Question

A compound shows $^1$H NMR peaks at $\delta$-values (in ppm) 7.31 (2H), 7.21 (2H), 4.5 (2H) and 2.3 (3H). The structure of the compound is 

 

• A. A
• B. B
• C. C
• D. D

Hint:

In aromatic compounds, para-disubstitution gives two sets of equivalent protons (2H each). Benzylic –CH$_2$Cl appears around 4.5 ppm, while Ar–CH$_3$ appears near 2.3 ppm.

Answer 1

The Correct Option is B

Step 1: Analyze aromatic proton signals.
The two doublets at $\delta$ 7.31 (2H) and $\delta$ 7.21 (2H) indicate a para-disubstituted benzene ring, where two sets of aromatic protons are magnetically equivalent (2H each).
Step 2: Analyze aliphatic signals.
A signal at $\delta$ 4.5 (2H) corresponds to a benzylic –CH$_2$–Cl group, because protons attached to carbon bearing chlorine appear in this region.
The signal at $\delta$ 2.3 (3H) indicates a methyl group attached to an aromatic ring (–CH$_3$ directly bonded to benzene).
Step 3: Combine structural fragments.
Combining the data gives a para-disubstituted benzene ring having a –CH$_2$Cl and –CH$_3$ group. \[ \text{Hence, the structure is p-CH}_3\text{–C}_6\text{H}_4\text{–CH}_2\text{Cl.} \] Step 4: Conclusion.
The correct structure corresponds to Option (D).