Question

The number of proton NMR signals for the compounds P and Q, respectively, is 

• A. 3 and 4
• B. 3 and 5
• C. 4 and 3
• D. 5 and 4

Hint:

Look for symmetry to reduce the number of unique proton environments in NMR.

Answer 1

The Correct Option is A

Step 1: Counting NMR signals for compound P.
Compound P contains:
– One aromatic ring with symmetry → 2 sets of aromatic protons. – One methyl (Me) group attached to the ring → 1 signal. Therefore total = 3 proton signals.
Step 2: Counting NMR signals for compound Q.
Q is a symmetric diester with two methoxy groups and a central CHMe group. Protons appear as:
– Two equivalent O–CH$_3$ groups → 1 signal.
– Two equivalent CH$_2$ groups → 1 signal.
– One central CH (unique) → 1 signal.
– One methyl attached to CH → 1 signal.
– Possibly splitting due to lack of full molecular symmetry gives a total of 5 distinct environments.
Step 3: Conclusion.
Thus the compounds show 3 signals (P) and 5 signals (Q).