Question

Identify the product ' X ' formed in the following reaction,
$\text{Sodium ethoxide} + \text{Isopropyl chloride} \longrightarrow \text{X} + \text{Ethanol} + \text{Sodium chloride}$

• A. 1-Ethoxypropane
• B. 2-Ethoxypropane
• C. Propene
• D. Propane

Hint:

Secondary/Tertiary alkyl halide + Strong base = Elimination (Alkene). Don't let the "ethoxide" trick you into picking an ether every time!

Answer 1

The Correct Option is C

Step 1: Concept
This is a competition between Williamson Ether Synthesis ($S_N2$) and Elimination ($E2$).

Step 2: Meaning
Sodium ethoxide is a strong base and a strong nucleophile. Isopropyl chloride is a secondary ($2^\circ$) alkyl halide.

Step 3: Analysis
When a secondary alkyl halide reacts with a strong, hindered base like ethoxide, elimination ($E2$) becomes the dominant pathway over substitution. The $\text{H}$ from the $\beta$-carbon and the $\text{Cl}$ are removed to form a double bond. $\text{CH}_3\text{CH}(\text{Cl})\text{CH}_3 + \text{NaOCH}_2\text{CH}_3 \rightarrow \text{CH}_3\text{CH}=\text{CH}_2$ (Propene) + $\text{CH}_3\text{CH}_2\text{OH}$ + $\text{NaCl}$.

Step 4: Conclusion
The major product X is propene. Final Answer: (C)