nmrshiftdb2 - Collection of "Problem of the Month" items

Strategy

• Fig. 13.1 shows a type of ¹³C NMR experiment which preserves coupling to ¹H. Thus, it is possible to identify the multiplicity of carbon atoms [note, however that this experiment is insensitive as compared to ¹J H,C correlations]
• In Fig. 13.2, you will see the 'modern' version of the same experiment [CH and CH3 groups give positive cross peaks (red), CH2 groups give negative cross peaks (blue). Quaternary carbons, as well as OH protons, do not yield any crosspeaks].
• As always, it is helpful to calculate the degree of unsaturation (DBE) from the molecular formula: C₇H₁₂O₂.

Hints

• There are two groups of spin systems in this molecule - they can be verified conveniently with a H,H COSY spectrum (Fig. 13.4).
• Even though there is no HMBC correlation provided here, chemical shifts and H,H correlations help to connect the fragments across the only existing quaternary carbon.

Solution

• The molecule of May could be almost solely elucidated through proton experiments. The only carbon that does not connect to any proton is typical for carboxylic acids and their derivatives.
• The two spin systems are divided into an olefinic AMX and an aliphatic A₂M₂P₂X₃ part. The olefinic couplings may be estimated, since 1 ppm corresponds to 600 Hz at the field strength employed for these experiments.
• On cheminfo, there is a nice tool to simulate the ¹H NMR spectrum: In the third row, select 'predict 1H' and simply draw the assumed structure with the editor and compare [if you got real spectra, you may even superimpose them as a jcamp file via drag-and-drop].
• If you are still in doubt - the correct structure is found at http://www.nmrshiftdb.org/molecule/60004995.

Strategy

• The degree of unsaturation (DBE) can be derived from the molecular formula: C¹⁰H⁸.
• Check the number of distinct signals for carbon and hydrogen NMR: This will give you a clue.
• Carefully analyse the structure of the multiplets visible in the 1D ¹H NMR spectrum - if you are struggling, there is a H,H COSY provided, too.

Hints

• It is most important to recognise the symmetry of the compound - all chemically equivalent atoms give rise to one common signal. Integration of proton signals is given below the corresponding peaks.
• In the HMBC, most of the cross peaks derive from ³J_CH coupling. In the COSY experiment, two spin systems may be identified - caution, there are also long range correlations visible (> ³J_HH).

Solution

• The molecule of February is quite symmetric - there are four positions that exists twice. Also, it is an aromatic compound for the chemical shifts in the proton spectrum, it contains 7 DBE.
• The two spin systems are divided into an AX₂ and an AA'MM'X part. You may simulate such a 5-spin system at icheminfo (middle box in the 'tools' section, choose ABCDE system and set δ 8.28 for A/B, 7.51 for C and 7.09 for D/E. Providing uniform ³J couplings of 7 Hz between A/D, B/E, C/D and C/E is sufficiently accurate).
• Connecting the annulated system is feasible through HMBC correlations. If youare still in doubt - enter the chemical shifts of the ¹³C signals (compare results with and without carbon multiplicities, i.e. S = no proton attached, D = proton attached) in nmrshiftdb2's search function.

• This month's problem is a mixture of two isomeric compounds with an elemental composition of C₇H₁₄O₂.
• It is always a good idea to determine the degree of unsaturation (double equivalents) first.
• The edited HSQC shows ¹H,¹³C coupling over one bond and also encodes CH₃ and CH groups as positive signals (red) and CH₂ groups as negative signals (blue).
• The mixture consists of a major component and a minor component. Try using the 1H integrals to determine which signals belong to which of the two components.
• Now use the HSQC to assign the carbon signals from the 13C spectrum to each component.
• Did you notice how the HSQC resolves the overlap of the multiplet at δ_H 0.91 ppm?
• The HMBC shows ¹H,¹³C coupling over two to four bonds.
• The signals δ_H 2.04/2.03 ppm are singlets in both compounds and must be connected to a quaternary carbon due to the absence of ¹H,¹H coupling. Can you determine which?
• Do you know which structural fragment you have with the singlet and the quaternary carbon? If not, search for the ¹³C shifts of the fragment on nmrshiftdb.org to get an idea.
• Now use the COSY spectrum to connect the aliphatic portion of the two molecules.
• Can you find out how they are connected to the fragment you identified before?
• Can you explain why the CH₂ groups give completely different signals in the two molecules?

• Hint: Use the HSQC to deduce the coupling patterns of the overlapping methyl signals at δ^H 0.91.
• If you cannot find a solution, try to search for the minor component by its ¹³C shifts on nmrshiftdb.org - the major component is not yet in the database but should be easy to identify once you know the minor compound!

Strategy

• The spectra for structure elucidation presented this month are a good example to show, how the QuickCheck tool might assist you. However, there is a catch in it!
• The experiments shown on the first page derive the C-H connectivities in the molecule: HMQC with cross peaks for ¹J_CH are visible in red, while HMBC peaks for long range correlations (²J_CH, ³J_CH and some dublets for ¹J_CH). In aromatic systems, the intensity of ²J cross peaks is weaker than the one of ³J_CH cross peaks.
• The molecular formula of this month's molecule is C₈H₉IO₂. First, derive the number of DBE (degree of unsaturation).

Hints

• The previous two experiments allow to evaluate the compound by chemical shifts, multiplicity of carbons (APT), signal intensity (integrals ¹H) and by analysis of the H-H coupling patterns. The absolute number of protons is printed in blue.
• There is one signal the ¹H NMR which differs from the other ones in its broad lineshape. Also, there is no correlation in the HMQC with this proton - that is why this area is not shown on the first page.
• After identification of the aromatic spin system and collecting information about individual fragments, try to identify which substituent is bound to which position by carefully re-investigating the HMBC correlations.

Solution

• There are only non-equivalent carbons in the target molecule‘s structure - which means that there is no higher symmetry. For the chemical shifts of proton signals in the left part, there must be aromatic moiety, which is supported by the degree of unsaturation (DBE = 4).
• Four spin systems exist in ¹H: One O-CH₃ unit (3.82), one O-CH₂ (4.64) unit, an "exchanging" OH proton (2.03) and an aromatic AMX spin system. The latter consists of two protons that are three bonds a part (7.68 and 6.61) and a third proton (7.08) that shares long-range coupling to the one of them (6.61).
• The substitution pattern of the aromat can be derived through the ³J_CH cross peaks. The location of iodine is easy to spot for the ¹³C chemical shift. If you are still in doubt - enter the chemical shifts of the ¹³C and ¹H signals with your suggested structure in nmrshiftdb2's QuickCheck. The solution can be found here.

Strategy

• In the H,H NOESY experiment (NOE = nuclear Overhauser effect) shown in Fig. 9.1, protons which are apart from each other no more than a distance of 4Å, show cross peaks. Here, through-space, as well as chemical exchange correlations can be analyzed.
• This experiment is phase-sensitive, i.e. cross peaks that are of the opposite phase (blue) as the diagonal peaks (red) stem from NOE. Signals, which are of the same phase as diagonal peaks (red) originate from chemical exchange.
• The molecular formula of this month's molecule is C₆H₁₁NO. What does this tell you about the number of DBE?

Hints

• In Fig. 9.2, H-H correlations through bonds can be inspected in the COSY experiment. Here, long-range couplings (> ³J_HH) are visible in two cases. The absolute number of protons is printed in blue. Carefully compare the cross peaks with the ones visible in the NOESY spectrum.
• Take a look at the chemical shifts in the multiplicity-edited APT spectrum (Fig. 9.3). Here, a C-H carbon stands out due its atypical shift. If you do a subspectrum search for a doublet (D) at this frequency, the type of carbons in question can be narrowed down. It might be interesting to search for the group of sp3 hybridized carbon signals, as well.
• To elucidate the 'skeleton' of the compound, there are H-C correlation spectra in Fig. 9.4. ¹J_CH correlations from the HMQC experiment are labelled in red.

Solution

• There are six non-equivalent carbons in the target molecule's structure - the same number as in the molecular formula. Also, there are - with one exception - no sp² hybridized carbons, which excludes the possibility of an aromatic/olefinic molecule. Calculation of DBE yields two.
• Connections between the CH₂ units are straight forward from the COSY spectrum: One coupled spin system consisting of 3.12-1.38-1.49-1.33-3.27 can be identified, additionally, long range correlations from 7.79 to 3.12 and 3.27 are present.
• From the NOESY experiment, exchange can be detected at least between 3.12/3.27. Also, there is only a NOE contact from 7.79 to 3.27, not to 3.12 - why? If you are still in doubt - enter the chemical shifts of the ¹³C signals in nmrshiftdb2 as a "spectrum search" (option "complete").

Strategy

• Given the molecular formula C₉H_{14O7, you can determine the number of double bond equivalents.}
• Determine the CH_x fragments in the molecule from the edited HSQC spectrum. CH and CH₃ groups give positive crosspeaks (red), CH2 groups give negative crosspeaks (blue). Quaternary carbons, as well as OH protons, will not yield any crosspeaks.
• Besides the correlation information, the inspection of the 1D ¹H NMR spectrum (projection on top, intensities and chemical shifts given in Fig. 8.2) provides only few couplings. Investigate the ¹H chemical shifts at 3.84 & 3.69 through a subspectrum search to get an idea of the type of possible functionalities.

Hints

• Do a subspectrum search for the three chemical shifts that appear at the highest frequencies (173.8, 170.2) in the ¹³C NMR spectrum displayed in Fig. 8.3. Check which are the most common fragments. Include the multiplicity (S) in your search. Repeat and then combine the search for the signals at 53.3, 52.1 (Q).
• Determine the connectivity of the fragments using the HMBC experiment (Fig. 8.4). It shows coupling between ¹H and ¹³C over more than one bond.
• You need to analyze the stereochemistry carefully to see why there is coupling between the protons 2.91 and 2.81.

Solution

• The target molecule contains, according to DBE (3) and carbon chemical shifts, three carbonyl double bonds, each of an ester moiety. There must be symmetry in the molecule, as is also reflected in the number of ¹³C signals and the relative intensities of e.g. signals C1 vs. C2, and C4 vs. C5, rsp.
• In the ¹H NMR spectrum, symmetry is also detectable: The signals C (δ_H 3.69) and A/B(δ_H 2.81/2.91) corresond to double sets of CH₃ and CH₂ groups. Only one multiplet (an AB system for signals A/B) with a geminal coupling (²J_HH = 15.7 Hz) is present, since methylene protons are diastereotopic. Also, there is one OH group, signal E (identified with the aid of HSQC).
• Try to summarize the fragments, before entering your structure in the editor using the prediction menu of nmrshiftdb2. No solution? Enter the chemical shifts of the ¹³C signals in nmrshiftdb2 as a "spectrum search" (option "complete").

Strategy

• The only NMR experiment required to solve this month’s problem is the H,C long range correlation (’HMBC’) shown in Fig. 7.1. It shows connectivities between proton and carbon ranging from one (in some cases) up to five bonds .
• Besides the correlation information, the inspection of the 1D ¹H NMR spectrum (projection on top, intensities given in blue numbers) presents a simple spin system.
• The molecular formula is C₅H₆O₂, which allows to determine the number of DBE (i.e. degree of saturation) for the structure.
• To facilitate assignment, ¹³C chemical shifts are labelled in red (CH or CH₃), blue (CH₂) and black (quaternary C).

Hints

• Do a subspectrum search for the three chemical shifts that appear at the highest frequencies (152.7, 74.7, 74.5) and check which are the most common fragments. Include the multiplicity (S, D, ...) in your search.
• The H,C long range correlation experiment needs to be assigned with special care - the expansion shown in Fig. 7.2 allows a clearer insight in the most overcrowded shift range in ¹³C. Note: The size of the ¹J couplings is explicitely printed on the spectrum. It helps to use the empirical rule valid for the degree of hybridization of C-H orbitals and the size of ¹J_CH.

Solution

• The target molecule contains, according to DBE (3) and carbon chemical shifts, a carbonyl double bond as well as an alkyne triple bond (result of subspectrum search).
• ¹H NMR allows the observation of an A₂X₃ spin system (ethyl group) which is bound to an oxygen atom (δ_H 4.26, 1.33). The remaining signal (δ_H 2.92) stems from the alkyne; extraordinarily large J^CH can be detected from the HMBC (²J ∼ 50 Hz, ¹J ∼ 250 Hz are typical for sp hybridized carbons).
• Try to compare the predicted ¹³C chemical shifts, when entering your structure in the editor using the prediction menu of nmrshiftdb2. This month’s molecule is not yet in our database, however, agreement of experimental and predicted shifts is very good. No solution? Enter the chemical shifts of the 13C signals in nmrshiftdb2 as a "spectrum search" (option "complete") after October 15th.

Strategy

• In Fig. 6.1, an H,H correlation experiment is presented which allows to derive three spin systems that are coupled over two or three bonds, respectively. Note, that the spectrum was recorded in DMSO-d₆. Thus, correlation can also be observed with so-called ‚exchanging‘ protons.
• It is always useful to determine the number of DBE (i.e. degree of saturation) for the molecule, if the molecular formula is given: C₉H₁₉NO₄.
• H,C connectivities can be assigned from the correlation experiment shown on the next page. As one can note, there are diastereotopic protons. Protons which are not bound to a carbon show no cross peak. Also, to facilitate assignment, cross peaks in red represent CH or CH₃ carbons, while blue peaks signify CH₂ groups.

Hints

• Quarternary carbon centers are printed in black in the ¹³C chemical shift list. Take a look at the four chemical shifts that appear at the highest frequencies (173.4 ... 59.1) and check the heteroatoms which are in the molecular formula.
• Another NMR experiment appears to be crucial to solve this problem: Only with H,C long range correlation experiment (HMBC, Fig. 6.2), the fragments identified so far can be connected. Note, that besides correlations for ²J and ³J, also ¹J (dublets) are visible in some cases. ‚Exchanging‘ protons do show l.r. correlations, too.
• As an additional hint, there is an expansion of the H,N correlation (¹J_HN) and a selective NOE experiment available in Fig. 6.3, which deliver information from correlations through bond and through space, rsp.

Solution

• The target molecule contains, according to DBE (1) and carbon chemical shift, a double bond (ester or amide, result of subspectrum search for δ_C 173.4). The signal at δ_H 7.75 correlates to the corresponding amide nitrogen.
• Also, three more exchanging proton signals can attributed to hydroxy groups (δ_H 5.36, 4.51, 4.49). Since they do show H,H correlations, the spin systems for two primary and one secondary alcohol can be derived; H,C long range correlation and NOE contacts help to integrate amide and diastereotopic methyl groups.
• Try to compare the predicted 13C chemical shifts, when entering your structure in the editor using the prediction menu of nmrshiftdb2. This month’s molecule is not in our database, however, the agreement of experimental and predicted shifts is very good. No solution? Enter the chemical shifts of the ¹³C signals in nmrshiftdb2 as a "spectrum search" (option "complete") after June 15th 2015.
• The solution has been added to the database.

Strategy

• First, determine the number of DBE for the molecule. Its molecular formula is C₈H₁₀FN.
• Various types of ¹³C NMR spectra are presented in Fig. 5.1: Without ¹H or ¹⁹F coupling (top trace), with only 19F coupling visible (middle) and with both, ¹H and ¹⁹F coupling present (bottom). With this information, you can assign the number of attached hydrogen or fluorine atoms.
• Now, assign protons to corresponding carbons from the ¹H NMR spectrum in Fig. 5.1. Try to assign the multiplets and derive the observable spin systems. Note, the upper spectrum is ¹⁹F decoupled.

Hints

• The ¹H signal at δ_H 1.38 will cease, once a drop of deuterated water is added to the sample!
• This information, together with the number of DBE, should be helpful.
• This month’s problem does not contain any through-bond correlation experiments. The only 2D experiment available as additional information is a heteronuclear Overhauser effect (Fig. 5.2). This type of spectrum gives information about nuclei which are (farely) close in space (maximum 4 Å as a rule of thumb).
• Also, you might verify the other part of the aromatic spin system with the 19F NMR spectrum given in Fig. 5.3.

Solution

• The target molecule should be, according to DBE and ¹H/ ¹³C chemical shifts in the sp² region of the spectra, an aromatic compound. From the sum formula and the hint with the perishing signal at δ_H 1.38 (intensity = 2H), an amine is feasible.
• There is an AA’XX’ system visible in the ¹H NMR spectrum, if ¹⁹F is decoupled. In combination with the ¹⁹F coupling observed in the ¹³C{¹H} spectrum, the substitution pattern can be identified. This is affirmed by the NOE spectrum.
• To derive the aliphatic amine part, the ¹H NMR spectrum is sufficient: Here, a AX₃ spin system is observed.
• No solution? Enter the chemical shifts of the 13C signals in nmrsshiftdb2 as a "spectrum search" (option "complete") or click here.

Strategy

• This month's molecule has the molecular formula C₁₃H₁₈O₂. H,C connectivities can be assigned from the edited HMQC (Fig. 4.1). Note: In this type of experiment, cross peaks in red represent CH or CHi₃ carbons, while blue peaks signify CH₂ groups. Two signals appear at almost identical shifts (multiplicities in brackets). Integrals of the proton signals are given below the 1D projection.
• Several ¹³C signals do not possess cross peaks. By comparison with the ¹H NMR spectrum (Fig. 4.2), it can be seen that there is also one proton signal without cross peak. In addition, there must be some symmetry in the target structure (number of signals vs. molecular formula). Although from a standard ¹³C{¹H} NMR spectrum no quantitative conclusions can be drawn, relative signal intensities (see 1D projection) should give you a clue.

Hints

• In total, three coupled spin system can be derived from ¹H and H,H COSY NMR spectra, amongst them a higher order aromatic pattern. Also, from the number of DBE, an aromatic compound should be considered.
• Do a "search by spectrum" in nmrshiftdb2 for one of the fragments (select option "subspectrum"): First, enter the carbons at δ_C 45.1, 30.1 and 22.4 with their corresponding multiplicity (e.g. "45.1T" for a carbon with a chemical shift of δ_C 45.1 with two protons attached). Since there are two carbons at δ_C 22.4, you have to enter this value twice! Observe the type of functional group, that results from your search.
• If additional information is needed, the fragments that you identified by now can be connected with the HMBC (Fig. 4.4).

Solution

• An AA'XX' system, made up from protons attached to sp² hybridized (aromatic) carbons, can be identified. Also, two aliphatic residues (A₂MX₆ and AX₃) are distinguished. The total number of DBE.s is 5.
• Subspectrum search for the ¹³C signals indicates an isobutyl group, the signal at δ_C 180.9 in combination with the broad proton signal at δ_H 10.58 a carboxylic acid. Assignment of quaternary aromatic carbons and substitution pattern can be deduced of the HMBC spectrum.
• Hint: Enter the chemical shifts of the ¹³C signals in nmrsshiftdb2 as a "spectrum search" (option "complete", multiply existing carbons that deliver only one signal due to symmetry need to be entered the corresponding number of times). Or click for the solution here.

Strategy

• The molecular formula of this month's molecule is C₁₀H₁₆O₂: As a first step, it is always important to determine the number of DBE.
• By inspection of the 13C{1H} NMR spectrum in Fig. 3.1, the shift at highest frequency clearly indicates one class of compound. If you are not sure, do a "search by spectrum" in nmrshiftdb2 (select option "subspectrum"): Enter the carbon chemical shift (219.6). From the chemical shifts present, an aromatic or olefinic molecule can be excluded. For easier discussion, ¹³C signals are labelled from left to right (C1-C10).
• Assign H,C connectivities from the HMQC (Fig. 3.2). Since this is an edited experiment, cross peaks in red represent CH or CH3 carbons, while blue peaks signify CH2 groups. An HMQC also allows to define diastereotopic protons. Integrals of the proton signals are given below the projection.

Hints

• An interesting candidate to look after is the ¹H signal at δ_H 2.1. Do a "search by spectrum" in nmrshiftdb2 (select option "subspectrum"), and enter the carbon shift corresponding to this signal (enter "43.1D" to indicate a shift at δ_C 43.1 with one proton attached). This brings up some types of molecules which, in combination with the number of DBE, should be helpful.
• This month's problem can best be solved with the assistance of the HMBC (H,C long range correlation) experiment (Fig. 3.3): E.g., all three methyl groups are connected to quarternary carbons (no splitting in 1H). Also, we are dealing with a bicyclic structure (DBE!). After summing up all H-C correlations, dissect the neighbours (e.g. by relating all correlations to the carbon at δ_C 219.6, etc.). To back up the observations, a COSY spectrum is also provided.

Solution

• The target molecule contains, according to DBE (3) and carbon chemical shifts, a double bond (ketone, result of subspectrum search for δ_C 219.6) and two rings. Also, three methyl groups, attached to the two residual quarternary carbons, are present (δ_H 0.95/0.82 to δ_C 46.8 and δ_H 0.91 to δ_C57.7).
• Subspectrum search for the 13C signal at δ_C 43.1 derives the option of a bridge head carbon. Thus, visible H,C l.r. correlations for this atom are mainly for ³J_CH connectivities. In combination with the methyl fragments mentioned above, this allows to distinguish fragments C10-C2, C4 and C5 as the next neighbours of C1. For the assignment of C6/C7, the COSY spectrum elucidates a chain of coupled spins H6-H7-H5-H4.
• No solution? Enter the chemical shifts of the 13C signals in nmrsshiftdb2 as a "spectrum search" (option "complete") or click here.

Strategy

• The molecular formula of this month's molecule is C₇H₅BrO₂: As a first step, determine the number of DBE.
• Assign H,C connectivities from the HMQC (Fig. 2.1). There is one 1H signal without cross peak. Identify the different types of protons in the target compound by comparison with the 1H NMR spectrum in Fig. 2.2 (upper trace). Note: The 1D projection of the 13C NMR spectrum is multiplicity-edited!
• Do a "search by spectrum" in nmrshiftdb2 (select option "subspectrum"), and enter the carbon shift corresponding to the 1H signal at 9.9 ppm (enter "195.5 D" for a shift at 195.5 ppm having one proton attached to it). This provides you with a valuable hint.
• Finally, evaluate the through-space information yielded from the 1D NOE spectrum (Fig. 2.2, lower trace).

Hints

• There is only one coupled spin system visible in the 1H NMR spectrum. Also, from the number of DBE, an aromatic compound should be considered.
• The information yielded from nmrshiftdb2 search derives a characteristic functional group which is present in the molecule. Its neighbour protons are visible in the 1D NOE experiment.
• Summing up these informations, an aromatic molecule is identified. The substitution pattern can be deduced from NOE, 1H coupling pattern and HMBC (Fig. 2.2, Fig. 2.3)

Strategy

• This month's molecule has the formula C₉H₆O₂. Check first the number of DBE.
• After assigning H,C connectivities (Fig. 1.3,1.4), try to identify existing spin systems from the ₁H projection of the HSQC. Relative intensities are indicated below the ₁H trace.
• Do a "search by spectrum" in nmrshiftdb2 (select option "subspectrum"), and enter the two leftmost carbon shifts with their corresponding multiplicity (e.g. enter "123.4 T" for a shift at 123.4 ppm with a triplet splitting ).

Hints

• There are two spin systems visible in the ₁H NMR spectrum.
• From your search for the two signals at δ_C 160.8 and 154.0, you should obtain an idea, which type of compound we are looking for.
• Additional information can be gained from the selective 1D NOE experiment provided in image 4.