A compound, C10H14, shows an IR peak at 825 cm-1. Its 1H NMR spectrum has peaks at delta 7.0 (4 H, broad singlet), 2.85 (1 H, septet, J=8 Hz), 2.28 (3 H, singlet), and 1.20 (6 H, doublet, J=8 Hz). Draw its structure in the window below.
The structure elucidated from the spectral data has been attached as a separate file. Spectral information explained below:
Explanation:
Taking into account the Quaternary state of all carbon atoms, the structure of this compound has been identified. Proton NMR has been sectioned into group A-D based on their chemical shift and environment.
IR peak at 825 cm-1 indicates presence of unsaturated hydrocarbon
The structure is that of p-Cymene as shown attached.
Explanation:
To solve the question, we note that at an IR peak at 825 cm⁻¹ we have either a tri-substituted alkenes or a para-di-substituted benzene.
The ¹H NMR spectrum peaks are at delta 7.0 which corresponds to an aromatic and with the 4 H broad singlet indicating that they are sharing two protons while being adjacent to a carbon with no protons. The peak at 2.85 with 1 H septet indicates methyl group with one proton attached to two equivalent alkyl groups with 3 protons each. The peak at 2.28 with three proton is equivalent to a methyl group attached to the aromatic ring while the peak at 1.2 with 6 protons are the 2 equivalent methly groups attached to a carbon with only one proton.
