Question

Draw the structural formula of the products formed when each alkene is treated with one equivalent of \(\mathrm{NBS}\) in \(\mathrm{CH}_{2} \mathrm{Cl}_{2}\) in the presence of light.

Short answer

Answer: The product formed is 3-bromopropene.

Step-by-step solution

Identify the allylic position

To do this, we first need to identify the carbon atoms adjacent to the double bond in the alkene. These carbons are known as allylic carbons and it is where the bromination will occur.

React the alkene with NBS

Next, we need to react the alkene with NBS in CH2Cl2. NBS is a source of bromine, which is able to selectively brominate the allylic position in the alkene in the presence of light. The light helps to generate bromine radicals, needed for the reaction to progress.

Draw the products of the reaction

Finally, draw the structural formula of the products formed by incorporating the bromine at the allylic positions. Keep in mind that the reaction may have multiple products if there is more than one possible allylic carbon. Since we don't have a specific alkene given in the exercise, let's consider an example alkene: propene (CH2=CHCH3).

Example: Propene

1. Identify the allylic position: In propene, the carbon adjacent to the double bond is the methyl group (CH3). 2. React propene with NBS: \(\mathrm{CH}_{2}=\mathrm{CHCH}_{3} \xrightarrow [\mathrm{CH}_{2}\mathrm{Cl}_{2}][]{\mathrm{NBS}, \mathrm h\nu}\) 3. Draw product: After bromination, the product will be 3-bromopropene: \(\mathrm{CH}_{2}= \mathrm{CHCH}_{2} \mathrm{Br}\) So, treating propene with one equivalent of NBS in dichloromethane in the presence of light produces 3-bromopropene as the final product.