This document discusses alkyl halides and their reactions, including nucleophilic substitution and elimination reactions. It covers industrial sources of alkyl halides like vinyl chloride, methods of preparing alkyl halides from alcohols using reagents like thionyl chloride, and reactions like halogenation of hydrocarbons and addition of hydrogen halides to alkenes and alkynes. It also discusses nucleophilic substitution reactions, factors that affect the rate and mechanism such as nucleophiles and leaving groups. The mechanisms of SN1 and SN2 reactions are explained along with effects of stereochemistry, carbocation rearrangements, solvent polarity, and reactivity based on structure.

1. © E.V. Blackburn, 2010
Alkyl halides
Nucleophilic substitution and
elimination reactions

2. © E.V. Blackburn, 2010
Alkyl halides - industrial
sources
H C C H
HCl
HgCl2
H2C=CHCl
vinyl chloride
H
H H
vinyl

3. © E.V. Blackburn, 2010
Alkyl halides - industrial
sources
H C C H
HCl
HgCl2
H2C=CHCl
vinyl chloride
H2C=CH2
Cl2
500o
H2C=CHCl
CH3Cl + Hg2F2
CH3F + Hg2Cl2
CCl4 + SbF3 CCl2F2
Freon-12

4. © E.V. Blackburn, 2010
Preparation from alcohols
R-OH
HX
or PX3
R-X
or SOCl2
SOCl2 - thionyl chloride
RCH2OH + SOCl2 RCH2Cl + HCl +SO2

5. © E.V. Blackburn, 2010
Halogenation of hydrocarbons
R-H RX
X2/h
CH3 CH2Br
Br2
h
lachrymatory

6. © E.V. Blackburn, 2010
Addition of HX to alkenes
C C
HX
C C
H X

7. © E.V. Blackburn, 2010
C C
X2
C C
X X
C C
2X2
C C
X X
X X
Addition of halogens to
alkenes and alkynes

8. © E.V. Blackburn, 2010
R-X + NaI R-I + NaX
acetone
soluble insoluble
Finkelstein reaction

9. © E.V. Blackburn, 2010
Nucleophilic substitution
reactions
When bonded to a carbon, the halogen is easily displaced
as a halide ion by stronger nucleophiles - it is a good
leaving group.
The typical reaction of alkyl halides is a nucleophilic
substitution:
R-X + Nu R-Nu +
the leaving
group
X-
The halide ion is the conjugate base of a strong acid. It
is therefore a very weak base and little disposed to share
its electrons.

10. © E.V. Blackburn, 2010
Nucleophiles
• reagents that seek electron deficient centres
• negative ions or neutral molecules having at least
one unshared pair of electrons
C C
H3C
-
+ CH3-Br C C
H3C CH3 + Br-
H3C
H
O + CH3-I H3C
H
O
CH3
+
+ I-
nucleophile leaving group

11. © E.V. Blackburn, 2010
Leaving groups
• a substituent that can leave as a weakly basic
molecule or ion
L L
Nu Nu
Nu + L:
+
Br Br
NC NC
CN-
+ Br-
 
OH2
Cl
-
OH2
Cl Cl + H2O
 
+
Br Br
PH3P Ph3P + Br
-
 
Ph3P:
+

12. © E.V. Blackburn, 2010
Nucleophilic substitution
CH3Br + CH3OH + Br-
OH-
A knowledge of how reaction rates depend on reactant
concentrations provides invaluable information about
reaction mechanisms. What is known about this
reaction?

13. © E.V. Blackburn, 2010
Nucleophilic substitution
rate = k[CH3Br][OH-]
CH3Br + CH3OH + Br-
OH-
rate a [CH3Br][OH-]
0.002 M 1.0 M 6 x 10-7 molL-1s-1
[CH3Br]I [OH-]I initial rate
0.001 M 1.0 M 3 x 10-7 molL-1s-1
0.002 M 2.0 M 1.2 x 10-6 molL-1s-1

14. © E.V. Blackburn, 2010
Order - a summary
The order of a reaction is equal to the sum of the
exponents in the rate equation.
Thus for the rate equation rate = k[A]m[B]n, the overall
order is m + n.
The order with respect to A is m and the order with
respect to B is n.

15. © E.V. Blackburn, 2010
Nucleophilic substitution
CH3-C-CH3
CH3
Br
CH3-C-CH3
CH3
OH
+ + Br-
OH-
rate a [(CH3)3CBr][OH-]0
rate = k[(CH3)3CBr]
0.002 M 1.0 M 8 x 10-7 molL-1s-1
[(CH3)3CBr]I [OH-]I initial rate
0.001 M 1.0 M 4 x 10-7 molL-1s-1
0.002 M 2.0 M 8 x 10-7 molL-1s-1

16. © E.V. Blackburn, 2010
References of interest:
E.D. Hughes, C.K. Ingold, and C.S. Patel, J. Chem. Soc., 526 (1933)
J.L. Gleave, E.D. Hughes and C.K. Ingold, J. Chem. Soc., 236 (1935)
CH3Br + CH3OH + Br-
OH-
rate = k[CH3Br][OH-]
The SN2 mechanism
OH
-
Br HO
+ Br-
HO Br
- -

17. © E.V. Blackburn, 2010
Stereochemistry of the SN2
reaction
OH
-
Br HO
+ Br-
HO Br
- -
Br
C6H13
H3C
H
(-)-2-bromooctane
[a] = -34.6o
OH
C6H13
H3C
H
(-)-2-octanol
[a] = -9.9o
(+)-2-octanol
[a] = +9.9o
HO
C6H13
CH3
H

18. © E.V. Blackburn, 2010
A Walden inversion.
P. Walden, Uber die vermeintliche optische Activät der
Chlorumarsäure und über optisch active Halogen-
bernsteinsäre, Ber., 26, 210 (1893)
Stereochemistry of the SN2
reaction
Br
C6H13
H3C
H HO
C6H13
CH3
H
(-)-2-bromooctane (+)-2-octanol
[a] = -34.6o
[a] = +9.9o
NaOH
SN2
optical purity = 100%

19. © E.V. Blackburn, 2010
The SN1 mechanism
rate =
CH3-C-CH3
CH3
Br
CH3-C-CH3
CH3
OH
+ +
OH-
Br-
k[(CH3)3CBr]
H3C C
CH3
Br
CH3
1. H3C C
CH3
CH3
+
+ Br-
2. H3C C
CH3
CH3
+
+ OH-
H3C C
CH3
CH3
OH
slow
fast

20. © E.V. Blackburn, 2010
Carbocations
G.A. Olah, J. Amer. Chem. Soc., 94, 808 (1972)
CH3 CH3CH2 CH3CHCH3 CH3CCH3
CH3
+ + + +
1o
2o
3o
sp2

21. © E.V. Blackburn, 2010
Carbocation stability
R C
R
R
+
3o
> R C
R
H
+
2o
> R C
H
H
+
1o
> H C
H
H
+
Hyperconjugation stabilizes the positive charge.
H
H
H
H
H

22. © E.V. Blackburn, 2010
Stereochemical consequences of
a carbocation
H3C C
CH3
Br
CH3
1. H3C C
CH3
CH3
+
+ Br-
slow
OH-
H2O
SN1
Br
C6H13
H3C
H
(-)-2-bromooctane
[a] = -34.6o
?

23. © E.V. Blackburn, 2010
Stereochemical consequences of
a carbocation
H3C C
CH3
Br
CH3
1. H3C C
CH3
CH3
+
+ Br-
slow
(+)-C6H13CHOHCH3
OH-
H2O
SN1
reduced optical
purity
Br
C6H13
H3C
H
(-)-2-bromooctane
[a] = -34.6o
Why?

24. © E.V. Blackburn, 2010
retention
Stereochemical consequences of
a carbocation
CH3
H
C6H13
X
-
+
inversion predominates
C6H13
CH3
H
HO
H2O

25. © E.V. Blackburn, 2010
Carbocation rearrangements
(CH3)3CCH2Br
C2H5O-
SN2
C2H5OH
SN1
(CH3)3CCH2OC2H5
Williamson ether synthesis
(CH3)2CCH2CH3
OC2H5
+
(CH3)2C=CHCH3
a rearrangement and
elimination

26. © E.V. Blackburn, 2010
Carbocation rearrangements
1,2 hydride and alkyl shifts
CH3CH2CH2CH2
+ +
1
o
2
o
CH3CH2CHCH3
C C
H
. .
+
C C
+ H
C C
R
. .
+
C C
+ R

27. © E.V. Blackburn, 2010
Carbocation rearrangements
(CH3)3CCH2Br (CH3)3CCH2
+
CH3
CH3
H3C
H
H
+
CH3
H3C CH2CH3
+
CH3
H3C CH2CH3
+
C2H5OH
CH3
H3C CH2CH3
OC2H5
H
+
CH3
H3C CH2CH3
OC2H5
H
+
-H+ CH3
H3C CH2CH3
OC2H5

28. © E.V. Blackburn, 2010
HO Br
- -
Steric effects in the SN2
reaction
OH
-
Br HO
HO Br
- -
+ Br-
Look at the transition state to see how substituents might
affect this reaction.

29. © E.V. Blackburn, 2010
Steric effects in the SN2
reaction
The order of reactivity of RX in these SN2 reactions is
CH3X > 1o > 2o > 3o
HO Br
- -

30. © E.V. Blackburn, 2010
Steric effects in the SN2
reaction
> (CH3)2CHBr
0.01
CH3Br >
150
reactivity
CH3CH2Br
1
RBr + I-
RI + Br-
- -
I Br - -
I Br - -
I Br
- -
I Br
> (CH3)3CBr
0.001

31. © E.V. Blackburn, 2010
Structural effects in SN1
reactions
3o > 2o > 1o > CH3X
R-X R X
+ -
R+
+ X-
RBr + H2O ROH + HBr
HCO2H
(CH3)3CBr > (CH3)2CHBr > CH3CH2Br > CH3Br
100,000,000 45 1.7 1

32. © E.V. Blackburn, 2010
Nucleophilicity
A base is more nucleophilic than its conjugate acid:
CH3Cl + H2O  CH3OH2
+ slow
CH3Cl + HO-  CH3OH fast
The nucleophilicity of nucleophiles having the same
nucleophilic atom parallels basicity:
RO- > HO- >> RCO2
- > ROH >H2O
Rates of SN2 reactions depend on concentration and
nucleophilicity of the nucleophile.

33. © E.V. Blackburn, 2010
Nucleophilicity
When the nucleophilic atoms are different, their relative
strengths do not always parallel their basicity.
In protic solvents, the larger the nucleophilic atom, the
better:
I- > Br- > Cl- > F-
In protic solvents, the smaller the anion, the greater its
solvation due to hydrogen bonding. This shell of solvent
molecules reduces its ability to attack.

34. © E.V. Blackburn, 2010
Nucleophilicity
Aprotic solvents tend to solvate cations rather than
anions. Thus the unsolvated anion has a greater
nucleophilicity in an aprotic solvent.

35. © E.V. Blackburn, 2010
Polar aprotic solvents
H N
O
CH3
H3C
N,N-dimethylformamide
DMF
H3C CH3
S
O
dimethyl sulfoxide
DMSO
P
O
hexamethylphosphoramide
HMPA
N(CH3)2
(H3C)2N
N(CH3)2
These solvents dissolve
ionic compounds.

36. © E.V. Blackburn, 2010
Solvent polarity
more polar transition state less
solvated than reagents
A protic solvent will decrease the rate of this reaction and
the reaction is 1,200,000 faster in DMF than in methanol.
Cl
-
I
H
H
H Cl I
- -

37. © E.V. Blackburn, 2010
Solvent polarity
R-X R X
+ -
R+
+ X-
less polar more polar
greater stabilization by
polar solvent
The transition state is more polarized.
Therefore the rate of this reaction increases with
increase in solvent polarity.
A protic solvent is particularly effective as it stabilizes
the transition state by forming hydrogen bonds with the
leaving group.

38. © E.V. Blackburn, 2010
Solvent polarity
Explain the solvent effects for each of the following second
order reactions:
a) 131I- + CH3I  CH3
131I + I-
Relative rates: in water, 1; in methanol, 16; in ethanol, 44
b) (n-C3H7)3N + CH3I  (n-C3H7)3N+CH3 I-
Relative rates: in n-hexane, 1; in chloroform, 13 000

39. © E.V. Blackburn, 2010
Leaving group ability
Weak bases are good leaving groups.
They are better able to accommodate a negative charge
and therefore stabilize the transition state.
Thus I- is a better leaving group than Br-.
I- > Br- > Cl- > H2O > F- > OH-

40. © E.V. Blackburn, 2010
SN1 v SN2
SN1 SN2
reactivity: 3o > 2o > 1o > CH3X CH3X > 1o > 2o > 3o
rearrangements no rearrangements
partial inversion inversion of configuration
eliminations possible
kinetics: 1st order second order

41. © E.V. Blackburn, 2010
Problems
Try problems 6.6 – 6.11 and 6.14 – 6.16 in chapter 6 of
Solomons and Fryhle.

42. © E.V. Blackburn, 2010
Functional group transformations
using SN2 reactions
CN-
R-CN nitrile
R C C R'
alkyne
'R
C
C-
R = Me, 1o, or 2o

43. © E.V. Blackburn, 2010
Problems
Try problems 6.12 and 6.17 in chapter 6 of Solomons
and Fryhle.

44. © E.V. Blackburn, 2010
ROH + HX - an SN reaction
ROH + HX RX + H2O
HX: HI > HBr > HCl
ROH: 3o
> 2o
> 1o
CH3CHCH3
OH
HBr or
NaBr/H2SO4
CH3CHCH3
Br

45. © E.V. Blackburn, 2010
Experimental facts
1. The reaction is acid catalyzed
H3C C
CH3
H
C
H
OH
CH3
HCl
H3C C
CH3
Cl
C
H
H
CH3
3. Alcohol reactivity is 3o > 2o > 1o < CH3OH
2. Rearrangements are possible

46. © E.V. Blackburn, 2010
The mechanism
1. ROH + HX
+
ROH2 + X-
2. ROH2
+
+
R + H2O
3. R +
+ -
X RX

47. © E.V. Blackburn, 2010
Reaction of primary alcohols
with HX
SN2
1. ROH + HX
+
ROH2 + X-
1o
2. ROH2
+
+ X-
X R OH2
- +
RX + H2O
HX: HI > HBr > HCl

48. © E.V. Blackburn, 2010