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

💡 Solution & Explanation

# Analysis of Nucleophile in $CH_3CHO + LiAlH_4$ Reaction **Step 1: Identify the reaction type** This is a reduction of an aldehyde ($CH_3CHO$) to a primary alcohol ($CH_3CH_2OH$) using lithium aluminum hydride. **Step 2: Determine what $LiAlH_4$ provides** $LiAlH_4$ dissociates to give $AlH_4^-$ (tetrahydridoaluminate ion), which is a powerful source of hydride ions ($H^-$). **Step 3: Identify the nucleophile** The hydride ion ($H^-$) is the actual nucleophile that attacks the electrophilic carbonyl carbon. Hydride is a strong nucleophile because: - It has a negative charge (electron-rich) - It can donate an electron pair to the electron-deficient carbonyl carbon - It initiates nucleophilic addition to the $C=O$ double bond **Step 4: Reaction mechanism** $$H^- \text{ attacks } C=O \rightarrow [CH_3CH(OH)^-] \rightarrow CH_3CH_2OH$$ **Answer: The nucleophile is $H^-$ (hydride ion)** While $LiAlH_4$ is the reagent, the actual nucleophile species performing the attack is the hydride ion derived from it.