H
CH3C
CH3
Cl
Qualitative test for Alcohol
Characterization
primary
OH
Cl
>10 minutes
(if at all)
OH
ZnCl2, HCl
secondary
Cl
<5 minutes
OH
Cl
t ertiary
1-2 seconds
1o and 2o Alcohols: best to use
SOCl2, PBr3, or P/I2
All are SN2 Reactions
SOCl2
pyridine
OH
PBr3
P , I2
(in situ prep.
of PI3)
Cl
Br
I
Thionyl chloride mechanism
O
Cl
S
Cl
SOCl 2
OH
Cl + SO2 + HCl
pyridine
O
O
O
H
S
Cl
O
Cl
H
N
S
+
Cl
-H
O
Cl
O
S
Cl
How would you prepare the
following RX from the appropriate
alcohols?
(A)
Cl
(B)
Br
(C)
Br
(D)
Cl
7.25] Draw and name the monochlorination products
of the reaction of 2-methylpentane with chlorine gas.
Which are chiral?
7.36] How would you prepare the ff cpds, starting from
cyclopentene and any other reagents needed?
a) Chlorocyclopentene
b) Cyclopentanol
c) Cyclopentylmagnesium chloride
d) cyclopentane
Reactions of Alkyl Halides
Grignard Reagents
What’s special with Grignard
reagents?
Br
H3CH2C
MgBr
Mg, Ether
H3CH2C
H
O
MgBr
H3CH2C
H
MgBr
H3CH2C
H
H
OH
1.
2. H3O+
H3CH2C
H
H
H 2O
H3CH2C
+ HOMgBr
H
Outline a synthetic scheme from
the appropriate alkyl halide
OH
OH
OH
OH
Reactions of Alkyl Halides
Substitution and Elimination
Substitution, Nucleophilic,
Bimolecular – SN2
Nuc :

C

X

Nuc
C

X
Nuc
C
+ X
transit ion st ate
Rate = k[Nuc: ][R-X]
Se cond O rde r Rate Kine tics
Reaction Profile for SN2 Reaction
(Wade)
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
T sCl
H
Bz
CH3
(S)(-)
[]D = -33o
CH3
(S)
KOAc
SO2Cl
p-t oluenesulfonyl chloride RO-H
(T s-Cl)
O
CH3
CH3
S O R
O
a tosylat e (ROT s)
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
OT s
Inversion on a Ring is often more
Obvious: Cis
Trans
Substrate Reactivity
Since the energy of the transition state is significant in determining
the rate of the reaction, a primary substrate will react more rapidly
than secondary (which is much more rapid than tertiary).
R
Rate : ~0
Br + Cl
R
Cl + Br
6
1
500
40,000
2 x 10
(CH3)3CBr
(CH3)3CCH2Br
(CH3)2CHBr
CH3CH2Br
CH3Br
tertiary
neopentyl
secondary
primary
methyl
1o > 2o >> 3o
Bulkiness of Substrate
Polar, Aprotic Solvents favor SN2
S olve n ts sh oul d be able to "cage " th e m e tal cati on
O
CH3SCH 3
DMSO
O
O
CH3CN
HCN(CH3)2 CH3CCH3
acetonitrile DMF
acetone
Polar, proti c solve n ts lowe r e n e rgy of nu cle oph i le
by solvati on
HOCH3
CH3OH
Br
CH3OH
HOCH3
Nucleophilicity
Nu cl e oph i l e stre n gth rou gh l y paral l e l s basi ci ty
-
-
-
-
CH3 > NH2 > OH > F
Nu cl e oph i l e stre n gth i n cre ase s goi n g down a grou p
OH < SH
-
-
-
-
F < Cl < Br < I
NH3 < P H3
A base i s al ways a stron ge r n u cl e oph i l e th an i ts con ju gate a
-
NH2 > NH3
-
OCH3 > CH3OH
Iodide vs. Fluoride as
Nucleophiles
Nucleophiles
(preferably non-basic)
basic
-
-
-
n on -basi c
-
-
-
-
-
HS > :P (CH3)3 > CN > I > OCH3 > OH > Br > Cl > NH3 > OAc
Good Leaving Groups are Weak
Bases
C
LG bon d is brok e n du rin g RDS
Q u al ity of le avin g grou ps is cru cial
Sulfonat es are excellent leaving groups
O
SO
CH3
O
tosylate
TsO -
O
CH3SO
O
mesylat e
MsO -
Common Leaving Groups
TsO - = MsO - > NH
3- > I- > H2O - = Br- > C l- >> F-
Sulfonates are easily prepared from alco
O
CH3OH + ClSR
in pyridine
O
CH3OSR + HCl
O
O
tosylate R =
mesylate R = CH3
CH3
SN2 and E2
SN2
H
R1 C
R2
Nuc:
C
H
Nuc
R1 C
R2
Br
C
+ Br
E2
H
R1 C
R2
C
B:
Br
-]
rat e = 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
-El im i nati on

Base
C
H
C

C


Br
SN2 Competes with E2
Depends on the Nature of the Nucleophile
CH3CO2
wk. base
Br
CH3CHCH3
CH3CH2O
st r. base
S u bstitu tion
OAc
CH3CHCH3
100%
CH2=CHCH3
0%
OEt
CH3CHCH3
CH2=CHCH3
20%
El im in atio
80%
SN2 Competes with E2
Depends on the Size of the Base
CH3CH2O
str. base
CH3CH2CH2CH2OEt
90%
CH3CH2CH=CH2
10%
CH3CH2CH2CH2Br
(CH3)3CO
CH3CH2CH2CH2OtBu CH3CH2CH=CH2
str. bulky base
85%
15%
SN2 Competes with E2
Depends on the Nature of the Substrate
CN
CH3CH2CH2CH2Br
CH3CH2CH2CH2CN
str. nuc.; wk. base
o
100% S N2
1
(CH3)3CBr
3o
CN
CH2=C(CH3)2
100% E2
Stereochemistry of E2
rate = k [R-X][base ]
se con d orde r rate k ine tics
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-3,4dimethylhexane
CH2CH3
Br
CH3
NaOCH 3
H
CH3
in CH3OH
heat
CH2CH3
H and Br 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 /
+ KOT s
OTs
OTs
has no ant i-coplanar H
H
OtBu
H
Zaitsev’s Rule
NaOCH3
in CH3OH
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-bute ne
trans 2-bute ne
C
C
C
C
mono-substitute d
disubstitute d
With Bulky Base,
Hofmann Product Forms
Which will react more rapidly?
CH3
Cl
NaOEt in Et OH
heat
CH(CH3)2
Me n th yl ch l ori de
CH3
Cl
NaOEt in Et OH
heat
CH(CH3)2
Ne om e n th yl ch l ori de
Reactive Conformations
Me n th yl ch l oride
(CH3)2CH
Ne om e n th yl ch lori de
Cl
CH3
CH3
(CH3)2CH
Cl
st able
H
H
st able andreact ive
flip
NaOEt
CH(CH3)2
CH3
CH3
CH(CH3)2
CH(CH3)2
H
NaOEt
Cl
react ive
CH3
E2 Reaction of
(R,R) 2-iodo-3-methylpentane
I
CH3CHCHCH2CH3
CH3
H
NaOCH2CH3
C
in et hanol
C
CH3
CH3
(R,R)
CH2CH3
OR
CH3
CH2CH3
H
CH2=CHCHCH2CH3
C
OR
CH3
C
CH3
Stereochemistry is Important
re active con form ation
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 CH3OH
CH3
CH3
CH3
C
CH3
SN 1
Rate = k [R-Br]
1st orde r rate k i n e ti cs
CH3
OCH3 +
C=CH2
CH3
+ HBr
E1
SN1 mechanism (Wade)
1st step is rate determining
Reaction Profiles (Wade)
SN1
S N2
SN1 Transition State
SN1 Solvent Effects
CH3
CH3
C
ROH
Cl
react .: 1
CH3
C
OR + HCl
CH3
CH3
EtOH
CH3
40% 2HO / 60% Et OH
100
80% 2H
O / 20% Et OH
14,000
O
2H
100,000
Tran si ti on state e n e rgy is l owe re d by polar proti c solve n ts
Partial Racemization in SN1
Carbocation Stability
more highly substituted, lower energy
Carbocation Stability
CH3
CH3
C
H
> CH3
CH3
te rti ary
>
C
= CH2=CH CH2 =
CH3
se condary = prim ary al lyl ic
CH2 > CH3CH2
=
pri m ary be n z yli c
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
H2SO 4 (aq) cat .
+ H2O
H
regenerat ed
H
O
HSO 4
or H2O
H
-H2O
H
Can R-X form a good LG?
No
Yes
no react ion cl assi fi cati on of
 carbon
1
o
3
2
o
o
nuc. hindered, st rong base?
st rong base? Yes
E2
nuc. a st rong base?
No
Yes
No
Yes
No
Yes
SN1*
polar
solvent
?
E2 good nuc., non-basic?
E1
good
nuc.,
nonbasic?
E2
Yes
No
(some SN2) Yes
No
SN2 (sl ow SN2)
S N2
solvent polar?Yes
SN1*
E1
* SN1 is favored over E1 unless high t emp. and t race amount s of base are used.
Draw the Major Product & Predict
the Mechanism
OH
CH3
6M H2SO4
120oC, distill
OH
CH3
6M H2SO4
120oC, distill
E1
CH3
NaNH 2 in liq. NH3
OTs
NaNH 2 in liq. NH3
OTs
E2
H
CH3
CH2CH3
OTs
KBr
in acetone, 20oC
H
CH3
CH2CH3
OTs
KBr
in acetone, 20 oC
SN2
Br
CH3
CH2CH3
H
Br
1% AgNO3
in CH3CH2OH
Br
1% AgNO3
in CH3CH2OH
SN1
CH3CH2O
+
AgBr
Br
CH3CH2CH2OH
warm
CH3CH2CH2OH
Br
OCH2CH2CH3
warm
SN1/E1
+
CH3
Br
NaSCH2CH3
in CH3CN
CH3
Br
NaSCH2CH3
CH3
in CH3CN
SN2
SCH2CH3
I
CH3
CH3
NaOCH 2CH3
in refluxing ethanol
I
CH3
CH3
NaOCH 2CH3
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
trans
I
re acti ve
I
(CH3)2CH
stable
CH(CH3)2
I
cis
(CH3)2CH
re acti ve & stable
H
react s more rapidly
What is an E1CB reaction?
• Poor LG 2 carbons away from a carbonyl
carbon (beta alcohol)
• Proceeds via a carbanion intermediate
stabilized by resonance.
• Step 1: Abstraction of alpha proton
• Step 2: Ejection of leaving group.