Interpreting a NMR Spectrum

Interpreting a NMR Spectrum The following ¹Hnmr spectrum of a C₁₀H₁₂O₂ compound was obtained on a 90 MHz spectrometer. The 1Hnmr spectrum of the unknown C10H12O2 compound is shown above. Seven structural formulas of some possible candidates for this compound are presented below. After reviewing the chemical shifts, coupling patterns and integration values shown in this spectrum, select the structure most likely to give this spectrum,  ( A through G )  THE FOLL POINTS WILL HELP YOU   1.  Ignoring the TMS signal at δ = 0, how many discrete groups of proton signals are present in this spectrum?
2.  What is the multiplicity ( s, d, t, q ) of the highest field signal from this sample?
3.  The sample has a singlet at δ = 3.8 ppm. In units of Hertz (Hz) how far is this signal from the TMS signal?
4.  What structural feature is suggested by the singlet at δ = 3.8 ppm?
          A CH₃–C=O     B –CH₂–     C –O–H     D –O–CH₃     E C–CH₃     F C=C–H 
5.  From multiplet line separations (Js), which of the other signals is coupled to the quartet at δ = 2.9 ppm?
          A δ = 1.2 ppm     B δ = 3.8 ppm     C δ = 6.9 ppm     D δ = 7.9 ppm

6.  Using the integrator trace and the formula of the sample, assign a whole number ratio to the sample signals as follows:
          7.9 ppm signal ;     6.9 ppm signal ;     3.8 ppm signal ;     2.9 ppm signal ;     1.2 ppm signal          Structure Analysis

The compound's formula, C₁₀H₁₂O₂, has less hydrogen than decane (C₁₀H₂₂). The difference of 10 hydrogens suggests the presence of five double bonds and/or rings in the structure. Since a benzene ring has three double bonds and one ring, and the nmr signals at δ = 7.9 and 6.9 ppm are consistent with aryl hydrogen resonances, we conclude that this compound has a benzene ring.

Integration indicates the presence of two methyl groups, a O–CH₃ at δ = 3.8 ppm, and a C–CH₃ at δ = 1.2 ppm. The former is of course a singlet, but the latter is a triplet (J = 8 Hz) due to coupling with a –CH₂– moiety, which in turn is split into a quartet (signal at δ = 2.9 ppm). From the chemical shifts, we conclude there is an ethyl group bonded to an sp² hybrid carbon atom.

If we examine the structural formulas of the seven candidate compounds, we can eliminate A, C, F & G because they do not have a –O–CH₃ group and cannot account for a singlet at δ = 3.8 ppm. Structure D does not have an ethyl group, so we are left with B & E as the most likely source of this spectrum.

It is not a simple job to distinguish B & E by ¹Hnmr alone. The simplicity of the aromatic proton region indicates para-disubstitution of the benzene ring. If the two substituents are similar in nature, the ring proton signals would have similar chemical shifts. If the substituents are very different in nature, these signals would also have significantly different chemical shifts. Since the aromatic doublets in this spectrum are separated by 90 Hz (1.0 ppm), we may tentatively select E (oxygen and carbon substituents) over B (two different carbon substituents).

A better means of distinguishing these compounds is by infrared spectroscopy. Ester carbonyl stretching frequencies are usually higher than equivalent ketone frequencies by roughly 25 cm^-1. Also the ¹³Cnmr carbonyl signal from esters (δ = ca. 170 ppm) is usually observed at higher field than ketone carbonyl signals (δ = ca. 200 ppm).

The spectrum is in fact that of E.

1H NMR

13 C NMR

IR

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