See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

# Synthesis of Cyclopentyl Ethyl Ether **Step 1: Identify the target molecule** Cyclopentyl ethyl ether has the structure: $C_5H_9-O-CH_2CH_3$ The ether linkage connects cyclopentyl group to ethyl group. **Step 2: Analyze Option (A)** $$\text{Cyclopentyl-ONa} + CH_3CH_2Br \rightarrow \text{Cyclopentyl-O-CH}_2CH_3 + NaBr$$ - Cyclopentanol is deprotonated to form cyclopentoxide anion (strong nucleophile) - Ethyl bromide undergoes $S_N2$ attack by the alkoxide oxygen - This directly forms the C-O-C ether linkage - ✓ Correct: Nucleophilic substitution on alkyl halide **Step 3: Why other options fail** **(B)** Cyclopentyl-Br with ethoxide: Incorrect because $S_N2$ on secondary alkyl halide (cyclopentyl-Br) is too slow; cyclopentyl cation is also too unstable for $S_N1$. **(C)** Cyclopentanol + ethyl chloride: Ethyl chloride is a poor electrophile in $S_N2$ and requires harsh conditions; cyclopentanol (neutral) is a weak nucleophile without activation. **Answer: (A)** ✓ The cyclopentoxide anion (strong nucleophile) attacks the unhindered primary carbon of ethyl bromide (good electrophile) via $S_N2$ mechanism, efficiently forming the desired ether.