Isotopes can be used as markers to determine the nature of intermediates in chemical reactions. The focus in such studies is primarily the location and distribution of the label in the product(s). There is a more subtle effect of isotopes which when studied quantitatively can give even greater details of chemical processes. Isotopic substitution can influence an equilibrium, sometimes dramatically. For example, the equilibria are shifted significantly to the right for the two cases given below.

HT + H_{2}O = H_{2} + HTO K_{a} = 6.3 ......(1)

2D_{3}O^{+} + 3 H_{2}O = 2H_{3}O^{+} + 3 D_{2}O K_{a} = 8.2 ......(2)

Isotopes also affect reaction rates. The magnitude of this kinetic isotope effect (KIE) would depend on the location of the isotope with respect to the 'scene of action'! Therefore the following types of effects are commonly encountered: (a) primary KIE, (b) secondary KIE ( primary and secondary types) and (c) steric KIE.

In this article we shall give examples of reactions which exhibit interesting primary kinetic isotope effects. We will see how such studies can help us understand the nature of the transition state in each case. The concepts needed for interpreting the KIE will also be developed.

**Primary Kinetic Isotope Effect**

*Primary kinetic isotope effect arises when the bond to the isotopic substitution is broken at or before the transition state*.

The existence of primary KIE can be easily seen from the following equations for the electrophilic substitution of benzene, and for the oxidation of a secondary alcohol.

C_{6}H(D)_{6} + HNO_{3} C_{6}H(D)_{5}-NO_{2} k_{H}/k_{D} = 1.0 ......(3)

C_{6}H(D)_{6} + Hg(ClO_{4})_{2} C_{6}H(D)_{5}-Hg^{+} k_{H}/k_{D} = 6.7 ......(4)

Me_{2}CH(D)-OH + CrO_{3} Me_{2}C=O k_{H}/k_{D} = 7.7 ......(5)

It is intriguing that equations 3 and 4 both represent electrophilic aromatic substitutions, but there is KIE in one, but not in the other.
Equally puzzling is the reaction of toluene with chlorine or bromine, in which toluene(d_{6}) shows a KIE of 1.5 with Cl^{.}. but 4.6 with
Br^{.}. How can one explain these variations?

**Theory**

In this section, we will look at a very elementary theory of primary KIE, and see how it helps us understand the reactions described above. We will find that this theory also allows us to calculate the equilibrium isotope effects shown in eq. 1 and 2.

Let us consider the potential energy diagram for a typical system such as a C-H bond. The function is described by the well-known
Morse curve (Figure 1). On either side of the minimum, the energy increases sharply. The value becomes constant at large internuclear
distance, corresponding to dissociation. To a very good approximation, the curve is unchanged on going to a C-D bond* (***note :**

The zero point energy can be estimated by assuming the bond to be a simple harmonic oscillator (a spring which obeys Hooke's law).
The energy is given by 0.5 hc(nu-bar), where h is Planck's constant, c is the velocity of light, and (nu-bar) is the vibrational frequency
(in cm^{-1}) given by equation 6:

(nu-bar) = [1/(2(pi)c)]*SQRT(k/µ) ... (6)

In the above equation, *k* is the force constant of the bond (second derivative or the curvature at the minimum), and µ is the reduced
mass of the system (endnote 1).

Since the potential energy diagram is unaltered on going from H to D, *k* remains constant. However, the reduced mass µ is different.
As a good approximation, on going from C-H to C-D, the reduced mass increases by a factor of approximately 2 (see box 1).
Therefore, the frequency for a C-D bond should be approximately 1/SQRT(2) or 0.71 times that of the corresponding C-H bond
(nu-bar)_{H}. In practice, however, this ratio is found to be closer to 0.74. What this means is that in any case the zero-point-energy for a
C-D system is *lower* than that of a C-H bond. Since the energy is the same at the dissociation limit, the bond energy is effectively
higher for the C-D bond. In other words, the C-D bond is *stronger*. Therefore, any reaction in which the C-H bond is broken during or
before the rate determining step will be slower if the hydrogen is replaced by a deuterium. The relative rates, k_{H}/k_{D} should be greater
than 1. This is known as the *normal* primary kinetic isotope effect.

But what should be the value of this ratio? That part is easy to calculate. Let us consider a situation in which the C-H or the C-D bond
is completely broken at the transition state. Then the relative activation energies is entirely determined by the zero point energy
difference in the reactant. This quantity corresponds to ZPE = 0.5 hc(nu-bar)_{H}(1- 0.74) = 0.13 hc(nu-bar)_{H}. The relative rates, i.e., the
KIE, will be given by:

k_{H}/k_{D} = (exp)[0.13 hc(nu-bar)_{H}/kT] ... (7)

[The eleventh letter of the English alphabet is used to represent different quantities: force constant, rate and equilibrium constants, Boltzmann constant etc. The authors who have suffered through this maze during their training urge the readers to figure out what the symbols stand for from the context!]

Using a typical C-H vibration frequency of

What about the KIE value of unity shown in equation 3? The likely mechanism of nitration of benzene is shown in Figure 2. The reaction involves two steps. First a sigma-complex is generated, which then undergoes C-H(C-D) bond cleavage to regenerate aromaticity. If the second step had been the rate determining step, a normal KIE of close to 7 should have been observed. The absence of any isotope effect in the reaction implies that the first step is rate determining. In other words, the bond to the isotopic atom is broken after the rate determining step.

This discussion still leaves one question unanswered. Why, for example, the KIE for chlorination and bromination of toluene are so
different? Remember, in our simple derivation for the estimation of the primary KIE, we have *assumed* that both isotopically
substituted transition states have the same zero point energy. In reality, this is far from true. While the C-H(C-D) bond may cleave
completely at the transition state, the H(D) must get attached to some other atom! For example, for chlorination of toluene, the Cl^{.}
attacks the C-H(C-D) bond, and forms a benzyl radical and H(D)-Cl. Clearly, the energies of HCl and DCl (or for that matter HBr and
DBr) are different! Hence, the transition state for the two isotopomers will necessarily have different zero point energy. The only
question is how different they are. It is clear from our discussion that if the H or the D atom is completely transferred to the X
(halogen) atom the zero point energy difference at the transition state will be maximum. This will lead to a small KIE. The same result
is also true in the opposite extreme case in which the H or D atom is hardly transferred at the transition state. On the other hand, if the
H(D) atom is "half-transferred" to X at the transition state, it will be "equally" bonded to both atoms. This will lower the force
constant, and therefore reduce the vibrational frequency. This will cause the zero point energy difference at the transition state to be
rather small, and, as a result, KIE will be high.

The extent to which an atom is transferred at the transition state can be guessed by qualitative considerations. For a highly exothermic
reaction, the transition state resembles the reactants. [margin note: This is known as Hammond postulate] Since the hydrogen
abstraction reaction with Cl** ^{.}** is more exothermic, the C-H bond cleavage in this case will be minimal. Therefore, the observed KIE of
only 1.5 with chlorination of toluene is clearly consistent with the above analysis. The value for the reaction with bromine is closer to
what one would expect for a greater degree of bond cleavage. In general, if the KIE is greater than 2, it usually implies that there is
considerable C-H bond cleavage at the transition state.

**Origin of Equilibrium Isotope Effects**

Let us now see if we can explain the equilibrium isotope effects shown earlier in equations 1 and 2. Substitution by a heavier isotope
always lowers the zero point energy. The magnitude of change is greater in systems with higher intrinsic vibrational frequency. For
example, the stretching vibrational frequencies for H_{2} and H_{2}O are 4370 and 3440 cm^{-1}, respectively. Introduction of tritium (^{3}H)
produces a large decrease in the frequencies. The values for HT and HOT (O-T bond) are approximately 3570 and 2100 cm^{-1},
respectively. Therefore, one can see that the combination of HT and H_{2}O will have higher ZPE than the combination of H_{2} and HOT,
resulting in the equilibrium isotope effect which is greater than unity. The precise numerical value will depend on the temperature. For
quantitative prediction, all the vibrational modes (including bending modes) of the species involved should be taken into account.
Similarly, the equilibrium isotope effect for equation 2 can be calculated if all the vibrational frequencies are known.

**Concluding Remarks**

We have shown how the magnitudes of primary KIE can be used to derive information about transition states of reactions. We have exclusively discussed examples of isotope effects resulting from the replacement of hydrogen by its isotopes. Isotope effects should also exist with heavier atoms, such as C, N, S, Cl etc. However, since the ratio of reduced mass approaches unity as the atomic weight increases, the KIE values become smaller for heavier atoms. Typically, such isotope effects are between 1 and 1.1. Since the rate variations are very small, special efforts are needed to measure them with high precision.

**Endnotes**

**1** The reduced mass µ of a system containing two masses M and m is given by:

1/µ** = **1/M + 1/m, or µ = Mm/(M+m)

We can see that typical values of reduced masses are ½ for H_{2}**, ** 2/3 for H-D and ¾ for H-T. ** **If the masses of the two atoms differ
significantly, the reduced mass will be closer to that of the lighter atom. For example, the reduced mass for the C-H bond is close to 1
(0.92), while that for C-D is 1.7.

**2 **A spectacularly large primary kinetic isotope effect is found in the intramolecular hydrogen abstraction reaction shown below.

The low temperature measurement contributes only partially to this large magnitude. A quantum mechanical process known as
*tunneling* is primarily responsible for this unusual effect. Tunneling usually involves light particles (electron, hydrogen atom). The
barrier should be low, and the width of the potential well should be small. Tunneling probability is highly sensitive to the mass. That's
why the rate drops dramatically on substitution of the hydrogen atom by deuterium. Extremely large KIE values are used to prove the
occurrence of tunneling.

**Suggested Reading**

Determination of Organic Reaction Mechanisms, B.K. Carpenter, John Wiley, New York, 1984.

Advanced Organic Chemistry, F.A. Carey and R.J. Sundberg, Part A, 3rd Ed., Plenum Press, New York, 1990.