Q48E - Organic Chemistry (Mcmurry)

Question

Propose a structure for a molecule ${\mathbf{C}}_{14}{\mathbf{H}}_{12}$ that has the following ${}^1\mathbf{H}$ NMR spectrum and has IR absorptions at 700, 740, and $\mathbf{890}{\mathbf{cm}}^{-1}$ :

Step-by-Step Solution

Answer

Formation of the given structure

Step-by-Step Solution Step 1: Determination of the structure

For the molecular formula of ${\mathbf{C}}_{14}{\mathbf{H}}_{12}$ , we have an NMR spectrum that has two peaks: One at 5.41 ppm and one at 7.31 ppm. Because of the lack of peaks in the spectrum we know that there is a high degree of symmetry in the molecule.

Because of the molecular formula, we know that there are 9 degrees of unsaturation in the molecule. Also, the peak at 7.31 ppm integrates to 5 relatives to the peak at 5.41. With this information, we can assume that there is at least one aromatic ring in the structure which contains 6 carbons and 5 hydrogens.

Step 2: Based on the relative shift in NMR

The peak at 5.41 is a sharp singlet with a relative shift that suggests that it is an alkene proton. The fact that it is a singlet indicates that either this proton is isolated or that the neighboring proton is chemically equivalent through symmetry.

We have used 7 carbons and 6 hydrogens. This is half the number we were given. If we extend the structure by doubling it, we arrive at the following structure.

Based on the IR absorptions. These are useful for determining whether the alkene is cis or trans since NMR cannot easily distinguish between the two. The most important peak is the 700 wavenumber absorption. This is indicative of a cis alkene. Therefore, the final structure is as follows:

Formation of the given structure