Háskóli Íslands/
University of Iceland
Raunvísindadeild
/Science faculty,
Eðlisefnafræði B / Physical chemistry B / EFN403G
Dæma og verkefnaskammtur III /problems and projects III, 13.04.10
(updated 16.04.10)
Tímadæmi:
NB: Students (see names in brackets) will solve the problems/projects during the lecture / problem
class (13.04.10) and hand in solutions in a computerised form for displaying on the web, no later
than one week after the problem class. ATH: Tilgreindir nemendur leysa viðkomandi dæmi í
tímanum (13.04.10) og skila úrvinnslum á tölvutæku formi til birtingar innan viku eftir það .
Tímadæmi A (J. Long): IR spectrum of H35Cl:
a) Make us e of relevant database to determine the rotational constants Bv, v = 0,1 (cm-1) for
H35Cl.
b) Determine the average internuclear distances (rv / nm)) for the two vibrational states v = 0 and 1
for H35Cl from the rotational constants in (a) and discuss the difference.
c) Show that the wavenumbers for rotational peaks in the IR spectrum of H35Cl for the P and R
lines respectively (~P and ~R ) can be expressed as:
~P ~0  (2 Be  2 e ) J "   e J "2
~R ~0  (2 Be  3 e )  (2 Be  4 e ) J "   e J "2
where ~0   e  2 e xe
d) Use relevant spectroscopic constants and a simulation program to calculate and plot the IR
spectrum for H35Cl at 25oC (present the spectrum).
Tímadæmi B (Niels) NMR:
a) At -89oC two 1H-NMR peaks are observed for partly deuterated cyclohexan (C6D11H) (see
Fig. 1 below). The spacing between the peaks is found to be 28.8 Hz when recorded by a
60 MHz NMR equipment. i) Explain the existance of the peaks. ii) What is the spacing
between the peaks in ppm? iii) What would be the spacing between the peaks in Hz if the
spectrum was recorded by a 400 Mz NMR equipment?
Fig. 1
b) Show (find) the position (in Hz) of all 1H-NMR peaks due to HA and HB in ethyl 3-(3indolyl)botanoat) (see Fig.2 below). Make a sketch of the spectrum. JAB = 14.8 Hz; JAC =
5.8 Hz; JBC = 8.8 Hz; A = 42.09 Hz; B = 28.03 Hz.
HC
HB
HA
N
C
O
CH2CH3
O
Fig. 2
Skiladæmi
ATH: Skilið eigi síðar en í verkefnatímanum, þriðjudaginn 13. april, 2010 /
Problems/projects to be handed in no later than 13.04.10.
1) Rotational spectroscopy
Rotational energy levels for the organic molecule acetylene (C2H2), derived from
rotational spectroscopic data, are listed in the table below. Assuming that the rotational
energy levels can be expressed by the 2nd order approximation equation
E(J) = B J (J +1) – D J 2(J +1)2
(1)
derive the rotational constants B and D (cm-1) by fitting expression (1) to the energy levels vs J
(see for example HERE, HERE and HERE)
E(J)/cm-1
J
0
1
2
3
4
5
6
7
8
9
10
0
2.35
7.06
14.12
23.53
35.30
49.41
65.88
84.71
105.88
129.41
(answer: B = 1.1766 cm-1) (Úrlausn/solution)
2) Styrkákvarðanir skv. gleypnimælingum / concentrations determined by absorption.
Virkni lífhvatans chymotrypsin (CT) er hindruð af rauðu litarefni (RL). Lausn sem inniheldur
1,00x10-4M heildarstyrk af lífhvatanum og 2,00x10-3 M heildarstyrk af litarefninu mælist með
gleypnina A = 1,4 við 280 nm.
CT + RL = CT-RL (komplex);
(1)
Mólar gleypni efna við 280 nm ( / M-1cm-1) er: (CT) = 12000; (RL) = 0; (CT-RL) = 15000.
a) Hver er styrkur komplexins (CT-RL) í lausninni.
b) Reiknið út jafnvægisfasta (K) hvarfsins (1) (svar: K = 1.03x103).
English:
The activity of the enzyme chymotrypsin (CT) is inhibited by a red dye (RL). An absorption of A
= 1,4 is measured for a solution which contains 1,00x10-4M total concentration of the enzyme
and 2,00x10-3 M total concentration of the dye at 280 nm.
CT + RL = CT-RL (complex);
(1)
Molar absorption coefficients at 280 nm ( / M-1cm-1) are found to be (CT) = 12000; (RL) = 0;
(CT-RL) = 15000
a) What is the concentration of the complex(CT-RL) in the solution?
b) Determine the equilibrium constant (K) for the reaction (1) (answer: K = 1.03x103).
(Úrlausn/solution)
3) Microwave spectrum for NO and spectral analysis:
* 26.75 cm-1
Microwave spectrum
Molecule = NO(g)
t = 25oC
Bandwidth = 1.5 cm-1
0
20
40
60
80
100
120
-1
 [cm
]
Enclosed (above) is a microwave spectrum of NO(g) derived at room temperature. The
wavenumber value for the * -marked peak is indicated.
a) What transition does the * -marked peak in the spectrum correspond to?
b) Derive the rotational constant (B / cm-1) for the molecule from the data.
(answer: 1.672 cm-1)
c) What is the average spacing (cm-1) between the spectral peaks?
d) What is the average internuclear distance (nm) between the atom nuclei?
e) What is the rotational energy (J mol-1) of the molecules which absorb at 26.75 cm-1 (*) ?
f) What is the rotational frequency (Hz) of the molecules in (e)? (Úrlausn/solution)
2) 4) IR spectroscopy
Enclosed are IR spectra of CO (Fig 1: low resolution, Fig 2: high resolution).
a) Based on the numerical values given in the text of figure 1, derive the vibrational
constants, e and exe for CO and compare with values given in a database. (answer:
calculated e = 2169 cm-1)
b) determine the zero energy for the molecule by using the e and exe constants from (a).
c) What transition do the * -marked peaks in the spectrum (Fig. 2) correspond to?
d) Determine the wavenumber values of the peaks in (c) (the *-marked peaks) by using an
appropriate program to calculate the IR spectrum for CO based on relevant spectroscopic
constants given in a database. (answer: left *: 2139.37 cm-1) (Úrlausn/solution)
Fig. 1
Fig. 2
5) Luminescence lifetime:
Luminescence from an organic molecule was measured as a function of time following excitation
(absorption) by a light pulse.
t /ns
0
Relative
luminescence 100
intensity
2
5
10
20
30
82
61
37
14
5
a-b) Determine the luminescence lifetime () and half-life (t1/2)
c) Argue whether the luminescence is fluorescence or phosphorescence? (Úrlausn/solution)
6) Ozon shielding
Enclosed (see below) are absorption spectra of ozon in the near-UV and Visible regions plotted as
cross sections (/cm2) as a function of wavelength. Cross sections () are closely related to molar
absorptivity (molar extinction coefficients; see p: 711 in RC) () and defined by the relationship:
I/I0 = exp(- *l*c) = 10-A
where I0 and I are the intensities before and after passing through a sample,  is the cross section
in cm2, l is the length of the absorbing sample in cm and c is the concentration of the sample in
molecules per cm3. A is the absorbance. In reality ozon is to be found everywhere in the
atmosphere to some extend, but its maximum concentration is close to 20 km above sea level (see
for example HERE and HERE). Therefore, it is customary to define the ozon layer close to that
altitude. The ozon concentration, in the atmosphere, on the other hand, is so little that if it was all
collected together around the earth at 1 bar pressure and room temperature, it would only form
about 3 mm thick gas layer! Judging from the near-UV absorption spectrum of ozon, the cross
section is about 1.15 x 10-17 cm2 for 255 nm but about 0.04x10-17 cm2 for 210 nm.
Determine a) the absorbance (A) and b) % transmission for i) 255 nm, ii) 210 nm and iii) 600 nm
sun radiations by assuming that he ozon layer equals 3 mm thick gas layer of ozon at 25oC and 1
bar pressure.
(svar: a, ii) : A = 1.27) (Úrlausn/solution)
7) Greenhouse effects
Enclosed is a figure showing total absorption of the solar radiation in the upper atmosphere (top)
as well as absorption spectra of (O2 + O3), CO2 and H2O (below). The two major absorption peaks
of CO2 in the infrared region are marked and indicated. Determine its wavenumber values (cm-1)
and search for relevant data (for example in a relevant database) to determine what energy
transitions these absorption peaks correspond to: specify the type of transitions involved as well
as relevant quantum numbers and quantum number changes.
(Úrlausn/solution)
8) NMR
a) What is the difference in spin frequency (Hz) of a 1H nucleus attached to an aromatic ring ( =
7) and a 1H nucleus in a TMS (tetra methyl silan) ( = 0) in a fixed magnetic field of a 60 MHz
NMR equipment?
b) What is the difference in magnetic fields in a 60 MHz CW NMR equipment needed to cause
absorption of the nuclei mentioned in (a)? (answer: 1.2 10-5 T)
(Úrlausn/solution)
9) NMR
Kjarnsýrusameind getur myndað tvenns konar spíral lögun (helical structure), A- og B- form.
Mismunurinn felst meðal annars í mismunandi lögun sykursameinda hluta keðjanna. Ribosa
sameindahluti RNA keðju, þar sem skipt hefur verið á H og D frumeindum að hluta, er sýndur hér
fyrir neðan:
Hliðrunargildi Hi´ róteinda beggja formana eru þau sömu, en viðkomandi kúplingsfastar eru hins
vegar frábrugðnir. Í meðfylgjandi töflu eru gefin upp gildi fyrir A formið:
Kjarnar
H1´
H2´
H3´
H4´
Hliðranir
(ppm)
6.0
5.0
4.5
4.0
kúplingsfastar (A-form)
Jij
Hz
J1´2´
0
J2´3´
5
J3´4´
9
Rissið upp 1H-NMR róf A-forms RNA og auðkennið klofnanir með því að tilgreina Jij. Tiltakið
hvort NMR merki einstakra róteinda séu “singlettar”, “doublettar” eða “triplettar” o.s.frv.
English:
Nucleic acid molecule can form two types of helical structures, A and B. This is partly due to
different structures of the suger units. Below you see the ribose part of a RNA chain where some
H atoms (1H) have been replaced by D atoms (2H).
The chemical shifts for the Hi´ atoms are independent of the helical forms, whereas the coupling
constants differ. In the table below you will ses coupling constants for the helical structures A:
Nuclei
H1´
H2´
H3´
H4´
Chemical shifts
(ppm)
6.0
5.0
4.5
4.0
coupling constants (A-form)
Jij
Hz
J1´2´
0
J2´3´
5
J3´4´
9
Make a drawing for a 1H-NMR spectrum of RNA ( the A form) and characterize couplings by
specifying Jij. Tell whether the signals are singlets, doublets, triplets etc. (Úrlausn/solution)
10) Röntgengreining kristalla / X-ray diffraction:
a) The layers of atoms in a crystal are separated by 325 pm. At what angle in a diffractometer will
diffraction occur using:
i)
– molybdenum K X rays ( = 70.8 pm)
ii)
– copper K X rays ( = 154 pm) (answer:  = 13.7o)
b) The smallest observed diffraction angle of silver taken with copper K X rays ( = 154 pm) is
19.076o. This angle is associated with the (111) plane in the cubic close-packed structure of silver.
i)
ii)
Determine the value of the unit cell length (answer: 408.6 pm)
If density,  = 10.500 g cm-3, calculate the number of atoms in the unit cell and assign
the cubic cell (sc, bcc or fcc). (Úrlausn/solution)