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The Acid–Base Chemistry of Nicotine: Extensions, Analogies, and a Generalization Addison Ault Department of Chemistry, Cornell College, Mount Vernon, IA 52314;
[email protected] The acid–base chemistry of nicotine provides a nice example of equilibria in aqueous solution where the major form in which the solute is actually present depends upon the pH of the solution. As Summerfield points out (1), nicotine can exist in aqueous solution either as the dication, N2+, the monocation N1+, or the neutral molecule, N.
N+ H N+
N
N+
CH3
H N
CH 3
CH3
N
H
dication of nicotine
monocation of nicotine
nicotine
N 2+ pKa = pK1 = 3
N 1+ pKa = pK2 = 8
N
N1+ H+ N2+
(1)
and then eq 2, the equilibrium expression for the ionization of N1+, to form the free base, N, and hydronium ion.
K2 =
N H+ N1+
(2)
Equation 1 can be rewritten as eq 3 to express the ratio of the concentrations of N2+ and N1+,
N2+ N1+
500
H+ = K1
+ N1+ H = N K2
(4)
You can then see that the ratio of N2+ to N, the product of the left sides of eqs 3 and 4, equals the product of the right sides of eqs 3 and 4, as shown in eq 5.
N2+ N1+
×
+ H+ N1+ N2+ H = = × N N K2 K1
(5)
The Fractional Concentration of N
The question is: Which is the predominant form in an aqueous solution at a particular pH? The intuitive answer is that the dication, N2+, should predominate in strongly acidic solutions, the neutral molecule, N, should predominate in strongly basic solutions, and the monocation N1+ could predominate in solutions of intermediate pH. More precisely, as we will show next, N2+ will predominate in solutions more acidic than a pH of 3, the monocation N1+ will predominate in solutions whose pH lies between 3 and 8, and the free base, N, will predominate in solutions that are more basic than a pH of 8. Furthermore, at a pH of 3 forms N2+ and N1+ will be present in equal concentrations, and at a pH of 8 forms N1+ and N will be present in equal concentrations. Following Summerfield’s lead we will write an expression for the fractional concentration of the free base, N, as a function of acid ionization constants and hydrogen ion concentrations, and then extend this treatment to the fractional concentrations of N1+ and N2+. We start by writing eq 1, the equilibrium expression for the ionization of the dication, N2+, to form N1+ and hydronium ion,
K1 =
and eq 2 can be rewritten as eq 4 to express the ratio of the concentrations of N1+ and N.
We now write in eq 6 the expression for the fractional concentration of the free base, N, as the concentration of N over the sum of the concentrations of N2+, N1+, and N, where α N is read “fractional concentration of N”. αN =
(6)
Dividing top and bottom by N gives eq 7. 1 αN = (7) 2+ N N1+ + +1 N N Substituting from eqs 4 and 5 then gives the fractional concentration of the free base, N, as eq 8. 1 αN = (8) + + H H+ H × + +1 K2 K1 K2 Equation 8, written in this form, provides the insight for understanding the variation of the fractional concentration of the free base, N, as a function of the acidity of the solution. For N to be the major form in solution, its fractional concentration must approach unity. The only way in which this can happen is for the first two of the three terms of the denominator of eq 8 to approach 0, so that the denominator approaches 0 + 0 + 1 = 1 and the value of the fraction approaches 1. This can happen only when [H+] is small; that is, this will be true only when the solution is somewhat basic. We will see next that the solution must be more basic than pH 8 for the free base to be the predominant form of nicotine. Substituting the values for K1 and K2 for the dication of nicotine into eq 8 gives eq 9,
αN = (3)
N N + N1+ + N 2+
H+ 108
×
1 H+ 103
+
Journal of Chemical Education • Vol. 78 No. 4 April 2001 • JChemEd.chem.wisc.edu
H+ 108
(9)
+1
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and from eq 9 we can see that when [H+] is exactly 108 the fractional concentration of N will be 1/(0.00001 + 1 + 1) or 1/2, and that as [H+] becomes substantially less than 108 the fractional concentration of N will approach 1.
Equation 16 indicates that the dicationic form of nicotine, N2+, can be the major species when [H+] is larger than both K1 and K2; that is, when the solution is strongly acidic. Substituting the values for K1 and K2 for the dication of nicotine into eq 16 gives eq 17,
The Fractional Concentration of N1+
N1+
α N1+ =
N2+ + N1+ + N Dividing top and bottom by N1+ gives eq 11. 1 α N1+ = 2+ N +1+ N 1+ N N1+
(10)
(11)
Substituting from eqs 3 and 4 then gives the fractional concentration of N1+ as eq 12.
1
α N 1+ =
(12)
+
K H + 1 + 2+ K1 H
You can see from eq 12 that the only conditions under which the fractional concentration of N1+ can approach unity (the denominator of eq 12 approaches 1) is for [H+] to be smaller than K1 and larger than K2. Substituting the values for K1 and K2 for the dication of nicotine into eq 12 gives eq 13, 1 α N 1+ = + (13) H 108 +1+ + H 103 and from eq 13 we seen that for N1+ to be the major species in solution [H+] must be between 103 and 108, or the pH must be between 3 and 8. The Fractional Concentration of N2+ Approaching the fractional concentration of N2+ in the same way we first write its fractional concentration as in eq 14.
N2+
α N2+ =
N2+ + N1+ + N Dividing top and bottom by N 2+ gives eq 15, 1
α N 2+ = 1+
N1+ N2+
(14)
(15)
+ N N2+
α N 2+ = 1+
K1 H+
+
H+
×
K2 H+
(16)
(17)
3
1+
10 103 108 + + × + + H H H
and from eq 17 we seen that for N 2+ to be the major species in solution [H+] must be greater than 103; the pH of the solution must be on the acidic side of 3. At a pH of exactly 3, N2+ and N1+ will be present in equal concentrations. The Other Conjugate Acid of Nicotine The monocation N1+ is the predominant conjugate acid of nicotine; the alternative monocation, N1+′ is also present, but in lower concentration. The reason for this is that the pyridine nitrogen, being sp2 hybridized, is more electronegative and therefore less basic than the pyrrolidine nitrogen, which is sp3 hybridized and therefore less electronegative and more basic than the pyridine nitrogen (2):
N
N+ H N
CH 3
N+
CH3
H N 1+ pKa = 8
N 1+' pKa = 3
Since the pKa values of N1+ and N1+′ differ by 5 powers of ten, their concentrations will always differ by 5 powers of ten, and the concentration of N1+ ′ will always be 1 × 105 times that of N1+. In solutions whose pH is between 3 and 8, when N1+ is the major form in which nicotine is present, the concentration of N1+ ′ will be 1 × 105. In solutions more acidic than a pH of 3 or more basic than a pH of 8, the concentration of N1+′ will be 1 × 105 times that of the lower fractional concentration of N1+. It is for these reasons that the concentration of N1+′ could be ignored in the preceding calculations. Summary and Generalization When a solute can exist in more than one form, say A, B, C, …, the fractional concentration of any one form, say A, can be expressed by an equation of the form of eq 18: fractional concentration of A =
and substituting from eqs 3 and 5 then gives the fractional concentration of N2+ as eq 16.
1 K1
1
α N 2+ =
We can think about the fractional concentration of N1+ in the same way. First, we write the fractional concentration of N1+, α N 1+ , as in eq 10.
1 B C 1+ + +… A A
(18)
When a single equilibrium separates A from another form, say B, the relative concentrations of A and B will be determined by the ratio of the equilibrium constant that relates A and B, and the concentration of another species that determines the relative amounts of A and B. This “other
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species” is typically hydronium ion but could be something else, such as the substrate or the inhibitor of an enzymatic reaction, or anything else that enters into a rapidly reversible equilibrium with A. When two equilibria separate A from another form, say C, the relative concentrations of A and C will be determined by the product of two ratios of equilibrium constant and concentration. Both of these possibilities were illustrated in the nicotine example. You can determine the sense of the ratio, whether it is equilibrium constant over concentration or concentration over equilibrium constant, either analytically, as in the nicotine example, or through your chemical intuition and experience. Application to Amino Acids
H
O
C
C
H
N
+
O
H
H
H
O
R
C
C
H
N
H
− O
H
+
H
O
R
C
C
H
N
+/-
A
A pKa = pK2 = 10
A pKa = pK1 = 2
− O
H
H +
-
Following the example of nicotine, we expect that A+ will be the predominant form of the amino acid present in solutions more acidic than pH 2, the neutral zwitterionic form A+/ will be the major form present in solutions whose pH lies between 2 and 10, and the basic form, A, will be the major form in solutions that are more basic than a pH of 10. The major neutral form will be the zwitterionic form A+/ because the carboxyl proton of A+, having a pKa of 2, is more acidic than the ammonium proton, with its pKa of 10. The alternative neutral uncharged form, A0, will never be present at a high concentration. H
O
R
C
C
H
N
H
+
− O
O
O H
H
O O O
+ NH A
H
O
R
C
C
H
N
H
O
H
H A+/-
A
O
O −
O− O
+ NH
OH
pKa = 10 aspartic acid; A+; major form at pH = 1
502
3
+/-
O −
O
O− O
3
NH2
A2 -
In solutions more acidic than pH 2, A+ is the predominant form; in solutions whose pH is between 2 and 4, A+/ is the predominant form (the isoelectric pH is 3); in solutions whose pH is between 4 and 10, A1 is the predominant form; and in solutions more basic than pH 10, A2 is the predominant form. Application to Enzyme Kinetics We can express the rate of a simple enzyme-catalyzed reaction by eq 19, in which kp is the rate constant for conversion of the enzyme–substrate complex, ES, to product, and E0 is the total enzyme concentration. rate = kp [E0](fraction of E0 present as ES)
(19)
The maximum rate is realized when the fractional concentration of ES approaches 1, and it is therefore essential to understand what factors determine the fractional concentration of ES. In general we can imagine that the enzyme is present as free enzyme, E, as the enzyme–substrate complex, ES, as an enzyme–inhibitor complex, IE, and as a complex of the enzyme with both substrate and inhibitor, IES. We can therefore express in a general way the fractional concentration of ES as shown in eq 20.
α ES =
ES ES + E + IE + IES
α ES =
1 IE IES E + + 1+ ES ES ES
(20)
pKa = 2
(21)
Proceeding now as in ref 3, we can reexpress this fraction in terms of concentrations and equilibrium constants as eq 22
O
+ NH 3
+ NH
Dividing top and bottom by [ES] gives eq 21.
HO O
O− O
3
+
A0
Since the two pKa values of A+ differ by 8 powers of ten, 0 A will always be 108 of the concentration of A+/. You can easily extend this line of reasoning to amino acids that contain either acidic or basic groups in the side chain. For aspartic acid, for example,
pKa = 4
O
H
A 1-
The acid–base chemistry of amino acids follows the same pattern. A typical “neutral” amino acid can exist in three forms in aqueous solution: the cationic or “acidic” form A+, the neutral zwitterionic form A+/, and the anionic or “basic” form A. R
the significant forms present in aqueous solutions are A+, A+/, A1, and A2:
1
α ES = 1+
KM S
+
KM S
×
I I + K2 K3
(22)
where KM, the Michaelis constant, is the substrate concentration at which the rate is half maximal in the absence of inhibitor; K2 is the dissociation constant for the dissociation
Journal of Chemical Education • Vol. 78 No. 4 April 2001 • JChemEd.chem.wisc.edu
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of I from IE; and K3 is the dissociation constant for the loss of I from IES. In the absence of inhibitor, I, the IE and IES terms in eq 21 are zero. When this is so, high substrate concentration can drive the E term to zero and the rate will be maximal. If the inhibitor binds only to free enzyme (the IES term is zero), the rate can still be driven to the original maximum by a high substrate concentration, since the E and IE terms in the denominator of eq 21 can both be driven to zero. This is one of the characteristics of competitive inhibition. If, however, I binds only to ES to give IES (IE does not form), high substrate concentration cannot raise the rate to the maximum that can be achieved in the absence of inhibitor because the
IES term in the denominator of eq 21 cannot be driven to zero by high [I]; the fractional concentration of ES never becomes equal to 1. This is one of the characteristics of uncompetitive inhibition. Formation of both IE and IES accounts for the kinetic behavior called noncompetitive inhibition. Reference 3 presents a more detailed analysis of these modes of inhibition. Literature Cited 1. Summerfield, J. H. J. Chem. Educ. 1999, 76, 1397. 2. Loudon, G. M. Organic Chemistry, 3rd ed.; Benjamin/ Cummings: Redwood City, CA, 1995; pp 664, 1182. 3. Ault, A. J. Chem. Educ. 1974, 51, 381.
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