Question

Decreasing order of (-I) effect

$^+NF_3 > ^+NR_3 > ^+SR_3 > ^+NH_3 > -NO_2 > -SO_3H > -CN > -CHO > -COR > -COX > COOCOR$ $>-COOR > -COOH > CONH_2 > -F > -Cl > -Br > -I > -OR > -OH > -C \equiv CH >$ $-NH_2 > -C_6H_5 > -CH = CH_2 > H$

Answer 1

$^+NF_3 > ^+NR_3 > ^+SR_3 > ^+NH_3 > -NO_2 > -SO_3H > -CN > -CHO > -COR > -COX > COOCOR > -COOR > -COOH > CONH_2 > -F > -Cl > -Br > -I > -OR > -OH > -C\equiv CH > -NH_2 > -C_6H_5 > -CH=CH_2 > H$

Answer 2

We need to arrange various substituents in order of decreasing (i.e. “most to least”) –I (inductive electron‐withdrawing) effect. One acceptable answer is as follows:

$$^+NF_3 > ^+NR_3 > ^+SR_3 > ^+NH_3 > -NO_2 > -SO_3H > -CN > -CHO > -COR > -COX > COOCOR > -COOR > -COOH > CONH_2 > -F > -Cl > -Br > -I > -OR > -OH > -C\equiv CH > -NH_2 > -C_6H_5 > -CH=CH_2 > H$$

Minimal Explanation

1. Key Idea: The groups with a formal positive charge (e.g. $^+NF_3, ^+NR_3$) are very strongly electron‐withdrawing via the inductive effect.

2. Next: Classic –I groups like $-NO_2$, $-SO_3H$, $-CN$ follow.

3. Then: Other carbonyl‐containing groups and halogens follow in the order of their –I strength.

4. Least: Groups like $-CH=CH_2$ and $H$ have little or no –I effect.

Thus, the above order is the decreasing order of the –I effect.