Show the mechanism of KOH and alcohol for forming double bond in an alkyl halide?

Question

HIX Tutor · Accepted Answer

undefined To form a double bond in an alkyl halide using KOH and alcohol, you would follow these steps:

Step 1: Mix the alkyl halide (containing the halogen atom) with KOH (potassium hydroxide) and the alcohol.

Step 2: The alkyl halide reacts with KOH to form an alkoxide ion and a potassium halide.

Step 3: The alkoxide ion then reacts with the alcohol to form an alkoxide ion of the alcohol and a new alkyl halide.

Step 4: Elimination occurs, where the alkoxide ion from the alcohol removes a hydrogen atom from the neighboring carbon, resulting in the formation of a double bond.

Remember, these steps are an overview, and each step has its own set of reactions and mechanisms that occur. If you need more detailed information about a specific step, let me know and I'll be happy to help!

Isabella Acosta · Answer

Here's what I get.

In a solution of alcoholic #"KOH"#, we have the equilibrium:

#"CH"_3"CH"_2"O-H" +"OH"^"-" ⇌ "CH"_3"CH"_2"O"^"-" + "H"_2"O"#

The position of equilibrium lies to the left, so there is not much ethoxide ion present.

However, ethoxide is a much stronger base than hydroxide, so it is the favoured reactant.

The ethoxide attacks an α-hydrogen atom in an #"E2"# elimination reaction.

The products are ethanol, an alkene, and a halide ion.

Jacob Anderson · Answer

undefined The mechanism of KOH (potassium hydroxide) and alcohol for forming a double bond in an alkyl halide involves the elimination reaction known as the E2 (bimolecular elimination) mechanism. In this reaction, the hydroxide ion ($OH^-$) from KOH acts as a strong base, abstracting a proton from the beta carbon (the carbon adjacent to the halogen) of the alkyl halide. Simultaneously, the halide ion ($X^-$) leaves, resulting in the formation of a double bond between the alpha and beta carbons.

This mechanism proceeds in a single step, involving the concerted removal of the proton and the leaving group, leading to the formation of the alkene product. The elimination occurs in a bimolecular fashion, meaning that both the alkyl halide substrate and the base (KOH) participate in the rate-determining step of the reaction.

The general mechanism can be represented as follows:

1. Deprotonation: The hydroxide ion ($OH^-$) abstracts a proton from the beta carbon of the alkyl halide, leading to the formation of a carbon-carbon double bond and the generation of water ($H_2O$).

2. Leaving Group Departure: The halide ion ($X^-$) departs from the molecule, resulting in the formation of the alkene product.

Overall, this process converts an alkyl halide into an alkene through the elimination of a halide ion and a proton, facilitated by a strong base such as KOH in the presence of an alcohol solvent.