See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

# C–C Bond Length Analysis **Step 1: Identify the C–C bonds in each compound** - **I**: $H_3O^+–CH=CH–NO_2$ → C=C (double bond) - **II**: $CH_2=CH–NO_2$ → C=C (double bond) - **III**: $CH_2=CH–Cl$ → C=C (double bond) - **IV**: $CH_2=CH_2$ → C=C (double bond, isolated alkene) **Step 2: Apply resonance and electron-withdrawing effects** All compounds have C=C bonds, but resonance and EWG effects shorten them: - **I**: $NO_2$ (strong EWG) + $H_3O^+$ (strong EWG) donate electron density from the π-bond through resonance → **shortest C=C** - **II**: $NO_2$ (strong EWG) withdraws electrons → **shortened C=C** - **III**: $Cl$ (weak EWG, primarily inductive) has minimal effect → **longer C=C than II** - **IV**: No substituents; plain alkene → **longest C=C** **Step 3: Order by bond length (shortest to longest)** $$\text{C=C length: I < III < II < IV}$$ Therefore, in decreasing order of C–C bond length: $$\boxed{\text{I} > \text{III} > \text{II} > \text{IV}}$$ **Answer: (C)**