Because chemical reactions depend principally on electrostatics, different isotopes of the same elem — Physical Chemistry — Kinetics Chemistry Question
Isotope Effects in Azo Coupling Reactions
Because chemical reactions depend principally on electrostatics, different isotopes of the same element generally have almost indistinguishable chemical properties. However, when the fractional difference in mass is large, the slight dependence of chemical properties on nuclear mass can result in perceptibly different reactivities. This is most commonly observed with isotopes of hydrogen, with compounds of protium (1H) often displaying quantitatively distinct reaction rates compared with those of deuterium (2H, abbreviated D) or tritium (3H, abbreviated T). In particular, the reduced masses of bonds to hydrogen, and thus the quantum mechanical zero-point energies of vibrations involving these bonds, E0 = 1/2 hν, where ν = 1/(2π) * sqrt(k/µ) with k being the force constant of the bond to H and µ = reduced mass = m1 m2 / (m1 + m2) with m1 and m2 the masses of the two bonded atoms, depend significantly on the mass of the hydrogen isotope. Heavier isotopes have larger reduced masses and lower zero-point energies. If a bond to hydrogen is broken during an elementary reaction, the vibrational frequency of the bond in the transition state, and hence its zero-point energy, is very low. Since compounds of all hydrogen isotopes therefore have similar or identical energies in the transition state, but heavier isotopes have lower energies in the reactants, compounds of protium will have a smaller activation energy and, therefore, react faster than those of deuterium or tritium.
The ratio (kH /kD), called a primary kinetic isotope effect when a bond to hydrogen is broken, is often in the range of 5 – 8 at ambient temperatures. Secondary kinetic isotope effects, where a bond remote to the site of isotopic substitution is broken, are typically much smaller, usually with kH / kD < 1.4.
Kinetic isotope effects have proven invaluable in the study of reaction mechanisms because of their ability to shed light on the details of processes that make or break bonds to hydrogen. A classic example is the study of the reaction between 2-naphthol-6,8-disulfonate and 4-chlorobenzenediazonium ion to form a highly colored azo dye:
[VISUAL]
Propose a synthesis of 4-chlorobenzenediazonium ion 2 from benzene.
Model Answer
To synthesize 4-chlorobenzenediazonium ion (2) from benzene:
1. Chlorination: React benzene with Cl2 in the presence of AlCl3 (or FeCl3) catalyst to obtain chlorobenzene.
2. Nitration: Treat chlorobenzene with HNO3/H2SO4 to yield 1-chloro-4-nitrobenzene as the major product (which can be separated from the minor 1-chloro-2-nitrobenzene isomer).
3. Reduction: Reduce the nitro group of 1-chloro-4-nitrobenzene using Sn and aqueous HCl to obtain 4-chloroaniline.
4. Diazotization: React 4-chloroaniline with NaNO2 and aqueous H2SO4 at low temperature (0-5 °C) to form 4-chlorobenzenediazonium ion (2).
Propose a structure for compound 3 (with H in 1), and explain the selectivity of the reaction.
Model Answer
Structure of 3 is the azo coupling product at position 1 of 2-naphthol-6,8-disulfonate: [VISUAL]
Explanation of selectivity:
The phenol group is strongly activating and ortho, para-directing. However, in 2-naphthol, the activation is directed exclusively to the 1-position, not the 3-position. Attack by an electrophile (such as the diazonium ion) can form a highly stabilized species when it adds at the 1-position, while addition at the 3-position is highly disfavored as it requires disrupting the aromaticity of the other benzene ring of the naphthalene system.
The kinetics of the reaction between compound 1H (compound 1 with hydrogen substitution) and compound 2 was studied in buffered aqueous solution (pH = 6.6) in the presence of variable amounts of pyridine. The reaction was found to be first order in both 1H and in 2 under all conditions. Describe in detail the experiments by which one could measure the second-order rate constants and determine the order of the reaction in each reagent.
Model Answer
The product of the reaction is an azo dye and is much more intensely colored in the visible region of the spectrum than either of the starting reagents. One could thus measure the absorbance of light at an appropriate wavelength as a function of time using a spectrophotometer to determine the concentration of product as a function of time.
To determine the reaction orders, it is easiest to run the reactions under 'pseudo-first-order conditions' where one reagent is present in massive excess over the other:
1. React 1 with 2 with one reagent (such as the diazonium ion 2) in > 10-fold excess over compound 1.
2. Under these circumstances, the concentration of the excess reagent (2) does not change significantly during the course of the reaction. A plot of ln(|At - A∞|) as a function of time t will be linear with a slope equal to -kobs, confirming a first-order dependence on compound 1.
3. The reaction is then re-examined at different excess concentrations of compound 2. If the reaction is first-order in 2, a plot of the observed rate constants kobs vs. [1] will be linear with a slope equal to the overall second-order rate constant k.
In the absence of pyridine, the reaction between 1H with 2 is faster than the reaction of 1D with 2 (k1H/k1D = 6.55). In contrast, the analogous reaction between 4 and 5 shows no discernible isotope effect (k4H/k4D = 0.97). Explain these results.
[VISUAL]
Model Answer
Electrophilic aromatic substitution is a two-step reaction involving an initial attack by the electrophile on the benzene/naphthalene ring and a subsequent removal of a proton by a base:
1 + 2 <-> intermediate I (k1, k-1) → 3 + base-H+ (k2)
Only the second step (deprotonation) involves breaking a bond to protium/deuterium, meaning only this step shows a primary kinetic isotope effect.
In the reaction of 1 with 2, the deprotonation step is rate-limiting, and thus a significant kinetic isotope effect is observed (6.55). One might plausibly imagine that the steric bulk of the 8-sulfonate group in 1 makes expulsion of the diazonium ion more facile (higher k-1) or impedes access to the proton by base (lower k2), either of which would contribute to step 2 becoming rate-limiting.
In contrast, in the reaction of 4 with 5, it is step 1 (electrophilic attack) that is rate-limiting. Since that step does not involve making or breaking a bond to H/D, no kinetic isotope effect is observed (0.97), which is the most typical behavior of electrophilic aromatic substitutions.
The second-order rate constants of reaction of 1H and 1D with 2 are tabulated as a function of pyridine concentration in the table below. Account for the variation of rate and isotope effect with , both qualitatively and quantitatively. (The pyridine concentrations listed are those of the free-base form of pyridine, they have been corrected for the protonation of pyridine at pH 6.6).
, mol dm–3 | k1H, dm3 mol–1 s–1 | k1D, dm3 mol–1 s–1 | kH/kD
0.0232 | 6.01 | 1.00 | 6.01
0.0467 | 11.0 | |
0.0931 | 22.4 | |
0.140 | 29.5 | |
0.232 | 46.8 | |
0.463 | 80.1 | |
0.576 | 86.1 | |
0.687 | 102. | |
0.800 | 106. | |
0.905 | 110. | 30.4 | 3.62
Model Answer
Qualitatively, pyridine acts as a base that accelerates the second step of the mechanism (general base catalysis). Since this step is rate-limiting, increasing accelerates the overall reaction, but the effect becomes smaller at high as the first step becomes increasingly rate-determining. Similarly, the isotope effect at low is close to the true kH/kD for the second step (~6.0), but as the first step becomes partially rate-limiting at high , the isotope effect is diminished to 3.62 because step 1 does not show a primary isotope effect.
Quantitatively, applying the steady-state approximation on intermediate I:
k1 [2] [1] = k-1 [I] + k2 [I]
[I] = k1 [2] [1] / (k-1 + k2 )
rate = k2 [I] = k1 k2 [2] [1] / (k-1 + k2 )
Thus the experimentally measured second-order rate constant k is given by:
k = k1 k2 / (k-1 + k2 )
This can be analyzed using a double-reciprocal plot:
1/k = 1/k1 + (k-1 / (k1 k2)) * (1/)
Plots of 1/k vs. 1/ for both 1H and 1D give linear fits:
For 1H: y = 0.0052 + 0.00379x, which gives k1 = 193 dm3 mol–1 s–1 and k2 / k-1 = 1.37 dm3 mol–1.
For 1D: y = 0.0075 + 0.0230x.
The intercepts of the two plots are close because k1 has little isotope effect. If the equilibrium isotope effect on the formation of intermediate I is neglected, the isotope effect on k2 is given by the ratio of the slopes: k2H / k2D ≈ 6.1, which is close to the experimental value observed at the lowest .
Predict the variation of the rate constant for the reaction of 4H with 5 as pyridine concentration is increased.
Model Answer
Since the reaction of 4 with 5 shows no isotope effect, step 1 must be rate-limiting. Acceleration of step 2 by pyridine will therefore have no effect on the observed rate, and this reaction should not be catalyzed by pyridine (which is consistent with experimental observations by Zollinger).
Explain the observed variation of the isotope effect of reactions of 1 with the structure of the diazonium salt used (all reactions in the absence of pyridine):
Diazonium ion: [VISUAL]
k1H/k1D: 6.55 | 5.48 | 4.78
Model Answer
As the diazonium ion becomes more electrophilic (as we go from left to right: e.g., p-methoxybenzenediazonium to p-chlorobenzenediazonium to p-nitrobenzenediazonium), its expulsion from the intermediate I (governed by k-1) becomes slower because of a more stable carbon-nitrogen bond. This increases the ratio k2/k-1, making the first step (which has little kinetic isotope effect) increasingly rate-limiting. As step 1 becomes more rate-determining, the overall observed kinetic isotope effect decreases from 6.55 to 4.78.