Question

Treatment of 1-bromo-2-deuterio-2-phenylethane with strong base leads to a mixture of deuterated and nondeuterated phenylethylenes in an approximately 7: 1 ratio. Explain.

Short answer

The product ratio is due to the Kinetic Isotope Effect favoring hydrogen removal over deuterium.

Step-by-step solution

Identify the Reaction Type

The reaction described involves the use of a strong base on 1-bromo-2-deuterio-2-phenylethane, which suggests an elimination reaction. In such reactions, the base abstracts a proton (or deuterium in this case), leading to the formation of a double bond, resulting in an alkene.

Determine Possible Products

The possible products from the elimination reaction are phenylethylenes that can either be deuterated (containing a deuterium atom) or nondeuterated (without a deuterium atom). The deuterium can be either retained in the alkene or replaced during the reaction.

Consider the Mechanism - E2

In an E2 elimination mechanism, the base removes the β-hydrogen or β-deuterium, leading to the formation of the double bond. If deuterium is specifically targeted, a deuterated alkene forms. Given that the nondeuterated product is prevalent in a 7:1 ratio, it's important to analyze the base's kinetic preference toward deuterium over hydrogen.

Analyze Kinetic Isotope Effect

The Kinetic Isotope Effect (KIE) explains that breaking a C-H bond is generally more favorable than a C-D bond due to lower zero-point energy in C-D bonds, making the reaction faster when hydrogen is removed. Hence, the major product is the nondeuterated phenylethylene.

Conclude the Product Ratio

The observed 7:1 ratio of nondeuterated to deuterated phenylethylenes can be attributed to the kinetic isotope effect. The strength of the C-D bond results in lower reaction rates for deuterium removal, leading to a predominance of the nondeuterated product.

Key concepts

Elimination Reaction

An elimination reaction is a type of organic reaction where certain atoms or groups are removed from a molecule, leading to the formation of a new product. In the context of the problem, the strong base induces the elimination reaction by abstracting a proton. This is typical in organic chemistry, where bases cause the removal of a proton, forming a carbon-carbon double bond, and thus generating an alkene.
Elimination reactions often lead to the formation of alkenes, which contain double bonds. Such reactions can be broadly categorized, based on the details of their mechanisms, into several types, including the E2 mechanism. Elimination reactions are essential because they introduce unsaturation into molecules, making them more reactive substrates for other chemical transformations.
In addition, elimination reactions are not just about removing atoms, they also involve the rearrangement of electrons, leading to the formation of new chemical structures. This makes them valuable in synthetic chemistry for constructing more complex molecules.

E2 Mechanism

The E2 mechanism is a bimolecular reaction, meaning it involves two reacting species. In this case, the strong base and the substrate, 1-bromo-2-deuterio-2-phenylethane, interact directly with each other. This is a single-step concerted mechanism where the base abstracts a β-hydrogen, resulting in the formation of a double bond and thus an alkene.
E2 mechanisms are characterized by a direct, simultaneous removal of hydrogen from the β-carbon and the leaving group (like bromine in this scenario) from the α-carbon. The reaction proceeds through a transition state without forming any intermediate products.
A key feature of the E2 mechanism is the stereochemistry: it often requires an anti-periplanar geometry, where the hydrogen and the leaving group must be on opposite sides of the molecule for optimal overlap of orbitals. This geometric feature influences which products are formed and in what ratio. Understanding this mechanism is crucial for predicting the outcomes of elimination reactions.

Deuterium

Deuterium is an isotope of hydrogen that contains one proton and one neutron, making it twice as heavy as hydrogen with no neutrons. This additional mass influences the behavior of deuterium in chemical reactions due to differences in bond strength and reaction kinetics.
In chemical reactions, particularly in this exercise, deuterium's presence in a molecule like 1-bromo-2-deuterio-2-phenylethane can alter the rate at which the reaction proceeds. Deuterium forms stronger bonds compared to hydrogen, resulting in a phenomenon known as the Kinetic Isotope Effect (KIE). This effect makes C-D bond breaking slightly less favorable and slower compared to C-H bond breaking.
This exercise highlights how the KIE impacts product distribution in elimination reactions. Because breaking the C-D bond requires more energy, reactions involving deuterium often progress at a slower rate compared to those involving regular hydrogen, favoring the formation of nondeuterated products.

Alkene Formation

Alkene formation is a crucial aspect of elimination reactions, where the removal of atoms leads to the creation of carbon-carbon double bonds. In the specific reaction we've been discussing, the transition from 1-bromo-2-deuterio-2-phenylethane to phenylethylenes is a prime example.
The formation of alkenes is influenced by many factors, including the strength of the bonds and the alignment of atoms and groups within the molecule. In the E2 mechanism, the correct alignment is critical for the spontaneous formation of the double bond.
The development of a double bond not only changes the chemical structure but also impacts the physical and chemical properties of the molecule. Alkenes are more reactive than their saturated counterparts due to the electron-rich nature of the double bond, making them vital intermediates in various chemical syntheses and industrial applications.