Optical Activity and Mechanism in 2-Bromobutane Synthesis
A student prepares 2-bromobutane by refluxing 5.0 g of enantiopure 2-butanol with 48% HBr. After workup, the crude product is distilled and 6.1 g of a colorless liquid is collected (bp 91 °C). The student measures the optical rotation in a 1.0 dm tube at 1.0 g/mL concentration and records [α]D = +9.4°. A control sample of the pure major enantiomer, prepared independently and run under identical conditions, gives [α]D = +23.5°. The student's report states: 'The product is optically active, so the reaction is stereospecific and proceeds by SN2.' The lab manual notes that secondary substrates in strongly acidic media can ionize and that the reaction mixture contains Br−, H2O, and H3O+. The student has PBr3, pyridine, and dry Et2O available for a repeat synthesis.
1. Calculate the enantiomeric excess and the percentage of each enantiomer, then evaluate the student's claim.
Read the worked answer
ee = observed rotation / pure rotation × 100 = 9.4/23.5 × 100 = 40%. A 40% excess means the mixture is 70% major enantiomer and 30% minor enantiomer, because (70 − 30)/100 = 40%.
The student's claim is wrong. A stereospecific SN2 reaction on enantiopure 2-butanol would give essentially 100% inversion and near 100% ee. A pure SN1 reaction through a planar carbocation would give a racemic mixture and 0% ee. A 40% ee means partial racemization, consistent with competition between SN1 ionization and SN2 bromide attack, or with ion-pair return. The acidic aqueous HBr favors SN1 for this secondary substrate, but Br− is also a good nucleophile, so both pathways contribute.
2. Propose a modification that would favor inversion and explain the mechanistic basis.
Read the worked answer
Use PBr3 in dry ether or pyridine at low temperature. PBr3 converts the alcohol into a bromophosphite intermediate that is displaced by Br− in an SN2-like step; this avoids a free carbocation and favors inversion. Keep the reaction dry and cool to suppress ionization.
Alternatively, convert the alcohol to a tosylate with TsCl/pyridine, then displace it with LiBr in acetone. That is a clean SN2 with inversion. Avoid aqueous HBr and heat, which promote SN1 and racemization. Starting from enantiopure 2-butanol, the SN2 route should give high ee.
3. If the starting alcohol were 3-methyl-2-butanol instead of 2-butanol, predict the major product and explain the reasoning.
Read the worked answer
Protonation and loss of water from 3-methyl-2-butanol gives a secondary carbocation at C2, adjacent to a tertiary carbon. A 1,2-hydride shift from C3 to C2 converts it into a more stable tertiary carbocation. Bromide then attacks the tertiary center.
The major product is 2-bromo-2-methylbutane, not the secondary bromide. This rearrangement is a signature of an SN1 pathway and shows that changing the substrate can change the carbon skeleton of the product as well as the stereochemistry. It also supports the idea that acidic HBr promotes ionization rather than a clean SN2.