Question 1

B  = o0  = (o0 2 ) =   470.516 MHz   = 11.74 T

0          Y      (Y 2 )    40.078 MHz T 1

B  = o0  = (o0 2 ) =    94.183 MHz    = 2.35 T

0          Y      (Y 2 )    40.078 MHz T 1

Unseen example

(1) Spectrum 1b has lower signal-to-noise ratio (SNR) than spectrum 1a      because NMR sensitivity is lower in weaker magnetic fields due to the         lower Larmor frequency and lower polarisation levels. (2) Peaks are            broader and the peaks in the multiplets are further apart in spectrum 1b        compared to 1a because the resolution of the NMR spectrum decreases with magnetic field. 1 ppm in spectrum 1 is 470.516 Hz, while 1 ppm in              spectrum 1b is only 94.183 Hz. (3) Roofing is observed in spectrum 1b due to strong coupling. Strong coupling is observed at lower magnetic fields      because the difference in chemical shift in Hz is reduced while the J            coupling values stay the same and so J becomes more significant relative to the difference in chemical shift in Hz.

[any combination of two of these three reasons is acceptable.]

Application of knowledge to unseen example

Peaks 3, 4 and 5 form a triplet and corresponds to the single fluorine F4 . F4    couples to n = 2 chemically equivalent neighbours (F3 and F5, which are in  the same chemical environment). According to the 2nI+1 coupling rule, this leads to a triplet (n = 2, I = ½) with amplitudes 1:2:1.  In this case F4 is the   active spin (the one that flips) and F3 and F5 are the passive spins. The         peaks in the triplet correspond to the different spin–up and spin-down states of the passive spins F3 and F5 . Both spin up gives m = +1 (up shifted peak), both spins down gives m = 1 (down shifted peak), or one up and one down  gives m = 0 (central peak). The central peak is twice the height of the outer  peaks because there are two ways to combine F3 and F5 to get a total spin of zero: (up,down) or (down,up).

Application of knowledge to unseen example

F3/F5 and F4 in spectrum (a) are weakly coupled while in (b) they are         strongly coupled. This can be seen by the more pronounced roofing effect for the doublet in spectrum (b). In (a) the triplet displays the classic 1:2:1  intensity pattern because the (spin-up,spin-down) and (spin-down,spin-up) states of the two passive spins (F3/F5) are degenerate. However in (b) the   central peak of the resonance is split because these states are no longer      degenerate.

To quantitatively justify the strong coupling in (b), the coupling constant can be calculated from Table 1 based on the doublet splitting:

6 =  152.38 ppm  (152.34 ppm) = 0.04 ppm

J = 6 vref   = 0.04 ppm  470.516 MHz = 18.8 Hz          The difference in chemical shift between F3/F5 and F4 is 2.1 ppm. In (a) this corresponds to 2.1 ppm x 470.516 MHz = 988 Hz.       In (b) this corresponds to 2.1 ppm x 94.183 MHz = 198 Hz.

Therefore in (a) J is 18.8 Hz/ 988 Hz = 1.9% of the difference in chemical shift, while in (b) J is 18.8 Hz/199 Hz = 9.4%. Therefore F5/F3 and F4 are strongly coupled in (b) but weakly coupled in (a).

Problem solving

F3 and F5 are chemically equivalent because they are in the same chemical environment and so have the same chemical shift. This is not affected by  decoupling. In the spectra in Figure 1, F3 and F5 are also magnetically       equivalent because they both couple to F4 with the same coupling constant due to the symmetry of the molecule. However, in the absence of 1H         decoupling, F3 and F5 would be magnetically inequivalent because they     would coupling differently to the 1H nuclei.

Unseen example

●    Transverse (T2) relaxation. Broad poorly resolved peaks come from the rapid dephasing of the nuclei during signal acquisition, which is caused by transverse relaxation. In this case the dephasing is caused by local field inhomogeneity around the bubble because of the         magnetic field susceptibility mis-match between the air and the       solvent. This will cause rapid dephasing. Some students might         correctly identify this as T2 * relaxation (effective T2 relaxation due to reversible dephasing) although this distinction is not required for full marks.

Application of knowledge to unseen example

●    T1 relaxation. T1 drives the return to equilibrium magnetization        along z following an RF pulse due to re-establishing the equilibrium population difference across the nuclear spin states. If a 30° RF        pulse is used, less magnetisation is rotated into the transverse plane  and so it takes less time to return to thermal equilibrium following   the pulse compared to the 90° pulse case where all magnetization is rotated into the transverse plane. After 1 spectrum the 90° pulse will give more signal (SNR) because it generates more transverse            magnetization. However, here the experiment is repeated 128 times and averaged. If the repetition time does not allow for full  T1                 relaxation between pulses, more signal (higher SNR) will be            observed in the 30° RF pulse case because more magnetization will be recovered between each experiment.

Application of knowledge to unseen example

Question 2

(a)         The longest wavelength absorption band arises from the lowest                                              (4)

energy transition that gives an observable band. The very high

absorption coefficient and oscillator strength of ca. 1 indicate that

the band arises from essentially a fully allowed transition

according to both spin and orbital selection rules. From an S0

singlet ground state, the lowest energy spin-allowed transition

promotes an electron from the HOMO to the LUMO and populates the S1 singlet        excited state of 1. A fully allowed transition according to the orbital selection rule      indicates that there is good spatial overlap of the HOMO and LUMO orbitals in 1 (the structure is asymmetric, so all transitions are symmetry allowed).

Application of principles covered in the course.

(b)         The shoulders at 465 or 440 nm arise from partially resolved vibrational fine structure         (3)

and excitation at these wavelengths populates higher vibrational levels of the S1 state.   Molecules in higher vibrational levels undergo vibrational relaxation to the v = 0 level  very rapidly in solution, via non-radiative collisional energy transfer to the solvent on a time scale that out-competes that for fluorescence. Hence, fluorescence is observed       essentially only from the v = 0 level regardless of which vibrational level is populated  initially on absorption.

Application of principles covered in the course.

(c)         The absorption band may be assigned to the S0 to S1 transition and the emission band to      (3)

the S1 to S0 transition of 2 by comparison with the relevant data given for 1; the         profiles, εmax, Φem and τobs values are all comparable and consistent with transitions to and from the lowest excited singlet state for a dye of similar structure.

The slightly longer wavelengths may be attributed to the different substituents on the  rings in 2 resulting in slightly different HOMO and LUMO orbitals that are closer in   energy and give a slightly smaller energy gap between the S0 and S1 states in 2 than in 1.

Application of principles covered in the course.

(i)          In the absence of 2, k = 1/τobs = 1/(4. 16 × 10-9 s) = 2.40 × 108 s- 1        In the presence of 2, kobs = 1/τobs = 1/(0.60 × 10-9 s) = 1.67 × 109 s- 1

kobs = k + kq [2], hence kq = (kobs - k)/[2]

= (1.67 × 109 s- 1 - 2.40 × 108 s- 1)/(4.00 × 10-3 mol dm-3)

= 3.58 × 1011 dm3 mol- 1 s- 1

Calculation; application ofprinciples covered in the course.

(ii)         Long-range Coulombic energy transfer is a non-radiative process by which a

donor in an excited state returns to the ground state by transferring its              excitation energy to an acceptor. It is described as long range because it can    operate over distances that are large in comparison with molecular size, and so contact between donor and acceptor is not required.

In this case, the fluorescence from 1 at

510 nm becomes weak when 2 is added

at this concentration because most of the

excited states of 1 will decay by the non-

radiative quenching mechanism, as

shown by the lifetimes. Efficient energy

transfer is consistent with a close match

in the energy gaps for 1 and 2, as shown

by the strong overlap of their respective emission and absorption spectra.

The very strong emission band at 550 nm can be attributed to fluorescence  from 2, as shown in Figure 2, because S1 excited states of 2 are produced in high yield by the efficient energy transfer process. (And the new emission  band is stronger than that which would be observed in the absence of 1       because 2 absorbs relatively weakly at this excitation wavelength.)

The calculated value of kq is above the diffusion-controlled limit, i.e. where  quenching occurs more rapidly than the time for 1 and 2 to diffuse together,  and is consistent with a long-range mechanism where contact is not required.

Application of principles covered in the course.

    The absorption of 3 does not overlap very strongly with the emission of 1,

which means that the energy gaps for 1 and 3 are not very well matched so         long-range Coulombic energy transfer quenching is likely to be weak. (Also, 3  absorbs only very weakly at 495 nm, so any emission from direct excitation will be very weak.) Hence, a strong emission band at 510 nm from 1 and a weak       emission band from 3 might be expected.

    The absorption band of 3 overlaps strongly with the emission band of 2, which

means that the energy gaps for 2 and 3 are well matched so long-range              Coulombic energy transfer quenching may be efficient from 2 to 3, occurring    sequentially after the energy transfer from 1 to 2. Hence, a weak emission band at 510 nm from 1 and moderately strong emission bands at 550 nm from 2 and  590 nm from 3 might be expected, with the emission band from 2 being weaker than it was from the sample of 1 and 2 in the absence of 3.

The concept of sequential energy transfer is unseen; the earlier parts of the question

are intended to help. This is a relatively open-ended question and credit is available for other plausible answers with appropriatejustification.

Question 3

Assignment of the methyl signals are shown below. There is only one methyl group that will not couple to anything and so this is assigned to the singlet at 1.79 ppm. The methyl group that is next to the CH2 group will appear as a     double doublet due to 3J coupling (7.2 Hz) to the two adjacent diastereotopic protons – this is assigned to the 0.79 ppm signal. The methyl group on the     alkene shows 4J coupling to the alkene proton (2.3 Hz) and so is a doublet –  this is assigned to the signal at 1.71 ppm.

Problem solving.

HA is an alkene proton and so the chemical shift should be around 5-6 ppm.   HA will be the signal at 5.48 ppm. The 4J coupling to the methyl protons        (quartet, 2.3 Hz) was previously identified from assigning the methyl group   above. As it is adjacent to two other protons (with 3J couplings), a ddq would be expected. However, only one of the 3J couplings (6.0 Hz) is observed        because the alkene proton and one of these two protons must have a dihedral angle close to 90 degrees (Karplus relationship). This is the reason that its     multiplicity is simpler than might have been expected.

Problem solving.

(iii)        HB is the signal at 2.61 (ddd, J = 12.7, 10.7, 3.8 Hz). This is a rather unusually       (3)

high chemical shift value and so J values are key to the assignments. It            contains two large Jvalues due to the two 3-bond axial-axial couplings and     one medium value due to a 3-bond axial-equatorial coupling. There are two     ways of reaching this assignment: (i) start with HA at 5.48 ppm, match the 6.0 Hz with the 2.01 ppm signal – this will be the axial proton of the adjacent CH2 group – 2.01 ppm signal will have a 2J = 16.7 Hz to the other proton on the     same carbon and an axial-axial coupling to HB which is the 10.7 Hz coupling; (ii) start with the CH proton adjacent to the OH group – this is also allylic and so would be expected at 3-4.5 ppm region. This proton will be the signal at      4.44 ppm which is a doublet due to axial-axial 3-bond coupling to the adjacent axial proton (3J = 8.6 Hz). This adjacent axial proton is the signal at 2.48 ppm which also has an axial-axial 3-bond coupling to HB of 12.7 Hz, identifying     HB .

Problem solving.

(iv)        For protons HC and HD, we are expecting double doublets, which will share a         (3)

2J value of 10- 18 Hz.  The axial proton will also have a 3J axial-equatorial coupling of 2-6 Hz and the equatorial proton will also have 3J equatorial-   equatorial coupling of 2-6 Hz. This means that these two protons cannot be assigned to a specific proton from the information provided. The only two  protons that fit the above Jvalues are those at 2.88 and 2.37 ppm.

[Not needed: To confirm these assignments, it is possible to match the J       values around from the already assigned methyl group at 0.79 ppm. The        diastereotopic pair of protons next to the methyl group are the signals at 1.15 and 0.98 ppm (both ddq). Matching the 11.6 and 9.8 Hz 3J values with the    signal at 3.07 ppm allows assignment of the equatorial proton on the 6-ring.  This has two other 3J values (4.8 and 3.1 Hz) to protons HC and HD .]

The best approach to assigning the signals at 2.88 and 2.37 ppm to specific    protons would be to carry out a NOE experiment. For example, irradiation of the methyl group signal at 1.79 ppm would give a NOE to the axial proton on the same side of the ring. NOESY can also be suggested.

Problem solving.

(v)         In a DEPT- 135 NMR experiment, quaternary carbons are absent, CH/CH3                     (4) groups point up and CH2 groups point down.

5 CH signals: 1 will be in 6C  150- 100 region (alkenes/aromatic), 1 will be in    the 6C  150- 100 region and 3 will be in 6C 50-0 region; 3 CH3 signals: all 3 will be in 6C 50-0 region; 3 CH2 signals: all will be in 6C 50-0 region

Problem solving.

(b)         This signal clearly appears as a quartet but it should not be called a quartet as there is           (4)

no proton in this molecule that couples to three identical other protons. However, the    same type of signal could be produced due to coupling to three different protons that all have the same J value by chance – this is what is occurring here. So, this signal is in      fact a double double doublet (ddd). Given that its chemical shift is 3.64 ppm, it can be   assigned to either of the two protons indicated by the arrows in the structure shown       below. These two protons would be expected to come in this range due to the adjacent  electron withdrawing C=O or electron deficient heteroaromatic group – this is why we  cannot assign it unequivocally based on the information provided.

The three identical J values can be calculated by taking the difference in ppm of any of the two adjacent lines and then multiplying by 400 (the frequency of the spectrometer   in MHz). Depending on which two lines are chosen, the J value could be 9.0, 8.5 or 8.0 Hz. The average of these values is 8.5 Hz, therefore it would be ddd, J = 8.5, 8.5, 8.5    Hz.

Problem solving.

Question 4

Part (i) comesfrom the lecture notes, but the remaining sections all require the application of knowledgefrom the course and are unseen.

(a)          Propan-2ol acts as a molecular source of H2, forming acetone as it is released                      (2)

according to the equation shown below.

(c)         Oxidative addition of 1-propanol forms the corresponding alkoxide hydride and fails to       (2)

undergo facile beta hydride transfer and will therefore poisons the catalyst. If beta         hydride transfer were to occur, an aldehyde would form which could then eliminate CO and this would also deactivate the system.

(d)         The mixture retains some, lower activity. Hence some of the surface is blocked by the         (4)

alkoxide which prevents H2 formation or acetophenone binding.  This is consistent      with the deactivation step being reversible and suggests that either some of the alcohol is physisorbed or that the chemisorption process leading to OH bond cleavage is          reversible. Or that CO binding is reversible. This allows the catalyst to be reactivated.

 is the fractional surface coverage and given by

9 =

Hence  = 50/Vm at 15 mbar and  = 62.5/Vm at 20 mbar.

Thus 50/Vm = 15b/(1+15b) or 62.5/Vm = 20b(1+20b)

Rearranging gives

1/b = (15Vm - 750)/50 and 1/b = (20Vm - 1250)/62.5

Solving for 1/b (60) we find that Vm the monolayer capacity of the surface is 250 cm3 .

In order to answer this it is necessary to recognise that the values of b for Kr and methanol must be the same. Thus if a monolayer of methanol were to bind, the   volume of gas adsorbed at 10 mbar pressure would be 35.7 cm3 . This comes       from = (250 x 10 x 0.0166667)/ (1 + 10 x 0.016667). The actual volume is 43     cm3 or ~20% higher than this. Hence, the surface occupancy is higher than a      monolayer and the Langmuir Isotherm is invalid.  If students take an approach   that demonstrates that the b values for the two materials must be different, credit will be given.