Given below are two statements :
Statement (I) : Alcohols are formed when alkyl chlorides are treated with aqueous potassium hydroxide by elimination reaction.
Statement (II) : In alcoholic potassium hydroxide, alkyl chlorides form alkenes by abstracting the hydrogen from the $$\beta$$-carbon.
In the light of the above statements, choose the most appropriate answer  from the option given  below : 

The behaviour of alkyl halides changes with the nature of the $$KOH$$ medium.

Case 1: Aqueous $$KOH$$ reacts with alkyl chlorides mainly through nucleophilic substitution ($$S_{N1}$$ or $$S_{N2}$$).
  • The hydroxide ion $$OH^-$$ attacks the carbon bearing the chlorine and replaces $$Cl^-$$.
  • This produces an alcohol: $$R{-}Cl + KOH_{(aq)} \rightarrow R{-}OH + KCl$$.
  • The mechanism is substitution, not elimination.
Therefore Statement (I) is incorrect because it labels the reaction as “elimination” instead of “substitution.”

Case 2: Alcoholic $$KOH$$ behaves as a strong base rather than a nucleophile.
  • The ethoxide/alkoxide ion ($$RO^-$$ formed in situ) abstracts a hydrogen from a $$\beta$$-carbon (carbon adjacent to the one carrying $$Cl$$).
  • Simultaneously the $$Cl^-$$ leaves, giving an alkene: $$R{-}CH_2{-}CH_2Cl \xrightarrow[KOH_{(alc)}]{} R{-}CH{=}CH_2 + KCl + H_2O$$.
  • This is an elimination (E2/E1) reaction, exactly as described in Statement (II).
Hence Statement (II) is correct.

Combining the conclusions:
  • Statement (I) - Incorrect.
  • Statement (II) - Correct.

Thus the most appropriate choice is Option B.