CHEM10007 Lecture Notes - Lecture 34: Nuclear Magnetic Resonance Spectroscopy, Analytical Chemistry, Infrared Spectroscopy
12 Jun 2018
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LECTURE 34
QUALITATIVE ORGANIC ANALYSIS
CHAPTER 16 - BIOCHEMISTRY
QUALITATIVE ANALYSIS
DETERMINING THE STRUCTURE OF AN UNKNOWN ORGANIC COMPOUND
•Scientists wanted to make compounds less/more toxic.
•Eg. For cancer, needed to make them more potent. Some cancer drugs kill cancer cells but
also kill healthy normal cells and this needed to be fixed. So scientists wanted to improve these
drugs.
•Approximately 70% of drugs on the market have come from natural sources.
•The techniques which are used routinely by the chemist to determine the structure of an unknown
compound include;
•Combustion Analysis; (empirical formula), the first step to identifying structure is to find the
ratios of elements in a compound.
•Mass Spectrometry (MS); (determination of molecular weight and molecular formula).
hUltraviolet Spectroscopy (UV)
•Infrared Spectroscopy (IR); (functional groups present)
•Nuclear Magnetic Resonance Spectroscopy (NMR) (gives info about the C/H framework; how it
is structured, gives meaningful info about the structures of organic compounds).
hX-Ray Diffractometry (X-Ray) most definitive tool. Uses X-Rays which have a wavelength
similar to the distance of the space
between atoms, and proves a precise 3D
structure from the light refractions.
hA limitation is that it requires growing
crystals, and not all organic compounds
are naturally solid.
•Spectrometry measures mass while
spectroscopy involves shining light on a
molecule.
COMBUSTION ANALYSIS
•Used to determine the empirical
formula of a pure organic compound by
combusting under conditions where the
resulting combustion products can be quantitatively analysed.
•A Lecture 10 Example:
•A 2.203 g sample of an organic compound (containing C, H and O) was extracted from a plant.
When it was burned in oxygen, 1.32 g of water and 3.23 g of carbon dioxide were collected.
•(a) Find the empirical formula of the compound.
•(b) Another sample was analysed in a mass
spectrometer. The mass spectrum produced showed
that the molar mass of the compound was 60.0 g mol–
1.!
What is its molecular formula?
•Answer a):
•n(H2O) = m/Mr = 1.32 /18 = 0.073 mol
•n(H) = 2 x n(H2O) = 0.15 mol
•m(H) = 0.15 x 1.0 = 0.15 g
•n(CO2) = 3.23 / 44 = 0.073 mol
•n(C) = n(CO2) = 0.073 mol
•m(C) = 0.073 x 12 = 0.88g
•m(O) = 2.203 – m(C) – m(H) = 2.203 – 0.88 – 0.15
•= 1.173g
•= 1.17g.
•Answer b):
•empirical formula mass CH2O!
= 12.00 + (1.00 × 2) + 16.00 = 30.00.
•The molar mass of the molecule =given molecular formula mass = 60.00
•Therefore the molecular formula mass/empirical formula mass = 60.00 / 30.00 = 2
•Therefore the molecular formula is double everything in the empirical formula = C2H4O2.
MASS SPECTROMETRY
•Used to determine the
molecular mass of an
organic compound based
on their mass to charge
ration (m/z).
•Eg. Trying to move a
bowling ball with water
squirting from a hose will
work less effectively than
trying to move a tennis
ball. In this example the
bowling ball represents a
large molecule and the
tennis ball represents a
small molecule.
•Inside of spectrometer is
a vacuum, there’s no gas.
•Under the conditions of
mass spectrometer, the
molecule is fragmenting.
