Download - Alkyl Halides
1
Alkyl HalidesOrgano halogen
Alkyl halide
Aryl halide
Halide vynilik C
CCl3
H
ClCl
OH2CHO
I
I
I
I
HC
NH2
C
O
OH
C Cl
H
H
Cl
C Cl
Cl
Cl
Cl
C Cl
Cl
H
Cl
C C
H
H Cl
H
Alkyl halide Reactions : Substitution : SN1 dan SN2 Elimination : E1 dan E2
NUCLEOPHILIC SUBSTITUTION R X Y- R Y X-
1. Leaving groups
++ XR YYR Xstronger
baseweaker
base
K > 1
Br F+ NaF + NaBrSB WB
Br I+ NaI + NaBr (s)
WB SB
acetone
weaker base = better leaving group
reactivity: R-I > R-Br > R-Cl >> R-Fbest L.G.
most reactiveworst L.G.
least reactive
precipitatedrives rxn(Le Châtelier)
2. Mechanisms SN
general: Rate = k1[RX] + k2[RX][Y–]
RX = CH3X 1º 2º 3ºk1 increases
k2 increases
k1 ~ 0
Rate = k2[RX][Y–](bimolecular)
SN2
k2 ~ 0
Rate = k1[RX](unimolecular)
SN1
SN2 MechanismA. 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:
CH3I + OH–
CH3OH + I–
[HO---CH3---I]–
B. Stereochemistry: inversion of configuration
H Br HO HNaOH
(R)-(–)-2-bromooctane (S)-(+)-2-octanol
Stereospecific reaction:Reaction proceeds withinversion of configuration.
Back-side attack:
HO C IH
HH
C
H
IHO
H H
CH
HOH
HI+
+ -
HO C I HO C I HO C I
inversion of configuration
C. Mechanism
D. Steric effects
e.g., R–Br + I– R–I + Br–
1. branching at the a carbon ( X–C–C–C.... )
a b g
Compound Rel. Rate
methyl CH3Br 150
1º RX CH3CH2Br 1
2º RX (CH3)2CHBr 0.008
3º RX (CH3)3CBr ~0
increasingsteric hindrance
C BrH
HH
I
H CH
C BrCH
H
H
H
H
HCH
I
minimal steric hindrance
maximum steric hindrance
Reactivity toward SN2: CH3X > 1º RX > 2º RX >> 3º RX
reactreadilyby SN2
(k2 large)
moredifficult
does notreact by
SN2(k2 ~ 0)
E. Nucleophiles and nucleophilicity
1. anions R X OH R OH X+ +
++ XR CNCNR X
R X R O H
H
+ +
++ R O R'
H
R X
H2O
R'OH
X
X
ROH + HX
ROR' + HX
hydrolysis
alcoholysis
2. neutral species
Summary:very good Nu: I–, HS–, RS–, H2N–
good Nu: Br–, HO–, RO–, CN–, N3–
fair Nu: NH3, Cl–, F–, RCO2–
poor Nu: H2O, ROHvery poor Nu: RCO2H
SN1 MechanismA. Kinetics
C
CH3
H3C
CH3
Br CH3OH CH3C
CH3
CH3
O CH3 HBr+ +e.g.,
3º, no SN2Find: Rate = k[(CH3)3CBr] unimolecular
RLS depends only on (CH3)3CBr
A. Kinetics
C
CH3
H3C
CH3
Br
CH3C
CH3
CH3
O CH3 HBr+
RLS: C
CH3
H3C
CH3
+ Br
C
CH3
H3C
CH3
HOCH3 C
CH3
H3C
CH3
OH
CH3
C
CH3
H3C
CH3
OH
CH3
-H+
A. Kinetics
Two-step mechanism:
RBr + CH3OH
R+
ROCH3 + HBr
B. Stereochemistry: stereorandom
CH3CH2
Br
H
CH3 CH3CH2
OH
H
CH3 CH3CH2
H
OH
CH3+H2O
racemic
CH3CH2 CH
CH3
+OH2
OH2
sp2, trigonal planar
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 SN2A. Solvent effects
nonpolar: hexane, benzenemoderately polar: ether, acetone, ethyl acetate
polar protic: H2O, ROH, RCO2Hpolar aprotic: DMSO DMF acetonitrile
CH
O
N(CH3)2
CH3 C NCH3
S
O
CH3
SN1 mechanism promoted by polar protic solvents
stabilize R+, X– relative to RX
RX
R+X–
in less polar solventsin more polar solvents
A. Solvent effects
SN2 mechanism promoted by moderately polar & polar aprotic solvents
destabilize Nu–, makethem 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)
RX + OH–
ROH + X–
in DMSO
in H2O
B. Summary
RX = CH3X 1º 2º 3º
rate of SN1 increases (carbocation stability)
rate of SN2 increases (steric hindrance)
reactprimarilyby SN2
(k1 ~ 0, k2 large)
reactsprimarilyby SN1
(k2 ~ 0, k1 large)
may goby either
mechanism
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
C
H
C
X
C C+ + BH +B X
strong base: KOH/ethanolCH3CH2ONa/CH3CH2OHtBuOK/tBuOH
Elimination
Follows Zaitsev orientation:
Br
+ +EtONaEtOH
61% 20% 19%
Br+EtONa
EtOH71% 29%
The E2 mechanism
• reaction is bimolecular, depends on concentrations of both RX and B–
Rate = k[RX][B–]
RLS must involve B–
• reactivity: RI > RBr > RCl > RF
RLS must also involve breaking the R—X bond
(and reaction doesn’t depend on whether RX is 1º, 2º, or 3º)
increasing R—X bond strength
elimination, bimolecular
1. Single step, concerted mechanism:
C C
X
H
C C
X
HB
C C
B H
X
B
Br
+ OH-
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)
HC C
X
HC C
X
anti periplanar-most moleculescan adopt thisconformation more easilyE2 eliminations usually occur when H and X are anti
syn periplanar-but eclipsed!
2. stereoelectronic effects: anti elimination
CH3
Br
Br
CH3
+EtONaEtOH
""
major minor
major
but
2. stereoelectronic effects: anti elimination
CH3
Br
HH
CH3
Br
HH
but
Br must be axial to be anti to any b-H’s:
Br is anti to both H’s normal Zaitsev orientation
Br is anti only to H that givesnon-Zaitsev orientation
Recall:
Rate = k[RBr][B–] E2Reactivity: RI > RBr > RCl > RF (and no effect of 1º, 2º, 3º)
However:
Rate = k[RBr]E1 (no involvement from B–)Reactivity: RI > RBr > RCl > RF (RLS involves R–X breaking)
and: 3º > 2º > 1º (RLS invloves R+)
3. the E1 mechanism
minormajor
EtONaEtOH +
Br
minormajor
EtOH
+Br
3. the E1 mechanism
Step 1:(RLS)
Step 2:
Br
H
+ Br
EtOH
+ EtOH2
EtOH + HBr
- and R+ can rearrange eliminations usually carried out with strong base
Substitution vs Elimination
A. Unimolecular or bimolecular reaction?(SN2, E2)(SN1, E1)
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?
1. substrate structure: steric hindrance
decreases rate of SN2, has no effect on rate of E2 E2 predominates
Br
Br
Br
Br
NaOEt O
O
O
+
+
+tBuOK
91% 9%
13% 87%
100%
15% 85%
"
"
sterichindranceincreases
sterically hindered nucleophile
Rate = kSN2[RX][Nu] + kE2[RX][B]
B. Bimolecular: SN2 or E2?
2. base vs nucleophile
• stronger base favors E2• better nucleophile favors SN2
tBuOK
Br I
OCH3
OtBu
+
+
NaI
NaOCH3
100%
40% 60%
5% 95%
good Nuweak B
good Nustrong B
poor Nustrong B
C. Unimolecular: SN1 or E1?
Br H2O
(weak B,poor Nu)
H
OH2
OH2
OH
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 mostly SN2 (except with tBuO–)2º RX both SN2 and E2 (but mostly E2)3º RX E2 only
b-branchinghinders SN2
2. unimolecular: SN1 & E1
Occurs in absence of good Nu or strong B
poor Nu, weak B: H2O, ROH, RCO2H
Substrate:
1º RX SN1 and E1 (only with rearrangement)2º RX3º RX
SN1 and E1 (may rearrange)
can’t controlratio of
SN1 to E1
1. Halogenation of Alkanes
R–H + X2 — R–X + HX a substitution reactionheat orlight
Reactivity: F2 > Cl2 > Br2 > I2
commontooreactive
toounreactive
(endothermic)
CH4 CH3Cl CH2Cl2 CHCl3 CCl4
+ HCl + HCl + HCl + HCl
Cl2
hnCl2
hnCl2
hnCl2
hn
Problem: mixture of productsSolution: use large excess of CH4 (and recycle it)
A. Free-radical chain mechanism
Step 1: Cl2 2Cl• (homolytic cleavage) Initiation
Step 2: Cl• + CH4 HCl + CH3•Step 3: CH3• + Cl2 CH3Cl + Cl•
net: CH4 + Cl2 CH3Cl + HCl
Sometimes: Cl• + Cl• Cl2 TerminationCH3• + CH3• CH3–CH3 (infrequent dueCH3• + Cl• CH3Cl to low [rad•])
Propagation-determines net reaction1000’s of cycles = “chain” reaction
B. Stability of free radicals: bond dissociation energies
BDECH3—H 104 kcalCH3CH2—H 98 kcalCH3CH2CH2—H 98 kcal (any 1º)(CH3)2CH—H 95 kcal (any 2º)(CH3)3C—H 91 kcal (any 3º)
easier to break bonds free radical more stable
R–H R• + H• H = BDE
CH3–CH2–CH3
CH3CH2CH2• CH3CHCH3• lower energy, more stable,
easier to form
98 kcal 95 kcalReactivity of C–H:
3º > 2º > 1º > CH3–H
C. Higher alkanes: regioselectivitySome alkanes give only one monohalo product:
CH3 CH3 CH3 CH2 ClCl2hn
Cl2hn
Cl2hn Cl
Cl
But:Cl2hn CH3CH2CH2Cl +CH3CH2CH3 CH3CHCH3
Cl
find: 43% 57%even though statistically: 75% 25%
(6 H) (2 H)
Syntheticallyuseful.
Not asuseful.
C. Higher alkanes: regioselectivity
Reactivity of C–H: 3º > 2º > 1º-for Cl2, relative reactivity is 5.2 : 3.9 : 1
Cl2hn CH3CH2CH2Cl +CH3CH2CH3 CH3CHCH3
ClPredicting relative amounts of monochloro product:
= x2º product1º product
reactivity of 2º Hreactivity of 1º H
number of 2º H’snumber of 1º H’s
= = =3.9 x 21 x 6
7.86
57%43%
C
CH3
CH3
H
C
H
H
C
CH3
CH3
H
Cl2
hvC
CH3
CH3
H
C
H
H
C
CH3
CH2
H
Cl C
CH3
CH3
H
C
H
H
C
CH3
CH3
Cl
C
CH3
CH3
H
C
H
Cl
C
CH3
CH3
H
1-chloro-2,4-dimethylpentane 2-chloro-2,4-dimethylpentane 3-chloro-2,4-dimethylpentane
# H 12 2 2reactivity factor x 1 x 5.2 x 3.9 12 10.4 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
12 H, primary 2H, tertiary 2H, secondary
Cl2hn
Bromine is much more selective:
Cl2hn
Br2hn
Cl
Cl
+43% 57%
3% 97%
Relative reactivities for Br2: 3º 2º 1º1640 82 1
Syntheticallymore useful.
Br2hn