Two Routes to a Bromide: Rearrangement versus Inversion
A second-year student must convert (R)-3-methylbutan-2-ol into a bromoalkane. Route A is concentrated HBr at room temperature; route B is PBr3 in dry ether at 0 °C. The student hoped to keep both the carbon skeleton and the stereochemical information intact, but TLC of the route A product shows two spots with very similar Rf, and a polarimeter reading of the isolated major product is almost zero although the starting alcohol was strongly optically active. A supervisor separately asks for a small sample of 2-bromo-2-methylbutane to use as a reference standard, and another student needs a single enantiomer of the unrearranged bromide. The student's write-up must explain which route gives which product, why the optical activity is lost in one case, and which route to choose for each target.
1. Explain the route A results mechanistically: identify both products, account for the rearranged skeleton, and explain the near-zero optical rotation of the major product.
Read the worked answer
Route A is an SN1 process under strongly acidic, ionising conditions:
- Protonation of the OH, then loss of water from C2 gives a secondary carbocation at C2, which is planar at that carbon.
- The adjacent C3 bears a hydrogen and two methyl groups. A 1,2-hydride shift moves that H to C2 and places the positive charge on C3, giving a tertiary carbocation (2-methylbutan-2-yl) - lower energy, so this is fast and strongly favoured.
- Bromide attack on the tertiary carbocation gives 2-bromo-2-methylbutane as the major product.
Direct capture of the secondary carbocation before rearrangement gives 2-bromo-3-methylbutane as the minor product. Those are the two close spots on the TLC plate.
Optical activity: the planar carbocation has lost the configuration at C2, so attack by Br⁻ occurs from either face and gives a racemic mixture. In the rearranged product C2 also becomes a new stereocentre, again formed racemically. A nearly zero rotation is exactly what a racemate plus a minor chiral impurity produces. The starting (R) configuration has been destroyed, not retained.
2. Explain why route B gives a different outcome, predict the configuration of its product from the (R) alcohol, and state the practical precautions that matter for this reaction.
Read the worked answer
PBr3 converts the alcohol to an alkoxyphosphonium species (via R–OPBr2 and Br⁻), and bromide then displaces the phosphite leaving group in a single concerted step at C2. No free carbocation is formed, so there is no opportunity for a hydride shift and the skeleton stays intact: the product is 2-bromo-3-methylbutane.
Stereochemistry: SN2 proceeds with inversion at the carbon bearing the leaving group. Because the leaving group (OPBr2/OH) and the incoming nucleophile (Br) are both the highest-priority substituent at C2, inversion flips the descriptor: (R)-3-methylbutan-2-ol gives (S)-2-bromo-3-methylbutane, essentially as one enantiomer.
Precautions: the ether must be dry and the alcohol dry, since water hydrolyses PBr3; keep the mixture cold (0 °C) to suppress E1/elimination to 2-methylbut-2-ene, which competes badly with a secondary substrate; add PBr3 slowly and quench carefully because HBr is liberated; and avoid prolonged storage of the product, which can solvolyse. Note the reaction consumes only one-third of a mole of PBr3 per mole of alcohol as written, since the by-product P(OH)Br2 can react further.
3. Recommend a route for each target - the 2-bromo-2-methylbutane reference sample and a single enantiomer of the unrearranged bromide - and state the general lesson about choosing between SN1 and SN2 conditions here.
Read the worked answer
Reference sample of 2-bromo-2-methylbutane: use route A, HBr. The rearrangement is the desired outcome, and SN1 conditions deliver the tertiary bromide as the major product because the tertiary carbocation is much more stable than the secondary one. Trying to reach this compound through PBr3 would give the unrearranged secondary bromide instead; forcing rearrangement from a secondary bromide would need added Lewis acid or a separate rearrangement step.
Single enantiomer of 2-bromo-3-methylbutane: use route B, PBr3 in dry ether at 0 °C. Only the SN2 pathway both avoids rearrangement and delivers a defined configuration (inversion from the (R) alcohol). HBr cannot deliver this: whatever the temperature, the ionising conditions cause partial or complete racemisation and partial rearrangement, so the best achievable is an enriched mixture, not a single enantiomer.
General lesson: match the conditions to the target's skeleton and stereochemistry rather than yield alone. Ask two questions first - does the target retain the carbon skeleton (if not, a carbocation route is required) and must the configuration be controlled (if yes, an SN2 route on a substrate that will not ionise). Secondary substrates are the hard case, because they sit where both mechanisms are competitive.