See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

# Analysis of -I Effect Order The **-I effect** (inductive effect) measures electron-withdrawing ability. Stronger electronegativity and greater charge density withdrawal = stronger -I effect. ## Evaluating Each Option: **(A) $-I < -Cl < -Br < -F$** ✓ Correct - Electronegativity: $F > Cl > Br > I$ - Order follows electronegativity perfectly **(B) $-NR_2 < -\overset{\oplus}{O}R_2$** ✓ Correct - The positively charged oxygen ($-\overset{\oplus}{O}R_2$) is far more electron-withdrawing than $-NR_2$ - Charge makes it strongly -I **(C) $-NR_2 < -OR < -F$** ✗ **Incorrect** - Actual order: $-NR_2 < -F < -OR$ (not < -F) - Why? $-OR$ has **oxygen with higher electronegativity AND formal negative charge** on the oxygen, making it extremely -I - $-F$ is less electron-withdrawing than a charged $-OR$ group - The given order violates this relationship **(D) $-SR < -OR < -\overset{\oplus}{O}R_2$** ✓ Correct - $S$ is less electronegative than $O$ - Positively charged oxygen is most -I - Order is correct **Answer: (C)** — The incorrect part is placing $-F$ as more -I than $-OR$. The negatively charged oxygen in $-OR$ makes it far more electron-withdrawing than fluorine.