Question

Mechanism involved and product obtained when 3 -iodo-3-ethylpentane is treated with sodium methoxide methanol is (A) Ether, \(\mathrm{S}_{\mathrm{N}} 1\) (B) Alkene, \(\mathrm{E}^{1}\) (C) Ether, \(\mathrm{E}^{\prime}\) (D) Alkene, \(\mathrm{E}^{2}\)

Short answer

The correct mechanism involved and product obtained when 3-iodo-3-ethylpentane is treated with sodium methoxide methanol is (D) Alkene, \(\mathrm{E}^{2}\).

Step-by-step solution

Identify the given reactants

We have the reactants: 1. 3-iodo-3-ethylpentane 2. Sodium methoxide methanol

Analyze option A: Ether, \(\mathrm{S}_{\mathrm{N}} 1\)

In this option, we are considering the \(\mathrm{S}_{\mathrm{N}} 1\) substitution mechanism. In this mechanism, the nucleophile attacks the substrate after the leaving group (iodine) has left, resulting in an ether product. However, our substrate is a tertiary substrate, and tertiary substrates generally favor \(\mathrm{S}_{\mathrm{N}} 2\) or \(\mathrm{E}^{2}\) mechanisms rather than \(\mathrm{S}_{\mathrm{N}} 1\). Based on this reasoning, option A is not likely the correct choice.

Analyze option B: Alkene, \(\mathrm{E}^{1}\)

In this option, we are considering the \(\mathrm{E}^{1}\) elimination mechanism. It involves a two-step reaction where the leaving group leaves to create a carbocation, and then a deprotonation occurs to produce the alkene. Similar to the reasoning for the \(\mathrm{S}_{\mathrm{N}} 1\) mechanism in option A, tertiary substrates favor other mechanisms over the \(\mathrm{E}^{1}\) mechanism. Therefore, option B is not likely the correct choice.

Analyze option C: Ether, \(\mathrm{E}^{\prime}\)

This option mentions the product as an ether with an ambiguous \(\mathrm{E}^{\prime}\) mechanism. Since this mechanism is not clearly defined, we cannot assess the validity of option C.

Analyze option D: Alkene, \(\mathrm{E}^{2}\)

In this option, we are considering the \(\mathrm{E}^{2}\) elimination mechanism. The \(\mathrm{E}^{2}\) mechanism is a one-step reaction where the nucleophile, in this case, sodium methoxide, removes a proton from the carbon next to the one that iodine is attached to. Simultaneously, the iodine leaves the substrate, and a double bond is formed between the two carbons, producing an alkene product. Since the substrate is a tertiary system and has a strong nucleophile, the \(\mathrm{E}^{2}\) mechanism seems to be the most reasonable choice. Given our analysis of the options, the correct mechanism involved and product obtained when 3-iodo-3-ethylpentane is treated with sodium methoxide methanol is:

Answer: Option D

(D) Alkene, \(\mathrm{E}^{2}\)

Key concepts

E2 mechanism

In organic chemistry, the E2 mechanism stands for bimolecular elimination. This mechanism involves a one-step process where both the base and the substrate simultaneously interact. In the context of our exercise, sodium methoxide acts as a base. It removes a proton from the β-carbon (the carbon adjacent to the carbon bonded with the iodine), while the iodine leaves as the leaving group. This simultaneous action results in the formation of a double bond, creating an alkene.

Key characteristics of the E2 mechanism include:

• A concerted reaction, meaning it happens in one step without the formation of intermediates.
• Requires a strong base because simultaneous bond formation and breakage occur.
• Favored by higher temperatures and polar aprotic solvents.

The elimination follows a stereospecific route, typically resulting in the trans-diastereomer if the carbon chain permits. Understanding E2 is crucial for predicting alkene products and is widely applied in synthetic organic chemistry.

tertiary substrate

Tertiary substrates are molecules where the carbon atom bonded to the leaving group (in this case, iodine) is further connected to three other carbon atoms. This makes them sterically hindered yet quite reactive under certain conditions, like the E2 mechanism.

The reasons tertiary substrates favor the E2 mechanism include:

• The hindered central carbon makes it difficult for nucleophiles to approach and attack directly, ruling out mechanisms like S_N2.
• E2 reactions are less affected by steric hindrance since elimination involves an external base removing a proton.
• Tertiary carbon centers are typically more stable, allowing easier removal of the leaving group.

In the exercise we analyzed, the tertiary substrate, 3-iodo-3-ethylpentane, is indeed highly prone to undergoing E2 reactions, resulting in the efficient formation of alkenes.

alkene formation

Alkene formation through elimination reactions like E2 is fundamental in organic synthesis. An alkene, characterized by a carbon-carbon double bond, is formed when a hydrogen atom and a leaving group (i.e., iodine) are removed from adjacent carbon atoms. This process converts saturated hydrocarbons into unsaturated ones.

The steps leading to alkene formation in an E2 reaction include:

• The base (sodium methoxide in this case) attacks a proton on the β-carbon.
• Simultaneously, the leaving group (iodine) departs from the α-carbon.
• A π-bond forms, resulting in an alkene.

Such resultant alkenes are vital building blocks in various chemical syntheses and can be used to create polymers, alcohols, and other complex molecules. The understanding of how conditions like solvent and temperature affect the regioselectivity and stereoselectivity of alkene formation is crucial for chemists aiming to maximize desired product yields.