(R)-3-Chloro-1-butene reacts with HCl by Markovnikov addition,and the products are separated by gas chromatography.How many total fractions would be obtained and how many would be optically active?

Question

Quizplus · Accepted Answer

The reaction of (R)-3-chloro-1-butene with HCl will result in the formation of two products: (R)-2-chlorobutane and (S)-2-chlorobutane. These two isomers are enantiomers, and they will be separated as two different fractions by gas chromatography. Therefore, two total fractions will be obtained. Since (R)-2-chlorobutane is optically active, one optically active fraction will be obtained. However, (S)-2-chlorobutane is also optically active, but it is a mirror image of (R)-2-chlorobutane, so it will rotate the plane polarized light in the opposite direction. Therefore, it is also an optically active compound, but it is considered as optically inactive with respect to (R)-2-chlorobutane. Thus, one optically active fraction and one optically inactive fraction will be obtained, making the best choice B.

(R)-3-Chloro-1-Butene Reacts with HCl by Markovnikov Addition,and the Products Are