Question

Compound K displayed a strong band at 1680 cm⁻¹ in its IR spectrum. Its ¹H-NMR spectral data are as follows:

δ (ppm):
7.30 (d, J = 7.2 Hz, 2H)
6.80 (d, J = 7.2 Hz, 2H)
3.80 (septet, J = 7.0 Hz, 1H)
2.20 (s, 3H)
1.90 (d, J = 7.0 Hz, 6H)

The correct structure of compound K is:

• A.
• B.
• C.
• D.

Hint:

Use IR to identify key functional groups (e.g., carbonyls), and match $^1$H NMR patterns to symmetry and splitting to deduce the substitution pattern. Para-substituted benzenes show two doublets; isopropyl groups show a septet and doublet.

Answer 1

The Correct Option is C

\[ \boxed{\text{Correct structure of K is given in option (C)}} \]