Question

When \(\mathrm{D}_{2}\) reacts with ethylene \(\left(\mathrm{C}_{2} \mathrm{H}_{4}\right)\) in the presence of a finely divided catalyst, ethane with two deuteriums, \(\mathrm{CH}_{2} \mathrm{D}-\mathrm{CH}_{2} \mathrm{D}\), is formed. (Deuterium, \(\mathrm{D}\), is an isotope of hydrogen of mass 2.) Very little ethane forms in which two deuteriums are bound to one carbon (for example, \(\left.\mathrm{CH}_{3}-\mathrm{CHD}_{2}\right)\). Use the sequence of steps involved in the reaction to explain why this is so.

Short answer

In the presence of a catalyst, the deuterium addition reaction proceeds by breaking the more reactive C=C double bond in ethylene and forming two new D-C single bonds with separate deuterium atoms, resulting in the CH₂D-CH₂D molecule. The formation of molecules with two deuterium atoms bound to the same carbon, such as CH₃-CHD₂, is less likely since it requires breaking a less reactive C-H bond, thus producing very little ethane with two deuteriums on the same carbon.

Step-by-step solution

Identify the reaction mechanism

The reaction between D₂ and ethylene (C₂H₄) in the presence of a catalyst is an example of a hydrogenation reaction, specifically, deuterium addition, where deuterium atoms replace hydrogen atoms in ethylene to form deuterated ethane. In this case, the catalyst helps to break the D-D bond and form new D-C bonds to produce CH₂D-CH₂D.

Understand the catalytic process

In the presence of the finely divided catalyst, the hydrogenation reaction proceeds through a series of steps. First, the catalyst adsorbs the reactants, ethylene, and D₂, onto its surface. This facilitates the breaking of the D-D bond and the formation of new D-C bonds with the ethylene molecule.

Deuterium bonding to the ethylene molecule

As the D₂ molecule interacts with the catalyst and breaks the D-D bond, two separate deuterium atoms are formed. These deuterium atoms can then bond with the ethylene molecule. The ethylene molecule has a double bond between the two carbon atoms, and each deuterium atom forms a bond with one of the carbon atoms, breaking the double bond and creating two new D-C single bonds. At this stage, we obtain the product: CH₂D-CH₂D.

Formation of CH₃-CHD₂ molecules

For the CH₃-CHD₂ molecule to form, the first deuterium addition would occur, comparable to step 3. However, the second deuterium atom would have to bond with the same carbon atom that the first deuterium atom bonded to. This process requires breaking a C-H bond, which is less likely to occur than breaking the C=C double bond, which is more reactive. #Conclusion# The sequence of steps involved in the reaction of D₂ with ethylene in the presence of a catalyst produces the molecule CH₂D-CH₂D primarily because the formation of this molecule involves breaking the more reactive C=C double bond in ethylene, leading to the formation of two new D-C single bonds. Forming molecules with two deuterium atoms bound to the same carbon, such as CH₃-CHD₂, would require breaking a C-H bond, which is less likely to happen due to the lower reactivity of this bond. Therefore, very little ethane with two deuteriums bound to the same carbon forms during this reaction.

Key concepts

Catalysis

Catalysis plays an essential role in chemical reactions by accelerating reaction rates and enabling pathways that might otherwise be inaccessible. When we consider the role of a catalyst in the reaction between deuterium (\(\mathrm{D}_2\)) and ethylene (\(\mathrm{C}_2\mathrm{H}_4\)), it becomes clear that the finely divided catalyst is crucial for breaking the strong D-D bond. This bond-breaking is a key step, allowing the deuterium atoms to add across the ethylene double bond.
The catalyst provides a surface onto which both reactants can adsorb. This adsorption primes the D-D bond for cleavage and facilitates the sequential formation of D-C bonds. By lowering the activation energy necessary to break these bonds, the catalyst significantly enhances the reaction rate.
Overall, the catalyst fundamentally changes how the reaction proceeds, ensuring that the formation of the product, typically in the form of \(\mathrm{CH}_2\mathrm{D}-\mathrm{CH}_2\mathrm{D}\), is favored over other possible products. Without the catalyst, the conditions needed for the reaction would be much more severe, potentially making the reaction impractical.

Deuterium Chemistry

Deuterium chemistry focuses on the interaction and bonding involving deuterium, an isotope of hydrogen with a nucleus containing one proton and one neutron, giving it atomic mass of 2. In the context of the given reaction, deuterium replaces hydrogen atoms in ethylene to form a deuterated product.
Understanding deuterium chemistry is essential for knowing why some products form in greater amounts than others. In this reaction, deuterium atoms preferentially add across the \(\mathrm{C=C}\) double bond of ethylene, resulting in the balanced product \(\mathrm{CH}_2\mathrm{D}-\mathrm{CH}_2\mathrm{D}\). This occurs because breaking the \(\mathrm{C=C}\) bond is energetically more favorable than breaking the \(\mathrm{C-H}\) bond, allowing deuterium atoms to seamlessly integrate themselves into the ethylene structure.
Deuterium's unique properties, such as its slightly different electron distribution compared to hydrogen, mean reactions involving deuterium can have subtly different dynamics. These differences are crucial when it comes to examining reaction mechanisms as they can lead to preferential product formation.

Hydrogenation Reaction

In organic chemistry, hydrogenation reactions involve the addition of hydrogen (\(\mathrm{H}_2\)) across double or triple bonds to form single bonds, often transforming unsaturated compounds into saturated ones. In the given scenario, this type of reaction can also involve deuterium, turning it instead into a deuteration reaction.
The process begins with the adsorption of ethylene and \(\mathrm{D}_2\) onto the catalyst surface. The catalyst then facilitates the breaking of the diatomic deuterium bond, enabling the individual deuterium atoms to add across the ethylene’s double bond. This turns the double bond into a single bond, producing a saturated compound, \(\mathrm{CH}_2\mathrm{D}-\mathrm{CH}_2\mathrm{D}\).
Hydrogenation reactions are widely used in the chemical industry to saturate hydrocarbons. When involving isotopes like deuterium, these reactions help in studying reaction mechanisms due to the isotope effect, which can reveal insights into the steps involved in bond cleavage and formation.