Question:
What is the main product obtained when sodium ethoxide reacts with isopropyl chloride?
What is the main product obtained when sodium ethoxide reacts with isopropyl chloride?
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A strong base with a secondary alkyl halide mainly causes elimination (E2) rather than substitution.
Updated On: Nov 12, 2026
- \(\text{C}_{2}\text{H}_{5}\text{-O-CH-(CH}_{3})_{2}\)
- \(\text{CH}_{3}\text{-CH=CH}_{2}\)
- \(\text{C}_{2}\text{H}_{5}\text{-O-C}_{2}\text{H}_{5}\)
- \((\text{CH}_{3})_{2}\text{-CH-O-CH-(CH}_{3})_{2}\)
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept
Isopropyl chloride, \((\text{CH}_{3})_{2}\text{CHCl}\), is a secondary halide. Sodium ethoxide is a strong base as well as a nucleophile.
Step 2: Competing reactions
With a secondary halide and a strong base, E2 elimination dominates over SN2 (Williamson ether synthesis). The ethoxide removes a \(\beta\)-hydrogen.
Step 3: Product
Loss of \(\text{HCl}\) gives \(\text{CH}_{3}\text{-CH=CH}_{2}\) (propene) as the main product. The ether is only a minor product.
Final Answer:
The main product is propene. \[ \boxed{\text{(B) }\text{CH}_{3}\text{-CH=CH}_{2}} \]
Isopropyl chloride, \((\text{CH}_{3})_{2}\text{CHCl}\), is a secondary halide. Sodium ethoxide is a strong base as well as a nucleophile.
Step 2: Competing reactions
With a secondary halide and a strong base, E2 elimination dominates over SN2 (Williamson ether synthesis). The ethoxide removes a \(\beta\)-hydrogen.
Step 3: Product
Loss of \(\text{HCl}\) gives \(\text{CH}_{3}\text{-CH=CH}_{2}\) (propene) as the main product. The ether is only a minor product.
Final Answer:
The main product is propene. \[ \boxed{\text{(B) }\text{CH}_{3}\text{-CH=CH}_{2}} \]
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