When 2-bromopropane is treated with sodium ethoxide, propene is produced. What molecule is lost from 2-bromopropane in this process?

Guida verificata passo dopo passo

Step 1: Recognize the type of reaction occurring. This is an elimination reaction (E2 mechanism), where a molecule is lost from the substrate to form a double bond.

Step 2: Identify the substrate, 2-bromopropane, and note that it contains a bromine atom attached to the second carbon of a propane chain.

Step 3: Understand the role of sodium ethoxide (NaOCH₂CH₃). It acts as a strong base, abstracting a proton (H⁺) from a β-carbon (the carbon adjacent to the carbon bonded to bromine).

Step 4: Determine the molecule lost during the reaction. The bromine atom leaves as a bromide ion (Br⁻), and a proton (H⁺) is abstracted from the β-carbon, resulting in the formation of propene (CH₃-CH=CH₂). The lost molecule is HBr (hydrogen bromide).

Step 5: Summarize the process: Sodium ethoxide facilitates the elimination of HBr from 2-bromopropane, leading to the formation of propene via the E2 mechanism.

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Elimination Reactions

Elimination reactions involve the removal of a small molecule from a larger one, resulting in the formation of a double bond. In the case of 2-bromopropane reacting with sodium ethoxide, an elimination reaction occurs, leading to the formation of propene. Understanding the mechanism of elimination is crucial for predicting the products of such reactions.

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Leaving Groups

A leaving group is an atom or group that can depart from the parent molecule during a chemical reaction. In the context of 2-bromopropane, the bromine atom acts as a leaving group when the sodium ethoxide facilitates the elimination process. The stability and ability of the leaving group to depart significantly influence the reaction's efficiency.

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The 3 important leaving groups to know.

Base-Induced Elimination

Base-induced elimination refers to the process where a strong base, such as sodium ethoxide, abstracts a proton from a β-carbon, leading to the formation of a double bond. This mechanism is essential in understanding how propene is generated from 2-bromopropane, as the base facilitates the removal of a hydrogen atom alongside the leaving group.

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