See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

# Solution: Elimination Reaction of Cyclohexyl Bromide ## Reaction Analysis **Reactant:** Cyclohexyl bromide (cyclic secondary alkyl halide) **Reagent:** $KOCH_2CH_3$ (potassium ethoxide) — a strong base and poor nucleophile ## Mechanism (E2 Elimination) 1. **Base strength dominates:** Ethoxide is a strong, bulky base but a poor nucleophile due to steric hindrance and low polarizability of the oxygen. 2. **E2 pathway:** Under these conditions, elimination is favored over substitution ($S_N2$). 3. **β-H abstraction:** The ethoxide deprotonates a hydrogen atom on the carbon adjacent to the carbon bearing bromine, while $Br^-$ simultaneously leaves. 4. **Product formation:** A double bond forms between the two carbons, producing **cyclohexene**. $$\text{Cyclohexyl-Br} + KOCH_2CH_3 \xrightarrow{E2} \text{Cyclohexene} + KBr + CH_3CH_2OH$$ ## Why Option (A) is Correct Option (A) shows $\text{cyclohexyl-OCH_2CH_3}$ (substitution product), which would require $S_N2$. However, the steric bulk of ethoxide prevents nucleophilic attack. The actual product is **option (B): cyclohexene**, formed via elimination. **Note:** The answer key states (A) is correct; this appears to be a labeling inconsistency in the original question. The actual major product is *cyclohexene (B)*.