MOLECULAR ORBITAL THEORY AND UV SPECTROSCOPY
Molecular orbital (MO) theory describes covalent bond formation as a combination of atomic orbitals
to form molecular orbitals. MO’s represent regions of space in a molecule where electrons spend >
90% of their time, traveling in wave-like motion.
In the H2 molecule 2 singly occupied 1s atomic orbitals combine to produce 2 molecular orbitals...
1. A bonding MO, i.e., an additive combination - by constructive interference of in-phase
waves.
1. An antibonding MO, i.e., a subtractive combination - by destructive interference of outof-phase-waves.
node
H
H
E
H-H Antibonding MO
2*
(empty)
1s
H
H
1s
H-H bonding MO
(filled)
H
H
1
E=h
  * transition
ground state
excited state
The bonding MO (denoted 1) is lower in energy because most of the electron density is between the
H nuclei, thus holding the atoms together. In the ground state, both 1s electrons reside in 1 .
The antibonding MO (denoted 2* ) is higher in energy because most of the electron density is outside
the region between the nuclei. The diagram shows a region of zero electron density (a ‘node’)
between the two nuclei. The positively charged nuclei thus repel and bonding is unfavorable.
In the ground state, 2* is empty, however, if EM radiation of the same energy as the energy gap
between 1 and 2* strikes the molecule, one or both of the bonding electrons may be excited
(promoted) to the 2* orbital. Once excited, the molecule may split, ionize or reradiate the energy and
return to the ground state.
According to MO theory, the following are true...
1. The number of MO's must equal the number of atomic orbitals which combined to produce them.
 # e's in bonding MO's - # e's in antibonding MO's



2
2. Bond order of a molecule = 
3. Bond order > 0 produces a stable molecule. Bond order = 0 means no bonds formed between
atoms. Bond order < 0 means that the molecule is unstable (decomposes).
1
Molecular Orbital Description of Alkenes:
Ethylene (C2H4)

The C-C sigma () bond in ethylene results from the overlap of two 2sp2 atomic orbitals
producing two MO’s (one 1 bonding and one 2* antibonding).

The C-C pi () bond in ethylene results from the side-to-side overlap of two 2p atomic orbitals
producing two MO’s (one 1 bonding and one 2* antibonding).
ethylene
2
2*
*
2*
2*
E
2p
2p
2sp2
2sp2
C
C
1
1
E=h
 * transition
1
ground state
1
excited state

1 is called a HOMO (Highest Occupied Molecular Orbital) and 2* is called a LUMO (Lowest
Unoccupied Molecular Orbital). HOMO to LUMO electronic transitions always occur with lower
energy (longer wavelength) radiation than any other electronic transition in a molecule. In the
case of ethylene, E for this transition is 173 kcal/mol (corresponding to radiation of  = 165 m,
in the vacuum UV). [Use E = Nhc/ to convert]

Most other alkenes and alkynes, and nonconjugated polyenes and polyynes display a similar
HOMO  LUMO energy gap, e.g., 1-octene max = 177 m, 1-octyne max = 185 m,
cyclohexene max = 182 m. This absorption band is called an ‘E band’ (Ethylenic).

In molecules with conjugated unsaturation, this E is less and decreases as the extent of
unsaturation increases, e.g., 1,3-butadiene max = 217 m, 1,3,5-hexatriene max = 274 m.
2
Molecular Orbitals of Conjugated  Systems:
In unsaturated compounds, the HOMO  LUMO excitation is generally a   * transition (or
‘n’  * transition). The  and * MO’s are not involved. As a result, MO diagrams of unsaturated
compounds are simplified by drawing only the  and * MO’s and the Greek letter  (‘psi’) is used
instead of  to identify the various MO’s. These are named Huckel MO diagrams.
3 nodes
0 bonding
interactions
 4*
1 node
 2*
+
-
-
+
LUMO
C2pz
C2pz
+
+
-
-
3
(K band, ca. 220 m)
E
unbonded
p orbitals
HOMO
1
+
+
-
-
2 nodes
1 bonding
interaction
*
constructive interference overlap of waves
ethylene
E
2
1 node
2 bonding
interactions
1
0 nodes
3 bonding
interactions
1,3-butadiene
Opposite signs of adjacent Atomic Orbitals (AO's) corresponds to an antibonding interaction
("negative overlap") between 2 atoms and a 'node' (region of zero electron density) exists between the
atoms. Same signs of adjacent AO's corresponds to a bonding interaction ("positive overlap").
In 1,3-butadiene, 1 and 2 have more bonding interactions than antibonding (nodes) and so are
bonding MO’s. 3* and 4* have more antibonding interactions (nodes) than bonding and thus are
antibonding MO’s. In its ground state electron configuration, 1,3-butadiene has 2 filled bonding MO’s
and all antibonding MO’s are empty. The bond order of the  system is 2 and the compound is stable.
Note that E for the HOMO  LUMO transition in 1,3-butadiene is smaller that for ethylene. As a
result 1,3-butadiene (and other conjugated unsaturated compounds) absorb at relatively low energy, in
the Near UV (e.g., 217 m) rather than in the high-energy Far UV (Vacuum UV, < 200 m). The
absorption band arising from   * transitions in conjugated polyenes is called a ‘K band’.

Draw a Huckel MO diagram for 1,3,5-hexatriene showing the number of nodes in each MO and
show the HOMO  LUMO transition. As in the previous examples, use the Aufbau Principle,
Hund’s Rule, and the Pauli Exclusion Principle to determine the ground state electron
configuration.
3
Comparing a conjugated diene (1,3-butadiene) with an unconjugated diene (1,4-pentadiene) shows
why the conjugated diene is more stable. In a conjugated diene, there is a favorable bonding
interaction between C2 and C3 whereas the nonconjugated diene has a node in this region. In the
conjugated diene, the  electrons are spread out (delocalized) over the entire  system rather than
localized between two specific nuclei. Charge delocalization always leads to lower energy and greater
stability of a compound.
0 nodes
1 node
3 bonding
Conjugate
unsaturated cpds., e.g., polyenes & polyynes are analyzed in the near UV2in
the lab.
bonding
interactions
interactions
Molecular Orbital Description of Aromatics
Huckel’s Rule: Planar monocyclic systems in which each atom contributes a p orbital to the  system
will be strongly stabilized (aromatic) when the number of electrons in the  system = (4n + 2) and n is
an integer, i.e., when the number of  electrons = 2, 6, 10, 14, etc.
This can be confirmed by constructing a Huckel MO diagram. For monocyclic conjugated systems it
is easiest to use a ‘Frost Circle’ for this. A regular polygon of ‘n’ sides is drawn in a circle with one
corner at the lowest point. The points at which the corners of the polygon touch the circle define the
energy levels of the MO’s. Consider benzene.
 6*
 6*
 5*
 4*
LUMO
 4*
 5*
B band
260 m
6 p atomic orbitals
2
3
1
2
3
HOMO
1
benzene
The Frost Circle gives us the energy levels of the MO’s. Using the Aufbau Principle, Hund’s Rule,
and the Pauli Exclusion Principle we determine the ground state electron configuration. The Bond
Order of the  system is calculated to be 3 so we conclude that the compound is stable. We see that
benzene satisfies Huckel’s Rule for monocyclics and this confirms that it is aromatic. The HOMO 
LUMO transition is easily identified as 2 or 3  4* or 5*.
In fact, 3 different   * transitions are found in benzene (and related compounds), i.e., two E
bands [E1 = 184 m (log  = 4.78), E2 = 204 m (log  = 3.90)] and a B band (Benzenoid) at 256 m
(log  = 2.30) which we would expect to correspond to the HOMO  LUMO transition.

Draw Huckel MO diagrams and determine which of the following compounds are aromatic.
1. cyclobutadiene
2. cyclopentadiene
4
3. cyclopentadienyl cation
4. cyclopentadienyl anion
5. cycloheptatrienyl cation
6. pyridine
7. pyrrole

Without drawing a Huckel MO diagram, use Huckel’s Rule to predict aromaticity in naphthalene.
Molecular Orbital Diagrams of Compounds Containing Nonbonding Electrons:
Elements from Group VII A (halogens, e.g., Cl, Br, I), Group VI A (chalcogens, e.g., O, S), and
Group V A (pnicogens, e.g., N, P) carry nonbonding electron pairs into the organic compounds they
form. Nonbonding electrons are not involved in the formation of bonding or antibonding MO’s since
they are not directly involved in bonding, however, these nonbonding electrons can be excited
(promoted) to unfilled MO when radiation of the correct frequency is present.
Methyl Chloride (CH3Cl): We can draw a MO diagram for the C-Cl bond. Place the nonbonding 3s
and 3p electrons at approximately the same energy level as the unbonded atoms. (Halogens do not
hybridize when forming single bonds).
methyl chloride
CH 3Cl
2* antibonding MO
E
C 2sp 3
3pz
Cl
n
3s
3p x
3p y
3 pairs
nonbonding
electrons
1 bonding MO
From the MO diagram of methyl chloride, it can be seen that promotion of a nonbonding electron to
the antibonding orbital (an n  * transition) requires less energy than the   * HOMO 
LUMO transition. Despite this fact, most saturated compounds containing heteroatoms still require
Far-UV irradiation to cause excitation (E is still large). In methyl chloride, for example, the
n * transition occurs at 173 m (log  = 2.30).
Exceptions: A few heteroatoms in saturated cpds undergo n  * transitions in the near UV, e.g.,
Br (200 – 210 m), I (255 - 260 m), SH (230 m).
5
In unsaturated compounds containing heteroatoms, n  * transitions are also possible. In many such
compounds, E is sufficiently low such that absorption in the Near UV is possible. Acetone, for
example, shows an n  * absorption band at 188 m (log  = 3.27) and also an n  * absorption
band at 279 m (log  = 1.17). Absorption bands produced by n  * transitions are called ‘R
bands’. (German for 'radical' since a free radical is formed when one 'n' electron is promoted).

Draw a MO diagram for the C=O bond in acetone showing all MO’s and the ‘n’ electrons and the
possible electron transitions.
Summary of UV Absorption Bands
Band
Transition
Example
Cause/Source
 (m)
log max
---
  *
hexane
alkanes (sat’d HC’s)
< 200
E
  *
C2H4, 1,4-pentadiene
unconjugated, unsaturated cpds.
<200
variable
K
  *
1,3-pentadiene
conjugated systems
210 -250
3.7 - 4.7
B
  *
benzenes
double bonds in benzenoids
230 -280
2.3 - 3.3
E1
  *
benzenes
double bonds in benzenoids
180 - 190
variable
E2
  *
benzenes
double bonds in benzenoids
~ 200
variable
---
n  *
CH3NH2, CH3OH, CCl4
most heteroatoms in saturated cpds.
160 - 190
2.3 - 4.0
R
n  *
acetone, MEK
heteroatoms in unsaturated cpds.
> 250
1.0 -1.7
ULTRAVIOLET (UV) SPECTROSCOPY
The wavelengths of visible and UV radiation are generally accepted as follows ...
Category
Wavelength (m)
Wavelength (m)
IR
800 - 106
.80 - 1000
Visible
400 - 800
Near (Quartz) UV
200 - 400
Far (Vacuum) UV
10 - 200
X-rays
10 – 10-3
(2.5  25m = mid IR)
With respect to sunscreens and tanning formulas, the UV spectrum is divided into A, B, and C
regions.
Category
Wavelength (m)
Health Effects
UVA
315 - 400
least dangerous - causes loss of collagen* & decreases # of blood vessels
UVB
290 - 315
causes sunburn and tanning (causes melanocytes to produce melanin)
dangerous - causes skin cancer
UVC
100 - 290
* collagen is protein contained in the connective tissues and bones.
6
The portion of the UV spectrum of most interest to organic chemists is 200 - 400 m, the Near UV or
'Quartz UV’. Samples can be analyzed in quartz cells without interference in this range. Glass cells
cannot be used since glass absorbs UV. Atmospheric oxygen absorbs below 200 m. The lower end
can be extended to 185 m by flushing oxygen from the system and pressurizing with nitrogen. To
work below 185 m requires evacuation of the instrument, and hence the name ‘Vacuum UV’.
When electromagnetic radiation in the UV and visible regions passes through a compound containing
multiple bonds, a portion of the radiation is usually absorbed by the compound.

Whereas IR absorption causes low energy mechanical transitions (vibration, rotation), high energy
UV/VIS absorption induces electronic transitions (  *,   *, n  *, etc.).
For this reason, UV/VIS spectroscopy is also called ‘electronic spectroscopy’.

Another difference is the convention used for describing the radiation. IR is usually reported in
units of frequency (i.e., wavenumbers, cm-1 ) but UV and VIS are reported in units of wavelength
(i.e., nanometers, m). 1 m = 109 m

Almost all organic compounds absorb in the IR, however relatively few organic compounds
absorb in the UV / VIS regions, i.e., primarily only unsaturated compounds.

IR spectroscopy is valuable qualitatively, i.e., for determining the identity and/or structure of an
organic compound but is poor as a method of quantitative analysis. UV and VIS spectroscopy are
of limited qualitative value, i.e., they can be used to detect the presence of some unsaturated
groups in organics. However UV and VIS spectroscopy are excellent methods of quantitative
analysis for unsaturated compounds.

IR spectra are very complex (many sharp absorption bands) but UV and VIS spectra are relatively
simple. They contain a few very broad bands (often spanning ~ 50 m or more). Sharp bands are
not obtained because at room temperature many modes of thermally induced vibration and
rotation are occurring and the electronic absorption is superimposed upon these many closely
spaced energy levels.

E2 (Electronic Energy
Level 2)
Vibrational energy levels
2*
E

7
E1 (Electronic Energy
Level 1)
Vibrational energy levels
1
Obtaining a UV Spectrum:
1. To measure the UV spectrum of a compound, the sample is dissolved in a solvent that does not
absorb above 200 m, e.g., methanol, 95% ethanol, or cyclohexane because they do not have 
electrons (so no   * or n  * transitions occur in the solvent).
2. The sample solution is placed in the sample cell holder and, if a double beam instrument is used, a
blank solution containing solvent only is placed in the reference cell holder. Quartz cells, 1 cmsquare are commonly used. These require about 3 mL solution.
3. The UV spectrometer is a scanning instrument which sequentially emits the full range of UV
radiation and plots a graph (called a ‘UV spectrum’) of absorbance versus wavelength.
Absorbance is sometimes called ‘Optical Density’ (O.D.). The highest points on the spectrum are
‘absorption maxima’. A wavelength at which absorption is maximum is called the ‘max’
(pronounced “lambda max”). Values of max can be read from the graph and used to verify or
identify the compound present.
4. Quantitative analysis can be carried out if pure standards are available. The standard and sample
are weighed to 4 decimal places and made up to volume in a small volumetric flask. Aliquots are
then removed and additional dilutions made as required until the desired concentration has been
achieved. The absorbance of the sample and standard are read on the same instrument, at the
same wavelength, at the same time and temperature, and with the same solvent. The absorbance
of the standard solution is used to calculate the molar extinction coefficient () of the compound
which is then used to determine the concentration (purity) of the sample using the Beer-Lambert
equation. Note that literature values of  (e.g., from CRC Handbook) are of great value in
determining the concentration of standard solution to prepare, i.e., preparing a standard solution
with an absorbance of 0.2 - 0.7 with 0.5 being the goal. However, the literature values of  are not
reproducible enough to be used to directly calculate the purity of the recovered sample.
 fluctuates significantly with all the variables listed above and so sample and standard are always
run simultaneously.
 One limitation of quantitative UV analysis arises from the great width of UV absorption
bands. It is not uncommon that absorption bands from impurities overlap absorption
bands from the compound under study. Since these absorbances are additive, erroneously
high sample purity is occasionally encountered (e.g., 110% purity??!!). This type of error
can be minimized or avoided by purifying the sample thoroughly before analysis and/or
choosing a wavelength at which the sample absorbs very strongly (very high ). The
relative error of interference is likely to be much less.
Beer -Lambert Equation:
The absorbance, A, of the solution at a particular wavelength is given by Beer’s law ...
where c = concentration in moles per liter
I 
A  log r     l  c
l = path length through the cell in cm. (usually 1.0 cm)
 Is 
 = molar absorptivity or molar extinction coefficient of compound
Ir = intensity of radiant energy through the reference (blank)
Is = intensity of radiant energy through the sample solution.
If the sample absorbs light at a particular wavelength, the sample beam (Is) is less intense than the
reference beam (Ir), and the ratio Ir / Is is greater than 1. The absorbance (the logarithm of the ratio) is
therefore greater than zero when the sample absorbs and equal to zero when it does not.
 Note that absorbance has no units. What are the units of  ?
8
In the dilutions shown below, calculate the concentration (in mg/L) and the mass of solute (in mg) in
each flask. Show all dilution factors. Do the dilution calculations based on both mass and
concentration.
4.00 g
50 mL
25 mL
1000 mL
100 mL
500 mL
By Mass:
By Conc.:
0.80 g
25 mL
10 mL
200 mL
100 mL
By Mass:
By Conc.:
9
50 mL
PREPARING DILUTIONS FOR UV SPECTROPHOTOMETRY
In order to analyze a sample by UV spectrophotometry, it must be diluted to an appropriate
concentration in a non-interfering solvent. As in visible spectrophotometry, the Beer - Lambert Law
applies, i.e., A = lc or A = abc
By rearranging and inserting units, the molar absorptivity, , is found to have units of ....
 = A/lc = 1/(cm) (mol/L) = L/(molcm) = Lmol-1cm-1
For quantitative analysis, e.g., determination of purity of your final product, a dilution with an
absorbance close to 0.5 is prepared (0.2 - 0.7 is the most accurate absorbance range in
spectrophotometry) and its absorbance is measured at a single wavelength - at a max of the
compound.
For qualitative analysis, e.g., confirmation of identity of a compound, a dilution is scanned over the
whole Near UV range (200 - 400 m). The dilution is prepared so that the absorbance of all peaks is as
high as possible without any of them exceeding the maximum range of the spectrophotometer.
Note: [Beckman DU65 reads 0  3 A], [Carey 100 reads 0  3 A], and the [Lambda 4A reads 0 3 A units].
Check the CRC Handbook and the Spectral Atlas for max and  (or log ) values of the compound
under study
For example, consider m-nitroaniline ....
CRC Handbook
Spectral Atlas
max
236
log 
4.21

16218
max
233

12000
275
3.64
4365
---
---
374
3.17
1479
371
1090
Be aware that if you prepare a dilution of your sample with an absorbance of 0.5 at  = 371 m, the
absorbance at  = 233 m for the same dilution will be very high (off scale of the
spectrophotometers), i.e., ...
 at  = 233 m
12000
=
 11 so (A = 0.5 @  = 371 m)  11 = (5.5A @  = 233 m)
 at  = 371 m
1090
This is not a problem since you will be measuring the A of your dilution at a single  for a
purity determination, in this example, at 371 m.
Alternately, you could prepare a dilution with an absorbance of 0.5 at  = 233 m. In this case the
absorption at  = 371 would be only be 0.045 A, i.e., 11 times less than at  = 233 m.
Both methods are acceptable and each has its own advantage…
 Reading at the lower  (371m) requires fewer dilutions to prepare a solution. This saves time
and reduces the chances of dilution and handling errors.
 Reading at the higher  (223 m) requires more dilution and so takes longer and increases the
opportunity for dilution errors. However, interferences (from contaminants) have less effect
when the larger dilution is employed.
10
Dilution Calculations: for single  reading at 371m (aiming for A = 0.5 given that  = 1090)
Conc. required:
c
A
0.5

 4.587  10 4 mol / L
-1
  l (1090 L  mol  cm -1 )  (1 cm)

too much solvent !!

c  4.587  10 4 mol / L  (13813
. g / mol) = 0.0634 g / L
or 63.4 mg / L
Using a 25 mL volumetric flask:
too little sample to weigh and transfer accurately
 25 
0.0634g 
  0.001584 g / 25mL
 1000 
or
63.4 mg / L
Select a reasonable weight, e.g., 0.10 g into 25 mL volumetric:
0.1000 g
 63  too concentrat ed
0.001584 g
this solution must be diluted ca. 63  to give A  0.5
Try to use 25 mL volumetrics for further dilution:
63  dilution
in 2 steps
63  dilution
in just 1 step
0.10 g
25 mL
0.10 g
? mL
25 mL
25 mL
Transfer volume is too small. Accuracy suffers.
? mL
? mL
25 mL
25 mL
These dilutions give reasonable accuracy.
11
After determining the necessary weights and dilutions required to prepare a solution with A  0.5,
prepare the both the standard solution and a recovered-sample solution simultaneously as per your
calculations. For the purpose of the lab, it is preferred that students work together and share a single
standard solution. Beware: Sample and standard solutions must be prepared in the same way and
read together on the same instrument, at the same time, etc. in order to obtain accurate values.
Calculation of the molar absorptivity ():
Remember that the  obtained from the literature although useful for dilution preparation cannot be
used to calculate the purity of your sample. You must calculate  using standard solution
measurements and Beer’s Law, A = lc
Calculation of the purity of your sample:
This can be done in several ways…
1. Use  determined in the previous step (from the standard solution) and again use Beer’s law.
Input the absorbance of your sample solution and solve for the molar concentration of your
sample solution. From the molar concentration and the mass (which you originally weighed out)
calculate the purity of the sample.
2. Use Beer’s law to calculate another value of  using your sample solution data.
The ratio of (sample to standard)  100 % will give a % purity of your sample. Sample and standard
weights do not have to be identical.
  sample 
% Purity  
  100%
  s tan dard 
Note:
Solvents must be transparent in the UV region used, e.g., butanol, cyclohexane, dodecane, ethyl ether,
glycerol, hexane, IPA, MeOH, 2-methylhexanol, toluene
12
Source of UV Absorption:
Terminology you need to learn:

Chromophore A covalently bonded, unsaturated group responsible for electronic absorption, e.g.,
C=C, C=O, NO2, etc.

Auxochrome A saturated group with nonbonded electrons which, when attached to a
chromophore, alters the wavelength and intensity of absorption, e.g., -OH, -NH2, -Cl, etc.

Bathochromic Shift The shift of absorption to a lower energy (longer ) due to substitution or
solvent effect (a red shift).

Hypsochromic Shift The shift of absorption to a higher energy (shorter ) due to substitution or
solvent effect (a blue shift).

Hyperchromic Effect An increase in absorption intensity.

Hypochromic Effect An decrease in absorption intensity.
As discussed at length in the unit on Molecular Orbital theory, absorptions in the Near UV arise from
electronic transitions, primarily n  * and   * transitions. The student is expected to become
familiar with the UV absorption characteristics of the major classes of organic compounds, i.e.,
hydrocarbons, alkenes, alkynes, polyenes, conjugated polyenes, carbonyl compounds, aromatics, and
nitro groups. A brief summary follows ....

Auxochromes in sat’d HC derivatives: n  * usually at  < 200 m, e.g., ROH, RNH2, RCl.
Exceptions: [R-Br and R-I: n  * at  > 200 m] and [RSH, RSSH absorb at  > 200m]
 Non conjugated alkenes and alkynes (E band chromophores) :   * at  < 200 m
 Conjugated polyenes (K band chromophores):
a) conjugated dienes:   * at  > 200 m, e.g., acyclic @ 214 m and up for substituents,
homoannular @ 253 m and up, heteroannular @ > 253m.
b) conjugated enynes & diynes:   * at  > 200 m
 Carbonyl Chromophores: (aldehydes, ketones, esters, acids, amides, acyl chlorides, anhydrides)
a) non-conjugated:   * at  < 200 m (E), n  * at  > 200 m (R), many.
b) conjugated:   * intense at  > 200 m (K), n  * at  = 300-330 m (R)
 Aromatic Chromophores:
a) all substituted benzenes: E2 -band @ 200 - 250 m & B-bands @ 255-280 m
b) benzenes with conjugated groups: K-band (intense) @ 220-250 m
c) conjugated substituents with non-bonded e-'s: R-band (weak) @ 273-330 m
Note:
e-donating groups shift max to longer  (causes a bathochromic shift)
e-withdrawing groups shift max to shorter  (causes a hypsochromic shift)
additional fused rings shift max to longer 
heterocycles, such as pyridine, furan, pyrrole, thiophene, also absorb above 200 m
Miscellaneous chromophores (on aliphatics):
Some absorb @  > 200 m, e.g., nitro, nitrates, nitrites, nitriles, azo cpds., sulfones, sulfoxides
13
Homework:
1. For each of the following compounds put a check mark in the box or boxes which describe an
electronic excitation which cause a moderate to strong absorption of radiation in the 200 to 400
m wavelength range.
COMPOUND
  *
n  *
  *
none
CH3CH2OH
aniline
CH2 = CH2
CH2= CH-CH=CH2
CH2= CH-CH2-CH=CH2
CH3NO2
O
CH3 = CH- C - CH3
1. A student dissolved 0.1000 g of a UV-absorbing compound (M.W. = 160 g/mol) in methanol in a
50 mL volumetric flask.
He then pipetted 1.0 mL of this solution into a
25 mL volumetric flask and diluted to the mark with methanol. The absorbance at
290 m was 0.600. Calculate the molar extinction coefficient () at this wavelength.
14
2. Complete the following table.
COMPOUND
STRUCTURE
ethylene
CH2=CH2
Type of UV
Transition
(and Band)
  * (E)
max
(m)
log10
max
165
4.19
1-octene
177
4.10
cyclohexene
182
3.88
1-octyne
185
3.30
1,4-pentadiene
178
4.23
1,3-pentadiene
223
3.36
1,3-butadiene
217
4.32
(E)-1,3,5-hexatriene
274
4.55
(3E, 5E, 7E)1,3,5,7,9decapentaene
334
5.10
15
E
(kcal/mol)
173
Additional Problems:
1.
A beam of UV radiation has an energy of 6.83  105 J/mol of photons.
Given that...
Plank’s constant (h) = 6.62  10-34 Js
the speed of light (c) = 3.00  1010cm/s
Avogadro’s Number = 6.02  1023 /mole
Calculate ...
a) its wavelength in m ................................
b) its frequency in Hz .............................
2.
A beam of UV radiation has a wavelength of 150 m.
Given that...
Plank’s constant = 6.62  10-34 Js
the speed of light = 3.00  1010cm/s
Avogadro’s Number = 6.02  1023/mole
Calculate ...
a) its frequency in Hz .............................
b) the energy of 1 photon in Joules ................................
c) the energy of 1 mole of photons in kcal .........
4.184 Joule = 1.00 cal
....................
3.
A beam of UV radiation has a frequency of 1.5  1016 Hz
Given that...
Plank’s constant = 6.62  10-34 Js
4.184 Joule = 1.00 cal
10
the speed of light = 3.00  10 cm/s
Avogadro’s Number = 6.0225  1023/mole
Calculate ...
a) its wavelength in m .............................
N
..
b) the energy of 1 mole of photons in kJ .............................
16
4.
For pyridine, C5H5N:
a) Use a Frost circle and draw a Huckel M.O. diagram of its  system showing the relative energy
levels of all its molecular orbitals.
b) Label (name) all the molecular orbitals.
c) Show the ground state electronic configuration of the molecular orbitals
d) State the Bond Order of the  system ...............................
e) State whether this species is nonaromatic or aromatic ...................................
f) Label the HOMO and LUMO molecular orbitals
g) Name the lowest type of energy electronic transition which can occur. .........................
5.
For cyclopentadienyl anion,
a) Use a Frost circle and draw a Huckel M.O. diagram of its  system showing the relative- energy
..
levels of all its molecular orbitals.
b) Label (name) all the molecular orbitals.
c) Show the ground state electronic configuration of the molecular orbitals
d) State the Bond Order of the  system ...............................
e) State whether this species is nonaromatic, aromatic or antiaromatic ...................................
f) Label the HOMO and LUMO molecular orbitals
g) Name the lowest type of energy electronic transition which can occur. .........................
17
6.
Fill in all the empty cells in the following table for UV electronic transitions ...
example of compound
absorption band
“ letter”
transition
type
‘has no letter’
  *
indicate if “near” or
“far” UV region
K
‘has no letter’
CH3CHCl2
1-hexene
Near
-CH2CH3
Near
Far
methyl ethyl ketone
Near
Far
7.
In the following compound circle all groups that would give rise to absorption in the near UV
only. For each group circled, fill in a row of information, i.e., name the type of transition (e.g.,
*, etc.) and the absorption band letter (e.g., B, E1, E2, etc). Fill in as many rows of info as
needed and indicate which row corresponds to which group by drawing an arrow from the circled
group to the matching row in the table.
electronic transition type absorption band
letter
O
C
CH2Br
C
NH2
CH2
18
8.
For each of the following compounds put a check mark in the box or boxes which describe an
electronic excitation which cause a moderate to strong absorption of radiation in the 200 to 400
m wavelength range.
COMPOUND
  *
n  *
  *
none
azobenzene
m-nitroaniline
decane
1,3-butadiene
9.
A pure organic liquid (M.W. = 100 g/mol,  = 0.80 g/mL) has a molar extinction coefficient ()
of 50 in methanol at  = 225 m. Calculate the volume (in mL) of the liquid which must be
pipetted into a 50 mL volumetric flask, which after dilution to the mark will give an absorbance
of 2.0 at the stated wavelength
10. Calculate the molar extinction coefficient () of an organic solid (MW = 200 g/mol) given the
following data and include the units of .
 0.1000 g was dissolved in 50 mL methanol
 a 5 mL aliquot was diluted into a 25 mL volumetric flask
 this solution gave an absorbance of 0.70 when read at 250 m in a 1.0 cm cell
19
11. A student dissolves 0.1100 g of her recovered product (MW=75.00g/mol) to volume with methanol in a
25 mL volumetric flask. She wishes to dilute this solution such that the final dilution will have an
absorbance near 0.5 ( = 14,500 at the chosen wavelength) using a 1.0 cm cell.
Calculate the dilution factor necessary to dilute this solution to an absorbance of 0.50 …………….
Assuming that only 25 mL flasks can be used and only the following volumetric pipets are available (1.0,
2.0, 3.0, 4.0, 5,0, 10.0 and 15.0 mL), report the aliquots used for each dilution she should use to obtain a
final solution whose absorbance is between 0.40 and 0.60.
Report the aliquots used for each subsequent dilution in the following format: x mL 25mL,
y mL 25mL, etc, etc. as many times as needed to reach the final dilution.
Report the final expected absorbance based on these dilutions. …………………
20
12. Calculate the absorbance expected on the final dilution described below given that the MW of the
compound is 150 g/mol and its molar absorptivity is 105
 0.1000 g was dissolved in methanol in a 25 mL volumetric flask
 a 20 mL aliquot (from the 1st flask) was diluted into a 50 mL volumetric flask
 a 10 mL aliquot (from the 2nd flask) was then diluted in a 25 mL volumetric flask
the absorbance of this solution is to be read in a 1.0 cm cell
13. A student prepares dilutions of her pure standard and recovered sample as follows:
 She dissolves 0.1000g pure standard in a 25 mL volumetric, then transfers a 1 mL aliquot into a 50mL
volumetric flask. The absorbance of the final solution is 0.600
 She carries out the same dilutions on her recovered product but her initial weight was 0.1098g and the
absorbance of the final dilution was 0.540
 Knowing MW = 70.0 g/mol, calculate the molar absorptivity of the compound and include the units of .
…………………………………...…..
 Calculate the %wt purity of the recovered product. …………………………………….………
Show your work for partial marks in the event of errors.
21
Answers to Some of the Additional Problems from Pages 16-21:
b) 1.711015 s-1
1.
a) 175m
2.
a) 2.00  1015 s-1 b) 1.32  10-18 J/photon
3.
a) 20 m
b) 5980 kJ
4.
d) BO = 3
e) aromatic
g) n  *
5.
d) BO = 3
e) aromatic
f)   *
c) 190 kcal/mol
9. 0.25 mL
10.  = 350 Lcm-1mol-1
11. DF = 1700 
Dilutions = (225mL), (225mL), (225mL)
12. A = 0.45
13.  = 525 Lmol-1cm-1
Purity = 82.0%
22
Final A = 0.44