See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

# Solution **Step 1: Propene reacts with MCPBA (m-chloroperoxybenzoic acid)** MCPBA performs epoxidation on the $C=C$ double bond of propene: $$CH_3CH=CH_2 \xrightarrow{MCPBA} CH_3-\underbrace{CH-CH_2}_{\text{epoxide}} + \text{m-ClC}_6H_4\text{COOH}$$ Product (A) is **propylene oxide** (methyloxirane). **Step 2: Reduction with $LiAlH_4$** $LiAlH_4$ is a strong reducing agent that cleaves epoxides by nucleophilic attack. The hydride ion attacks the epoxide ring, opening it and forming an alcohol. Due to steric factors: - The hydride preferentially attacks the **less hindered carbon** (the terminal $CH_2$) - This breaks the C-O bond and forms a secondary alcohol at the central carbon $$CH_3-CH-CH_2 + LiAlH_4 \xrightarrow{\text{ring opening}} CH_3-CH(OH)-CH_3$$ $$\text{(epoxide)} \hspace{2cm} \text{(isopropanol)}$$ **Step 3: Workup with $H_3O^+$** Aqueous acidic workup neutralizes excess $LiAlH_4$ and protonates the alkoxide intermediate to give the final alcohol. **Answer: (C) $CH_3CHOHCH_3$** (isopropanol/2-propanol) This forms a secondary alcohol with the OH group on the central carbon, which is the thermodynamically favored product from epoxide ring opening.