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alkylgghhhhhhhyyggghyyyyyuuu_halides.pptx
- 2. Alkyl halides - industrial
sources
H C C H
HCl
HgCl2
H2C=CHCl
vinyl chloride
H H
H
vinyl
© E.V. Blackburn, 2010
- 3. Alkyl halides - industrial
sources
H C C H
HCl
HgCl2
H2C=CHCl
vinyl chloride
H2C=CH2
Cl2
500o
H2C=CHCl
CH3Cl + Hg2F2
CCl4 + SbF3
CH3F + Hg2Cl2
CCl2F2
Freon-12
© E.V. Blackburn, 2010
- 7. C C
© E.V. Blackburn, 2010
X2
C C
X X
C C
2X2
X X
C C
X X
Addition of halogens to
alkenes and alkynes
- 8. R-X + NaI
© E.V. Blackburn, 2010
acetone
R-I + NaX
soluble insoluble
Finkelstein reaction
- 9. Nucleophilic substitution
reactions
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.
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 + X-
the leaving
group
© E.V. Blackburn, 2010
- 10. Nucleophiles
• reagents that seek electron deficient centres
• negative ions or neutral molecules having at least
one unshared pair of electrons
-
3 3
H C C C + CH -Br 3
H C C C CH3 + Br-
H3C O
H
+ CH3-I H3C O
CH3
+ H
+ I-
nucleophile leaving group
© E.V. Blackburn, 2010
- 11. Leaving groups
• a substituent that can leave as a weakly basic
molecule or ion
Nu L
Nu + L Nu + L:
Br
NC Br NC
CN-
+ Br-
2
OH Cl OH2
Cl-
Cl 2
+ H O
+
Br
© E.V. Blackburn, 2010
PH3P
Br
+
Ph3P + Br
-
3
Ph P:
- 12. Nucleophilic substitution
CH3Br + CH3OH + Br-
© E.V. Blackburn, 2010
OH-
A knowledge of how reaction rates depend on reactant
concentrations provides invaluable information about
reaction mechanisms. What is known about this
reaction?
- 13. Nucleophilic substitution
rate [CH3Br][OH-]
rate = k[CH3Br][OH-]
CH3Br + CH3OH + Br-
© E.V. Blackburn, 2010
OH-
[CH3Br]I
0.001 M
[OH-]I
1.0 M
initial rate
3 x 10-7 molL-1s-1
0.002 M 1.0 M 6 x 10-7 molL-1s-1
0.002 M 2.0 M 1.2 x 10-6 molL-1s-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. Nucleophilic substitution
Br
© E.V. Blackburn, 2010
CH3
CH3-C-CH3
OH
CH3
CH3-C-CH3 + OH-
+ Br-
rate [(CH3)3CBr][OH-]0
rate = k[(CH3)3CBr]
[(CH3)3CBr]I
0.001 M
[OH-]I
1.0 M
initial rate
4 x 10-7 molL-1s-1
0.002 M 1.0 M 8 x 10-7 molL-1s-1
0.002 M 2.0 M 8 x 10-7 molL-1s-1
- 16. + Br-
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
-
HO
© E.V. Blackburn, 2010
-
Br
- 17. Stereochemistry of the SN2
reaction
OH- Br
-
HO
-
Br
Br
C6H13
H
H3C
(-)-2-bromooctane
[] = -34.6o
OH
C6H13
H
H3C
(-)-2-octanol
[] = -9.9o
(+)-2-octanol
[] = +9.9o
© E.V. Blackburn, 2010
HO
HO
+ Br-
C6H13
H
CH3
- 18. 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
© E.V. Blackburn, 2010
C6H13
H
H3C
HO
C6H13
H
CH3
(-)-2-bromooctane
[] = -34.6o
(+)-2-octanol
[] = +9.9o
optical purity = 100%
NaOH
N
S 2
- 19. The SN1 mechanism
rate = k[(CH3)3CBr]
Br
CH3
CH3-C-CH3 + OH-
CH3
CH3-C-CH3
OH
+ Br-
CH3
1. H3C C CH3
Br
CH3
+
H3C C CH3 + Br-
2.
© E.V. Blackburn, 2010
CH3
+
H3C C CH3 + OH-
CH3
H3C C CH3
OH
slow
fast
- 20. Carbocations
G.A. Olah, J. Amer. Chem. Soc., 94, 808 (1972)
CH3
CH3 CH3CH2
+ +
1o
CH3CHCH3 CH3CCH3
+ +
2o
3o
sp2
© E.V. Blackburn, 2010
- 21. Carbocation stability
R
R C +
R
3o
H
> R C +
R
2o
H
> R C + >
H
1o
H
H C +
H
Hyperconjugation stabilizes the positive charge.
H
© E.V. Blackburn, 2010
H
H
H
H
- 22. Stereochemical consequences of
a carbocation
CH3
1. H3C C CH3
Br
CH3
H3C C CH3
+
+ Br-
slow
OH-
© E.V. Blackburn, 2010
H O
2
SN1
Br
C6H13
H
H3C
(-)-2-bromooctane
[] = -34.6o
?
- 23. Stereochemical consequences of
a carbocation
CH3
H3C C CH3
+
+ Br-
slow
6 13
© E.V. Blackburn, 2010
(+)-C H CHOHCH3
OH-
H O
2
SN1
reduced optical
purity
Br
CH3
1. H3C C CH3
Br
C6H13
H
H3C
(-)-2-bromooctane
[] = -34.6o
Why?
- 28. HO Br
- -
Steric effects in the SN2
reaction
OH- Br HO
+ Br-
-
HO
© E.V. Blackburn, 2010
-
Br
Look at the transition state to see how substituents might
affect this reaction.
- 29. Steric effects in the SN2
reaction
The order of reactivity of RX in these SN2 reactions is
CH3X > 1o > 2o > 3o
HO
© E.V. Blackburn, 2010
Br
- -
- 30. 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
-
© E.V. Blackburn, 2010
-
I Br
> (CH3)3CBr
0.001
- 31. Structural effects in SN1
reactions
3o > 2o > 1o > CH3X
R-X
+ -
R X R+
+ X-
HCO2H
RBr + H2O ROH + HBr
(CH3)3CBr > (CH3)2CHBr > CH3CH2Br > CH3Br
100,000,000 45 1.7 1
© E.V. Blackburn, 2010
- 32. © E.V. Blackburn, 2010
Nucleophilicity
Rates of SN2 reactions depend on concentration and
nucleophilicity of the nucleophile.
A base is more nucleophilic than its conjugate acid:
CH3Cl + H2O CH3OH2+
CH3Cl + HO- CH3OH
slow
fast
The nucleophilicity of nucleophiles having the same
nucleophilic atom parallels basicity:
-
RO- > HO- >> RCO2 > ROH >H2O
- 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. Polar aprotic solvents
H
O
N
CH3
H C
3
N,N-dimethylformamide
DMF
O
H3C
S
CH3
dimethyl sulfoxide
DMSO
These solvents dissolve
ionic compounds.
O
(H3C)2N P N(CH3)2
N(CH3)2
hexamethylphosphoramide
HMPA
© E.V. Blackburn, 2010
- 36. 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
© E.V. Blackburn, 2010
- -
I
- 37. Solvent polarity
R-X
© E.V. Blackburn, 2010
+ -
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:
131I-
3
a) + CH3I CH 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
SN2
second order
CH3X > 1o > 2o > 3o
no rearrangements
inversion of configuration
reactivity: 3o > 2o > 1o > CH3X
rearrangements
partial inversion
eliminations possible
kinetics:
SN1
1st 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.
- 43. © E.V. Blackburn, 2010
Problems
Try problems 6.12 and 6.17 in chapter 6 of Solomons
and Fryhle.
- 44. ROH + HX - an SN reaction
CH3CHCH3
© E.V. Blackburn, 2010
OH
ROH + HX RX + H2O
HX: HI > HBr > HCl
ROH: 3o
> 2o
> 1o
HBr or
NaBr/H2SO4
CH3CHCH3
Br
- 45. Experimental facts
1. The reaction is acid catalyzed
2. Rearrangements are possible
HCl
© E.V. Blackburn, 2010
H3C C C CH3 H3C C
H OH Cl
CH3 H CH3 H
C CH3
H
3. Alcohol reactivity is 3o > 2o > 1o < CH3OH
- 46. The mechanism
1. ROH + HX
+
ROH2 + X-
2. ROH2
+
+
R + H2O
+
© E.V. Blackburn, 2010
-
3. R + X RX
- 47. Reaction of primary alcohols
with HX
SN2
HX: HI > HBr > HCl
1. ROH + HX
+
ROH2 + X-
1o
+
© E.V. Blackburn, 2010
2. ROH2 + X-
- +
X R OH2 RX + H2O