Question

A compound, C10H14, shows an IR peak at 745 cm-1. Its 1H NMR spectrum has peaks at delta 7.18 (4 H, broad singlet), 2.70 (4 H, quartet, J=7 Hz), and 1.20 (6 H, triplet, J=7 Hz). Draw its structure in the window below.

You do not have to consider stereochemistry.

You do not have to explicitly draw H atoms.

In cases where there is more than one answer, just draw one.

Answer 1

1. M.F. C10H14. Corresponding saturated hydrocarbon formula is C10H22. This indicate deficiency of 8 H i.e. 4 unsaturation sites.
2. A single IR band at 745 cm^-1 indicate para-substituted ring which is confirmed by PMR peak (7.14, 4H, broad singlet).
3. PMR peaks: (2.70, 4H, quartet, J=7Hz) indicate benzylic –CH2- group, 2 such groups present and (1.20, 6H, triplet, J=7Hz) indicate 2 –CH3 groups. These protons are mutually coupled protons hence 2 –CH2-CH3 groups are present.
4. Hence 1,4-Diethylbenzene is the only proposed structure.