Chemistry 125: Lecture 43
January 24, 2011
Solvation & Water Dissocation
Brønsted Acidity
Nucleophilic Substitution
and its Components
This
For copyright
notice see final
page of this file
The Importance of Solvent for Ionic Reactions
E±Coulomb = -332.2 / dist (Å) [long-range attraction; contrast radical bonding]
kcal/mol
400
300
H+
+
OH-
392
(g)
H+ :OH2 bonding
plus close proximity
164 !
of + to eight electrons e transfer
OH- (aq)
similar
(polarizability shifts e-cloud)
H3O+ (g)
etc,
etc,
etc
106
28
Sum = 370
18
200
100
H3O+ (aq)
100
0
-(3/4  386)  10-290
K  10
BDE HO-H 120
H2O (g)
6.3
H2O (aq)
From small difference of
large numbers!
pKa = 15.8
21.5
H+(aq) + OH-(aq)
Brønsted Acidity
Substitution at Hydrogen
ABN
ABN AON
ABN
F
F
InMake
a Solvent!
Two
CH2
F H :OH2
+
CH2 F H H:OH
Break Two
OH2
CH2
CH2
H OH "E2 Elimination"
Make & Break
(Cf. Lecture 16)
Fortunately solvation energies of
analogous compounds are similar
enough that we can often make
reasonably accurate predictions
(or confident rationalizations)
of relative acidities in terms
of molecular structure.
When pH = pKa
Why should organic chemists bother about pH and pKa, which seem like
topics for general chemistry?
a) Because whether a molecule is ionized or not is important for
predicting reactivity (HOMO/LUMO availability), conformation, color,
proximity to other species, mobility (particularly in an electric field), etc.
b) Because the ease with which a species reacts with a proton might
predict how readily it reacts with other LUMOs (e.g. *C-X or *C=O).
[H+] [B-]
Ka =
[HB]
[B-]
pH = pKa + log
[HB]
= pKa, when HB is half ionized
With known pKa, measure pH by measuring [B-] / [HB].
Single indicators work best over ~2.5 pH units (95:5 - 5:95).
Bootstrap with overlapping indicators for wide coverage.
Factors that Influence
Brønsted Acidity
16
HOH
15.7
(BDE 119)
12
+
H3NH
9.2
HSH
7.0
(BDE
FH
3.2
(BDE 136)
pKa
8
91)
4
0
+
H2OH
-4
-1.7
Learning from
pKa Values
E-Mismatch
Decrease
of
Overlap
(e-negativity difference)
Ease of Heterolysis, Ka
Ease of Homolysis
pKa
BDE
H CH3
~55
105
H NH2
~35
108
H OH
H F
16
119
3
136
H SH
H Cl
7
91
~ -3
103
H Br
~ -5
88
H I
~ -9
71
16
HOH
Learning from
pKa Values
15.7
12
+
O
H3NH
9.2
HSH
7.0
9
pKa
8
4
3.2
0
+
H2OH
-4
-1.7
CH3-C-CH-C-CH3
H
O
4.8
FH
O
2.9
CH3-COH
O
ClCH2-COH
CH3 O
H3NCH-COH
+
Titration of Alanine
CH3 O
H2NCH-CO-
12
slow
10
(buffered)
8
pH
CH3 O
HH
H23NCH-CO+
6
4
It requires 0.50 equivalents
to change the ratio 9-fold
(from 75/25 to 25/75)
slow
2
CH3 O
H3NCH-COH 0
(buffered)
And only 0.03 equivalents
to change the ratio 9-fold
(from 3/100 to 1/300)
But only 0.22 equivalents to
change the ratio 9-fold
(from 25/75 to 3/97)
+
0.5
1.0
1.5
Equivalents of OH- added
2.0
Titration of Alanine
12
Then proximity of negative charge should make it ~300 times harder
to remove H+ from alanine “zwitterion” than from H3N+-CH2CH3 (
).
Actually it is 5 times easier!
pK2 9.87 10
8
Ar O
H3NCH-COCH3
But it is 400 times
harder than the
corresponding ester.
pH
+
CH3 O
HH
H23NCH-CO+
6
4
pK1 2.35
(pKa 7.3)
Apparently the CO2 group
without charge is sufficiently
electron withdrawing to
destabilize the cation more
than the negative charge
stabilizes it.
Reasonable that proximity of positive charge
makes it ~300 times easier to remove H+ from
alanine cation than from acetic acid (pKa 4.5)
2
CH3 O
H3NCH-COH 0
CH3 O
H2NCH-CO-
+
0.5
1.0
1.5
Equivalents of OH- added
2.0
Approximate “pKa” Values
50
CH3-CH2CH2CH2H ~ 52
pKa *
40
CH3-C C-CH2H
CH3-CH=C=CHH
(best E-match C-H)
_
2
sp C (no  overlap)
:
CH3-CH2CH=CHH ~ 44
_
3
sp C
(allylic)
_
C HOMO -  overlap
(better E-match N-H)
~ 34 H2NH
~ 38
30
CH3-CH2C CH : ~ 25
_
sp C
(no  overlap)
20
16 HOH
10
(bad E-match O-H)
* Values are approximate because HA1 + A2- = A1- + HA2 equilibria
for bases stronger that HO- cannot be measured in water. One must
“bootstrap” by comparing acid-base pairs in other solvents.
1st of 6 pages from
http://evans.harvard.edu/pdf/evans_pKa_table.pdf
Cf. http://research.chem.psu.edu/brpgroup/pKa_compilation.pdf
Problems for Wednesday:
1) List factors that help determine pKa for an acid.
2) Choose a set of several related acids from one of
the pKa Tables or from your text (inside back cover of J&F),
and explain what they teach about the relative
importance of these factors.
2) Explain your conclusions to at least one other class
member and decide together how unambiguous
your lesson is.
Feel free to consult a text book and its problems
or the references at the end of the Tables.
Hint: this could provide a good exam question.
Nucleophilic Substitution
and -Elimination
Chapter 7
ABN
(Cf. Lecture 16)
ABN AON
ABN
Make Two
F CH2
F H :OH
F CH3 :OH
CH2 H :OH
Break
Two
CH2
F FH OH
F
CH3 OH
"E2 Elimination"
CH
H
OH
"SN2 Substitution"
2
"Acid-Base"
Same
Make & Break
All are Nucleophilic
Generalization
Substitution
Williamson Ether Synthesis (1852)
*
EtBr
+ +
O- Na
HOMO
+
OEt + Na Br
LUMO
Finkelstein Reaction (1910)
acetone
*
+
+ RCl
Na I
also RBr
RI + Na+ Cl- ()
Na+ Br
-
Exchange
Ions
(Double Decomposition)
Menschutkin Reaction (1890)
+
*
Et3N + RI
Et3N-R + I-
Create
Ions
Meerwein Reagent (1940s)
* +
(CH
)
O
BF
+
+
3 3
4
RO Na
Destroy
Ions
Solvolysis
*
(CH3)3C-Br
EtOH
ROCH3 + (CH3)3O
+ Na+ BF4
(CH3)3C-OEt + HBr
Breaking apart
by solvent
Generality of Nucleophilic Substitution
Solvent
Nu:
R-L
Nucleophile
Substrate
But there
are different
mechanisms!
+
METHIONINE
Substitute
NHR2 for SR2
at C
L
Product
Leaving
Group
Substitute
SR2 for “OH”
at C
+
OH
CH3
ARGININE
H
Nu-R
-
:
+
(-)
(+)
RIBOSE
Substitute
Biological Methylation
Base for NR3
(Protein Modification by Methyl
etc.)
at Transferase,
H
H OH
ADENINE
Substitute
ADENOSINE
NR2
for “OH”
at C
S-Adenosylmethionine
End of Lecture 43
Jan. 24, 2011
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J. M. McBride, Chem 125. License: Creative Commons BY-NC-SA 3.0