Q15-40E - Organic Chemistry (Mcmurry)

Question

What is the structure of a hydrocarbon that has ${\mathbf{M}}^{+}\mathbf{=}\mathbf{120}$ in its mass spectrum and has the following ${}^1\mathbf{H}$NMR spectrum? 7.25 d (5 H, broad singlet); 2.90 d (1 H, septet, J 5 7 Hz); 1.22 d (6 H, doublet, J 5 7 Hz).

Step-by-Step Solution

Answer

Isopropyl benzene

Step-by-Step Solution Step 1: NMR spectroscopy

The proton NMR spectroscopy is used to study the characteristics of the proton and its chemical shift value defines, how many protons are attached by comparing the chemical shift value of the standard one.

Step 2: NMR spectroscopy of given compound

As the mass of the compound is 120 and from the NMR data, 12 protons are present so the molecular formula of the compound is ${\mathrm{C}}_9{\mathrm{H}}_{12}$. Six protons doublet and one proton septet indicate the presence of an isopropyl group, from the chemical shift value of the one proton signal we can say that it is present on the aromatic ring and a five proton broad singlet indicates the presence of substituted benzene ring. So, the structure of the compound is 

Isopropyl benzene