Chapter 14 Spectroscopic methods of structure determination ?????????

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Chapter 14 Spectroscopic methods of structure
determination ?????????

• 14.1 Introduction
• 1 Ultraviolet Spectroscopy (uv)
  ????----?????????????????????
• 2. Infrared spectroscopy (IR)????----??????????

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3. Nuclear Magnetic Resonance Spectroscopy(NMR)
??????----??????????? ?????????(1H
NMR and 13C NMR)

• 4. Mass Spectroscopy (MS)??----??????????

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The characteristic methods of determination

1. Microscale sample (1 5 mg)
2. It need short time to determine sample
3. Identify structure very fast.

1. ???????????

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14.1 The electromagnetic spectrum ????

• A wave is usually described in terms of its
  wavelength (??? ) or its frequency (???)
• The energy of quantum of electromagnetic energy
  (E) is directly related to its frequency.

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Since v c /?, the energy of electromagnetic
radiation is inversely proportional to its
wavelength.
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??(Energy)???(frequency)?????
????,???????? ????,????????
?????????????????????? (?
)?????,??????,????????
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The different regions of the electromagnetic
spectrum are shown in Fig.14.2.
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14.2 Visible and Ultraviolet Spectroscopy

• 1. What is it ultraviolet spectroscopy?

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A typical UV absorption spectrum, that of
2,5-dimethyl-2,4-hexadiene, is shown in Fig. 14.3.
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In addition to reporting the wavelength of
maximum absorption (?max ), chemists often report
another quantity that indicates the strength of
the absorption, called the molar absorptivity, e
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In the chemical literature this would be
reported as
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UV ???????????
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We can understand how conjugation of multiple
bonds brings about absorption of light at longer
wavelengths
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The greater the number of conjugated multiple
bonds a compound contains, the longer will be the
wavelength at which the compound absorbs light.
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ß Carotene ????
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Lycopene ????
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Compounds with carbon-oxygen double bonds also
absorb light in the UV region.
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200 400 nm (near UV)(????)(???????)

• ?????????? (??,?,?,??,???????

????????????? (- NH2, -NR2, -OH, -OR, -SR, -Cl,
-Br, -I), ? max ???????-----???????????
???
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Ultraviolet Spectroscopy (uv)?????
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• UV??????CH3OH, CH3CH2OH

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14.3 Infrared spectroscopy????

• Organic compounds also absorb electromagnetic
  energy in the infrared (IR) region of the
  spectrum. Infrared radiation does not have
  sufficient energy to cause the excitation of
  electrons, but it does cause atoms and groups of
  atoms of organic compounds to vibrate faster
  about the covalent bonds that connect them. These
  vibrations are quantized (???). And as they
  occur, the compounds absorb IR energy in
  particular regions of the spectrum.

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• Infrared spectrometers operate in a manner
  similar to that of visible-UV spectrometers. A
  beam of IR radiation is passed through the sample
  and is constantly compared with a reference beam
  as the frequency of the incident radiation is
  varied. The spectrometer plots the results as a
  graph showing absorption versus frequency or
  wavelength
• The location of an IR absorption band (or peak)
  can be showed representation of wavenumber

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IR???????????

• IR?????????

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The frequency of a given stretching vibration and
thus its location in an IR spectrum can be
related to two factors. These are the masses of
the bonded atoms---light atoms vibrate at higher
frequencies than heavier ones---and the relative
stiffness of the bond.

• ???????,???????,???????????,??, ???????????

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The stretching frequencies of groups
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Aldehydes, ketones, Esters, and carboxylic acids
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Frequency range of functional groups
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The peaks of frequency range for hydroxyl group
and triple bond
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The peaks of frequency range for carbonyl group
and nitrile group
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????????

• ?????????????????,????????,??????????????
• ????,????????,????,????,??????,?,?,??????????

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????Degrees of unsaturations (O)????
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3-pentanone CH3CH2COCH2CH3 Diethyl ketone
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Degree of unsaturation is 4
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C8H8O
Degree of unsaturation is 5
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• 3400 cm-1 no OH or NH peak present 3100 cm-1
  strong peak suggesting sp2 CH.
• 2900 cm-1 moderate peak suggesting sp3 CH . 2200
  cm-1 strong unsymmetrical triple bond peak
• 1690 cm-1 no carbonyl absorbance. 1610 cm-1
  peaks suggesting Ar CC

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Degree of unsaturation is 5
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O1

• 3400 cm-1 strong OH or NH peak present 3100
  cm-1 minor peak suggesting sp2 CH
• 2900 cm-1 minor peak suggesting sp3 CH 2200
  cm-1 no unsymmetrical triple bond peak
• 1650 cm-1 strong carbonyl absorbance 1550 cm-1
  weak peak suggesting NH bending

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• 3400 cm-1 no OH or NH peak present. 3100 cm-1
  no peak suggesting sp2 CH.
• 2900 cm-1 minor peak suggesting sp3 CH. 2200
  cm-1 no unsymmetrical triple bond peak.
• 1760 cm-1 strong carbonyl absorbance. 1600
  cm-1 no peak suggesting CC.

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A Nicolet 800 FT-IR, used to record infrared
spectra in the range of 400 to 4000 cm-1) 
??????1. KBr??? 2. ???(NaCl??) 3 ????
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14.4 Nuclear Magnetic resonance
spectroscopy?????

• 1H NMR---??????????????????

The nuclei of certain elements and isotopes
behave as though they were spinning about an
axis. The nuclei of ordinary hydrogen (1H) and
those of carbon-13 (13C) have this property.
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????????????????????????,?1H, 13C, 19F, 31P,
??????????????????(Ho)?, ???????????,?????????,???
??????
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• ?????????????,?????????,????????????,???????????,?
  ?????????,???????????(1H NMR)

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Fig. 14.9 Essential parts of a nuclear magnetic
resonance spectrometer.
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Fig. 14.10 Shows the 1H NMR spectrum of p-xylene
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We have an example of signal splitting.
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14.6 Shielding and Deshielding of Protons
(?????????)

• ?3-???, ?????????,????????????,??????????????????,
  ??????????,???????????,??????,??????H ??,
  ????????,d ??????????,???,?????????d ????

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Fig. 14.15 The circulations of the electrons of
a C-H bond under the influence of an external
magnetic field
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The electron circulations generate a small
magnetic field ( an induced field) that shields
the proton from the external field

• ?????????????,?????,??????????,d?????????????????,
  ???????,??????????,d????

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Fig. 14.17 the induced magnetic field of the
pelectrons of benzene deshields the benzene
protons
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18 Annulene. The internal protons are highly
shield. The external protons are highly
deshielded.
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Explanation
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14.7 The chemical shift ????d

• ???????????????,????????,????????,???????d
  (chemical shift)

Chemical shifts are measured with reference to
the absorption of protons of reference compounds.
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• The signal from the 12 equivalent protons of TMS
  is used to establish the zero point.
• A very small amount of TMS gives a relatively
  large signal.
• They give a single signal.
• Since silicon is less electronegative than
  carbon, the protons of TMS are in regions of high
  electron density.
• Tetramethylsilane, like an alkane, is relatively
  inert.
• It is volatile, its boiling point is 27 oC. After
  the spectrum has been determined, the TMS can be
  removed easily by evaporation

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Table 14.3 Approximate proton chemical shifts.
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14.8 Chemical shift equivalent and nonequivalent
protons

• Two or more protons that are in identical
  environments have the same chemical shift and,
  therefore, give only one 1H NMR signal. How do we
  know when protons are in the same environment?
  For most compounds, protons that are in the same
  environment are also equivalent in chemical
  reactions. That is, chemically equivalent protons
  are chemical shift equivalent in 1H NMR.

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Problem 14.5 How many different sets of
equivalent protons do each of the following
compounds have? How many signals would each
compound give its 1H NMR spectrum?
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14.9 Signal splitting(????) spin-spin
coupling(??-????)

• Signal splitting is caused by magnetic fields of
  protons on nearby atoms.
• ??????????,????????,??????????????????????????????
  ,?????????????????,??????????(n 1)??(n
  ????????)????????????(d)?????????

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Fig. 14.20 Signal splitting in 1,1,2-trichloroetha
ne
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Tert-Butyl methyl ether ?-???
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1,1,2-Trichloroethane
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Fig. 14.26 the 1H NMR spectrum of ethyl bromide.
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Explanation
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????????

• ???a?n?????b??,???a??????n?????b?????(n1)??,?????
  ???????(ab)n??????????

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P 584 Fig. 14.27 The 1H NMR spectra
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B. C2H4Br2
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C, C3H6Cl2
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Proton NMR spectra have other features

• 1. Signals may overlap
• 2. Spin-spin couplings between the protons of
  nonadjacent atoms may occur.
• 3. The splitting patterns of aromatic groups are
  difficult to analyze. Because of long-range
  couplings, the phenyl group appears as a very
  complicated multiplet.

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Signals may overlap----The singlet from the
protons of (b) falls on one of the outmost peaks
of the quartet from (C).
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A monosubstituted benzene ring has three
different kinds of protons.
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Fig 14.3 The 1H NMR spectrum of 1-nitropropane
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Fig.14.3 The 1H NMR spectrum of ordinary ethanol
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For example
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C4H8O2 From the molecular formula, the compound
has "1degrees of unsaturation" (one double bond
or ring).

• The proton NMR has three peaks a quartet at d
  4.1 (2H), a triplet at d 1.2 (3H) and a singlet
  at d 1.95 (3H). The quartet and triplet suggest a
  CH2 coupled to a CH3 in an ethyl group. The peak
  at d 4.1 is in the area generally observed for CH
  groups adjacent to electronegative groups, i.e.,
  oxygen. The peak at d 2 is in the region for a
  methyl group adjacent a carbonyl.

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From the molecular formula (C8H8O), the compound
has "5 degrees of unsaturation" (four double
bonds or rings), suggesting the possibility of an
aromatic compound (benzene has four degrees of
unsaturation) attached to another unsaturated
center.
Explanation for (C8H8O)
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• The proton NMR has three peaks a singlet at d
  2.2 (3H), and a singlet at d 10 (1H) and two
  doublets centered around d 7.6. The doublets
  centered at d 7.6 are in the aromatic region the
  fact that two doublets are observed (2H each)
  suggests a 1,4-disubstituted aromatic compound.
  The peak at d 2.2 is in the region for a methyl
  group adjacent a mildly electronegative group.
  The singlet at d 10 is in the region observed for
  aldehydic protons. The presence of two doublets
  in the aromatic region is highly characteristic
  of 1,4-disubstitution.

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From the molecular formula (C10H14O), the
compound has "4 degrees of unsaturation" (four
double bonds or rings), suggesting the presence
of an aromatic compound (benzene has four degrees
of unsaturation).

• Explanation for C10H14O

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• The proton NMR has four sets of peaks a singlet
  at d 3.6 (3H), two sets of doublets centered
  around d 6.9 (4H), a septet at d 2.7 (1H) and a
  doublet at d 1.6. The singlet at d 3.6 is
  consistent with an isolated CH3 adjacent to an
  electronegative center, such as an oxygen. The
  septet and doublet strongly suggest an isopropyl
  group CH(CH3) 2 in which the carbon is bonded to
  something mildly electronegative and the two
  doublets centered at d 6.9 strongly suggest a
  1,4-disubstituted aromatic compound.

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The instrument of 500M NMR
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Using Deuterium? solvents dissolve sample for NMR
determinations
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Special Topic D---Mass Spectrometry (??)

• In a mass spectrometer molecules in the gaseous
  state under low pressure are bombarded with a
  beam of high-energy electrons. The energy of the
  beam of electrons is usually 70 ev (electron
  volts) and one of the things this bombardment can
  do is dislodge one of the electrons of the
  molecule and produce a positively charged ion
  called the molecular ion.
• M e- M . 2e-

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Fig.D.1 Mass spectrometer
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• The instrument of Mass spectra

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Sample molecules enter here
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• Magnetic field

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Collector for ions and spectrum recorder
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?-??????(GC-Mass)
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The mass spectrum of bromomethyl benzene is shown
below.
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C9H10O
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?(IR, 1H NMR, MS) ?????????????

• The molecular formula for an unknown compound is
  C5H10O. Data for the infrared, 1H NMR and mass
  spectra for this compound are as follows

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• 3400 cm-1no OH or NH present
• 3100 cm-1no peak to suggest sp2 CH
• 2900 cm-1strong peak indicating sp3 CH
• 2200 cm-1no unsymmetrical triple bonds
• 1710 cm-1strong carbonyl absorbance
• 1610 cm-1no peak to suggest CC

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• The proton NMR has a septet at d 2.8 coupled to a
  doublet at d 1.1, strongly suggesting the
  presence of an isopropyl group, and a singlet at
  d 2.2 suggesting an isolated methyl group next to
  a mildly electronegative group (a carbonyl?).

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