State the number and kind of stereoisomers formed when \((R)\)-s-methyl-1-pentene is treated with these reagents. Assume that the starting alkene is enantiomerically pure and optically active. Will each product be optically active or inactive?

First, we need to identify the structure of \((R)\)-3-methyl-1-pentene. This is a 1-pentene, meaning that it is a five-carbon chain with a double bond between the first and second carbons. Additionally, there is a methyl group on the third carbon. The fact that it is the \((R)\) enantiomer means that it has a specific configuration at the chiral center (the third carbon).

In this exercise, we are not given specific reagents. However, since we are dealing with an alkene, we can consider reactions that typically involve alkenes, such as electrophilic addition reactions, which include: 1. Halogenation (X2) - addition of halogens like Cl2 or Br2. 2. Hydrohalogenation (HX) - addition of a hydrogen halide (like HCl, HBr, or HI). 3. Hydration (H2O/H2SO4) - addition of water, under acidic conditions. Considering these reaction types will help us determine the potential stereoisomers that can be produced.

We will now analyze each reaction type listed in Step 2 and determine the number and kind of stereoisomers formed. 1. Halogenation - In this reaction, stereochemistry of the product is anti, meaning that the two halogen atoms (from the X2 reagent) will be on opposite sides of the alkene. This can lead to the formation of two enantiomers (mirror images), as the newly formed stereocenter can have two different configurations. Therefore, for this reaction, 2 stereoisomers will be produced, and they will both be optically active as they are enantiomers. 2. Hydrohalogenation - This reaction can involve either regioselectivity (Markovnikov or anti-Markovnikov) or stereoselectivity (syn or anti). The key thing to consider is the formation of the new stereocenter at the point where the hydrogen halide is added. This new stereocenter, combined with the existing chiral center in 3-methyl-1-pentene, means that there are two diastereomeric outcomes each with two enantiomers. Thus, there are a total of 4 stereoisomers for this reaction. All of these products will be optically active as both the existing and new stereocenters will have different configurations in each stereoisomer. 3. Hydration - This reaction involves the addition of a hydroxyl group (from H2O) and a hydrogen (from the acidic environment), with regiochemistry (Markovnikov) and stereoselectivity (typically anti). This results in the formation of a new stereocenter, which can have two different configurations, leading to two diastereomers each with two enantiomers. Again, a total of 4 stereoisomers will be produced, and all of these will be optically active.

Overall, the number and kind of stereoisomers produced by the treatment of \((R)\)-3-methyl-1-pentene with various reagents are: 1. Halogenation - 2 enantiomers, both optically active 2. Hydrohalogenation - 4 stereoisomers (2 pairs of enantiomers that are diastereomeric to each other), all optically active 3. Hydration - 4 stereoisomers (2 pairs of enantiomers that are diastereomeric to each other), all optically active. Since every possible stereoisomer is optically active, regardless of the reagent used, we can conclude that each product of these reactions will also be optically active.