CSUMB
ESSP 311 Organic Chemistry I
Ronald W. Rinehart, Ph.D.

Chapter 13  Spectroscopy

This page has Rod Oka's modified outline and some general web references.
See headings below for workshop pages with loads of additional web references.

Chapter 13.  Spectroscopy 

I.  Purpose of Spectroscopy

            A.  The main purpose of spectroscopy in organic chemistry is to determine the structure of a compound.
                        1.   Types of spectroscopy include [but are not limited to]
                                    a)  Infrared Spectroscopy, IR.
                                    b)  Nuclear Magnetic Resonance, NMR.
                                    c)   Mass Spectrometry.
                                    d)   UV-Visible Spectroscopy.
                        2.   Other types of spectroscopy have been developed, such as esr. 

II.   Infrared Spectroscopy
            A.   Theory of IR
                        1.  Molecules can vibrate in many ways.
                        2.  The energy required to excite these vibrations corresponds to electromagnetic radiations in the
                                     infrared region.
                        3.  Different types of bonds absorb specific frequencies of IR radiation.
                                    a)  The frequencies are often measured in wavenumbers (cm^-1), the number of waves which would
                                                fit in a 1 cm line.
                                    b)  For example, the C=O bond absorbs energy in the 1660-1770 cm^-1 range while the C-O bond
                                                 absorbs between 1050-1300 cm^-1.
                        4.  The information IR gives the chemist is the presence (or lack of) specific functional groups in the
                                     compound being analyzed.
                        5.  IR cannot tell you how many atoms of each type are in the compound. 

            B.  Components of the IR Spectrometer
                        1.  A simple IR spectrometer consists of an IR energy source which sends energy through a prism.
                                    a)  The prism separates the energy into specific frequencies, just like it separates visible light into
                                         the colors of the rainbow.
                                                i)  In the newer Fourier Transform IR instruments, prisms are not used; an improved  
                                                   technology is used instead.
                                    b)  The individual frequencies are sent through the sample cell which holds the compound being
                                                 studied.
                                                i)  The sample cell can hold gases, liquids, or solids.
                                                ii)  Gases and liquids can be held in the sample cell as the pure compound ("neat").
                                                iii)  Solids can be mixed with solid KBr (invisible to IR), ground into a fine powder, and
                                                       then subjected to pressure to make  a transparent window.
                                                iv)  Solids can also be ground with mineral oil to make a “Nujol mull”  
                                    c)  When energy is absorbed due to the presence of a specific bond type vibrating, a detector notes
                                           the sudden decrease of energy reaching it (a reference beam bypassing the sample cell is used for
                                                 comparison).
                                    d)  The detector sends the information to a recording device.
                                                a)  Energy absorptions are seen as "absoprtion bands" on the IR spectrum. 

            C.  Interpretation of IR Spectra
                        1.  Not every absorption band in an IR spectrum is significant since certain frequency ranges of IR are
                                     absorbed by several different types of bonds.
                                    a)  Only certain regions of the IR spectra are meaningful.
                        2.  Some of the important regions of the IR spectrum to look at include:  
                                    a)  3200-3600 cm^-1:  O-H from alcohols.
                                    b)  3000-3100 cm^-1:  C-H from sp² C.
                                    c)  2850-3000 cm^-1:  C-H from sp³ C.
                                    d)  1660-1770 cm^-1:  C=O
                                    e)  1600 cm^-1:  C=C from benzene.
                                    f)  1050-1300 cm^-1:  C-O

                                                            See the summary chart below

                        3.  The primary clues page gives more precise information on spectral interpretation.
 

II.  Nuclear Magnetic Resonance
            A.  Theory
                        1.  The nuclei of some atoms behave as though they spin on their axis.
                                    a)  Electrons have the same behavior.
                        2.  The ¹H nucleus (a proton) has two nuclear spin states with quantum numbers +1/2 and -1/2.
                        3.  Spinning charged particles create a magnetic field, so the proton has a magnetic moment.
                                    a)  The proton acts like a small bar magnet.
                        4.  The two spin states are identical in energy in the absence of an applied magnetic field.
                                    a)  In absence of an applied magnetic field, the nuclear magnetic moments are randomly oriented.
                        5.  When placed in a magnetic field the magnetic moments align either with or against the field.
                                    a)  The two spin states are no longer of equal energy.
                                                i)  The +1/2 spin state aligns with the applied magnetic field and is of lower energy.
                                                ii)  The −1/2 spin state aligns against the applied magnetic field and is of higher energy.
                                    b)  Example: think of the Earth's magnetic field as an applied field in which a compass is placed.
                                                i)  The compass needle aligns with the magnetic field of the Earth.
                                                ii)  This is of lower energy than if you were to align the needle in the opposite direction.
                        6.  The nucleus in the lower (+1/2) spin state can be elevated to the higher-energy spin state by the
                                    absorption of energy (electromagnetic radiation) equal to the energy difference of the two spin states.
                                    a)  This change from the +1/2 to −1/2 spin state is referred to as "spin flipping".
                                    b)  The amount of energy required for this transition is dependent upon the magnitude of the applied
                                                 magnetic field.
                                    c)  The greater the applied field, the higher the required energy to flip the spin.
                        7.  Many nuclei lack the property of spin.
                                    a)  Nuclei having an even number of protons and an even number of neutrons lack spin.
                                                i)  Pairs of protons and pairs of neutrons in such nuclei have opposed spins which cancel each
                                                   other out, leaving the nucleus with no net spin.     

            B.  NMR Spectroscopy
                        1.  The NMR spectrometer consists of a  powerful magnet.
                        2.  A sample is placed in the magnetic field of the spectrometer.
                        3.  A radio-frequency source irradiates energy through the sample until the frequency of the energy source
                                     is equal to the energy difference of the two spin states.
                                    a)  At this time the two are said to be in resonance with each other.
                        4.  The absorption of energy is detected by a radio-frequency receiver and recorded as a peak on an NMR
                                    spectrum.
                        5.  The fact that radio waves are used to cause the spin flip demonstrates that the energy difference
                                     between the two spin states is of very low energy.
                                    a)  In a very strong magnetic field of 14,100 gauss, the energy difference for the two spin states of a
                                                 proton is only 2.5 x 10^-5 kJ/mole.
                        6.  NMR instruments can operate either with a constant magnetic field and changing the frequency, or
                                    with constant frequency and changing magnetic field.
                                    a)  Either way, the spectra are identical.
                        7.  Not long ago 60 MHz instruments were in common usage, then 100 MHz, 220 MHz, and 360 MHz
                                     instruments were in use; now 400, 500 and 600 MHz instruments are in use.
 

            C.  Chemical Shift
                        1.  If all hydrogen nuclei spin flipped at the same frequency, NMR spectroscopy would have no value in
                                     compound structure determination since all NMR spectra would look the same:  a single   peak at the
                                    same place.

                        2.  Fortunately, the chemical environment of  hydrogen atoms affects the frequency of the energy
                                    absorbed for resonance to take place.
                        3.  In a constant frequency instrument, where the magnetic field is slowly changed to cause the spin flip, it
                                    is observed that the applied magnetic field required to spin flip a proton in an organic compound is
                                     greater than that required for a bare proton.
                                    a)  This is because the electrons in a molecule "shield" the nucleus from the applied magnetic field.
                                                i)  This is referred to as "diamagnetic shielding".
                                    b)  This shielding occurs because the application of the magnetic field results in the electrons moving
                                                and setting up their own magnetic field which opposes the applied field.
                                    c)  The nucleus of the bound proton "feels" less magnetic field than is actually applied.
                                    d)  In order to achieve the spin flip, a greater magnetic field must be applied as compared to the field
                                                required to flip a bare proton.
                        4.  The change in the resonance position of a nucleus (due to the chemical environment of the atom) is
                                    referred to as a chemical shift.
                        5.  A peak at a higher field is said to be "upfield".
                                    a)  A peak at a lower field is said to be "downfield".

                         6.  The chemical shift is measured relative to a substance standard
                                    a)  The standard used is TMS, tetramethylsilane.
                                    b)  The NMR signal from TMS is set at 0 (on the delta scale) since protons in TMS are highly
                                                 shielded due to the low electronegativity of Si.
                                                i)  There is therefore a relatively high electron density around the hydrogen atoms.
                                                ii)  Shielding in TMS is greater than that of most organic compounds, making it an ideal standard.
                                                iii)  The spectrum is adjusted electronically on the chart so that the TMS signal is set to 0.
                        7.  The placement of an NMR signal relative to TMS varies with magnetic field strength.
                                    a)  Because NMR instruments use many different magnetic field strengths, a unitless measure of
                                                chemical shift independent of field strength has been developed.
                                    b)  The delta scale is the ratio of the chemical shift of a given signal to the total radio frequency used:
            Chemical shift (d) =  (position of signal - position of TMS peak)  x 10⁶/   radio frequency of spectrometer
                                    c)  The values in the above equation are in Hz.
                                    d)  Because the difference in the two positions is so small, it is multiplied by 10⁶.
                                                i)  The delta scale is therefore expressed in ppm.

                        8.  The chemical shift yields important information.
                                    a)  By the magnitude of the shift, some details of chemical structure can be deduced.
                                                i)  The text has a more complete listing of chemical shifts; below are some representative values.

                                     b) Highly electronegative elements deshield the protons, resulting in downfield resonances. 

Chemical Shifts of CH₃Z          

 Substitution Effects on Chemical Shifts

                        9.  The chemical shift for benzene is around 7 ppm and that for the protons of alkenes is around 5.
                                    a)  Resonance occurs far downfield than might be expected for hydrocarbons.
                                    b)  The effect is due to anisotropy.
                                                i)  The pi cloud of electrons of these systems circulate due to the induced magnetic field,
                                                            creating their own magnetic field.
                                    c)  In the case of benzene, the induced magnetic field opposes the applied field in the middle of the
                                                 ring, but is with the applied field outside the ring.
                                                i)  The protons of benzene are in the space where the secondary magnetic field generated by
                                                            the pi electrons of the pi cloud circulating deshields the protons.
                                                ii)  A strong clue to the presence of the benzene ring in a compound is a NMR signal at ~7 ppm.

                                    d)  In the case of alkenes, the induced field opposes the applied field in the middle of the double
                                                bond, but are with the applied field in the area of the vinyl hydrogens.

             D.  Splitting Patterns 

                        1.  The NMR spectrum of ethane is a single peak since there is only one chemical environment for H                  
                        2.  The NMR spectrum of 1,1,2-trichloroethane should show how many NMR signals, and what would
                                     you expect their approximate chemical shifts?

                                    Answer:  Two.   

                                    The H at C-1:  ~6 ppm.  It is further downfield than the H’s at C-2 since two Cl are attached to its C.
                                    The H's at C-2:  ~4 ppm           

                                    a)  Two single peaks are NOT observed for the above compound.
                                    b)  A triplet is observed at 5.8 ppm and a doublet is observed at 3.9 ppm. 

                         3.  Multiplets are observed in place of singlets on the NMR spectrum when hydrogen atoms exist on
                                     adjacent carbon atoms.
                                    a)  Multiplets arise from "spin-spin splitting".
                                    b)  The chemical shift is in the center of the multiplet.
                        4.  Splitting occurs because the magnetic field of a proton of a hydrogen atom on carbon 1 adds or
                                     subtracts to the applied field felt by the hydrogen atom on carbon 2.
                                    a)  The "effective field" felt by a proton can be greater, less than, or equal to the applied field,
                                                depending on the orientation of spin on neighboring hydrogens.
                        5.  The two hydrogens on C-2 of 1,1,2-trichloroethane experience two different effective magnetic fields
                                     depending upon if the hydrogen on C-1 is in the spin up   or spin down state.
                        6.  The H at C-1 is near two hydrogens on C-2 and as a result can feel 3 different effective fields.
                                    a)  Both C-2  H-spins can be up, both could be down, or one could be up while the other is down.

                                     b)  Notice that the central peak of the triplet is twice the size of the other two since there are twice
                                          as many spin combinations possible than for the other two.
                                    c)  The hydrogens on the adjacent carbons are said to be "coupled".
                        7.  In simple cases, the observed multiplet is equal to the number of H on adjacent carbon atoms + 1.
                          This is called the "(n+1) rule"
                        8.  The splitting pattern for the ethyl group is quite distinctive. 
                                    a)  Predict what the proton NMR spectrum would look like for CH­₃CH₂F, and show the spin
                                                combinations and net spins below each signal. 

                        9.  The following chart shows the type of multiplet observed as a function of adjacent hydrogens, and the
                                     area ratios of the peaks of the multiplet. You may recognize this pattern as Pascal’s Triangle

                        10.  Protons of the hydroxyl group on alcohols usually appear as a singlet and are not split due to rapid
                                     intermolecular exchange of the hydroxyl protons. The same is true for amine hydrogens. 

            E.  Coupling Constants

                        1.  The distance between peaks in a multiplet, measured in Hz, is referred to as the coupling constant, J.
                        2.  The coupling constant is independent of the magnetic field strength of the NMR instrument.
                        3.  For the example fill in the blanks:
                                    The peak for H_a appears as a doublet.
                                    The peak for H_c appears as a doublet.   
                                    a)  Because the coupling constants J_ab and J_bc are not equal (J_ab = 3.6 Hz while J_bc = 6.8 Hz) the peak
                                                for H_b is not a quintet but instead a doubled quartet (or quadrupled doublet.)
                                                i)  This makes the multiplet rather messy rather than being a nice and easy-to-decipher peaks. 

                         4. Unequal coupling constants can result in complex spectra which are more difficult to interpret.
                        5. Coupling constants can yield important information about conformations in cyclohexane rings.
                                    a)  The coupling constant for adjacent axial hydrogens in cyclohexanes is 10-13 Hz.
                                    b)  The coupling constant for adjacent equatorial hydrogens in cyclohexanes is 3-5 Hz.   

                         6.  Coupling constant values can be used for the determination of stereostructure in alkenes.
                                    a)  J is always greater for trans H’s than for cis H’s for a pair of isomeric cis and trans alkenes.
                                                i)  J_trans = 11-19 Hz.
                                                ii)  J_cis = 5-14 Hz.
                                    b)  Notice there is overlap in the J values.
                                                i)  If the vicinal coupling constant is < 10, one can infer the alkene is the cis isomer.
                                                ii)  If the vicinal coupling constant is > 14, one can infer the alkene is the trans isomer.
                                                iii)  If the J value is between 10 and 14, then the spectra of both isomers must be available
                                                            to make a conclusion (J_cis is always less than J_trans for isomeric alkenes.)
                                    c)  The NMR spectrum of (E)- and (Z)-3-chloropropenenitrile illustrate how the coupling constants
                                                 can be used to assign structure.  

F.   Integration 

                        1.   In proton NMR the area of the peaks for each proton environment is proportional to the number of
                                    protons in that environment.
                        2.  When the NMR spectrometer is set to the integration mode it draws a step-like line whose height is
                                    directly proportional to the number of protons responsible for the signal.
                        3.  Integrations do not actually indicate the precise number of hydrogens at each chemical environment.
                                    a)  They instead yield a ratio of protons at each environment.
                        4.  The example below illustrates how integration yields a ratio of protons at each chemical environment. 

                                     a)  In the above example notice how the integral ratios are calculated.
                        5.  The information given by the chemical shifts, splitting patterns, coupling constants, and the
                                    integration can often lead to the structure of the compound in question.
                                    a)  In addition to this we also have information given by IR (functional groups present), mass
                                                spectrometry (molecular weight and other information), elemental analysis, and
                                                other spectroscopic methods. 

            G.  C-13 NMR (CMR)
                        1.  NMR experiments may be done with any atom which has a nucleus with a magnetic spin.
                        2.  Examples of various isotopes with nuclear magnetic spin states are listed below.
                                    a)  The chart shows the frequency and magnetic field strength for which the nuclei have their
                                                resonance. Note that for a given isotope, resonant frequency is proportional to applied field.

                        3.  A significant difference in proton NMR vs CMR is that ¹H is the most abundant isotope of hydrogen.
                                    a)  In CMR the lesser abundant isotope is observed. About 1% of all carbon atoms are C-13.
                                    b)  The lower abundance means a larger sample must be used in the instrument. 

                                                 Proton NMR requires about .1 mg of sample. CMR requires about 1-5 mg of sample.
                        4.  The ¹H NMR spectrum ranges typically from 0-12 ppm; the CMR spectrum ranges from 0-210 ppm.
                                    a)  Even slight differences in a carbon's chemical environment results in a significant chemical
                                                shift so that the number of carbon atoms (of different chemical environment) in the compound
                                                can be determined simply by adding up the number of signals in the spectrum.
                        5.  A great advantage of CMR is that the NMR experiment can be run in two different modes. 
                                    a)  The "off-resonance decoupling"  experiment results in the splitting of each signal for each carbon
                                                 atom in the molecule by the hydrogen atoms attached to it.
                                                i)  This means that the number of hydrogen atoms attached to the carbon atom can be quickly
                                                            deduced simply by counting the peaks of the multiplet and subtracting 1.
 

                                     b)  In the proton-decoupled mode each carbon nucleus appears as a singlet.
                                                i)  The protons are kept from coupling with the C-13 nuclei by irradiating the sample with
                                                            radio waves which flip the protons:  they do not spend enough time in either spin state to
                                                            couple with the C-13 nuclei. 

                                                 ii)  Adjacent carbon atoms do not split the signal in C-13 NMR. Why not?
                                                            Answer:  Since there are so few C-13 atoms (~1%), the chances of having two C-13                                                          atoms next to each other in a molecule are small.                                                                         

                        6.  In CMR as in NMR, electronegative atoms attached to the carbon atom results in a downfield shift.
                                    a)  In addition, replacement of a hydrogen atom on a carbon with an alkyl group results in a
                                                downfield shift.

                       7.  In the CMR spectrum of 2-butanol above, assign which carbons are responsible for each observed
                                    peak.  
                        8.  Coupling constants could be used in proton NMR to distinguish cis and trans isomers of alkenes.
                                    a)  The chemical shift of the allylic carbons in CMR are lower for the cis isomer than the trans
                                                isomer by about 5 ppm, so this is another way of deciding upon which isomer is being studied.
                                    b)  Notice the chemical shifts for the isomers cis- and trans-2-butene. 
                                             cis: 11.4 ppm                       trans: 16.8 ppm

                        9.  Unlike proton NMR, CMR integrals do not yield ratios of nonequivalent carbons   since the peak
                                     intensities in CMR are subject to distortion.
                        10.  Below are shown some CMR chemical shift correlations (ppm).

             H.  Spectral Interpretation
                        1.  It is not necessary to memorize the various chemical shifts for the different types of chemical
                                    environments.
                                    a)  Tables can be consulted.
                                    b)  It is helpful to remember some generalities such as, thinking of the proton NMR spectrum as a                                                             football field:
                                                i)  Alkanes resonate around the 10-20 yard line on the right side.
                                                ii)  Hydrogens bonded to carbons in turn bonded to highly electronegative groups (OH, Cl, Br)                                                             resonate near mid-field.
                                                iii)  Inductive shielding effects  are cumulative, so the greater the number of Cl's, O's, etc
                                                            bonded to a carbon, the more the C-H resonance will move downfield (past mid-field to
                                                            the other end zone.)
                                    c)  In CMR remember that double-bonded carbons resonate far downfield, C-O and C-X
                                                (X = halogens) moderately far downfield, and C=O very far downfield.
                        2.  From CMR one can find the number of nonequivalent carbons and in the off-resonance mode the
                                    number of hydrogens on each carbon.
                        3.  From proton NMR probably the most useful aspect is the spin-spin splitting patterns.  
                        4.  The integrations in proton NMR yield proton ratios in each chemical environment and should be
                                    evaluated.
                        5.  The chemical shift of ~7 is usually indicative of the presence of the benzene ring.
                                    a)  A signal between 9.5 and 10 usually indicates the presence of an aldehyde
                                    b)  A signal between 10 and 12 usually indicates the presence of a carboxylic acid.
            I. Sample Problems
                        1.  3-hexanol is treated with boiling sulfuric acid forming a mixture of isomeric alkenes:

                                           Identify each isomer. 

            Answer.  Isomers A and B have six signals and therefore must have six nonequivalent carbons (III and IV).  Isomer B has two chemical shifts about 5 ppm greater than Isomer A (17.5 vs 13.5, 35.1 vs 29.3), which indicates isomer B must be compound III (the trans isomer) and A must be compound IV.

            By similar analysis, Isomer C must be compound I (the cis isomer) and Isomer D must be compound II. 

                        2.  A ¹H NMR experiment is done on a compound of formula C₄H₇Cl₃.  Use the following proton NMR data to determine the structure of the compound.  (s =  singlet, d = doublet;  t = triplet;  m = multiplet). 

                                    d   (ppm):    0.9 (t);  1.7 (m); 4.3 (m);  5.8 (d) 
                           Integration ratio:       3  :      2      :       1    :      1 

                        Answer. The triplet at .9 (upfield) with an integration of 3 (3 H's) is a methyl group next to a methylene group.  CH₃CH₂-
                        The doublet at 5.8 is so downfield that it is most likely -CHCl₂.  Being a doublet there must be only one hydrogen on the adjacent carbon.
                        The multiplet at 4.3 with an integration of 1 could account for this adjacent carbon with 1              hydrogen atom;  the fact that it is downfield           can be explained by the presence of a Cl atom and being next to a carbon with two Cl atoms.

                        -CHCl- 

                        The structure of the compound is likely to be 1,1,3-trichlorobutane 

                        3.  A compound of formula C₃H₇NO­₂ (an -NO₂ group is present – thank IR for that) has the
                            following  proton NMR spectrum.  Determine the structure of this compound. 

                                     Answer:  CH₃CH₂CH₂NO₂

    Ha: 1.0 ppm (t);  Hb: 2.0 ppm (m);  Hc: 4.4 ppm (t)

                        4.  A compound has the formula C₄H₇O₂Br. The signal at 10.97 ppm was moved onto the chart since the
                            chart paper only runs from 0-10 ppm.  Determine the structure based on the proton NMR spectrum. 

                         Answer.           CH­₃CH₂CHBrCO₂H                                                             

   Ha: 1.1 ppm (t);  Hb: 2.0 ppm (m);  Hc: 4.2 ppm (t);  Hd: 10.97 ppm (s)

III.  Mass Spectrometry 

            A.  Theory
                        1.  Molecules are bombarded with a beam of  high energy electrons.
                        2.  The molecules are ionized and fragmented into many smaller pieces.
                        3.  Most of the particles carry a +1 charge.
                                    a)  This means that their mass/charge  ratio, the m/e value, is the mass of the ion.
                                    b)  For the case of the molecular cation, the m/e value is the molecular weight of the compound.
                                                i)  This is referred to as the M⁺ peak, and M⁺ is the molecular ion or parent ion.
                        4.  The different ions produced is sensed by a detector, which sends the information to a recorder which
                                    plots the signals in relationship to their intensities.
                                    a)  The largest signal is referred to as the base peak and is given a relative intensity of 100%.
                                                i)  The M⁺ is not necessarily the base peak.
                                                ii)  The M⁺ peak can sometimes be extremely small.
                                    b)  The intensities of the other peaks are then expressed relative to the base peak.
                                    c)   The mass spectrum is highly characteristic of a given compound and can be used to prove the
                                                identity of a compound or used to determine the structure of a compound.
                        5.  The mass spectrum helps in structure determination by
                                    a)  giving a molecular weight for the compound.
                                    b)  indicating certain structural units existing in the molecule, and
                                    c)  sometimes can give a molecular formula for the compound.

            B.  The M + 1 Peak
                        1.  The M⁺ peak is not the peak of highest m/e value.
                        2.  This is because of the existence of isotopes.
                        3.  Since about 1.1% of all carbon atoms is C-13, some of the molecular ions have a m/e value 1 unit
                                    higher than the parent peak.
                                    a)  The (M + 1) peak can also be due to the presence of deuterium, although the % abundance of ²H
                                                is only   0.015%.
                        4.  In the case of benzene, it would be expected that the (M + 1) peak would be about 6.6% as intense as  
                                    the M⁺ peak.
                                    a)  Since there are six carbons in benzene, the % intensity expected is 6 x (1.1%).
                        5.  In compounds containing chlorine a significant (M + 2) peak would be observed since the
                                     % abundance of ³⁷Cl (2 amu's more massive than the more abundant ³⁵Cl) is about 24.23%.
                        6.  In compounds containing Br, an (M + 2) peak of about equal intensity as the parent peak is observed
                                    since the two isotopes of Br, ⁷⁹Br and ⁸¹Br, exist in a 100:98 [~1:1] ratio.
                                    a)  “Twin” peaks of about equal intensity but two units different in mass is a good indication of Br in
                                                 the compound. 

            C.  Example

                        1.  Neopentane has been analyzed by mass spectrometry and found to break into several fragments
                                    including the following.
                                    a)  Notice that the fragments formed are not obvious fragmentation products of neopentane.
                                                i)  Rearrangements obviously must occur.
                                                ii)  Stability of carbocations is a driving force for rearrangements in mass spectrometry.
                                    b)  Only about 10^-6 gram of sample is needed for the experiment.

                         2.  Fragmentation patterns and types of   rearrangements are well-known and are a key to the full
                                    utilization of mass spectrometry as a tool for structure determination.

IV.  UV-Visible Spectroscopy

            A.  Theory
                        1.  Ultraviolet light is of much higher energy than that used in IR and NMR.
                        2.  UV spectroscopy uses electromagnetic radiation in the range from 200 to 400 nm.
                        3.  Visible spectroscopy uses light in the 400 to 800 nm range.
                        4.  Both are used in the investigation of  the electronic structures of unsaturated compounds and
                                    measuring the extent of conjugation.
                        5.  The absorbance A of a sample is proportional to its concentration in solution and the path length of
                                    the sample cell.
                                    a)  To correct for the above factors, the absorbance is converted to molar absorptivity, e.
                                                [formerly called the “molar extinction coefficient”]

                                    b)  When molar absorptivity is measured at  l_max, it is cited as e_max.
                                                i)  It is usually expressed without units.
                        6.  Solvents used in UV-Visible spectroscopy are ethanol, methanol, and cyclohexane.
                                    a)  These solvents do not absorb in the uv/visible range.
                                    b)  Solvents have an effect on the wavelength absorbed and the molar absorptivity, so they should be                                                             listed in electronic spectra data.
                        7.  Absorption of the energy occurs by the excitation of electrons to higher energy levels.
                                    a)  The electrons are promoted from a filled bonding molecular orbital (p) or a nonbonding
                                                molecular orbital (n) to an antibonding molecular orbital (p*).
                                    b)  These transitions are written as:
                                                i)     p à  p*
                                                ii)    n à  p*
                                    c)  The compound has been changed from its ground electronic state to an excited electronic state.
                        8.  The excitations are recorded as electronic spectra.  
                        9.  Sigma bonds require too great an energy to be used in this technique. 

            B.  Electronic Spectra and Delocalization

                        1.  Electronic spectra can indicate the size and degree of delocalization in an extended pi system.
                        2.  The greater the number of double bonds conjugated together in a molecule, the lower the energy of
                                    excitation will be.
                                    a)  There will also be a greater number of peaks in the spectrum.
                        3.  Examples of some molecules and their l _max follow.

                            4.  Notice that the first two compounds above are not conjugated and have a much higher energy (shorter
                                    wavelength) than the conjugated compounds.
                        5.  Beta-carotene has 11 double bonds in conjugation.
                                    a)  The wavelength absorbed is so long that it is in the visible range, giving beta-carotene its
                                                characteristic orange color.    
                        6.  Larger conjugated pi systems have lower-energy excited states because with more p-orbitals
                                    combining to form molecular orbitals, the energy gap between bonding and antibonding molecular
                                    orbitals gets smaller.
                                    a)  Also, more bonding and antibonding orbitals are available to rise to more electronic transitions.
                        7.  The structural unit associated with the electronic transition is referred to as a chromophore.
                        8.  In alkenes and polyenes the electronic transitions occur from the HOMO (highest occupied molecular
                                    orbital, a pi molecular orbital) to the LUMO (lowest unoccupied molecular orbital), the pi
                                                antibonding orbital.
                                    a)  In carbonyl compounds, it is the n à  p*  transition of the C=O group which occurs.
                                                i)  Here the electron being promoted is a nonbonding electron from the lone pair on oxygen.
                        9.  UV-visible spectroscopy does not offer as much info as NMR or IR, but is still an important tool.

Many thanks to Rod Oka of MPC for generously sharing his "Lecture Companion" outline,
reproduced here in extensively modified form by permission, with web references and other goodies added by me.
Structures drawn using  MDL IsisDraw™,  ACDLabs ChemSketch™, and  CS ChemOffice ChemDraw™.
Other graphics drawn with MS Excel™ [my favorite program]