Spectroscopy

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 groups → 1 signal.
– Two equivalent CH 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).

Step 1: Use the formula for converting ppm to Hz.
Chemical shift in Hz = (chemical shift in ppm) × (spectrometer frequency in MHz).
Step 2: Substitute values.
Chemical shift = 2 ppm.
Operating frequency = 350 MHz.
Thus, shift = 2 × 350 = 700 Hz.
Step 3: Conclusion.
Therefore, the resonance is shifted by 700 Hz from TMS.

Step 1: IR activity condition.
A molecule is IR active if it shows a change in dipole moment during vibration. Symmetric linear molecules with no change in dipole moment are IR inactive.
Step 2: Analyzing each molecule.
(A) CO : Though linear and symmetric, its asymmetric stretching mode is IR active → correct.
(B) CS : Linear and completely symmetric; no dipole change → IR inactive → incorrect.
(C) OCS: Asymmetric linear molecule with net dipole moment → IR active → correct.
(D) N : Homonuclear diatomic molecule with no dipole → IR inactive → incorrect.
Step 3: Conclusion.
Thus, CO and OCS are IR active molecules.

Step 1: Analyze aromatic proton signals.
The two doublets at 7.31 (2H) and 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 4.5 (2H) corresponds to a benzylic –CH –Cl group, because protons attached to carbon bearing chlorine appear in this region.
The signal at 2.3 (3H) indicates a methyl group attached to an aromatic ring (–CH directly bonded to benzene).
Step 3: Combine structural fragments.
Combining the data gives a para-disubstituted benzene ring having a –CH Cl and –CH group. Step 4: Conclusion.
The correct structure corresponds to Option (D).

Step 1: Relation between rotational constant and moment of inertia. and for diatomic molecules, where is reduced mass.
Step 2: Compute reduced mass. Step 3: Solve for bond length. Substitute: cm m
Step 4: Conclusion. The bond length of CO = 1.13 Å.

Step 1: Check molecular symmetry and dipole moment.
- BF : Trigonal planar ( symmetry), no permanent dipole moment → microwave inactive.
However, its asymmetric B–F stretching vibrations change dipole moment → infrared active.
- CH : Tetrahedral ( symmetry), no permanent dipole moment → microwave inactive.
Some vibrational modes (asymmetric stretches) change dipole moment → infrared active.
Step 2: Conclusion.
Both BF and CH are microwave inactive but infrared active. Hence, the most appropriate choice is **(B)**.