CH₃-CH₂-CH₂-Br+NaOH →(C₂ H₅ OH) Product 'A'
Product A →(H₂O) Product “B”; →(Diborane) Product “C”
Consider the above reactions, identify product B and product C.

Solution

Chemistry ma’am · 3 steps

Make propene first, then remember: acid puts OH on the middle carbon, borane puts it on the end.

1. Find product A. In ethanol, OH⁻ acts as a base and removes HBr from 1-bromopropane. This dehydrohalogenation gives propene, so A is CH₃-CH=CH₂.

   Reaction on propene	Rule followed	Where OH goes	Product
   H₂O / H⁺	Markovnikov: H⁺ adds to CH₂ to give the more stable 2° carbocation	Middle carbon	B = 2-propanol, CH₃CH(OH)CH₃
   B₂H₆, then H₂O₂ / OH⁻	Anti-Markovnikov: boron adds to the less hindered end carbon	End carbon	C = 1-propanol, CH₃CH₂CH₂OH

2. Why they differ. Acid hydration goes through a carbocation, so OH lands where the cation is most stable. Hydroboration has no carbocation: boron takes the end carbon, and oxidation replaces that boron with OH in the same position.
3. So. B is 2-propanol and C is 1-propanol, which is option (d).

The idea to keep. Acid-catalysed hydration follows Markovnikov's rule. Hydroboration-oxidation gives the anti-Markovnikov alcohol, with OH on the less substituted carbon.

Answer from the NTA final answer key. Solution by PrepWiser.