Alkyl Halides
Organo halogen
Alkyl halide
Cl
H
H
C
Cl
Cl
Cl
C
Cl
CCl3
H
C
Cl
Cl
Aryl halide
H
C
Cl
Cl
Halide vynilik
H
C
Cl Cl
C
H
H
I
HO
I
H2 H
C C
O
I
Alkyl halide Reactions :
Substitution : SN1 dan SN2
Elimination : E1 dan E2
Cl
O
C
OH
NH2
I
1
NUCLEOPHILIC SUBSTITUTION
R
Y-
X
R
Y
X-
1. Leaving groups
weaker base = better leaving group
reactivity: R-I > R-Br > R-Cl >> R-F
best L.G.
worst L.G.
most reactive
least reactive
R X + Y
stronger
base
R Y + X
weaker
base
Br + NaF
SB
Br
+ NaI
WB
K>1
F + NaBr
WB
acetone
I
+ NaBr (s)
SB
precipitate
drives rxn
(Le Châtelier)
2. Mechanisms SN
general: Rate = k1[RX] + k2[RX][Y–]
k1 increases
RX = CH3X
1º
2º
3º
k2 increases
k1 ~ 0
k2 ~ 0
Rate = k2[RX][Y–]
(bimolecular)
Rate = k1[RX]
(unimolecular)
SN2
SN1
SN2 Mechanism
A. Kinetics
e.g., CH3I + OH–  CH3OH + I–
find: Rate = k[CH3I][OH–], i.e., bimolecular
 both CH3I and OH– involved in RLS
and recall, reactivity: R-I > R-Br > R-Cl >> R-F
 C-X bond breaking involved in RLS
 concerted, single-step mechanism:
[HO---CH3---I]–
CH3I + OH–
CH3OH + I–
B. Stereochemistry: inversion of configuration
Stereospecific reaction:
H Br
Reaction proceeds with
inversion of configuration.
HO H
NaOH
(R)-(–)-2-bromooctane
(S)-(+)-2-octanol
C. Mechanism
HO
H + 
C I
H
H
H
HO C
H
HO C
I
H
HH
Back-side attack:
HO
C
+ I
H
inversion of configuration
I
HO
C
I
HO
C
I
D. Steric effects
e.g., R–Br + I–  R–I + Br–
1. branching at the a carbon
( X–C–C–C.... )
a b g
minimal steric hindrance
H
Compound
Rel. Rate
methyl
CH3Br
150
1º RX
CH3CH2Br
1
2º RX
(CH3)2CHBr
0.008
3º RX
(CH3)3CBr
~0
I
increasing
steric
hindrance
I
Reactivity toward SN2:
CH3X > 1º RX > 2º RX >> 3º RX
react
readily
by SN2
(k2 large)
more
difficult
does not
react by
SN2
(k2 ~ 0)
H
C
Br
H
H H
H C
H
C Br
H C
H C
H
H
H
maximum steric hindrance
E. Nucleophiles and nucleophilicity
1. anions
R X + OH
R OH + X
R X + CN
R CN + X
2. neutral species R X + H2O
R O H + X
H
R X + R'OH
R O R' + X
H
Summary:
very good Nu:
good Nu:
fair Nu:
poor Nu:
very poor Nu:
I–, HS–, RS–, H2N–
Br–, HO–, RO–, CN–, N3–
NH3, Cl–, F–, RCO2–
H2O, ROH
RCO2H
ROH + HX
hydrolysis
ROR' + HX
alcoholysis
SN1 Mechanism
A. Kinetics
e.g.,
CH3
H3C C Br + CH3OH

CH3
H3C C O CH3 + HBr
CH3
3º, no SN2
Find: Rate = k[(CH3)3CBr]
CH3
unimolecular
 RLS depends only on (CH3)3CBr
A. Kinetics
CH3
CH3
RLS: H3C C Br
H3C C
CH3
CH3
CH3
H3C C
+ Br
HOCH3
CH3
CH3 H
H3C C O
CH3 CH3
-H+
CH3 H
H3C C O
CH3 CH3
CH3
H3C C O CH3 + HBr
CH3
A. Kinetics
Two-step mechanism:
R+
RBr + CH3OH
ROCH3 + HBr
B. Stereochemistry: stereorandom
Br
CH3CH2
OH
CH3
H2O
CH3CH2
H
CH3 + CH3CH2
H
OH2
CH3CH2
+ H
C
CH3
OH2
sp2, trigonal planar
H
racemic
CH3
OH
C. Carbocation stability
R+ stability: 3º > 2º >> 1º > CH3+
R-X reactivity toward SN1: 3º > 2º >> 1º > CH3X
CH3+
1º R+
2º R+
3º R+
rearrangements
possible
SN1 vs SN2
A. Solvent effects
nonpolar:
moderately polar:
polar protic:
polar aprotic:
hexane, benzene
ether, acetone, ethyl acetate
H2O, ROH, RCO2H
DMSO
DMF
acetonitrile
O
O
CH3 C N
C
S
N(CH3)2
CH3
CH3 H
SN1 mechanism promoted by polar protic solvents
stabilize R+, X– relative to RX
R+X–
RX
in less polar solvents
in more polar solvents
A. Solvent effects
SN2 mechanism promoted by moderately polar & polar aprotic solvents
destabilize Nu–, make
them more nucleophilic
e.g., OH– in H2O: strong H-bonding to water makes OH– less reactive
OH– in DMSO: weaker solvation makes OH– more reactive (nucleophilic)
in DMSO
in H2O
RX + OH–
ROH + X–
B. Summary
rate of SN1 increases
RX =
CH3X
1º
2º
(carbocation stability)
3º
rate of SN2 increases
(steric hindrance)
react
may go
reacts
primarily
by either
primarily
by SN2
mechanism
by SN1
(k1 ~ 0, k2 large)
(k2 ~ 0, k1 large)
SN2 promoted good nucleophile (Rate = k2[RX][Nu])
-usually in polar aprotic solvent
SN1 occurs in absence of good nucleophile (Rate = k1[RX])
-usually in polar protic solvent (solvolysis)
ELIMINATION REACTIONS
Dehydrohalogenation of alkyl halides
H X
C C
+ B
C C
+ BH + X
Elimination
strong base: KOH/ethanol
CH3CH2ONa/CH3CH2OH
tBuOK/tBuOH
Follows Zaitsev orientation:
EtONa
EtOH
Br
+
61%
EtONa
EtOH
Br
+
20%
+
71%
29%
19%
The E2 mechanism
elimination, bimolecular
• reaction is bimolecular, depends on concentrations of both RX and B–
Rate = k[RX][B–]
 RLS must involve B–
• reactivity: RI > RBr > RCl > RF
increasing R—X bond strength
 RLS must also involve breaking the R—X bond
(and reaction doesn’t depend on whether RX is 1º, 2º, or 3º)
1. Single step, concerted mechanism:
X
X
C C
C C
H
H
B
B
Br
+ OH-
X
C C
B H
Zaitsev
2. stereoelectronic effects: anti elimination
spatial arrangement of electrons (orbitals)
In the E2 mechanism, H and X must be coplanar:
(in order for orbitals to overlap in TS)
H
H
C
C
X
C
C
X
syn periplanar
anti periplanar
-but eclipsed!
-most molecules
can adopt this
conformation more easily
E2 eliminations usually occur when H and X are anti
2. stereoelectronic effects: anti elimination
CH3
Br
EtONa
EtOH
+
major
CH3
Br
but
"
"
major
minor
2. stereoelectronic effects: anti elimination
Br must be axial to be anti to any b-H’s:
H
H
CH3
CH3
H
H
but
Br
Br is anti to both H’s
 normal Zaitsev orientation
Br
Br is anti only to H that gives
non-Zaitsev orientation
3. the E1 mechanism
Recall:
EtONa
EtOH
Br
+
major
Rate = k[RBr][B–] E2
Reactivity: RI > RBr > RCl > RF
However:
Br
minor
(and no effect of 1º, 2º, 3º)

EtOH
+
major
Rate = k[RBr]
E1
Reactivity: RI > RBr > RCl > RF
and: 3º > 2º > 1º
minor
(no involvement from B–)
(RLS involves R–X breaking)
(RLS invloves R+)
3. the E1 mechanism
Step 1:
(RLS)
+ Br
Br
EtOH + HBr
Step 2:
+ EtOH2
H
EtOH
- and R+ can rearrange
 eliminations usually carried out with strong base
Substitution vs Elimination
A. Unimolecular or bimolecular reaction?
(SN1, E1)
(SN2, E2)
Rate = k1[RX] + k2[RX][Nu or B]
• this term gets larger as [Nu or B] increases
 bimolecular reaction (SN2, E2) favored by
high concentration of good Nu or strong B
• this term is zero when [Nu or B] is zero
 unimolecular reaction (SN1, E1) occurs in
absence of good Nu or strong B
B. Bimolecular: SN2 or E2?
Rate = kSN2[RX][Nu] + kE2[RX][B]
1. substrate structure: steric hindrance
decreases rate of SN2, has no effect on rate of E2
 E2 predominates
Br
NaOEt
O
+
91%
Br
steric
hindrance
increases
"
O
9%
+
13%
Br
Br
"
tBuOK
100%
O
15%
sterically hindered nucleophile
87%
+
85%
B. Bimolecular: SN2 or E2?
2. base vs nucleophile
• stronger base favors E2
• better nucleophile favors SN2
NaI
I
Br
100%
good Nu
weak B
+
good Nu
strong B
NaOCH3
OCH3
60%
40%
tBuOK
poor Nu
strong B
OtBu +
5%
95%
C. Unimolecular: SN1 or E1?
OH2
Br
OH
H2O

(weak B,
poor Nu)
H
OH2
for both, Rate = k[R+][H2O]
 no control over ratio of SN1 and E1
D. Summary
1. bimolecular: SN2 & E2
Favored by high concentration of good Nu or strong B
good Nu, weak B: I–, Br–, HS–, RS–, NH3, PH3
favor SN2
good Nu, strong B: HO–, RO–, H2N–
SN2 & E2
poor Nu, strong B: tBuO– (sterically hindered)
favors E2
Substrate:
1º RX
2º RX
3º RX
mostly SN2 (except with tBuO–)
b-branching
both SN2 and E2 (but mostly E2)
hinders SN2
E2 only
2. unimolecular: SN1 & E1
Occurs in absence of good Nu or strong B
poor Nu, weak B: H2O, ROH, RCO2H
Substrate:
1º RX
2º RX
3º RX
SN1 and E1 (only with rearrangement)can’t control
ratio of
SN1 and E1 (may rearrange)
SN1 to E1
1. Halogenation of Alkanes
or R–X + HX a substitution reaction
R–H + X2 heat
—
light
Reactivity: F2 > Cl2 > Br2 > I2
too
reactive
common
too
unreactive
(endothermic)
Cl2
Cl2
Cl2
Cl2
CH4 
CH3Cl 
CH2Cl2 
CHCl3 
CCl4
hn
hn
hn
hn
+ HCl
+ HCl
+ HCl
Problem: mixture of products
Solution: use large excess of CH4 (and recycle it)
+ HCl
A. Free-radical chain mechanism
Step 1:
Cl2  2Cl•
Step 2:
Step 3:
Cl• + CH4  HCl + CH3•
CH3• + Cl2  CH3Cl + Cl•
(homolytic cleavage)
1000’s of cycles = “chain” reaction
net:
Initiation
Propagation
-determines
net reaction
CH4 + Cl2  CH3Cl + HCl
Sometimes: Cl• + Cl•  Cl2
CH3• + CH3•  CH3–CH3
CH3• + Cl•  CH3Cl
Termination
(infrequent due
to low [rad•])
B. Stability of free radicals: bond dissociation energies
R–H  R• + H•
H = BDE
BDE
CH3—H
104 kcal
CH3CH2—H
98 kcal
CH3CH2CH2—H 98 kcal (any 1º)
(CH3)2CH—H
95 kcal (any 2º)
(CH3)3C—H
91 kcal (any 3º)
CH3CH2CH2•
•
CH3CHCH3
98 kcal
CH3–CH2–CH3
95 kcal
easier to break bonds
 free radical more stable
lower energy, more stable,
easier to form
Reactivity of C–H:
3º > 2º > 1º > CH3–H
C. Higher alkanes: regioselectivity
Some alkanes give only one monohalo product:
CH3
CH3
Cl2
hn
CH3
Cl2
hn
CH2 Cl
Cl
Synthetically
useful.
Cl
Cl2
hn
Cl
But:
CH3CH2CH3
Cl2
hn
CH3CH2CH2Cl + CH3CHCH3
find:
even though statistically:
43%
75%
(6 H)
57%
25%
(2 H)
Not as
useful.
C. Higher alkanes: regioselectivity
Reactivity of C–H: 3º > 2º > 1º
-for Cl2, relative reactivity is 5.2 : 3.9 : 1
Predicting relative amounts of monochloro product:
Cl
Cl2
hn
CH3CH2CH3
CH3CH2CH2Cl + CH3CHCH3
2º product
reactivity of 2º H
number of 2º H’s
=
x
1º product
reactivity of 1º H
number of 1º H’s
=
3.9 x 2
7.8
57%
=
=
1x6
6
43%
Cl2
hn
CH3 H CH3
CH3
C C C CH3
CH3 H CH3
Cl2
hv
CH3
C C C CH2
CH3 H CH3
Cl
H H H
H H H
1-chloro-2,4-dimethylpentane
12 H, primary
#H
reactivity factor
12
x1
12
CH3
CH3 H CH3
C C C CH3 CH3
C C C CH3
H H Cl
H Cl H
2-chloro-2,4-dimethylpentane 3-chloro-2,4-dimethylpentane
2H, tertiary
2
x 5.2
10.4
2H, secondary
2
x 3.9
7.8
sum = 12+10.4+7.8 = 30.2
percent
12/30.2 x 100 = 39.7%
10.4/30.2 = 34.4%
7.8/30.2 = 25.8
Bromine is much more selective:
Cl
Cl2
hn
Br2
hn
Cl +
43%
57%
3%
97%
Relative reactivities for Br2: 3º 2º 1º
1640 82 1
Br2
hn
Synthetically
more useful.