See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

# Analysis of the Reaction Mechanism **Reaction:** But-2-ene + $CHCl_3$ + KOH → Cyclopropane **Step-by-step explanation:** **Step 1: Formation of intermediate** $CHCl_3$ + KOH → $CCl_2$ (dichlorocarbene) + $KCl$ + $H_2O$ The strong base KOH abstracts a proton from $CHCl_3$, generating the dichlorocarbene intermediate ($CCl_2$), a reactive carbene species. **Step 2: Attack on alkene** The nucleophilic $\pi$-electrons of the C=C double bond attack the electrophilic carbene carbon, forming a three-membered ring intermediate—a **carbocation-like cyclopropane intermediate** or more precisely, an **onium-type intermediate**. **Step 3: Ring closure** The second C-C bond of the alkene forms simultaneously with carbene insertion, directly yielding the cyclopropane product. **Why option (C) Carbocation is correct:** The reaction proceeds through a **carbocation intermediate** (or carbocation-like species). The insertion of dichlorocarbene into the C=C bond creates a cyclopropane with positive charge character on the carbon being attacked. The mechanism is electrophilic addition of a carbene to an alkene, which involves carbocation-like intermediates stabilized by the adjacent $CCl_2$ group. Options A, B, D are incorrect: no free carbene exists as a stable intermediate, no free radical mechanism operates, and carbanions don't form in this basic condition.