Analyze Statement (I)

Statement (I): Alcohols are formed when alkyl chlorides are treated with aqueous potassium hydroxide by elimination reaction.

• When an alkyl chloride reacts with aqueous KOH, the major reaction is nucleophilic substitution, not elimination.
• The hydroxide ion $\left(\mathrm{OH^-}\right)$ replaces the chloride ion: $\mathrm{R-Cl + KOH(aq) \rightarrow R-OH + KCl}$
• Thus, alcohol is indeed formed, but by substitution reaction, not by elimination.

Therefore, Statement (I) is incorrect.

Analyze Statement (II)

Statement (II): In alcoholic potassium hydroxide, alkyl chlorides form alkenes by abstracting the hydrogen from the $\beta$-carbon.

• In alcoholic KOH, elimination is favored.
• The base removes a hydrogen from the $\beta$-carbon, and the halogen leaves from the adjacent carbon, leading to formation of a double bond: $\mathrm{R-CH_2-CH_2-Cl \xrightarrow[alc.\ KOH]{} R-CH=CH_2 + KCl + H_2O}$
• This is $\beta$-elimination (dehydrohalogenation).

Therefore, Statement (II) is correct.

Evaluate the options

• A: Both incorrect $\rightarrow$ False
• B: Both correct $\rightarrow$ False
• C: Statement I incorrect but Statement II correct $\rightarrow$ True
• D: Statement I correct but Statement II incorrect $\rightarrow$ False

Final Answer

The correct option is: $\boxed{\text{C}}$

Comparison with stored answer

• Derived answer: C
• Stored correct answer: C
• They match.