How many different alkenes result when 3-bromo-3-methylheptane is treated with a strong base?Select answer from the options below6  5243

The reaction of 3-bromo-3-methylheptane with a strong base will result in an elimination reaction, specifically E2 elimination. This reaction will remove the bromine and a hydrogen on a beta-carbon, forming a double bond (alkene).

Here are the steps to determine the number of different alkenes:

1. Identify the alpha-carbon: The alpha-carbon is the carbon that is attached to the bromine. In this case, it is the third carbon in the chain.

2. Identify the beta-carbons: The beta-carbons are the carbons that are adjacent to the alpha-carbon. In this case, there are two beta-carbons: the second carbon and the fourth carbon in the chain.

3. Determine the possible alkenes: The double bond can form on either side of the alpha-carbon, resulting in different alkenes.

   • If the double bond forms at the second carbon (between the second and third carbons), the result is 2-methyl-2-heptene.

   • If the double bond forms at the fourth carbon (between the third and fourth carbons), the result is 3-methyl-2-heptene.

So, there are 2 different alkenes that can result from this reaction. Therefore, the answer is 2.