See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

# Analysis of –I Effect Order The **inductive effect (–I effect)** measures how strongly a group withdraws electron density through the σ-bond framework. Groups with higher electronegativity and lower polarizability exert stronger –I effects. ## Checking Each Option: **Option (A): $-F > -Cl > -Br > -I$** - Correct. Electronegativity decreases down the halogen group: F (4.0) > Cl (3.0) > Br (2.8) > I (2.5) - ✓ Valid ordering **Option (B): $-NR_3^{\oplus} > -NH_3^{\oplus} > -NO_2$** - Correct. Positively charged groups ($-N^+$) withdraw electrons much more strongly than neutral groups - The N-alkyl substituent in $-NR_3^{\oplus}$ provides additional –I effect - ✓ Valid ordering **Option (C): $-OCH_3 > -OH > -NH_2$** - Correct. Electronegativity: O (3.4) > N (3.0) - Methoxy withdraws more than hydroxyl due to both –I and resonance donation - Hydroxyl withdraws more than amino - ✓ Valid ordering **Option (D): $\text{(benzene)} > -C \equiv CH > -H$** - **False.** The phenyl group $-C_6H_5$ has **weak or negligible –I effect** (aromatic stabilization reduces electron withdrawal) - $-C \equiv CH$ (alkynyl) has **significant –I effect** due to high electronegativity of sp-hybridized carbon - **Correct order should be:** $-C \equiv CH > -C_6H_5 > -H$ - ✗ Invalid ordering **Answer: (D)** is false because the phenyl group should rank lower than the alkynyl group in –I effect.