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

💡 Solution & Explanation

# Analysis of Nucleophile in $CH_3CHO + LiAlH_4 \rightarrow CH_3CH_2OH$ **Step 1: Identify the reactants** - $CH_3CHO$ (acetaldehyde) — electrophile with a polar C=O bond - $LiAlH_4$ (lithium aluminum hydride) — strong reducing agent **Step 2: Determine what acts as nucleophile** $LiAlH_4$ dissociates to produce $H^-$ (hydride ion): $$LiAlH_4 \rightarrow Li^+ + AlH_4^-$$ The $AlH_4^-$ complex releases $H^-$ ions, which are the actual nucleophiles. **Step 3: Mechanism of reduction** The hydride ion ($H^-$) acts as a **nucleophile** because it: - Has excess electron density (negative charge) - Attacks the electrophilic carbonyl carbon ($\delta^+$) - Forms a C-H bond while the C=O π-bond breaks $$CH_3CHO + H^- \rightarrow CH_3CH_2O^-$$ **Step 4: Conclusion** The nucleophile is **$H^-$ (hydride ion)** or equivalently **$AlH_4^-$ (tetrahydridoaluminate ion)** from which hydride is transferred. Without seeing option D, it should read: **"$H^-$ (hydride ion)"** or **"$AlH_4^-$"** — the species donating electron density to the electrophilic carbonyl carbon.