Consider the light-initiated chlorination of (S)-2-chlorobutane followed by careful fractional distillation (or separation by GLC)of all of the products with the formula C₄H₈Cl₂.How many fractions (in total)would be obtained and how many of these fractions would be optically active? A)Three fractions total; all optically active B)Four fractions total; three fractions optically active C)Five fractions total; all optically active D)Five fractions total; four fractions optically active E)Five fractions total; three fractions optically active

Question

Quizplus · Accepted Answer

The light-initiated chlorination of (S)-2-chlorobutane can result in the substitution at any of the four carbon atoms, leading to a total of five possible products due to the creation of a new chiral center at the 3-position (two enantiomers at this position) and the retention of the original chiral center at the 2-position. However, not all of these products are optically active. The product where chlorination occurs at the 1-position or 4-position will not create a new chiral center, and thus each will be optically inactive. The original compound and the two enantiomers formed by chlorination at the 3-position are optically active, making a total of three optically active fractions.

Consider the Light-Initiated Chlorination of (S)-2-Chlorobutane Followed by Careful Fractional