nmrshiftdb2 - Collection of "Problem of the Month" items

All material is copyrighted by the authors

Table of Contents

February 2017
November 2016
May 2016
January 2016
September 2015
May 2015
February 2015
September 2013
May 2013
Novmber 2012
Septemberr 2012

Problem of the Month - February 2017 (as pdf)


Fig. 11.1: 1H-NMR in CDCl3 at 600 MHz (click to enlarge).
  • 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 δH 0.91 ppm?
  • The HMBC shows 1H,13C coupling over two to four bonds.
  • The signals δH 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?
Still no idea?
  • Hint: Use the HSQC to deduce the coupling patterns of the overlapping methyl signals at δH 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!
(jl)

Problem of the Month - November 2016 (as pdf)


Fig. 10.1: H,C HMBC (blue) and H,C HMQC (red) at 300/75 MHz (click to enlarge).

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 1JCH are visible in red, while HMBC peaks for long range correlations (2JCH, 3JCH and some dublets for 1JCH). In aromatic systems, the intensity of 2J cross peaks is weaker than the one of 3JCH 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.

Fig. 10.2: APT 13C{1H} NMR spectrum recorded at 75 MHz (click to enlarge).

Fig. 10.3: 1H NMR spectrum recorded at 300 MHz (click to enlarge).

Solution

  • There are only non-equivalent carbons in the target moleculeā€˜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 3JCH 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.
(nes)

Problem of the Month - May 2016 (as pdf)


Fig. 9.1: H,H NOESY, 600 MHz, List of 1H chemical shifts: 7.79, 3.27, 3.12, 1.49, 1.38, 1.33 (click to enlarge).

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 (> 3JHH) 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. 1JCH correlations from the HMQC experiment are labelled in red.

Fig. 9.2: H,H COSY, 600 MHz (click to enlarge).

Fig. 9.3: APT 13C{1H} NMR spectrum recorded at 150 MHz (click to enlarge).

Fig. 9.4: H,C HMBC (black) and HMQC spectra (red) at 600/150 MHz (click to enlarge).

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").
(nes)

Problem of the Month - January 2016 (as pdf)


Fig. 8.1: 1H/13C HSQC (600/151 MHz, CDCL3 (click to enlarge).

Strategy

  • Given the molecular formula C9H14O7, you can determine the number of double bond equivalents.
  • Determine the CHx 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.

Fig. 8.2: 1H NMR (600 MHz, CDCL3 (click to enlarge).

Fig. 8.3: 13C NMR (151 MHz, CDCL3 (click to enlarge).

Fig. 8.4: 1H/13C HMBC (600/151 MHz, CDCL3 (click to enlarge).

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 (δH 3.69) and A/B(δH 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 (2JHH = 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").
(jl)

Problem of the Month - September 2015 (as pdf)


Fig. 7.1: H,C HMBC spectrum, signal intensities 1H in blue (600 MHz, CDCl3). 1H chemical shifts: 4.26 (q), 2.92 (s), 1.33 (t), 13C chemical shifts: 152.7, 74.7, 74.5, 62.4, 13.8 (click to enlarge).

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 1JCH.

Fig. 7.2: H,C HMBC spectrum (expansion) (click to enlarge).

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 (δH 4.26, 1.33). The remaining signal (δH 2.92) stems from the alkyne; extraordinarily large JCH 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.
(nes)

Problem of the Month - May 2015 (as pdf)


Fig. 6.1: H,H COSY spectrum, signal intensities in blue (500 MHz, DMSO-d6(click to enlarge).

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 (1JHN) and a selective NOE experiment available in Fig. 6.3, which deliver information from correlations through bond and through space, rsp.

Figure 6.2: H,C HMBC with expansion, optimized for JHC of 10 Hz, 500/125 MHz (click to enlarge).

Figure 6.3: Selective 1H NOE (upper trace), and 1H NMR (below, 500 MHz) (click to enlarge).

Solution

  • The target molecule contains, according to DBE (1) and carbon chemical shift, a double bond (ester or amide, result of subspectrum search for δC 173.4). The signal at δH 7.75 correlates to the corresponding amide nitrogen.
  • Also, three more exchanging proton signals can attributed to hydroxy groups (δH 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.
(nes)

Problem of the Month - February 2015 (as pdf)


Fig. 5.1: 1H NMR spectra, 400 MHz (click to enlarge).

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 δH 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.

Figure 5.2: H,F HOESY, 400/376 MHz (click to enlarge).

Figure 5.3: 19F NMR spectra, 376 MHz (click to enlarge).

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 δH 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.
(nes)

Problem of the Month - September 2013 (as pdf)


Fig. 4.1: Edited H,C HMQC, optimized for JCH of 145 Hz, 400/100 MHz. List of 13C chemical shifts: 180.9, 140.8, 136.7, 129.4, 127.3, 45.1 (T), 45.0 (D), 30.1, 22.4, 18.1 (click to enlarge).

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 δC 45.1, 30.1 and 22.4 with their corresponding multiplicity (e.g. "45.1T" for a carbon with a chemical shift of δC 45.1 with two protons attached). Since there are two carbons at δC 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).
Figure 4.2: 1H NMR spectrum, recorded at 400 MHz (CDCl3). List of 1H chemical shifts and multiplets (brackets): 10.58 (s, broad), 7.26 (,d'), 7.13 (,d'), 3.74 (q), 2.48 (d), 1.88 (d hept), 1.54 (d), 0.93 (d) (click to enlarge).Figure 4.3: H,H COSY, 400 MHz (click to enlarge).

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 δC 180.9 in combination with the broad proton signal at δH 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.
Fig. 4.4: H,C HMBC, optimized for JHC of 10 Hz, 400/100 MHz (click to enlarge).
(rs)

Problem of the Month - May 2013 (as pdf)


Fig. 3.1: 13C{1H} NMR spectrum, 100 MHz. List of 13C chemical shifts: 219.6, 57.7, 46.8, 43.3, 43.1, 29.9, 27.1, 19.8, 19.1, 9.2 (click to enlarge).

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 δH 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 δC 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 δC 219.6, etc.). To back up the observations, a COSY spectrum is also provided.
Figure 3.2: Edited H,C HMQC, optimized for JHC of 145 Hz, 400/100 MHz. 1H chemical shifts: 2.36, 2.10, 1.96, 1.85, 1.69, 1.40, 1.35, 0.95, 0.91, 0.82 (click to enlarge).Figure 3.3: H,C HMBC (section!), optimized for JHC of 10 Hz, 400/100 MHz (click to enlarge).

Fig. 3.4: H,H COSY (section!), 400 MHz (click to enlarge).

Solution

  • The target molecule contains, according to DBE (3) and carbon chemical shifts, a double bond (ketone, result of subspectrum search for δC 219.6) and two rings. Also, three methyl groups, attached to the two residual quarternary carbons, are present (δH 0.95/0.82 to δC 46.8 and δH 0.91 to δC57.7).
  • Subspectrum search for the 13C signal at δC 43.1 derives the option of a bridge head carbon. Thus, visible H,C l.r. correlations for this atom are mainly for 3JCH 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.
(rs)

Problem of the Month - November 2012 (as pdf)


Fig. 2.1: H,C HMQC, optimized for JCH of 145 Hz, 500/125 MHz (click to enlarge).

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)
    Figure 2.2: 1H NMR spectrum, recorded at 500 MHz (CDCl3 ) (top), 1D 1H NOE spectrum (bottom) (click to enlarge).Figure 2.3: H,C HMBC, optimized for JHC of 10 Hz, 500/125 MHz (click to enlarge).

    Solution

    (nes)

    Problem of the Month - September 2012 (as pdf)


    Fig. 1.1: DEPT135 (top) and 13C{1H} NMR spectra, 125 MHz (click to enlarge).

    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 δC 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.

      Fig. 1.2 (left): H,C HSQC, optimized for JHC of 145 Hz, 500/125 MHz, Fig. 1.3 (right): H,C HMBC, optimized for JHC of 10 Hz, 500/125 MHz (click to enlarge).

      Figure 1.4: 1D 1H NOE spectrum, irradiation frequency indicated (arrow), 500 MHz (click to enlarge).

      Solution

      (nes)