Propose structures for alcohols that have the following ${}^{\mathbf{1}}\mathit{H}$ NMR spectra:

(a) ${\mathbf{C}}_{\mathbf{5}}{\mathbf{H}}_{\mathbf{12}}\mathbf{O}$

(b) ${\mathbf{C}}_{\mathbf{8}}{\mathbf{H}}_{\mathbf{10}}\mathbf{O}$

The hydrogen situated next to the oxygen is deshielded and has a value in the range of 3.4 to 4.5 ppm in the${}^{\mathbf{1}}\mathbf{H}$NMR spectrum. Spin-spin splitting is not seen between the alcohol’s O-H proton and the neighbouring protons on the carbon.

(a)The formula ${\mathrm{C}}_5{\mathrm{H}}_{12}\mathrm{O}$has zero degrees of unsaturation. There is a peak at 0.93. This peak integrates to six hydrogens. This peak is a triplet which indicates that the protons responsible for this splitting pattern is adjacent to two protons.

There are six hydrogens adjacent to a carbon with two hydrogens. This carbon must also be part of the rest of the molecule and this carbon is bonded to one additional carbon. This configuration is impossible without forming five bonds to the aforementioned carbon. By elimination, it is known that the signal is due to two methyl groups each of which is bonded to methylene. These groups are symmetrically oriented on our molecule.

A peak is noticed that integrates to 4.00 hydrogens at 1.42 ppm. The splitting pattern is of a quintuplet which indicate that these protons are adjacent to four other protons. There are two methyl groups in the molecule comprising three protons each. Also, with these four protons and the 6 of the peak at 0.93 ppm, we know that there are two hydrogens left to build the molecule. Only one of the hydrogens left can be connected to the carbon. This is because one of the hydrogens must be alcohol. This indicates that these protons are the protons of two methylene groups that is proposed in the previous part.

Therefore, there is a methine in the molecule at the center of the plane of symmetry at which both of the methylene groups are attached. The structure of the molecule can be given as:

Structure of compound (a)

(b)The formula ${\mathrm{C}}_8{\mathrm{H}}_{10}\mathrm{O}$comprises four degrees of unsaturation. There is a peak at 7.32 ppm. This peak integrates at 5.00 hydrogens and is in the aromatic region. This peak is a complex multiplet.

This indicates that the compound has a monosubstituted benzene ring. Lets consider the quartet at 4.80 ppm. This signals integrates to 1.00 and is very downfield. The chemical shift indicates that the carbon of this proton is connected to an oxygen and an aromatic ring. The fact that it is a quartet indicates that the proton responsible for this signal is adjacent to three protons. The structure of the compound can be given as:

Structure of compound (b)

The analysis can be confirmed by looking the peak at 2.43 ppm, which is the range of alcohol peak and a singlet. A signal which is a doublet is seen at 1.42 ppm. This signal integrates to 3.00 hydrogens and the splitting pattern indicates that these protons are near to only one hydrogen atom.