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7/30/2019 Explicaciones Del pH
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5/24/2013 MEDC 501 Fall 2007 1
pKA and Ions in Solution
Chemical Equilibrium
A + B C + Dk1
k2
At equilibrium, k1 [A] [B] = k2 [C] [D]
EK
BA
DC
k
k
]][[
]][[
2
1
Le Chateliers Principle: When a stress is applied on a system at equilibrium,
the system will re-adjust to diminish the stress or counteract the change.
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pKA and Ions in Solution
H2O Equilibrium
EKOH
OHH
][
]][[
2
H2O H + OHk1
k2
+ _
142 105.55][]][[
WEE KKOHKOHH
1410log]][log[ OHH
14]log[]log[ OHH
14]log[]log[ OHH
14 pOHpH
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pKA and Ions in Solution
Ionization of a weak acid
CH3COOH + H2O H3O+
+ CH3COO-
]][[
]][[
32
33
COOHCHOH
COOCHOHKE
][
]][[][
3
332
COOHCH
COOCHOHKOHK AE
][
][log]log[log
3
33
COOHCH
COOCHOHKA
][
].[log
][
][log
3
3
acid
baseconjpH
COOHCH
COOCHpHpKA
ApKpHacid
baseconj
][
].[log Henderson-Haselbach Equation
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pKA and Ions in Solution
What is the physical meaning of pKA?
11
1][
].[log ApKpHacidbaseconj
22
2
][
].[log ApKpH
acid
baseconj
}{}{
][
].[log
][
].[log 21
2
2
1
1AA pKpHpKpH
acid
baseconj
acid
baseconj
212
2
1
1
].[
][
][
].[log AA pKpK
baseconj
acid
acid
baseconj
122
1
].[
].[log AA pKpK
baseconj
baseconj [Unionized Acid1] =
[Unionized Acid2]
If pKA1 = 4 and pKA2 = 1. 001.01010
].[
].[ 3)41(
2
1
baseconj
baseconj
].[001.0].[ 21 baseconjbaseconj ][
].[001.0
][
].[
2
2
1
1
acid
baseconj
acid
baseconj
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Consequences of Bond Polarity
Inductive Effect - Influence on pKa
CH3
O
O
HF
3C O
H
O
pKa 4.5 0.23
CH3COOH ClCH2COOH
pKa 4.5 2.9
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Consequences of Bond Polarity
Inductive Effect - Additive Property
pKa
CH3COOH 4.5
ClCH2COOH 2.9
Cl2CHCOOH 1.3
Cl3CCOOH 0.7
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Consequences of Bond Polarity
Inductive Effect - Distance Dependent Property
pKa
CH3CH2CH2COOH 4.8
ClCH2CH2CH2COOH 4.5
CH3CHClCH2COOH 4.1
CH3CH2ClCH2COOH 2.8
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Consequences of Bond Polarity
Inductive Effect - Group Electronegativities
pKaNO2CH2COOH 1.7
ClCH2COOH 2.9COOHCH2COOH 2.8CNCH2COOH 3.5
Electron-withdrawing groups -C=N 0.5+1.0 = 1.5
-COOH 0.25+1.25+1.0+0 = 2.5
-CHO 0.25+1.25+0 = 1.5
-NO2 0.5+1.25+1.25 = 3.0
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pKa 6.1 5.7
Consequences of Bond Polarity
Influence on pKa . through space effect
ClCl
COOHCOOH
Cl
Cl
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pKA and Solubility in Water
NN
NH
O
F COOH
Ciprofloxacin
(piperazine)
Antibacterial
(anthrax)
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A molecule may have more than one pKA value!!
Consequences of Bond Polarity
Inductive Effect Multiple pKA values
OH O
OO
OH OH
OO
O O
OO
_ _ _
pKA = 2.8 pKA = 5.7
ONH
3
O
OHNH
3
O
ONH
2
OpKA = 2.35 pKA = 9.78
_ _+ +
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Consequences of Bond Polarity
Natural Amino Acids Structure and pKA
R
NH2
HCOOH Structure of Amino Acid Residues
All L-configuration or R geometry(except for glycine)
Non-Polar R = Polar R =
Glycine H Serine HOCH2
Alanine CH3 Threonine CH3CH(OH)
Valine (CH3)2CH Cysteine HSCH2
Leucine (CH3)2CH2CH Tyrosine HOC6H4CH2
Isoleucine CH3CH2CH(CH3) Asparagine H2NC(O)CH2
Phenyalanine PhCH2 Glutamine H2NC(O)CH2CH2
Methionine H3CSCH2CH2
Tryptophan
NH
CH2
__
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Consequences of Bond Polarity
Natural Amino Acids Structure and pKA
R
NH2
HCOOH Structure of Amino Acid Residues
All L-configuration or R geometry(except for glycine)
Acidic R = Basic R =
Aspartic Acid HOOCCH2 Lysine H2N(CH2)4
Glutamic Acid HOOCCH2CH2
Arginine H2NC(NH)NH(CH2)3
HistidineN
NH
H2C
__
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pKA Values of Natural Amino AcidsResidues pKA1 pKA2 Side-Chain pKA (group identification)
Non-Polar
Glycine 2.35 9.78
Alanine 2.35 9.87Valine 2.29 9.74
Leucine 2.33 9.74
Isoleucine 2.32 9.76
Phenyalanine 2.16 9.18
Methionine 2.13 9.28
Tryptophan 2.43 9.44
Hydrogen-bondingSerine 2.19 9.21
Threonine 2.09 9.10
Cysteine 1.92 10.78 8.33 (thiol group)
Tyrosine 2.20 9.11 10.13 (phenol group)
Asparagine 2.10 8.84
Glutamine 2.17 9.13
Acidic
Aspartic Acid 1.99 9.90 3.90 (g-COOH group)Glutamic Acid 2.10 9.47 4.07 (g-COOH group)Basic
Lysine 2.16 9.18 10.79 (-amino group)Arginine 1.82 8.99 12.48 (guanidino group)
Histidine 1.80 9.33 6.04 (imidazole group)
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pKA Values of Natural Amino Acids
and the Dominant Form
Serine
Cysteine
Tyrosine
Asparagine
Arginine
Aspartic Acid
Amino Acid pH 3.0 pH 7.0 pH 10.0
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Consequences of Bond Polarity
Influence on pKa of bases
pKA = pH - log[RNH2]
[RNH3+]
RNH3+ + H2O H3O+ + RNH2
CH3CH2NH2 FCH2CH2NH2
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Consequences of Bond Polarity
Inductive Effect
depend on the eW atom or eN atom is distance dependent may be re-inforced or cancelled is additive can affect the acidity or basicity of molecules can affect the physical properties of molecules
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Consequences of Bond Polarity
Resonance EffectNH
2NH
2
pKa 9-11 ~5
Resonance, Aromaticity & Conjugation
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Consequences of Bond Polarity
Resonance EffectNH
2NH
2
pKa 9-11 ~5
NH3+
NH3+
NH2
NH2
+ H2O + H3O+
+ H2O + H3O+
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Consequences of Bond Polarity
Resonance Effect
pKa 4.2 3.4 4.5
COOH COOH
NO2
COOH
OMe
N
OOH
O O
OMe
OOH
N
O O
OO
OMe
OO
+ H2O + H3O+
+ H2O + H3O+
_
_
__+ +
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Consequences of Bond Polarity
Resonance Effect
Electron-donating groups . OH, OMe, NH, NH2, NCH3
Electron-withdrawing groups . NO2, COOH, CHO, CN,SO3H, SO2NH2, SO2Cl
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Group Acid form Conjugate base form pKA
Amines RCH2NHR2+ RCH2NR2 8-11
Acids RCOOH RCOO- 3-5
Amides RCONH2 .. Does not ionize (neutral) ..
Alcohols ROH .. Does not ionize (neutral) ..
Phenols PhOH PhO- 9-11
Anilines PhNH3+ PhNH2 3-5
Imidazole 6.9
Pyridines 4-6
Imides 8-10
1,3-dicarbonyls 11-13
Imines ~7
Ketones R-C(=O)-R .. Does not ionize (neutral) ..
Aldehydes R-C(=O)-H .. Does not ionize (neutral) ..
Esters R-C(=O)-OR .. Does not ionize (neutral) ..
Ethers ROR .. Does not ionize (neutral) ..
NH
NH
+
NH
N
NH
+N
NH
O O
N
O O
_
O O
H H
O O
H
_
NNH
+
Group Acid form Conjugate base form pKA
Amines RCH2NHR2+ RCH2NR2 8-11
Acids RCOOH RCOO- 3-5
Amides RCONH2 .. Does not ionize (neutral) ..
Alcohols ROH .. Does not ionize (neutral) ..
Phenols PhOH PhO- 9-11
Anilines PhNH3+ PhNH2 3-5
Imidazole 6.9
Pyridines 4-6
Imides 8-10
1,3-dicarbonyls 11-13
Imines ~7
Ketones R-C(=O)-R .. Does not ionize (neutral) ..
Aldehydes R-C(=O)-H .. Does not ionize (neutral) ..
Esters R-C(=O)-OR .. Does not ionize (neutral) ..
Ethers ROR .. Does not ionize (neutral) ..
NH
NH
+
NH
N
NH
NH
+
NH
N
NH
+NNH
+N
NH
O O
N
O O
_NH
O O
N
O O
_
O O
H H
O O
H
_
O O
H H
O O
H
_
NNH
+
NNH
+
pKA Values of Common Organic Functional Groups
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Application of Henderson-Hasselbach Equation
1. Amphetamine has pKA of 9.8. Whatform will dominate at pH 2.8, 6.8,and 9.8?
CH3
NH3
CH3
NH2
+ H2O + H3O+
+
CH3
NH3
CH3
NH2
+ H2O + H3O+
+
log = pH - pKA[RNH2]
[RNH3+]
At pH 2.8
log = 2.8 9.8 = -7.0
= 10-7
[RNH2] = 10-7 [RNH3+] % [RNH2] = 10-7 0.00001%
% [RNH3+] = 100 - 10-5 99.99999 %RNH3
+ is an ion, therefore % ionization of
amphetamine is nearly 100% at pH 2.8
[RNH2][RNH3
+]
[RNH2][RNH3
+]
100
(1+10-7)
At pH 6.8
log = 6.8 9.8 = -3.0
= 10-3
[RNH2] = 10-3 [RNH3+] % [RNH2] = 10-3 0.1 %
% [RNH3+] = 100 - 10-1 99.9 %RNH3
+ is an ion, therefore % ionization of
amphetamine is nearly 100% at pH 6.8
[RNH2][RNH3
+]
[RNH2][RNH3
+]
100
(1+10-3)
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Application of Henderson-Hasselbach Equation
1. Amphetamine has pKA of 9.8. Whatform will dominate at pH 2.8, 6.8,and 9.8?
CH3
NH3
CH3
NH2
+ H2O + H3O+
+
CH3
NH3
CH3
NH2
+ H2O + H3O+
+
log = pH - pKA[RNH2]
[RNH3+]
At pH 9.8
log = 9.8 9.8 = 0.0
= 100 = 1
[RNH2] = 1[RNH3+] % [RNH2] = 1 50 %
% [RNH3+] = 100 - 50 50 %RNH3
+ is an ion, therefore % ionization of
amphetamine is 50% at pH 9.8
[RNH2][RNH3
+]
[RNH2][RNH3
+]
100
(1+1)
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Application of Henderson-Hasselbach Equation
2. Meperidine is a narcotic analgesic.Its pKA is 8.7. It needs to beinjected (iv). What pH will you need
to make the solution of meperidine?
log = pH - pKA[RNH2]
[RNH3+]
N
COOEt
H CH3
N
COOEt
CH3
++ H
2O + H
3O+
For meperidine to go into solution fully and easily, it
should be fully ionized. Therefore, protonated species
(the conjugate acid form, [A]), and not the base form
{[B]} should dominate.
[RNH2]
[RNH3+]
< 0.01 or 10-2
log < 2.0[RNH2][RNH
3
+]
pH pKA < 2.0 pH 8.7 < 2.0 pH < 2.0 + 8.7 pH < 6.7
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Application of Henderson-Hasselbach Equation
3. Hydrocortisone hemisuccinate has apKA of 5.1? It is a pro-drug form ofhydrocortisone, a steroidal anti-
inflammatory molecule. Why washemisuccinate derivative made?
O
O
O
OHOH
O
OH
O
OH
O
OHOH
Hydrocortisone hemisuccinate hydrocortisone
Hydrocortisone has poor solubility in water/blood. However, the hemisuccinate form
with a pKA of 5.1 generates good solubility (~100% ionized) at pH 7.2 (blood). Thus, it
can be injected directly in the vein for rapid anti-inflammatory activity. On the other
hand, hydrocortisone or its ester derivatives (acetate, valerate, etc) cannot be injected
into the vein directly because they will precipitate. If a solubilized form of
hydrocortisone or its ester is prepared and injected into a vein, dangerous possibilitiesof cutting of blood flow (and possibly death) exist!
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