The kinetic isotope effect

Up to now you have probably (and rightly) assumed that isotopes of an element are chemically identical. They differ only in the number of neutrons in their nuclei: chemistry generally depends on charge, orbitals, and electrons. It may come as a surprise to find that this is not quite true. Isotopes may differ chemically, because some chemical properties do depend on atomic mass. However, this difference is only significant for hydrogen—no other element has one isotope twice as massive as another! Kinetic isotope effects are the changes in rate observed when a (1H) hydrogen atom is replaced by a (2H) deuterium atom in the same reaction. For any reaction, the kinetic isotope effect (KIE) is defined as

KIE = k_H / k_D

where kH is the rate with a 1H atom in the molecule and kD is the rate with a 2H (deuterium, D) atom in the molecule. How do kinetic isotope effects come about? Even in its lowest energy state a covalent bond never stops vibrating. If it did it would violate a fundamental physical principle, Heisenberg’s uncertainty principle, which states that position and momentum cannot be known exactly at the same time: a non-vibrating pair of atoms have precisely zero momentum and precisely fixed locations. The minimum vibrational energy a bond can have is called the zero-point energy, and the zero-point energy depends on the mass of the atoms attached to the bond— heavier atoms have a lower zero-point energy than lighter ones.

In order to break a covalent bond, a certain amount of energy is required to separate the nuclei from their starting position. This energy has to raise the vibrational state of the bond to the point where it breaks. For the sake of argument, imagine taking a C–H bond in its low estenergy state and breaking it—the diagram shows the amount of energy required, which we can call ΔG‡H. Now do the same for a C–D bond: because the zero point energy of a C–D bond is smaller than that for a C–H bond, the C–D bond needs that little bit more energy ΔG‡D to break: in other words a C–D bond is marginally stronger than a C–H bond. This means reactions in which C–H bonds break go faster than reactions in which C–D bonds break, providing the bond to H (or D) is involved in the rate-determining step. The theoretical maximum value of the KIE is about 7 for reactions at room temperature in which a bond to H or D is being broken. For example, the rates of these two eliminations can be compared, and kH/kD turns out to be 7.1 at 25 °C.

In this case the fact that the KIE is non-zero tells us that the C–H (or C–D) bond is being broken during the rate-determining step, and so the reaction must be an E2 elimination. In E1 eliminations, the rate-determining step does not involve a breaking C–H bond. we told you that the rate-determining step in the nitration of benzene was the attack of the electrophile on the benzene ring. This is easily verified by replacing the hydro gen atoms round the benzene ring with deuteriums. The rate of the reaction stays the same, so the C–H (or C–D) bonds cannot be involved in the rate-determining step. If the second step, which does involve the breaking of a C–H bond, were the rate-determining step it would go more slowly if the H were replaced by D.

By contrast, for the iodination of phenol in basic solution there is a deuterium isotope effect of kH/kD = 4.1. Clearly, the loss of the proton from the intermediate must now be the rate determining step—the phenolate ion reacts so rapidly that the first step is faster than the second.

The deuterium isotope effect can add to the information from Hammett plots in building up a picture of a transition state. Three separate Hammett ρ values can be measured for the elimination reaction and this information is very valuable. In addition, a large KIE k_H/k_D = 7.1 is observed for the hydrogen atom under attack.

It is no surprise that the base (ArO⁻) donates electrons and the leaving group (ArSO₃⁻) accepts them, as the ρ values indicate. The large deuterium isotope effect tells us that the reaction is E2, but additional information comes from the moderate positive ρ(Y) value for the aromatic ring adjacent to the proton being lost. It might have been expected that this ring is merely a spectator, but in fact the reaction must involve a build-up of negative charge, which can be stabilized by an electron-donating substituent Y. This can be explained if we assume that the removal of the proton is slightly more advanced at the transition state than loss of the leaving group.

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شعبة متابعة الأداء تنظم محاضرة تربوية لمتطوعات مجمع العلقمي استعدادًا لزيارة الأربعين

شعبة الحرم الشريف تكثف استعداداتها لتقديم خدمات متنوعة لزائري الأربعين

قسم الشؤون الفكرية يفتتح محطتين ثقافيتين في ذي قار لخدمة زائري الأربعين

العتبة العباسية المقدسة تعقد ملتقى خاصًا لمنتسبيها العاملين في مراكز إرشاد التائهين خلال زيارة الأربعين

المزيد

الآخبار الصحية

علامات فارقة بين النوبة أو السكتة القلبية.. ماهي؟

الفلفل الأحمر.. سعرات حرارية منخفضة وفوائد صحية عديدة

8 أعراض إذا شعرت بها.. اذهب للطوارئ فورا !

تعرف على أكثر 8 أماكن اتساخا في المنزل

المزيد

الآخبار التكنلوجية

إنتاج الهيدروجين الأخضر في البحر.. هذه أحدث الدراسات

"ناسا" تخطط لبناء مفاعل نووي على سطح القمر !

ابتكار ومواهب وأموال.. مارك زوكربيرغ يعلن الحرب على أبل

عودة التعاون الفضائي بين روسيا وأميركا.. وبحث ملفات "هامة"

المزيد

مواضيع ذات صلة

Bonding and reactions in transition metal complexes

Ligands can be attached in many different ways

The 18-electron rule

Transition metals extend the range of organic reactions

The Ritter reaction and the Beckmann fragmentation

Specific base catalysis

Specific acid catalysis

Entropy of activation

Non-linear Hammett plots

A complete picture of the transition state from Hammett plots

Using the Hammett ρ values to uncover mechanisms

Reactions with small Hammett ρ values