Q.Pent-1-ene, CH3-CH2-CH2-CH=CH2, is treated with

Step 1. Pent-1-ene is CH3-CH2-CH2-CH=CH2, with the double bond between C1 (terminal, =CH2) and C2.

Step 2. BH3 adds across the double bond with boron going to the LESS hindered, less substituted carbon (C1) -- the anti-Markovnikov regiochemistry characteristic of hydroboration.

Step 3. H2O2/OH- then oxidises the C-B bond to C-OH with RETENTION of configuration/position, so -OH ends up at C1, the terminal carbon.

Step 4. The product is therefore the straight-chain PRIMARY alcohol pentan-1-ol, CH3-CH2-CH2-CH2-CH2-OH -- option (a); option (b) (a secondary alcohol) would be the Markovnikov/acid-hydration product, and option (c) (a diol) does not arise from a simple hydroboration-oxidation of a mono-ene.

Identify hydroboration-oxidation's anti-Markovnikov rule (OH to the less substituted alkene carbon) and apply it to pent-1-ene's terminal double bond.

• Confusing hydroboration-oxidation's anti-Markovnikov outcome with the Markovnikov outcome of acid-catalysed hydration -- they give DIFFERENT alcohols from the same alkene.
• Assuming hydroboration installs two -OH groups (that is dihydroxylation with cold alkaline KMnO4, a different reagent entirely).