Skip to main content
Ch. 7 - Structure and Synthesis of Alkenes; Elimination
Wade - Organic Chemistry 9th Edition
Wade9th EditionOrganic ChemistryISBN: 9780135213728Not the one you use?Change textbook
All textbooksWade 9th EditionCh. 7 - Structure and Synthesis of Alkenes; EliminationProblem 75b
Chapter 7, Problem 75b

Deuterium (D) is the isotope of hydrogen of mass number 2, with a proton and a neutron in its nucleus. The chemistry of deuterium is nearly identical to the chemistry of hydrogen, except that the C―D bond is slightly (5.0 kJ/mol, or 1.2 kcal/mol) stronger than the C―H bond. Reaction rates tend to be slower if a C―D bond (as opposed to a C―H bond) is broken in a rate-limiting step. This effect on the rate is called a kinetic isotope effect. (Review Problem 4-57)
b. When the following deuterated compound reacts under the same conditions, the rate of formation of the substitution product is unchanged, while the rate of formation of the elimination product is slowed by a factor of 7.
Deuterated compound reaction showing substitution and elimination rates, with elimination seven times slower.
Explain why the elimination rate is slower, but the substitution rate is unchanged.

Verified step by step guidance
1
Step 1: Understand the reaction mechanism. The reaction involves two competing pathways: substitution and elimination. Substitution replaces the bromine atom with a hydroxyl group, while elimination removes a proton and bromine to form a double bond.
Step 2: Analyze the substitution pathway. In substitution, the C―D bond is not broken during the rate-limiting step. The hydroxide ion directly replaces the bromine atom, which explains why the substitution rate remains unchanged.
Step 3: Analyze the elimination pathway. In elimination, the rate-limiting step involves breaking a C―D bond to remove a deuterium atom (D) along with bromine. Since the C―D bond is stronger than the C―H bond, breaking it requires more energy, slowing down the elimination reaction.
Step 4: Relate the kinetic isotope effect to the observed rates. The kinetic isotope effect arises because the stronger C―D bond leads to a slower rate of bond cleavage in the elimination pathway, while the substitution pathway is unaffected as it does not involve breaking the C―D bond.
Step 5: Conclude the explanation. The elimination rate is slower due to the kinetic isotope effect caused by the stronger C―D bond, while the substitution rate remains unchanged because the C―D bond is not involved in the rate-limiting step of substitution.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
5m
Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Kinetic Isotope Effect

The kinetic isotope effect (KIE) refers to the change in reaction rate that occurs when an atom in a molecule is replaced by one of its isotopes. In this case, the C―D bond is stronger than the C―H bond, leading to a slower reaction rate when breaking the C―D bond during elimination reactions. This effect is significant in understanding how isotopes influence reaction mechanisms and rates.
Recommended video:
2:02
Directing Effects in Substituted Pyrroles, Furans, and Thiophenes Concept 1

Substitution vs. Elimination Reactions

Substitution and elimination are two fundamental types of organic reactions. In substitution reactions, one atom or group is replaced by another, while in elimination reactions, elements are removed from a molecule, resulting in the formation of a double bond. The unchanged rate of substitution despite the presence of deuterium suggests that the C―D bond does not significantly affect the transition state of the substitution pathway, unlike in elimination.
Recommended video:
Guided course
00:38
Intro to Substitution/Elimination Problems

Transition State Theory

Transition state theory posits that during a chemical reaction, reactants pass through a high-energy transition state before forming products. The stability of this transition state is crucial in determining the reaction rate. In the case of the elimination reaction involving deuterium, the stronger C―D bond leads to a higher energy transition state, thus slowing the reaction compared to the substitution pathway, which may not be as sensitive to the bond strength.
Recommended video:
Guided course
06:55
Intermediates vs. Transition States
Related Practice
Textbook Question

Write a mechanism that explains the formation of the following product. In your mechanism, explain the cause of the rearrangement, and explain the failure to form the Zaitsev product.

962
views
Textbook Question

When the following compound is treated with sodium methoxide in methanol, two elimination products are possible. Explain why the deuterated product predominates by about a 7:1 ratio (refer to Problem 7-75).

3653
views
Textbook Question

The following reaction is called the pinacol rearrangement. The reaction begins with an acid-promoted ionization to give a carbocation. This carbocation undergoes a methyl shift to give a more stable, resonance-stabilized cation. Loss of a proton gives the observed product. Propose a mechanism for the pinacol rearrangement.

1787
views
Textbook Question

Deuterium (D) is the isotope of hydrogen of mass number 2, with a proton and a neutron in its nucleus. The chemistry of deuterium is nearly identical to the chemistry of hydrogen, except that the C―D bond is slightly (5.0 kJ/mol, or 1.2 kcal/mol) stronger than the C―H bond. Reaction rates tend to be slower if a C―D bond (as opposed to a C―H bond) is broken in a rate-limiting step. This effect on the rate is called a kinetic isotope effect. (Review PROBLEM 4-57)

a. Propose a mechanism to explain each product in the following reaction.

1453
views