Question

The IR stretching frequency of the carbonyl (C=O) group of a typical saturated ketone is 1715 cm\(^{-1}\). The IR stretching frequencies for the carbonyl groups present in three different acetophenone derivatives are given. Match the molecules in Group I with their corresponding frequencies in Group II. \[ \begin{array}{c|c} \text{Group I} & \text{Group II (C=O stretching frequency cm}^{-1}\text{)}
\hline P:\; p\text{-amino acetophenone} & 1:\; 1677
Q:\; p\text{-nitro acetophenone} & 2:\; 1700
R:\; p\text{-methoxy acetophenone} & 3:\; 1684
\end{array} \]

• A. P-1, Q-2, R-3
• B. P-2, Q-3, R-1
• C. P-3, Q-1, R-2
• D. P-3, Q-2, R-1

Hint:

Rule: EDG → lower C=O frequency; EWG → higher C=O frequency.

Answer 1

The Correct Option is A

Step 1: Recall substituent effects.
- Electron-donating groups (EDG) decrease the C=O stretching frequency by stabilizing the carbonyl resonance.
- Electron-withdrawing groups (EWG) increase the C=O stretching frequency by reducing resonance stabilization, making the bond stronger.
Step 2: Apply to given groups.
- p-amino (EDG): Strongly donates electrons → lowers frequency → 1677 cm\(^{-1}\).
- p-nitro (EWG): Strongly withdraws electrons → raises frequency → 1700 cm\(^{-1}\).
- p-methoxy (EDG but weaker than amino): Slight lowering effect → 1684 cm\(^{-1}\).
Step 3: Match.
P-1, Q-2, R-3.
Final Answer: \[ \boxed{P-1,\; Q-2,\; R-3} \]