Question

Identify each of the following compounds from its molecular formula and its ¹³C NMR spectrum:

a.C₄H₁₀O

b.C₆H₁₂O

Short answer

The compounds that fit best in the 13C NMR spectra are shown below.

Step-by-step solution

Step 1:13C NMR spectroscopy

Carbon-13 nuclear magnetic resonance is the application of nuclear magnetic resonance spectroscopy to carbon. It is analogous to proton NMR and allows the identification of carbon atoms in an organic molecule just as proton NMR identifies Hydrogen atoms.

The possibility of carbon NMR appears surprising at first glance.

The most abundant isotope of carbon ¹²C has no net nuclear spin, hence it is not detectable by NMR. The far less abundant isotope of carbon ¹³C has a nuclear spin of ½ and is detectable by NMR. ¹³C NMR absorptions occur over a range of 250 ppm, while ¹H NMR absorptions generally occur over a range of only 10 ppm.

Degree of unsaturation

The degree of unsaturation is also known as double bond equivalent (DBE). If the molecular formula is given, plug the number into the following formula.

\(DBE = \frac{{2C + 2 + N - X - H}}{2}\)

Where C is the number of Carbon.

N is the number of nitrogen.

H is the number of hydrogen.

X is the number of halogens.

Determine the degree of unsaturation in the following compound

(a) For C₄H₁₀O

The actual formula C₄H₁₀O , corresponds to zero degrees of unsaturation. It means the compound that has no double bond or ring, maybe simple aliphatic alcohol.

\(\begin{aligned}{}DBE &= \frac{{2(4) + 2 + 0 - 0 - 10}}{2}\\\mathop {}\limits_{} \mathop {}\limits_{} \mathop {}\limits_{} \mathop {}\limits_{} &= \frac{0}{2}\\\mathop {}\limits_{} \mathop {}\limits_{} \mathop {}\limits_{} \mathop {}\limits_{} &= 0\end{aligned}\)

(b) For C₆H₁₂O

\(\begin{array}{}DBE = \frac{{2(6) + 2 + 0 - 0 - 12}}{2}\\\mathop {}\limits_{} \mathop {}\limits_{} \mathop {}\limits_{} \mathop {}\limits_{} = \frac{2}{2}\\\mathop {}\limits_{} \mathop {}\limits_{} \mathop {}\limits_{} \mathop {}\limits_{} = 1\end{array}\)

The actual formula C₆H₁₂O , corresponds to one degree of unsaturation. It may be due to a ring or a double ring present in the compound.

Identification of compound by 13C NMR Spectroscopy

(a) Compound A has zero degree of unsaturation and is simple aliphatic alcohol. The ¹³C NMR spectrum has only three peaks at signal 19, 30, and 70 ppm which indicates that 2-methylpropan-1-ol is the best fit in the given NMR spectrum. The structure of the compound is shown below.

Here, the numbers shown in above structure indicate the ppm value of the signal given by that carbon.

(b) Compound (b) has one degree of unsaturation and iscyclic alcohol. The ¹³C NMR spectrum has only four peaks at signal 71, 36, 26, and 24 ppm which indicates that cyclohexanol is the best fit in the given ¹³C NMR spectrum. The structure of the compound is shown below.

Here, the numbers shown in above structure indicate the ppm value of the signal given by that carbon.