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 Question:
We need to determine the major product formed when a secondary alkyl halide (2-bromobutane) is reacted with an aqueous solution of a strong base ($NaOH$).

Step 2: Detailed Explanation:
The nature of the solvent used with $NaOH$ (or $KOH$) strictly dictates the reaction pathway for an alkyl halide:
1. Aqueous $NaOH$ or $KOH$: The hydroxide ion ($OH^-$) acts purely as a nucleophile because it is highly solvated by water, preventing it from acting as a strong base. This heavily favors Nucleophilic Substitution reactions ($S_N2$ or $S_N1$). The halogen is simply replaced by the -OH group.
2. Alcoholic $NaOH$ or $KOH$: The presence of alcohol forms alkoxide ions ($RO^-$), which are very strong bases. This heavily favors Elimination reactions, removing a proton and a halide to form a double bond (alkene).
Because the problem specifies an aqueous solution, the reaction proceeds via substitution.
Reactant: 2-bromobutane ($CH_3-CH(Br)-CH_2-CH_3$)
Reagent: $OH^-$ (nucleophile)
Reaction: $CH_3-CH(Br)-CH_2-CH_3 + OH^- \rightarrow CH_3-CH(OH)-CH_2-CH_3 + Br^-$
The resulting molecule has a hydroxyl group on the second carbon, making it Butan-2-ol.

Step 3: Final Answer:
The substitution product is Butan-2-ol, which matches option (d).