CH3-CH2-CH2-Br + NaOH C2H5OH Product A Product A undergoes the following two reactions: Consider the above reactions,…

Step 1: n-Propyl bromide + NaOH/ethanol → Product A

$\ce{CH3CH2CH2Br + NaOH ->[ C2H5OH ] CH3CH=CH2 + NaBr + H2O}$

Alcoholic KOH (or NaOH) promotes E2 elimination → propene (prop-1-ene). Product A = propene.

Step 2: Propene + H2O/H+ → Product B (Markovnikov addition)

Acid-catalysed hydration follows Markovnikov's rule:

• $\ce{H+}$ adds to terminal CH2 (less substituted end)
• Forms secondary carbocation at C2
• Water attacks C2
• B = 2-Propanol (propan-2-ol, isopropanol) ✓

Step 3: Propene + BH3·THF then H2O2/OH- → Product C (Anti-Markovnikov)

Hydroboration-oxidation:

• BH3 adds boron to terminal C (less substituted end, anti-Markovnikov)
• Oxidation with H2O2/OH- replaces B with -OH at terminal C
• C = 1-Propanol (propan-1-ol, n-propanol) ✓

Key Points to Remember:

• NaOH/C2H5OH → E2 elimination → alkene.
• Markovnikov hydration (H+/H2O): OH at more substituted C → secondary alcohol.
• Anti-Markovnikov hydration (BH3/H2O2): OH at less substituted C → primary alcohol.
• Same alkene can give two different alcohol isomers depending on hydration method.