Question

The NMR spectra on the right are for the organic compounds C₆H₁₂ and C₄H₁₀O. Deduce the structures for these compounds. See Exercise 74 for a discussion of the bonding in organic compounds. The structure of C₆H₁₂ has one double bond, with the rest being single bonds, and the structure of C₄H₁₀O has only single bonds. Note that the TMS reference has been omitted in each spectrum.

Short answer

The structure of C₆H₁₂is given below:

The structure of C₄H₁₀O is given below:

Step-by-step solution

 Explanation for C6H12  

In the NMR spectra of the given molecule, there is only one peak present in the spectrum, which means all the hydrogen atoms must be equivalent in their environment. Now, if we calculate the index of hydrogen deficiency, we find that there will be one double bond or a ring in the structure. Thus, a structure can be deduced by trial and error. Therefore, 2, 3-dimethylbut-2-ene is the correct compound for the given data.

 Explanation for C4H10O

In the NMR spectra for this molecule, we have a quartet and a triplet peak in the spectrum. The given quartet peak indicates that hydrogen atoms neighbor three H atoms, and the triplet peak indicates that hydrogen atoms neighbor two H atoms. Now, if we calculate the hydrogen deficiency, there is no hydrogen deficiency in this molecule. Thus, all the bonds will be single bonds and the structure will not be cyclic.

In the given spectrum, only two peaks are present which further indicates that there is a presence of symmetry in the compound. Therefore, the probable structure for the molecule is drawn below: