Question

Above conversion is carried out using: 

• A. (I) KOH (alc.), (II) oxymurcuration-demercuration
• B. (I) KOH (aq.), (II) Conc. H\(_2\)SO\(_4\)
• C. (I) KOH (alc.), (II) Hydroboration-oxidation
• D. (I) Nucleophilic addition-elimination, (II) oxymurcuration-demercuration

Hint:

The oxymurcuration-demercuration sequence involves hydration of alkenes using mercuric acetate followed by reduction with sodium borohydride.

Answer 1

The Correct Option is A

Step 1: KOH (alc.). Potassium hydroxide in alcohol (KOH (alc.)) causes elimination of the bromine atom, resulting in the formation of an alkene.

Step 2: Oxymurcuration-demercuration. The alkene subsequently undergoes oxymurcuration-demercuration, yielding an alcohol at the original site of bromine attachment. Therefore, option (1) is correct.

Final Answer: \[ \boxed{\text{(1) (I) KOH (alc.), (II) oxymurcuration-demercuration}} \]