Rearrangement During HBr Conversion of a Hindered Secondary Alcohol
You need 2-bromo-3,3-dimethylbutane, a secondary bromide, as an intermediate. A colleague treats 10.0 g (97.9 mmol) of 3,3-dimethyl-2-butanol with 30 mL of 48% aqueous HBr and heats at reflux for 2 h. GC-MS of the crude shows 6% of the desired secondary bromide, 88% of a tertiary bromide identified as 2-bromo-2,3-dimethylbutane, and 5% of an alkene. Repeating the reaction at 0 °C for 6 h gives 18% desired secondary bromide, 70% tertiary bromide, and 10% alkene. The alkene fraction increases when the mixture is heated. You need at least 85% selectivity to the secondary bromide and cannot use silver salts. The alcohol is limiting; HBr is in excess.
1. Explain the product distribution mechanistically. Why does cooling improve the desired product only slightly?
Read the worked answer
Under strongly acidic conditions the alcohol is protonated and water leaves, generating a secondary carbocation at C2. That carbocation is adjacent to a quaternary carbon at C3. A 1,2-methyl shift from C3 to C2 converts the secondary carbocation into a more stable tertiary carbocation. Bromide then attacks the tertiary cation, giving 2-bromo-2,3-dimethylbutane as the major product. The small amount of secondary bromide comes from direct bromide attack on the secondary carbocation before rearrangement, and the alkene comes from E1 loss of a proton from the carbocation.
Cooling to 0 °C slows all carbocation-mediated steps and shortens the carbocation lifetime, so direct trapping becomes slightly more competitive. However, the reaction still proceeds through a carbocation, and rearrangement is intramolecular and fast. Cooling cannot prevent rearrangement once the cation forms; it mainly reduces the overall rate and changes the balance between substitution and elimination. The protic aqueous HBr medium also stabilizes ions and favors ionization. To reach at least 85% secondary bromide, the carbocation must be avoided entirely rather than merely slowed.
2. Propose a two-step sequence that avoids carbocation rearrangement and gives the secondary bromide with high selectivity. Include expected stereochemistry.
Read the worked answer
Convert the alcohol to a sulfonate ester under conditions that do not ionize it, then displace the sulfonate with bromide in an SN2 reaction. A practical sequence is: (1) dissolve the alcohol in dry pyridine or dichloromethane with a catalytic amount of DMAP, cool to 0 °C, add p-toluenesulfonyl chloride (about 1.1 equiv), and stir until the tosylate forms; (2) isolate the tosylate, then treat it with excess anhydrous LiBr in acetone or DMF at moderate temperature. Bromide attacks the carbon bearing the tosylate from the back side, displacing tosylate. Because no carbocation is formed, the adjacent quaternary center does not rearrange.
The stereochemical outcome is inversion at the carbon bearing bromine. If the starting alcohol is racemic, the product is racemic because each enantiomer is inverted; if a single enantiomer of the alcohol is used, the bromide is the inverted enantiomer, provided the tosylation and displacement are clean. The main risks are elimination from the hindered tosylate and slow SN2 caused by the adjacent tert-butyl group. Keep the displacement temperature as low as practical, use excess LiBr, and monitor by GC to stop before elimination dominates. PBr3 can also convert secondary alcohols to bromides with inversion, but the tosylate/LiBr route is easier to control and avoids aqueous acid.
3. Calculate the theoretical mass of 2-bromo-3,3-dimethylbutane (M = 165.07 g/mol) from 10.0 g of alcohol (M = 102.18 g/mol). What mass would correspond to 85% selectivity and 80% recovery?
Read the worked answer
Moles of alcohol = 10.0 g / 102.18 g/mol = 0.0979 mol = 97.9 mmol. Because the alcohol is limiting and one molecule of product contains one carbon skeleton, the theoretical moles of bromide are also 97.9 mmol. Theoretical mass = 0.0979 mol × 165.07 g/mol = 16.16 g.
If the reaction gives 85% selectivity to the desired secondary bromide and the isolation recovery is 80%, the isolated mass is 16.16 g × 0.85 × 0.80 = 11.0 g. This calculation assumes the 85% selectivity is measured before isolation and that the recovery applies to the desired product. The current HBr route gives only 6-18% desired product, so its isolated yield would be far below this target. The alternative sequence must therefore be judged not only by conversion but by the selectivity ratio of secondary bromide to tertiary bromide and alkene.