Statement-1: Alcohol is prepared from alkyl halide in the presence of aqueous KOH by elimination.
Statement-2: Alkenes are prepared from alkyl halide with alcoholic KOH by $\beta$-elimination.
Which of the following options is correct?
1. Statement-1 and Statement-2 are correct.
2. Statement-1 and Statement-2 are incorrect.
3. Statement-1 is true and Statement-2 is false.
4. Statement-1 is false and Statement-2 is true.
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- Aqueous KOH typically leads to substitution reactions (formation of alcohols), while alcoholic KOH leads to elimination reactions (formation of alkenes).
Analysis of statements:
- Statement 1:
Alcohol is prepared from alkyl halide in the presence of aqueous KOH by elimination.
This statement is incorrect. Reaction of alkyl halide with aqueous KOH proceeds via $S_N1$ or $S_N2$ substitution, yielding alcohol, not elimination. Elimination occurs with alcoholic KOH.
- Statement 2:
Alkenes are prepared from alkyl halide with alcoholic KOH by $\beta$-elimination.
This statement is correct. Alcoholic KOH promotes $\beta$-elimination of alkyl halides, forming alkenes.
Conclusion: Statement-1 is false and Statement-2 is true. Thus, the correct option is (4) Statement-1 is false and Statement-2 is true.
