JEE MainChemistryAlcohols Phenols and Ethers
When 1-chlorobutane is heated with alcoholic KOH , it gives a major hydrocarbon product 'P'. Hydrocarbon 'P' reacts with dilute H₂SO₄ to form compound 'Q'. Separately, hydrocarbon 'P' undergoes hydroboration-oxidation (reaction with B₂H₆ followed by H₂O₂ / OH^- ) to form compound 'R'. Identify the IUPAC names of products 'Q' and 'R' respectively.
Options
- A'Q' is Butan-1-ol and 'R' is Butan-2-ol
- B'Q' is Butan-1-ol and 'R' is Butan-1-ol
- C'Q' is Butan-2-ol and 'R' is Butan-1-ol
- D'Q' is Butan-2-ol and 'R' is Butan-2-ol
Correct answer
C. 'Q' is Butan-2-ol and 'R' is Butan-1-ol
Step-by-step solution
When 1-chlorobutane ( CH₃-CH₂-CH₂-CH₂-Cl ) is heated with alcoholic KOH , it undergoes dehydrohalogenation (an elimination reaction) to form 1-butene ( CH₃-CH₂-CH=CH₂ ) as the major hydrocarbon product 'P'. When 1-butene ('P') reacts with dilute H₂SO₄ , acid-catalyzed hydration takes place according to Markovnikov's rule. The OH group attaches to the more substituted carbon, yielding butan-2-ol as product 'Q'. When 1-butene ('P') undergoes hydroboration-oxidation, hydration occurs with anti-Markovnikov regioselecti