Meeting 9 - Condensed Answer Key

¹H-NMR Spectroscopy

1. a. All three compounds in group 1 display three singlets in the aliphatic range (δ in ppm): compound 1: 2.40 (3 H), 3.53 (3 H), 4.75 (2 H); compound 2: 2.25 (3 H), 2.32 (3 H), 4.60 (2 H); compound 3: 2.35 (3 H), 3.79 (3 H), 3.57 (2 H). In compound 1, the signals are well separated, while in compound 2 and compound 3, two signals are close. In compound 1, methylene group is shifted the most because it is located between an oxygen atom and a carbonyl group. In compound 2, the two methyl groups displays similar chemical shifts because they are attached to a benzene ring or a carbonyl group, respectively. The methylene group is shifted the most again because of the oxygen atom and the carbonyl group. In compound 3, the methyl group of the methoxy function and the methylene functions are close because the effect of the carbonyl group and the benzene ring add up for the carbonyl group.

b. The molecules in group 2 display the following coupling pattern in the aromatic range: compound 1: d, t, d, t (1 H each); compound 2: s, d, t, d (1 H each); compound 3: d, d (2 H each). This analysis assumes that only J₃-couplings are taken in consideration. If J₄- and J₅-couplings are considered as well, the multiplet structures will be more complex (see below).

c. The simulated spectra look somewhat different: The signal for the OH group shifts downfield in acetone due to the hydrogen bond with the oxygen atom of acetone (R-OH...O=C(CH₃)₂). The signals are closer together in case of the acetone, showing a partial overlap. One of the most obvious changes are the locations of the methyl group, which is at δ=2.26 ppm in acetone and at δ=2.63 ppm in chloroform.

d. All coupling patterns are more complicated than expected due to the NMR activity of the fluorine atom (I=1/2). The H-F couplings tend to be larger than the H-H couplings i.e., J₃(H-H)= ~6-10 Hz, J₃(H-F)= ~20-30 Hz. Compound 1 in group 3 displays a triplet of doublets and a quartet of doublets for the ethyl group. The protons of the methyl group couple very weakly with the fluorine atom due to the larger number of bonds in between. Compound 2 displays two doublets of triplets, the coupling on the CH₂F group being twice (J₂=47 Hz) as large as the one observed in the adjacent CH₂-group. Compound 3 shows a doublet of doublets for the methyl group (1.64 ppm) and a doublet of quartets for the CH-function (5.70 ppm). The large shift of the CH-group can be explained by the presence of the flourine atom and the benzene ring.

2. The 60 MHz spectra look partially very messy (confusing) and generally display a significant signal overlap for the non-symmetric compounds because the lines are more separated. The reason is that the coupling constants are frequency independent. Thus, spectra acquired using a higher frequency (400 MHz) provide more resolution because of the tighter grouping of the multiplets, which can be expanded using the software. The higher field allows for information to be extracted from the spectra.

3. The step height indicates how many of these protons are present in the molecule. Note that the program only provides one continuous integration line, Thus, the integration line accounts for the number of protons in the compound. The experimental spectra usually display a separation of the integration lines if the signals are reasonably separated.

4. If J₄ and J₅ are considered, the multiplet structures are getting much more complicated. Since these coupling constants (J₄= ~1.2 Hz, J₅= ~0.5 Hz) are usually smaller than J₃-couplings (J₃= ~8.20 Hz), they are often not clearly visible unless the spectrum is expanded i.e., the proton adjacent to the hydroxy function should show as a doublet, but it is really a "ddd", which is not clearly resolved in this case because the coupling constants are too small. These long-range couplings explain why the experimental H-NMR spectra have so many numbers on the top.

¹³C-NMR Spectroscopy

1. The molecules in group 2 display different number of carbon signals for the benzene ring because of symmetry (para isomer) or the lack thereof (compound 1: 6 signals, compound 2: 6 signals, compound 3: 4 signals). The situation becomes more difficult in group 3 because of the fluorine, which couples with the carbon atom as well. Compound 1 displays fifteen carbon signals with more or less splitting due to the fluorine atom, the ipso-carbon the largest one. Compound 2 displays twelve signals, the carbon that bears the fluorine displaying the largest splitting (J₂=167 Hz). Compound 3 has ten signals due to the splitting with fluorine, the carbon that bears the fluorine displaying the largest splitting (J₂=168 Hz).

2. The spectra display the C-H couplings as well. The spectra show multiplets now, because the C-H couplings but the intensity of the signals decrease significantly. Overall, the spectra contain more information, but they are also more difficult to analyze due to the overlap of the multiplets.

3. a. The signals of the compound are relatively large compared to the solvent signal that appears as septet at δ=40 ppm due to the C-D coupling (CD₃: 2*1*3+1). Note that the deuterium atom displays a spin of I=1.

b. Due to the lower concentration of the solute, the signals due to the compound are very small compared to the solvent lines, which appear as triplet at δ=77 ppm.

Chromatography Simulations

1. The log P values are

Compound	Log P
o-hydroxyacetophenone	1.96
m-hydroxyacetophenone	1.39
p-hydroxyacetophenone	1.42
m-methoxyacetophenone	1.84

The m-isomer and the p-isomer of the hydroxyacetophenone are more polar than the o-isomer because the hydroxy groups are available for hydrogen bonding to water. In addition, these compounds also display a larger dipole moment. If the hydroxy group is converted to a methoxy group, the solubility of the compound decreases because the hydrogen bond donor ability is lost.

a. Based on the log P-values, the compounds would elute in the following sequence on a non-polar stationary phase (increasing retention times):

m-hydroxyacetophenone < p-hydroxyacetophenone < m-methoxyacetophenone < o-hydroxyacetophenone.

Since the first two compounds (meta and para-isomer) display very similar log P values, they will probably overlap in the chromatogram.

b. The situation would different on a polar stationary phase like silica (increasing R_f-values):

m-hydroxyacetophenone < p-hydroxyacetophenone < m-methoxyacetophenone < o-hydroxyacetophenone.

The most polar compound would migrate the least (meta isomer), while the least polar compound (ortho isomer) would migrate the most. Again, it is doubtful if the meta and para isomer would separate on the TLC plate.