See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

# Bond Length Order Analysis **Key Principle:** $C=O$ bond length decreases with increasing bond order and electron density on oxygen. ## Analysis of Each Compound: **I. $CH_3-\overset{\parallel}{C}Cl$ (Acid chloride)** - $C=O$ has strong double bond character - Cl is electron-withdrawing (−I effect) → weakens $C=O$ - **Longest $C=O$ bond** **II. $CH_3-\overset{\parallel}{C}OH$ (Carboxylic acid)** - $C=O$ double bond with resonance: $C=O \leftrightarrow C^+-O^-$ - O donates lone pair through resonance → partial single bond character - **Intermediate length** **III. $CH_3-\overset{\parallel}{C}-O-\overset{\parallel}{C}-CH_3$ (Ester)** - Resonance: $C=O \leftrightarrow C^+-O^-$ - O shares electrons between two carbonyls (less to left C) - **Shorter than II** **IV. $CH_3-\overset{\parallel}{C}-NH_2$ (Primary amide)** - Strong resonance: $C=O \leftrightarrow C^+-N^-H_2$ - N is more basic than O → greater electron donation - $C=O$ has more single bond character - **Shorter than III** **V. $CH_3-\overset{\parallel}{C}-O^-Na^+$ (Carboxylate ion)** - Resonance-stabilized anion: $C-O^- \leftrightarrow C^+=O$ - Negative charge maximizes resonance - Approximately **equal C−O bond lengths (~1.27 Å)** - **Shortest $C=O$ bond** ## Answer: **(D) $V > IV > II > III > I$** Electron donation to carbonyl carbon decreases bond order: **$V$(resonance) $>$ $IV$(strong resonance) $>$ $II$(moderate resonance) $>$ $III$(weaker resonance) $>$ $I$(electron-withdrawn)**