Question

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.

Answer 1

The structure is that of p-Cymene as shown attached.

Step-by-step 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.

The structure is attached.

Answer 2

The structure elucidated from the spectral data has been attached as a separate file. Spectral information explained below:

Step-by-step 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

H NMR

• A- 1.20 ppm (6H d): RCH2R
• B- 2.85 (1H, septet): RC≡C-H
• C- 7.0ppm (4H, s): vinyl hydrogens
• D- 2.28 (3H, s): alkyl hydrogens