Q 13-13-22 P - Organic Chemistry (Mcmurry)

Question

Propose a structure for an aromatic hydrocarbon, C₁₁H₁₆, that has the following ¹³C NMR spectral data:

Broadband decoupled: 29.5, 31.8, 50.2, 125.5, 127.5, 130.3, 139.8

DEPT-90: 125.5, 127.5, 130.3 

DEPT-135: positive peaks at 29.5, 125.5, 127.5, 130.3 d; negative peak at

50.2 

Step-by-Step Solution

Answer

A proposed structure for the unknown compound that satisfies all the data:

13 C NMR Spectroscopy

NMR stands for Nuclear Magnetic Resonance Spectroscopy. This is the technique used for determining the structure of the organic compounds. It is a non-destructive technique. It is of two types ¹H NMR and¹³C NMR.

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 carbo atom 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 carbon¹³C has a nuclear spin of ½ and is detectable by NMR.

Determine degree of unsaturation

Degree of unsaturation is also known as double bond equivalent (DBE). If molecular formula is given plug in the number into these formula.

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

Where, C is the number of Carbons

N is the number of nitrogen’s

H is the number of hydrogens

X is the number of halogens

For 

C₁₁H₁₆

_{$$\begin{gathered} DBE=\frac{2\left(11\right)+2+0-0-16}{2} \\ =\frac{8}{2}=4 \end{gathered}$$}

The actual formula, corresponds to four degree of unsaturation. Since the unknown hydrocarbon is aromatic, a benzene ring accounts for all four degree of unsaturation.

Although molecular formula indicates 11 carbons, only seven peak appear in the 13C NMR spectrum, indicating a plane of symmetry. Four of the seven peaks are due to aromatic carbons indicating a benzene ring that is probably monosubstituted.

DEPT 13 C NMR spectra

A modern technique called Distortionless Enhancement by Polarization Transfer, better known as DEPT ¹³C NMR spectra has been developed to distinguish between groups. It shows four spectra of the same compound each spectra provides different information.

Broadband decoupled Spectrum: It shows signal for all type of carbon atoms.

DEPT-45 spectrum: It shows signals for all carbon that are covalently bonded to hydrogens $({\mathrm{CH}}_3,{\mathrm{CH}}_{2 }\mathrm{and} \mathrm{CH})$

DEPT-90 spectrum:It shows signals for all carbon that are covalently bonded to one hydrogen (CH).

DEPT-135 spectrum:It shows signals for all carbon that are covalently bonded to hydrogens $({\mathrm{CH}}_3,{\mathrm{CH}}_{2 }\mathrm{and} \mathrm{CH})$but the phase of the signal will be different, depending on whether the number of attached hydrogen to each atom are odd or even. Signal arising from $({\mathrm{CH}}_3,\mathrm{CH})$odd groups will give positive peak and signal arising from (CH₂) even group will give negative peak.

Proposed a structure by given 13C NMR data

We start with a 139.8 ppm peak. It only shows up in the broadband-decoupled spectra, so it must be a quaternary carbon.

The DEPT-90 spectrum shows that 3 of the kinds of the carbon in the aromatic ring are CH carbons. The positive peak in the DEPT-135 spectrum include these three peaks, along with the peak at 29.5, which is due to a carbon. The negative peak in the DEPT-135 spectrum is due to a CH₂ carbon.

Two peaks remain unidentified and are thus quaternary carbons; one of them is aromatic.

At this point the unknown structure is a monosubstituted benzene ring with a substituent that contains ${\mathrm{CH}}_2,\mathrm{CH} \mathrm{and} {\mathrm{CH}}_3$carbons. A structure for the unknown compound that satisfies all the data is shown below:

                                Structure of unknown