Question

The group which shows \(-I\) and \(-R\) effect is:

• A. \(-\mathrm{Cl}\)
• B. \(-\mathrm{OR}\)
• C. \(-\mathrm{NO}_2\)
• D. \(-\mathrm{NHCOR}\)

Hint:

For substituent effects:
- \(-I\): Groups with electronegative atoms (e.g., \(\mathrm{NO}_2\), halogens).
- \(-R\): Groups with pi-acceptor ability (e.g., \(\mathrm{NO}_2\)).
- Check both inductive (sigma) and resonance (pi) effects.

Answer 1

The Correct Option is C

Substituents on aromatic rings affect electron density via: - Inductive effect (\(I\)): Electron withdrawal (\(-I\)) or donation (\(+I\)) through sigma bonds. - Resonance effect (\(R\)): Electron withdrawal (\(-R\)) or donation (\(+R\)) through pi bonds.
We need a group with both \(-I\) (electron-withdrawing via induction) and \(-R\) (electron-withdrawing via resonance).
Step 1: Analyze each group
- (A) \(-\mathrm{Cl}\): Chlorine is electronegative, causing a \(-I\) effect. It has lone pairs that can donate electrons via resonance (\(+R\)) in aromatic systems, so not \(-R\).
- (B) \(-\mathrm{OR}\): Alkoxy groups (e.g., \(-\mathrm{OCH}_3\)) are \(-I\) due to oxygen’s electronegativity but \(+R\) due to lone pair donation via resonance.
- (C) \(-\mathrm{NO}_2\): The nitro group is \(-I\) due to electronegative oxygens pulling electrons through sigma bonds. It is also \(-R\) because it withdraws electrons via resonance (e.g., resonance structures place positive charge on the aromatic ring). This matches both criteria.
- (D) \(-\mathrm{NHCOR}\): The amide group has a nitrogen lone pair, leading to \(+R\) (electron donation via resonance). It may have a weak \(-I\) effect due to the electronegative oxygen in \(-\mathrm{COR}\), but the resonance effect is dominantly \(+R\).
Step 2: Conclusion
Only \(-\mathrm{NO}_2\) shows both \(-I\) and \(-R\) effects.