The reaction of 2-Bromopentane with alcoholic KOH (potassium hydroxide) is an example of an elimination reaction, specifically E2 elimination.

Here are the steps to predict the major product:

1. Identify the leaving group: In this case, the leaving group is the bromine atom attached to the second carbon in the 2-Bromopentane.

2. Identify the beta carbon(s): The beta carbon is the carbon adjacent to the carbon attached to the leaving group. In this case, there are two beta carbons - one on the 1st carbon and one on the 3rd carbon.

3. Identify the hydrogen on the beta carbon: The beta hydrogen is the hydrogen atom(s) attached to the beta carbon.

4. The alcoholic KOH acts as a base and abstracts a proton from the beta carbon, causing the electrons in the C-H bond to move and form a double bond with the alpha carbon (the carbon attached to the leaving group). At the same time, the leaving group leaves, resulting in the formation of a double bond.

5. The major product is determined by Zaitsev's rule, which states that the most substituted product will be the major product. In this case, the double bond will form on the side of the 3rd carbon, as it is more substituted than the 1st carbon.

So, the major product of this reaction will be 2-Pentene.

Question

The reaction of 2-Bromopentane with alcoholic KOH (potassium hydroxide) is an example of an elimination reaction, specifically E2 elimination.

Here are the steps to predict the major product:

1. Identify the leaving group: In this case, the leaving group is the bromine atom attached to the second carbon in the 2-Bromopentane.

2. Identify the beta carbon(s): The beta carbon is the carbon adjacent to the carbon attached to the leaving group. In this case, there are two beta carbons - one on the 1st carbon and one on the 3rd carbon.

3. Identify the hydrogen on the beta carbon: The beta hydrogen is the hydrogen atom(s) attached to the beta carbon.

4. The alcoholic KOH acts as a base and abstracts a proton from the beta carbon, causing the electrons in the C-H bond to move and form a double bond with the alpha carbon (the carbon attached to the leaving group). At the same time, the leaving group leaves, resulting in the formation of a double bond.

5. The major product is determined by Zaitsev's rule, which states that the most substituted product will be the major product. In this case, the double bond will form on the side of the 3rd carbon, as it is more substituted than the 1st carbon.

So, the major product of this reaction will be 2-Pentene.

Knowee AI · Accepted Answer

The reaction of 2-Bromopentane with alcoholic KOH (potassium hydroxide) is an example of an elimination reaction, specifically E2 elimination.

Here are the steps to predict the major product:

1. Identify the leaving group: In this case, the leaving group is the bromine atom attached to the second carbon in the 2-Bromopentane.

2. Identify the beta carbon(s): The beta carbon is the carbon adjacent to the carbon attached to the leaving group. In this case, there are two beta carbons - one on the 1st carbon and one on the 3rd carbon.

3. Identify the hydrogen on the beta carbon: The beta hydrogen is the hydrogen atom(s) attached to the beta carbon.

4. The alcoholic KOH acts as a base and abstracts a proton from the beta carbon, causing the electrons in the C-H bond to move and form a double bond with the alpha carbon (the carbon attached to the leaving group). At the same time, the leaving group leaves, resulting in the formation of a double bond.

5. The major product is determined by Zaitsev's rule, which states that the most substituted product will be the major product. In this case, the double bond will form on the side of the 3rd carbon, as it is more substituted than the 1st carbon.

So, the major product of this reaction will be 2-Pentene.

Predict the major product formed when 2-Bromopentane reacts with alcoholic KOH.

Question

Predict the major product formed when 2-Bromopentane reacts with alcoholic KOH.

Solution

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