
Above conversion is carried out using:
Step 1: KOH (alc.).
Potassium hydroxide in alcohol (KOH (alc.)) induces elimination of the bromine atom, producing an alkene.
Step 2: Oxymurcuration-demercuration.
The alkene then undergoes oxymurcuration-demercuration to form an alcohol at the position where the bromine was originally attached. Hence, option (1) is correct.
Final Answer: \[ \boxed{\text{(1) (I) KOH (alc.), (II) oxymurcuration-demercuration}} \]
The reaction is carried out by:
