Question

When isopropyl bromide is treated with sodium ethoxide in ethanol, propylene and ethyl isopropyl ether are formed in a \(3: 1\) ratio. If the hexadeuteroisopropy1 bromide, \(\mathrm{CD}_{3} \mathrm{CHBrCD}_{3}\) is used, \(\mathrm{CD}_{3} \mathrm{CH}=\mathrm{CD}_{2}\) and \(\left(\mathrm{CD}_{3}\right)_{2} \mathrm{CHOC}_{2} \mathrm{H}_{5}\) are formed in a ratio of \(1: 2\). Explain.

Short answer

In summary, when isopropyl bromide is treated with sodium ethoxide in ethanol, the elimination reaction (E2) is favored over the nucleophilic substitution (SN2), resulting in a 3:1 ratio of propylene to ethyl isopropyl ether. However, upon deuterium substitution, the ratio changes to 1:2 for the hexadeuteroisopropyl bromide, due to the stronger C-D bond that slows down the E2 reaction and allows the SN2 reaction to compete more effectively. This leads to a higher ratio of the substitution product.

Step-by-step solution

Identify the reaction mechanism

The isopropyl bromide reacts with sodium ethoxide in ethanol, which is a classical example of nucleophilic substitution (SN2) reaction, where the bromide ion is replaced by the ethoxide ion. Alternatively, it could also undergo elimination (E2) reaction, where the bromide ion and a proton from a neighboring carbon are removed, leading to the formation of an alkene (propylene).

Identify the two possible products of the reaction

Upon reacting with sodium ethoxide, isopropyl bromide can yield two different products: 1. Substitution product - Ethyl isopropyl ether: This product is formed when the ethoxide ion replaces the bromide ion. 2. Elimination product - Propylene: This product is formed when a proton from the adjacent carbon is removed along with the bromide ion, and a double bond is formed between the carbons.

Discuss the effect of deuterium substitution on ratios

When hexadeuteroisopropyl bromide (\(CD_{3}CHBrCD_{3}\)) is used, the product ratio changes to: 1. \(CD_{3}CH=CD_{2}\): This is the elimination product, where a hydrogen from the adjacent carbon is removed along with the bromide ion, and a double bond is formed between the carbons. 2. \((CD_{3})_{2}CHOC_{2}H_{5}\): This is the substitution product, where the ethoxide ion replaces the bromide ion. Deuterium substitution of the isopropyl bromide has the effect of changing the relative rates of the nucleophilic substitution (SN2) and elimination (E2) reactions. This is because the C-D bond is stronger than the C-H bond, so the removal of a deuterium atom in the E2 reaction is slower than the removal of a hydrogen atom.

Explain the product ratios

In the case of isopropyl bromide, the 3:1 ratio in the formation of propylene and ethyl isopropyl ether suggests that the elimination (E2) reaction is favored over the substitution (SN2) reaction. When hexadeuteroisopropyl bromide is used, the ratio of elimination to substitution product changes to 1:2. The slower E2 reaction due to the stronger C-D bond allows for nucleophilic substitution (SN2) to compete more effectively, resulting in a higher ratio of the substitution product. So, the change in ratio of the products can be attributed to the effect of deuterium substitution on the relative rates of the E2 and SN2 reactions.

Key concepts

SN2 Reaction

In organic chemistry, the SN2 reaction is a form of nucleophilic substitution where a nucleophile directly displaces a leaving group on an sp3 hybridized carbon atom in a single, concerted step. The '2' in SN2 denotes the bimolecular nature of the reaction, indicating that both the substrate (such as isopropyl bromide) and the nucleophile (such as sodium ethoxide) are involved in the rate-determining step.

This reaction mechanism is characterized by a backside attack on the carbon atom, leading to an inversion of stereochemistry. It's a common mechanism for alkylation reactions in synthesis, especially when using a strong, unhindered nucleophile and a primary alkyl halide. Competitive experiments with deuterium-labeled compounds help to confirm the SN2 mechanism, as they can affect the reaction rate through the isotope effect.

E2 Reaction

On the other hand, an E2 reaction is a type of elimination reaction where a proton is removed from the β-carbon, and a leaving group (like a bromide ion) departs from the α-carbon, resulting in the formation of a carbon-carbon double bond. The '2' indicates that this is also a bimolecular process; the base (nucleophile) and the substrate are involved in the rate-limiting step.

The E2 mechanism typically requires a strong base, such as sodium ethoxide, and the reaction tends to favor more substituted alkene as the product due to the Zaitsev's rule. It can compete with the SN2 reaction, particularly with secondary substrates like isopropyl bromide, and the choice of base and the reaction conditions can influence which pathway will predominate. When deuterium effects are observed, they give insight into the reaction mechanism by altering the reaction rates.

Isopropyl Bromide

Isopropyl bromide, also known as 2-bromopropane, is a secondary alkyl halide commonly used as a reagent in organic synthesis. The presence of the bromide ion, which is a good leaving group, makes it prone to both nucleophilic substitution (SN2) and elimination (E2) reactions. The structure of isopropyl bromide allows for these reactions to take place at the secondary carbon, with potential competing reaction pathways.

Due to its reactivity, isopropyl bromide serves as an excellent example to study the competition between SN2 and E2 reactions. Depending on the conditions, such as solvent and temperature, isopropyl bromide can yield an ether via an SN2 reaction or an alkene via an E2 reaction, allowing students to observe how reaction conditions influence the outcome.

Deuterium Effect

The deuterium effect is an isotope effect observed in chemical reactions where deuterium, a heavier isotope of hydrogen, is involved. When hydrogen atoms in a molecule are replaced by deuterium, such as in hexadeuteroisopropyl bromide, this can significantly affect the rate of chemical reactions, particularly those involving the breaking of carbon-hydrogen or carbon-deuterium bonds.

Since deuterium-carbon bonds (C-D) are stronger and have a lower zero-point energy than the corresponding carbon-hydrogen (C-H) bonds, reactions that involve breaking these bonds, like E2 eliminations, occur more slowly. This results in a shift in the product distribution when deuterium is present at the reactive site, as can be observed in the different ratios of elimination and substitution products when comparing reactions with isopropyl bromide and its deuterated counterpart.