Transcript Periodic Table 5/24/2016 MEDC 501 Fall 2007 1

Periodic Table
5/24/2016
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1
Orbitals
Shapes of Orbitals
s - orbital
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p - orbital
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Electronic Structure
Ionic Bond - NaCl
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Electronic Structure
Can 6C form ionic bonds?
6C
::
1s22s22px12py1
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Hybridization
s -orbital
+
sp - orbital
p -orbital
‘n’ atomic orbitals give ‘n’ hybrid orbitals
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Hybridization in Carbon
sp3 hybridization …… CH4
6C
6C
::
1s22s22px12py1
:: 1s22s12px12py12pz1
sp3 orbitals
6C
:: 1s2(2sp3)1 (2sp3)1 (2sp3)1 (2sp3)1
4
1H
:: 1s1
CH4 :: 1s2(2sp3)2 (2sp3)2 (2sp3)2 (2sp3)2
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Hybridization in Carbon
sp2 hybridization …… CH2=CH2
6C
6C
::
1s22s22px12py1
6C
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:: 1s22s12px12py12pz1
:: 1s2(2sp2)1 (2sp2)1 (2sp2)1 2pz1
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Hybridization in Carbon
sp2 hybridization …… CH2=CH2
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Hybridization in Carbon
sp hybridization …… CH=CH
6C
6C
::
1s22s22px12py1
:: 1s22s12px12py12pz1
sp hybrid
orbital
p orbitals
s orbitals
p orbitals
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Hybridization in Carbon
sp2 hybridization …… benzene C6H6
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Bond Polarity
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Electronegativity
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Quantitative Measure of Electronegativity
Sanderson Scale
F 4.000, Cl 3.475,
O 3.654, N 3.194,
C 2.746, H 2.592
Pauling Scale
F 4.0, Cl 3.0,
O 3.5, N 3.0,
C 2.5, H 2.1
Kier-Hall Scale
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Br 3.219,
S 2.957
Br 2.8,
S 2.5
I 2.778
I 2.5
(column # - # of s bonds)
 =
(row #)2
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Calculating Kier-Hall Electronegativity
(column # - # of s bonds)
 =
(row #)2
f
O
HO c
a
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i
Cl
b
h NH h
e
OH
d
N
g
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2
3
4
5
4
6C
5
7N
15P
6
8O
16S
34Se
7
9F
17Cl
35Br
53I
14
Consequences of Bond Polarity
Dipolar Bonding
H3C
O
H3C
H3C
H3C
H3C H CH3
C
H 2C
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O
O
O
MW
bp
H3C
H3C
H3C
H3C
H 3C
CH3
72
28 OC
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O
H 2C
CH3
72
80 OC
15
Consequences of Bond Polarity
Hydrogen Bonding
H 2O
H 2S
MW 18
bp 100 OC
34
-60 OC
O = (6-2)/4 = 1.0
H
O
H
H
S
S = (6-2)/9 = 0.44
H
H
H
O
...
O
H
O
...
H
H
H
H
H
O
H
O
H
O
O
...
H
H
O
H
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H
H
O
H
H
H
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Consequences of Bond Polarity
Hydrogen Bonding
•
Each H in a hydrogen-bond is shared by two eN atoms
•
Linear arrangement of three atoms (H plus 2 eN) is strongest
•
Each linear arrangement is ~ 4 -5 kcal/mol
Typically form H-bond
N-H….OO-H….N=
O-H….F-
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Typically do not form H-bond
S-H OP-H N=
O-H Cl-
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Consequences of Bond Polarity
Hydrogen Bonding - Alcohols
CH3CH2CH3
CH3CH2CH2OH
MW 44
bp -45 OC
H
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O
H
….
….
O
H
46
78 OC
O
H
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O
O
H
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Consequences of Bond Polarity
Hydrogen Bonding - Acids
COOH
COOH
COOH
OH
OH
MW
mp
138
159
OH
138
214
138
201
OC
O H
HO
O
O
H
O
O
O
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H
O ...H O
H
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H ... O
O
O
H
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Consequences of Bond Polarity
Hydrogen Bonding
- Structure of Proteins and Nucleic Acids
O
N
H
H
N
R10
O
N
H
H
N
R12
O
N
H
O
R7
O
R15
H
N
R14
O
N
H
O
O
….
R13
N
H
….
R5
R8
O
H
N
….
R11 O
O
N
H
….
O
R3
N
H
R6
O
H
N
….
R9
O
R4
….
R1
O
H
N
….
N
H
….
H
N
H
N
R2
O
H
N
R16
Hydrogen Bonding in b-strands
(3D structure)
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Consequences of Bond Polarity
Inductive Effect
…the transfer of electronegative effect of polar s bonds
to neighboring s bonds, causing them to be more or
less polarized.
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Consequences of Bond Polarity
Change in Acidity or Basicity of Groups
O
H3C
O
H
H3C
a carboxylic acid
an alcohol
O
O
H3C
O
H
F3C
a carboxylic acid
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O
H
O
H
a trifluoro-carboxylic acid
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Consequences of Bond Polarity
Inductive Effect - Influence on pKa
1.25
O
H3C
pKa
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O
H
H3C
1.0
4.5
O
H
1.0
16.5
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pKA and Ions in Solution
Chemical Equilibrium
A
+
B
At equilibrium,
k1
k2
C + D
k1 [A] [B] = k2 [C] [D]
k1 [C ][ D]

 KE
k2 [ A][ B]
Le Chatelier’s 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
H2O
k1
k2
+
_
H + OH
[ H  ][OH  ]
 KE
[ H 2O]
 [ H  ][OH  ]  K E  [ H 2O]  K E  55.5  KW  1014

log[ H  ][OH  ]  log1014

log[ H  ]  log[OH  ]  14

 log[ H  ]  log[OH  ]  14

pH  pOH  14
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pKA and Ions in Solution
Acidity, basicity and neutrality
and pH conditions
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pKA and Ions in Solution
Ionization and Strength of Acids and Bases
Property of Ionization: The degree of
ionization of a particular compound in water is
dependent on its structure and is
characteristic for that compound.
O
Ph
C2H5
HN
O
N
H
Phenobarbital
O
O
O
CH3
OH
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Warfarin
O
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pKA and Ions in Solution
Ionization of a weak acid
CH3COOH +
H3O+ + CH3COO-
H 2O
[ H3O  ][CH3COO ]
KE 
[ H 2O][CH3COOH ]

[ H3O ][CH3COO ]
K E  [ H 2O]  K A 
[CH3COOH ]

[CH3COO ]
 log K A   log[ H3O ]  log
[CH3COOH ]


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
[CH3COO ]
[conj.base]
pK A  pH  log
 pH  log
[CH3COOH ]
[acid ]
log
[conj.base]
 pH  pK A Henderson-Haselbach Equation
[acid ]
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pKA and Ions in Solution
What is the physical meaning of pKA?
log
log
[conj.base1 ]
 pH  pK A1
[acid1 ]
log
[conj.base2 ]
 pH  pK A2
[acid2 ]
[conj.base1 ]
[conj.base2 ]
 log
 { pH  pK A1}  { pH  pK A2 }
[acid1 ]
[acid2 ]
log
[conj.base1 ]
[acid2 ]

  pK A1  pK A2
[acid1 ]
[conj.base2 ]
log
If pKA1 = 4 and pKA2 = 1
….
[conj.base1 ]
 pK A2  pK A1
[conj.base2 ]
[conj.base1 ]
 10 (14)  10 3  0.001
[conj.base2 ]
[conj .base1 ]  0.001  [conj .base 2 ]
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[Unionized Acid1] =
[Unionized Acid2]
[conj.base1 ]
[conj.base2 ]
 0.001 
[acid1 ]
[acid2 ]
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Consequences of Bond Polarity
Inductive Effect - Influence on pKa
O
O
H3C
pKa
O
H
F3C
4.5
0.23
CH3COOH
pKa
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O
H
ClCH2COOH
4.5
2.9
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Consequences of Bond Polarity
Inductive Effect - Additive Property
pKa
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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
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CH3CH2CH2COOH
4.8
ClCH2CH2CH2COOH
4.5
CH3CHClCH2COOH
4.1
CH3CH2ClCH2COOH
2.8
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Consequences of Bond Polarity
Inductive Effect - Group Electronegativities

Electron-withdrawing groups
-C=N
0.5+1.0
-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
NO2CH2COOH
ClCH2COOH
COOHCH2COOH
CNCH2COOH
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= 1.5
pKa
1.7
2.9
2.8
3.5
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Consequences of Bond Polarity
Influence on pKa …. through space effect
Cl
Cl
Cl
pKa
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Cl
COOH
COOH
6.1
5.7
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pKA and Solubility in Water
O
F
N
COOH
N
HN
Ciprofloxacin
(piperazine)
Antibacterial
(anthrax)
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Consequences of Bond Polarity
Inductive Effect – Multiple pKA values
A molecule may have more than one pKA value!!
O
OH
HO
O
HO
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pKA = 2.8
O
+
NH3
O
_
O
_
O
+
NH3
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O
_
O
HO
pKA = 2.35
pKA = 5.7
O
O
_
O
O
O
pKA = 9.78
_
O
NH2
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Consequences of Bond Polarity
Natural Amino Acids – Structure and pKA
R
H
COOH Structure of Amino Acid Residues
NH2
Non-Polar
All L-configuration or R geometry
(except for glycine)
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—
__
CH
2
Tryptophan
N
H
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Consequences of Bond Polarity
Natural Amino Acids – Structure and pKA
R
H
COOH Structure of Amino Acid Residues
NH2
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—
H
N
Histidine
__ H C
2
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N
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pKA Values of Natural Amino Acids
Residues
pKA1
pKA2
Non-Polar
Glycine
Alanine
Valine
Leucine
Isoleucine
Phenyalanine
Methionine
Tryptophan
2.35
2.35
2.29
2.33
2.32
2.16
2.13
2.43
9.78
9.87
9.74
9.74
9.76
9.18
9.28
9.44
Hydrogen-bonding
Serine
Threonine
Cysteine
Tyrosine
Asparagine
Glutamine
2.19
2.09
1.92
2.20
2.10
2.17
9.21
9.10
10.78
9.11
8.84
9.13
8.33 (thiol group)
10.13 (phenol group)
Acidic
Aspartic Acid
Glutamic Acid
1.99
2.10
9.90
9.47
3.90 (g-COOH group)
4.07 (g-COOH group)
Basic
Lysine
Arginine
Histidine
2.16
1.82
1.80
9.18
8.99
9.33
10.79 (-amino group)
12.48 (guanidino group)
6.04 (imidazole group)
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Side-Chain pKA (group identification)
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pKA Values of Natural Amino Acids
and the Dominant Form
Amino Acid
pH 3.0
pH 7.0
pH 10.0
Serine
Cysteine
Tyrosine
Asparagine
Arginine
Aspartic Acid
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Consequences of Bond Polarity
Influence on pKa of bases
RNH3+ +
H 3O+ +
H2O
pKA = pH - log
[RNH2]
[RNH3+]
CH3CH2NH2
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RNH2
FCH2CH2NH2
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Consequences of Bond Polarity
Inductive Effect






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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 Effect
NH2
pKa
NH2
9-11
~5
Resonance, Aromaticity & Conjugation
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Consequences of Bond Polarity
Resonance Effect
NH2
pKa
NH2
9-11
NH3+
~5
NH2
+ H2O
NH3+
H3O+
+
H3O+
NH2
+ H2O
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+
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Consequences of Bond Polarity
Resonance Effect
COOH
pKa
4.2
COOH
COOH
NO2
OMe
3.4
_O
O
HO
4.5
O
+ H2O
O
HO
_
O
O
+
H3O+
O
+ H2O
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H3O+
_ O N +O
+O _
N
OMe
+
OMe
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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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pKA Values of Common Organic Functional Groups
Group
Acid form
Conjugate base form
Amines
RCH2NHR’2+
RCH2NR’2
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
H
N
Imidazole
H
N
NH
6.9
N
+
Pyridines
4-6
NH
N
+
O
Imides
O
O
O
Imines
8-10
N_
N
H
O
O
O
H
H
H
1,3-dicarbonyls
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pKA
O
_
NH
+
N
11-13
~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) …..
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Application of Henderson-Hasselbach Equation
CH3
1. Amphetamine has pKA of 9.8. What
form will dominate at pH 2.8, 6.8,
and 9.8?
CH3
+
NH3
NH2
+ H3O+
+ H2O
[RNH2]
= pH - pKA
+
[RNH3 ]
log
At pH 6.8
At pH 2.8
log
[RNH2
= 2.8 – 9.8 = -7.0
[]RNH3+]
log
[RNH2
= 6.8 – 9.8 = -3.0
[]RNH3+]

[RNH2
-7
][RNH3+] = 10

[RNH2
-3
][RNH3+] = 10

[RNH2] = 10-7  [RNH3+]

[RNH2] = 10-3  [RNH3+]

% [RNH2] = 10-3 

% [RNH3+] = 100 - 10-1  99.9 %
100
 % [RNH2] = 10-7 
 0.00001
(1+10-7)
%

% [RNH3+] = 100 - 10-5  99.99999 %
RNH3+ is an ion, therefore % ionization of
amphetamine is nearly 100% at pH 2.8
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100
 0.1 %
(1+10-3)
RNH3+ is an ion, therefore % ionization of
amphetamine is nearly 100% at pH 6.8
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Application of Henderson-Hasselbach Equation
1. Amphetamine has pKA of 9.8. What
form will dominate at pH 2.8, 6.8,
and 9.8?
CH3
CH3
+
NH3
+ H2O
NH2
+ H3O+
[RNH2]
= pH - pKA
+
[RNH3 ]
log
At pH 9.8
log
[RNH2
= 9.8 – 9.8 = 0.0
[]RNH3+]

[RNH2
0
][RNH3+] = 10 = 1

[RNH2] = 1  [RNH3+]

% [RNH2] = 1 

% [RNH3+] = 100 - 50  50 %
100
(1+1)
 50 %
RNH3+ is an ion, therefore % ionization of
amphetamine is 50% at pH 9.8
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Application of Henderson-Hasselbach Equation
2. Meperidine is a narcotic analgesic.
Its pKA is 8.7. It needs to be
injected (iv). What pH will you need
to make the solution of meperidine?
log
COOEt
COOEt
N
N
H
+ H3O+
+ H2O
+
CH3
CH3
[RNH2]
= pH - pKA
+
[RNH3 ]
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.

 log
[RNH2] <
[RNH3+]
[RNH2]
[RNH3+]
 pH – pKA
<
0.01
<
10-2
–2.0
–2.0
 pH – 8.7
<
–2.0
 pH
<
–2.0 + 8.7
 pH
<
6.7
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or
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Application of Henderson-Hasselbach Equation
3. Hydrocortisone hemisuccinate has a
pKA of 5.1? It is a pro-drug form of
hydrocortisone, a steroidal antiinflammatory molecule. Why was
hemisuccinate derivative made?
O
O
O
HO
HO
OH O
OH OH
OH
O
O
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 possibilities
of cutting of blood flow (and possibly death) exist!
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