Order of Electron Withdrawing Power of Functional Groups

To determine the increasing order of electron-withdrawing power (inductively and mesomerically) for the given functional groups, we analyze the electron-withdrawing strength of each group:

1. Halogen ($-\text{I}$): Halogens possess a weak inductive electron-withdrawing ($-I$) effect compared to carbonyl, cyano, and nitro groups, and they do not exert a $-M$ (mesomeric) effect. Thus, $-\text{I}$ has the weakest electron-withdrawing power among the given groups.

2. Carboxylic Acid ($-\text{COOH}$): The carboxyl group exhibits a moderate $-I$ effect (due to two electronegative oxygen atoms attached to carbon) and a $-M$ effect. It is a stronger electron-withdrawing group than halogens.

3. Cyano Group ($-\text{CN}$): The cyano group has an $sp$-hybridized carbon atom bonded to a triply-bonded nitrogen atom, imparting a strong $-I$ effect and a strong $-M$ effect, making it more electron-withdrawing than the carboxylic acid group.

4. Nitro Group ($-\text{NO}_2$): The nitro group carries a formal positive charge on the nitrogen atom ($-\text{N}^+=\text{O}(\text{O}^-)$), which creates an extremely strong $-I$ effect along with a very strong $-M$ resonance effect. Thus, it is the strongest electron-withdrawing group among the options.

Comparing the electron-withdrawing power of the given groups: $-\text{I} < -\text{COOH} < -\text{CN} < -\text{NO}_2$

Using the labels provided in the question:

• a. $-\text{CN}$
• b. $-\text{COOH}$
• c. $-\text{NO}_2$
• d. $-\text{I}$

The increasing order of electron-withdrawing power is: $\text{d} < \text{b} < \text{a} < \text{c}$

This corresponds to Option C.