Q2(a). Hydroboration–oxidation of 1-methylcyclopentene
By taking 1-methylcyclopentene as an example, explain hydroboration-oxidation reaction. Justify that it is stereoselective and regioselective.
Answer in brief
1-Methylcyclopentene gives trans-2-methylcyclopentan-1-ol (racemic). Regioselective: B, then OH, on the less substituted carbon (anti-Markovnikov). Stereoselective: syn addition of H–B, then replacement of B by OH with retention.
Key points
- the empty p orbital of boron accepts the π electrons; B–C and C–H bonds form together through a four-centre transition state (no carbocation, no rearrangement). B goes to C-2 (the CH carbon), H to C-1 (the carbon carrying CH₃).
- The monoalkylborane adds to two more alkenes to give the trialkylborane R₃B.
- HOO⁻ adds to boron; an alkyl group migrates from B to O with loss of OH⁻. This 1,2-shift keeps the configuration of the migrating carbon (retention). Repeated three times, then hydrolysis gives three ROH and borate.
The full model answer
Written out in full, as it should go in the answer booklet. The solution page with its diagrams follows it.
Introduction. Hydroboration–oxidation converts an alkene into an alcohol with anti-Markovnikov orientation and syn addition of H and OH. Step 1: BH₃ (as BH₃*THF) adds across C=C; step 2: alkaline H₂O₂ replaces boron by OH.
Scheme: Product racemic: (1R,2R) shown, with its (1S,2S) enantiomer.
Mechanism
- Hydroboration: the empty p orbital of boron accepts the π electrons; B–C and C–H bonds form together through a four-centre transition state (no carbocation, no rearrangement). B goes to C-2 (the CH carbon), H to C-1 (the carbon carrying CH₃).
- The monoalkylborane adds to two more alkenes to give the trialkylborane R₃B.
- Oxidation: HOO⁻ adds to boron; an alkyl group migrates from B to O with loss of OH⁻. This 1,2-shift keeps the configuration of the migrating carbon (retention). Repeated three times, then hydrolysis gives three ROH and borate.
Regioselective
Boron goes to the less substituted carbon. (a) Steric: the bulky boron group approaches the less hindered CH end. (b) Electronic: in the transition state C-1 (tertiary) carries a partial positive charge and the hydride is delivered to it; the H is the nucleophilic end of B–H. The OH therefore ends on C-2: 2-methylcyclopentan-1-ol, not 1-methylcyclopentan-1-ol (the Markovnikov product of acid-catalysed hydration).
Stereoselective (stereospecific)
H and B add to the same face (syn addition). The H on C-1 and the B on C-2 are therefore cis, which places the CH₃ (on C-1) trans to B. Oxidation with retention puts OH exactly where B was: OH trans to CH₃. Attack on either face of the planar alkene is equally likely, so both enantiomers form: a racemic mixture of (1R,2R) and (1S,2S)-trans-2-methylcyclopentanol; no cis isomer.
Conclusion. 1-Methylcyclopentene gives trans-2-methylcyclopentan-1-ol (racemic). Regioselective: B, then OH, on the less substituted carbon (anti-Markovnikov). Stereoselective: syn addition of H–B, then replacement of B by OH with retention.
Application. Hydroboration–oxidation is used to make primary alcohols from terminal alkenes (e.g. 1-octanol from 1-octene) without rearrangement; chiral boranes (Ipc₂BH) give optically active alcohols.
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All parts of Question 2
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