Chemistry 324 - Spring 2008

One of you wrote “this sucks” next to your NMR assignments. I have to agree. A complete explanation of the 1H and 13C spectra is beyond me.

On other hand, while I wish I had chosen a more accessible set of data, these spectra are typical. Mother Nature is rarely kind. We just have to deal with that.

Here is my quick take on the NMR data along with some subtle relationships that should not be overlooked …

First, two useful ways of looking at compound 9:

1H NMR

First, group protons by chemical shift range:

• aromatics (b-e) - 4H
• alkene (i, j) - 2H
• alkane, O neighbor (p) - 3H
• alkane, unsat’d neighbor (h-l) - 3H
• alkane (n) - 2H

The data contain:

• four 1H signals between 6.86-8.24 ppm. These are b-e. Alkene protons could fall in this region, but only if the alkene is substituted with strong deshielding groups, e.g., C=O. Two signals are doublets with J=8 Hz. These are probably b (8.24) and e (6.86). c and d can’t be assigned reliably to 6.93/7.27 (both multiplets) without spectra to compare with.
• two 1H signals 6.67 and 5.71 ppm. These are i and j. These signals appear in all of the versions of this compound (O bridged, unbridged, Me- and Ph-substituted).
• a 3H singlet at 3.85 ppm (q).
• three 1H signals at 5.32 (d, J=4 Hz), 3.73 (d, J=1.5 Hz), and 3.37 (d, J=2.5 Hz). These are h-l. Two protons, h and l have two unsat’d neighbors and should be extra deshielded, but only one signal (5.32) appears at an extreme shift. The model shows that h and l are nearly coplanar and perpendicular with respect to the aromatic ring, respectively. This geometric difference must cause one proton to be extra deshielded. Other geometric differences: h is near O, while l lies behind C=O. It is hard to assess the shielding and deshielding effects of these geometric factors (note: Spartan can calculate chemical shifts, but I didn’t attempt it - it’s a little time-consuming and I wasn’t expecting you to use this tool).
• two 1H signals at 2.26 (m) and 2.17 (d, J=11 Hz). These are n1 and n2. 11 Hz is a standard geminal coupling constant. The n proton that points towards the aromatic ring is nearly perpendicular to its vicinal neighbors, so it might not couple noticeably with either of them (2.17).

13C NMR

First, group carbons:

• aromatics (a-f) - 6C
• alkenes (gmij) - 4C
• ester carbonyl (o) - 1C
• ester methyl (p) - 1C
• alkyl (hkln) - 4C

Because they give you CH coupling information, some assignments are simple:

• 51.5 CH3 - p
• 41.9 CH2 - n

Fortunately, these assignments are also consistent with the chemical shifts expected for these carbons. Speaking of expectations, Hans Reich’s 13C tables show that benzofuran should have signals at 106 (m), 144 (g), 121-127 (abcd), 111 (e), 155 (f). What you should notice here are the numbers and two patterns, namely, O has a deshielding effect on f and g and a shielding effect on e and m. Another table shows that ester C=O are typically located between 155-165 ppm and attaching an alkyl group to an arene or alkene C has a modest (10 ppm?) deshielding effect (this will apply to g and m).

Now for the observations:

• 166.1 (C) & 163.2 (C) & 145.2 (C) - Based on the above considerations, o must be either 166 or 163, but which? The other two signals must be f and g.
• 131.1 (C) & 124.9 (C) - a and m, but can’t be certain which is which.
• 145.9, 131.7, 128.2, 127.3, 121.2, 110.5 (all CH) - bcdeij. It is tempting to assign 110.5 to e based on the above shift comparisons.
• 83.0, 44.1, 41.9 (all CH) - hkl. First, consider branching - no help, all three C are tertiary. Next consider de/shielding neighbors. Again (see 1H NMR) we have the puzzle of two C attached to two unsaturated (deshielding) neighbors and one C attached to one unsaturated neighbor, but only one unusually deshielded signal.