Stereochemical Outcome of HBr Addition to a Chiral Alkene
A sample of (3R)-3-methyl-1-pentene is 98% enantiomerically pure. It is treated with HBr in dichloromethane at 0 °C. Markovnikov addition places Br at C2 and creates a new stereocenter there; the original stereocenter at C3 is not part of the double bond. GC analysis shows two product fractions. Fraction 1 has [α]D = +12° and fraction 2 has [α]D = −8°, both measured at the same concentration and path length. A student concludes that the reaction must have racemized C3, because two optically active products cannot both retain the original (R) configuration. The student also predicts that if the starting alkene were racemic, the same two fractions would appear with rotations equal in magnitude and opposite in sign. The instructor asks for a stereochemical analysis of the products, including the number of stereoisomers and the expected optical activity of the mixture.
1. Evaluate the student's conclusion that C3 must have racemized. How many stereoisomers are formed from the 98% enantiomerically pure starting material, and why are both fractions optically active?
Read the worked answer
The conclusion is incorrect. The reaction creates a new stereocenter at C2, but it does not break either bond at C3. The starting material is mostly (3R), so the C3 configuration remains R in the products. Addition at C2 can occur from either face, giving two diastereomers: (2R,3R) and (2S,3R). Diastereomers are not mirror images, so they can have different optical rotations. That is exactly what the two fractions show: +12° and −8° are not equal in magnitude and opposite in sign, so they are not an enantiomeric pair. Both fractions retain C3 as R. The small amount of (3S) starting material would give a tiny amount of the corresponding (2R,3S) and (2S,3S) products, but the major products are the two (3R) diastereomers. No racemization at C3 is needed to explain the data.
2. If the starting alkene were racemic instead of 98% enantiomerically pure, what products would form and what would be the net optical activity of the mixture?
Read the worked answer
A racemic starting material contains equal amounts of (3R) and (3S) alkene. Each enantiomer reacts to give two diastereomers at C2. The (3R) alkene gives (2R,3R) and (2S,3R). The (3S) alkene gives (2R,3S) and (2S,3S). The result is four stereoisomers, arranged as two enantiomeric pairs: (2R,3R) with (2S,3S), and (2S,3R) with (2R,3S). Because the starting enantiomers are equal in amount and the reaction conditions are achiral, each member of an enantiomeric pair forms in equal amount. The net optical rotation of the whole mixture is therefore zero. On an achiral GC column, enantiomers coelute, so the two observed fractions would each be racemic and optically inactive, not optically active with equal and opposite rotations as the student predicted.