Question

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.

• A.
• B.
• C.
• D.

Hint:

- Aqueous KOH typically leads to substitution reactions (formation of alcohols), while alcoholic KOH leads to elimination reactions (formation of alkenes).

Answer 1

The Correct Option is D

Let's analyze both statements: - Statement 1: Alcohol is prepared from alkyl halide in the presence of aqueous KOH by elimination. This statement is incorrect. When an alkyl halide is treated with aqueous KOH, the reaction follows an $S_N1$ or $S_N2$ mechanism, leading to the substitution of the halogen by a hydroxyl group, resulting in the formation of an alcohol. This is not an elimination reaction. In fact, elimination occurs when alcoholic KOH is used. - Statement 2: Alkenes are prepared from alkyl halide with alcoholic KOH by $\beta$-elimination. This statement is correct. When an alkyl halide is treated with alcoholic KOH, a $\beta$-elimination reaction occurs, resulting in the formation of an alkene. Alcoholic KOH favors the elimination reaction, where a hydrogen atom and a halogen are removed from adjacent carbons to form a double bond. Thus, the correct option is (4) Statement-1 is false and Statement-2 is true.