1. Alkyl Halides Organo halogen Alkyl halide Aryl halide Halide vynilik
Alkyl halide Reactions :
Substitution : SN1 dan SN2
Elimination : E1 dan E2

2. NUCLEOPHILIC SUBSTITUTION
1. Leaving groups
weaker base = better leaving group
reactivity: R-I > R-Br > R-Cl >> R-F
best L.G.
most reactive
worst L.G.
least reactive
precipitate
drives rxn
(Le Châtelier)

3. 2. Mechanisms SN
general: Rate = k1[RX] + k2[RX][Y–]
k1 increases
RX = CH3X 1º 2º 3º
k2 increases
k1 ~ 0
Rate = k2[RX][Y–]
(bimolecular)
SN2
k2 ~ 0
Rate = k1[RX]
(unimolecular)
SN1

4. 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–

5. B. Stereochemistry: inversion of configuration
Stereospecific reaction:
Reaction proceeds with
inversion of configuration.
(R)-(–)-2-bromooctane
(S)-(+)-2-octanol
C. Mechanism
Back-side attack:
inversion of configuration

6. 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
Compound Rel. Rate
methyl CH3Br 150
1º RX CH3CH2Br 1
2º RX (CH3)2CHBr 0.008
3º RX (CH3)3CBr ~0
increasing
steric
hindrance
Reactivity toward SN2:
CH3X > 1º RX > 2º RX >> 3º RX
maximum steric hindrance
react
readily
by SN2
(k2 large)
more
difficult
does not
react by
SN2
(k2 ~ 0)

7. E. Nucleophiles and nucleophilicity
anions
2. neutral species
hydrolysis
alcoholysis
Summary:
very good Nu: I–, HS–, RS–, H2N–
good Nu: Br–, HO–, RO–, CN–, N3–
fair Nu: NH3, Cl–, F–, RCO2–
poor Nu: H2O, ROH
very poor Nu: RCO2H

8. SN1 Mechanism A. Kinetics e.g., 3º, no SN2
Find: Rate = k[(CH3)3CBr] unimolecular
 RLS depends only on (CH3)3CBr

9. A. Kinetics

10. A. Kinetics
Two-step mechanism: R+
RBr + CH3OH
ROCH3 + HBr

11. B. Stereochemistry: stereorandom

12. 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

13. SN1 vs SN2 A. Solvent effects nonpolar: hexane, benzene
moderately polar: ether, acetone, ethyl acetate
polar protic: H2O, ROH, RCO2H
polar aprotic: DMSO DMF acetonitrile
SN1 mechanism promoted by polar protic solvents
stabilize R+, X– relative to RX
in less polar solvents
in more polar solvents
R+X– RX

14. them more nucleophilic
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–

15. B. Summary
rate of SN1 increases (carbocation stability)
RX = CH3X 1º 2º 3º
rate of SN2 increases (steric hindrance)
react
primarily
by SN2
(k1 ~ 0, k2 large)
may go
by either
mechanism
reacts
primarily
by SN1
(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)

16. ELIMINATION REACTIONS
Dehydrohalogenation of alkyl halides
Elimination
strong base: KOH/ethanol
CH3CH2ONa/CH3CH2OH
tBuOK/tBuOH
Follows Zaitsev orientation:

17. 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
 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

18. 1. Single step, concerted mechanism:
Zaitsev

19. 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)
anti periplanar
-most molecules can adopt this conformation more easily
E2 eliminations usually occur when H and X are anti
syn periplanar
-but eclipsed!

20. 2. stereoelectronic effects: anti elimination
but

21. 2. stereoelectronic effects: anti elimination
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 gives
non-Zaitsev orientation

22. 3. the E1 mechanism
Recall:
Rate = k[RBr][B–] E2
Reactivity: 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+)

23. 3. the E1 mechanism
- and R+ can rearrange
 eliminations usually carried out with strong base

24. 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

25. 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
steric
hindrance
increases
sterically hindered nucleophile

26. B. Bimolecular: SN2 or E2?
2. base vs nucleophile
stronger base favors E2
better nucleophile favors SN2
good Nu
weak B
good Nu
strong B
poor Nu
strong B

27. C. Unimolecular: SN1 or E1? for both, Rate = k[R+][H2O]
 no control over ratio of SN1 and E1

28. 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-branching
hinders SN2

29. 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 RX
3º RX
can’t control
ratio of
SN1 to E1
SN1 and E1 (may rearrange)

30. 1. Halogenation of Alkanes
heat or
light
R–H + X2 — R–X + HX a substitution reaction
Reactivity: F2 > Cl2 > Br2 > I2
common
too
reactive
too
unreactive
(endothermic)
Cl2 hn
Cl2 hn
Cl2 hn
Cl2 hn
CH4  CH3Cl  CH2Cl2  CHCl3  CCl4
+ HCl + HCl + HCl + HCl
Problem: mixture of products
Solution: use large excess of CH4 (and recycle it)

31. 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 Termination
CH3• + CH3•  CH3–CH3 (infrequent due
CH3• + Cl•  CH3Cl to low [rad•])
Propagation
-determines
net reaction
1000’s of cycles = “chain” reaction

32. B. Stability of free radicals: bond dissociation energies
R–H  R• + H• DH = 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)
easier to break bonds
 free radical more stable
CH3CH2CH2• •
CH3CHCH3
lower energy, more stable,
easier to form
98 kcal
95 kcal
Reactivity of C–H:
3º > 2º > 1º > CH3–H
CH3–CH2–CH3

33. C. Higher alkanes: regioselectivity
Some alkanes give only one monohalo product:
Synthetically
useful.
Not as
useful.
But:
find: 43% 57%
even though statistically: 75% 25%
(6 H) (2 H)

34. 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:
2º product
1º product
reactivity of 2º H
reactivity of 1º H
number of 2º H’s
number of 1º H’s = x
3.9 x 2
1 x 6
7.8 6
57%
43% = = =

35. 12 H, primary 2H, tertiary 2H, secondary
reactivity factor x x x 3.9
sum = = 30.2
percent /30.2 x 100 = 39.7% 10.4/30.2 = 34.4% /30.2 = 25.8

36. Bromine is much more selective:
Synthetically
more useful.
Relative reactivities for Br2: 3º 2º 1º