퐶퐻3퐶퐻푂+ 퐿푖퐴푙퐻4 →퐶퐻3퐶퐻2푂퐻 Nucleophile added in this reaction is — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

# Analysis of $CH_3CHO + LiAlH_4 \rightarrow CH_3CH_2OH$ ## Identifying the Nucleophile **Step 1: Understand the reaction type** This is a **reduction of an aldehyde** to a primary alcohol using lithium aluminum hydride. **Step 2: Identify what $LiAlH_4$ provides** $LiAlH_4$ dissociates in solution to give **hydride ions** ($H^-$): $$LiAlH_4 \rightarrow Li^+ + AlH_4^-$$ The $AlH_4^-$ complex provides hydride ($H^-$) species. **Step 3: Recognize the nucleophile** In this reduction: - The **hydride ion ($H^-$)** acts as the **nucleophile** - It attacks the electrophilic **carbonyl carbon** (C=O) of the aldehyde - The oxygen, being more electronegative, becomes negatively charged (forming an alkoxide intermediate) **Step 4: Mechanism confirmation** $$CH_3CHO + H^- \rightarrow CH_3CH(O^-)H \rightarrow CH_3CH_2OH$$ The hydride is electron-rich and seeks the electron-deficient carbonyl carbon, making it a classic nucleophile. ## Answer **The nucleophile is $H^-$ (hydride ion)** — Option **D** is correct as it identifies hydride as the nucleophilic species that attacks the carbonyl group.