Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Assuming that the mechanism for the hydrogenation of \(\mathrm{C}_{2} \mathrm{H}_{4}\) given in Section \(12.7\) is correct, would you predict that the product of the reaction of \(\mathrm{C}_{2} \mathrm{H}_{4}\) with \(\mathrm{D}_{2}\) would be \(\mathrm{CH}_{2} \mathrm{D}-\mathrm{CH}_{2} \mathrm{D}\) or \(\mathrm{CHD}_{2}-\mathrm{CH}_{3} ?\) How could the reaction of \(\mathrm{C}_{2} \mathrm{H}_{4}\) with \(\mathrm{D}_{2}\) be used to confirm the mechanism for the hydrogenation of \(\mathrm{C}_{2} \mathrm{H}_{4}\) given in Section \(12.7\) ?

Short Answer

Expert verified
The product of the reaction between \(\mathrm{C}_{2}\mathrm{H}_{4}\) and \(\mathrm{D}_{2}\) would be \(\mathrm{CH}_{2}\mathrm{D}-\mathrm{CH}_{2}\mathrm{D}\), based on the hydrogenation mechanism given in Section 12.7. The reaction can confirm the mechanism, as the formation of this product is consistent with the transfer of deuterium atoms to the double-bonded carbon atoms in the ethene molecule.

Step by step solution

01

Write down the mechanism

The reaction mechanism for the hydrogenation of ethene (\(\mathrm{C}_{2}\mathrm{H}_{4}\)) involves the formation of a pi complex between ethene and the hydrogen molecule, followed by the transfer of hydrogen atoms to the double-bonded carbon atoms, leading to the formation of ethane. In this case, we will consider the same mechanism with deuterium (\(\mathrm{D}_{2}\)) instead of hydrogen.
02

Apply the mechanism to the given reaction

Using the known mechanism, let's apply it to the reaction between \(\mathrm{C}_{2}\mathrm{H}_{4}\) and \(\mathrm{D}_{2}\). In the first step, a pi complex forms between the ethene molecule and the deuterium molecule. In the second step, the deuterium atoms will transfer to the ethene molecule.
03

Determine the product

According to the known mechanism, both deuterium atoms will transfer to the double-bonded carbon atoms in the ethene molecule, leading to the formation of \(\mathrm{CH}_{2}\mathrm{D}-\mathrm{CH}_{2}\mathrm{D}\) as the major product.
04

Discuss the mechanism confirmation

If the reaction between \(\mathrm{C}_{2}\mathrm{H}_{4}\) and \(\mathrm{D}_{2}\) yields a major product of \(\mathrm{CH}_{2}\mathrm{D}-\mathrm{CH}_{2}\mathrm{D}\), it confirms the mechanism for the hydrogenation of \(\mathrm{C}_{2}\mathrm{H}_{4}\). This is because the mechanism involves the transfer of deuterium (or hydrogen) atoms to the double-bonded carbon atoms, and the observed product formation is consistent with this mechanism.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Pi Complex
In the context of hydrogenation, a pi complex plays a significant role. It is formed when a molecule with a double bond, like ethene (\(\mathrm{C}_{2}\mathrm{H}_{4}\)), interacts with another molecule, such as hydrogen (\(\mathrm{H}_{2}\)) or deuterium (\(\mathrm{D}_{2}\)).

This interaction involves the electrons in the pi bond of ethene aligning with the particles of \(\mathrm{H}_{2}\) or \(\mathrm{D}_{2}\), creating a temporary complex.

This formation is crucial because it primes the ethene molecule for the subsequent steps in the reaction.
  • The pi complex stabilizes the intermediate state.
  • It facilitates electron density transfer from the double bond to the reacting molecule.
When deuterium is involved instead of regular hydrogen, the process is akin to hydrogenation but with key differences observed in isotopic substitution. This alters the properties slightly due to the heavier mass of deuterium.
Deuterium
Deuterium is a stable isotope of hydrogen, denoted as \(\mathrm{D}\), and contains one proton and one neutron in its nucleus. This makes it double the mass of a hydrogen atom, which only contains a single proton.

In reactions involving deuterium, such as the hydrogenation mechanism with ethene, deuterium behaves similarly to hydrogen in terms of forming and breaking bonds.

However, its added mass can influence the reaction rate and transition state energies slightly.
  • Deuterium is used to track reactions due to its distinguishable mass.
  • It helps confirm reaction pathways without changing the outcome significantly.
In the reaction with \(\mathrm{C}_{2}\mathrm{H}_{4}\), deuterium acts as a hydron donor. When added across the double bond, it forms \(\mathrm{CH}_{2} \mathrm{D}- \mathrm{CH}_{2} \mathrm{D}\), indicating that both deuterium atoms from \(\mathrm{D}_{2}\) successfully incorporated into the product.
Ethene
Ethene, also known as ethylene, is a simple alkene with the formula \(\mathrm{C}_{2}\mathrm{H}_{4}\). It has a double bond (\(\mathrm{C} = \mathrm{C}\)) making it highly reactive in chemical processes such as hydrogenation.

The pi bonds in ethene are responsible for its reactivity, allowing it to form pi complexes during reactions. These double bonds open the door for the incorporation of hydrogen or deuterium, converting \(\mathrm{C}_{2}\mathrm{H}_{4}\) into a single-bonded structure like ethane.

  • Ethene serves as an important feedstock in chemical industries for producing polymers.
  • It acts as a prototype for understanding the behavior of alkenes in addition reactions.
When ethene reacts in a hydrogenation mechanism, the \(\mathrm{D}_{2}\) interaction leads to the formation of a dibasic alkane, where each carbon atom in the original double bond bonds with deuterium.
Reaction Confirmation
Reaction confirmation refers to verifying the proposed mechanism using empirical evidence. In the case of hydrogenation involving deuterium, it is essential to ensure that the mechanism is consistent with the experimental results.

When ethene reacts with deuterium, the formation of \(\mathrm{CH}_{2}\mathrm{D}-\mathrm{CH}_{2}\mathrm{D}\) as the main product helps confirm that:
  • The pi complex correctly forms between deuterium and ethene, allowing for proper hydrogenation.
  • The deuterium atoms are being transferred correctly to the molecule’s double bonds.
Supporting the observed product aligns with the expected mechanism and rules out alternative reaction pathways. Additionally, isotopic labeling using deuterium offers an insightful perspective and highlights the actual position where atoms are incorporated during the reaction process.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Table \(12.2\) illustrates how the average rate of a reaction decreases with time. Why does the average rate decrease with time? How does the instantaneous rate of a reaction depend on time? Why are initial rates used by convention?

Cobra venom helps the snake secure food by binding to acetylcholine receptors on the diaphragm of a bite victim, leading to the loss of function of the diaphragm muscle tissue and eventually death. In order to develop more potent antivenoms, scientists have studied what happens to the toxin once it has bound the acetylcholine receptors. They have found that the toxin is released from the receptor in a process that can be described by the rate law Rate \(=k[\) acetylcholine receptor-toxin complex \(]\) If the activation energy of this reaction at \(37.0^{\circ} \mathrm{C}\) is \(26.2 \mathrm{~kJ} / \mathrm{mol}\) and \(A=0.850 \mathrm{~s}^{-1}\), what is the rate of reaction if you have a \(0.200 M\) solution of receptor-toxin complex at \(37.0^{\circ} \mathrm{C}\) ?

The reaction $$ \left(\mathrm{CH}_{3}\right)_{3} \mathrm{CBr}+\mathrm{OH}^{-} \longrightarrow\left(\mathrm{CH}_{3}\right)_{3} \mathrm{COH}+\mathrm{Br}^{-} $$ in a certain solvent is first order with respect to \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CBr}\) and zero order with respect to \(\mathrm{OH}^{-}\). In several experiments, the rate constant \(k\) was determined at different temperatures. A plot of \(\ln (k)\) versus \(1 / T\) was constructed resulting in a straight line with a slope value of \(-1.10 \times 10^{4} \mathrm{~K}\) and \(y\) -intercept of \(33.5\). Assume \(k\) has units of \(\mathrm{s}^{-1}\). a. Determine the activation energy for this reaction. b. Determine the value of the frequency factor \(A\). c. Calculate the value of \(k\) at \(25^{\circ} \mathrm{C}\).

Consider the following statements: "In general, the rate of a chemical reaction increases a bit at first because it takes a while for the reaction to get 'warmed up.' After that, however, the rate of the reaction decreases because its rate is dependent on the concentrations of the reactants, and these are decreasing." Indicate everything that is correct in these statements, and indicate everything that is incorrect. Correct the incorrect statements and explain.

Enzymes are kinetically important for many of the complex reactions necessary for plant and animal life to exist. However, only a tiny amount of any particular enzyme is required for these complex reactions to occur. Explain.

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free