Question

Identify the product formed when 2-Bromobutane is heated with aqueous solution of sodium hydroxide.

• A. But-1-ene
• B. But-2-ene
• C. Butan-1-ol
• D. Butan-2-ol

Hint:

This is a guaranteed exam trick question. Always circle the solvent!
Aqueous KOH/NaOH $\rightarrow$ Substitution $\rightarrow$ Product is an Alcohol.
Alcoholic KOH/NaOH $\rightarrow$ Elimination $\rightarrow$ Product is an Alkene.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
Heating an alkyl halide with aqueous NaOH leads to a nucleophilic substitution reaction, whereas alcoholic KOH/NaOH would lead to elimination (alkene formation).

Step 2: Formula Application:
$R-X + NaOH(aq) \rightarrow R-OH + NaX$.

Step 3: Explanation:
In 2-bromobutane ($CH_3-CH(Br)-CH_2-CH_3$), the hydroxide ion ($OH^-$) acts as a nucleophile and replaces the Bromine atom. The structure of the carbon skeleton remains unchanged, resulting in the formation of Butan-2-ol.

Step 4: Final Answer:
The product formed is Butan-2-ol.