Elemental analysis & Structure Elucidation
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Alkhair Adam Khalil,
B. Pharm., M. Pharm.
Department of Pharmaceutical Chemistry
College of Pharmacy – Karary University
There are four data required for structure
There are four data required for structure
elucidation:
elucidation:
•
Elemental
analysis
•
Mass spectrometry
(MS)
•
Infrared
spectroscopy
(IR)
•
Nuclear magnetic spectroscopy
(NMR)
Elemental analysis
Elemental analysis
•
Calculation of the Empirical formula and molecular formula
Calculation of the Empirical formula and molecular formula
•
The empirical formula
The empirical formula
of a compound is the
simple
ratio
of atoms present in
such compound.
•
The molecular formula
The molecular formula
is the
actual
number
of
atoms present in a
compound.
•
Suppose
we
have
a new compound whose
elemental
analysis gave the following:
•
Carbon 67.95%
•
Hydrogen 5.69%
•
Nitrogen 26.2%
•
Total 99.84
•
Since the total of these
percentage
is close
to
100% the small difference is
experimental error that is there is
no other element
like
oxygen.
To calculate the empirical formula:
1.
Convert the % mass of each element to moles by dividing the % mass
with atomic weight of the particular element
•
C = 67.95/
12
=
5.66 mole
•
H = 5.69/
1
=
5.69 mole
•
N
= 26.2/
14
=
1.87 mole
2.
Divide the mole ratios with the smallest number and then round ratio
to the nearest whole number
•
C: 5.66/
1.87
= 3.03 which ≈
3
•
H: 5.69/
1.87
= 3.02 which ≈
3
•
N: 1.87/
1.87
=
1
•
Thus the empirical formula of the compound is
C
3
H
3
N
To find the
molecular formula it is necessary to know the molecular weight.
•
Histidine an amino acid isolated from protein gave
the following elemental analysis: C,
46.38%
, H,
5.90%
, N,
27.01%
.
•
Total is
79.29
in this case the difference is
oxygen
i.e.
100- 79.29 =
20.71%
•
To get the ratios as moles divide by atomic weight:
•
C: 46.38/
12
=
3.86,
H:5.9/
1.0
=
5.90,
N: 27.29/
14
=
1.94,
O: 20.71/
16
=
1.29
Example:
Example:
•
Now divide with the smallest number which is
1.29
•
If the ratios numbers are too far multiply with factor
(2 or 4).
•
In the above example we need to multiply by 2 and
the empirical formula is
C
6
H
9
N
3
O
2
•
Mass spectrometry showed that histidine has the
molecular weight
155
•
The
empirical
formula
weight
155
, therefore
molecular formula of histidine is also
C
6
H
9
N
3
O
2
•
The way in which these atoms are joined together to
give histidine is called the molecular structure of
histidine this is determined by both physical and
chemical methods.
Problem 1
Problem 1
•
Aspartame (artificial sweetener agent) shows an
elemental analysis as follows:
•
57.14%
C
,
•
6.16%
H
,
•
9.52
N
•
calculate the empirical formula and find the
molecular formula (
M.wt = 294.3 g/mol
).
•
57.14+6.16+9.52 = 72.82%
•
The remaining is oxygen = 27.18
•
57.14%
C
, = 57.14/12 = 4.762 moles
•
6.16%
H
, = 6.16/1 = 6.16 moles
•
9.52%
N
, = 9.52/14 =
0.68
mole
•
27.18%
O
= 27.18/16 = 1.7 moles
•
Divide by 0.68
C
7
H
9
NO
2.5
•
Multiply by 2
C
14
H
18
N
2
O
5
Molecular mass =
294
•
Therefore the empirical formula is the molecular formula
•
C
14
H
18
N
2
O
5
294.3 gram/mole
Degree of unsaturation = 7 (
(aromatic ring = 4)
+
3 C=O
)
Aspartame
Problem 2
Problem 2
•
Citral is a chemical responsible for the odour
of lemon. Combustion analysis shows that
citral to contain 78.9%
C
, 10.6%
H
.
•
Assuming that the reminder is
oxygen
what is
the empirical formula of citral.
•
If citral has a molecular wt of 152 what is its
molecular formula
Steps to be followed for structure elucidation
Steps to be followed for structure elucidation
1.
Determine the possible molecular formula
MF
from data
provided by elemental analysis and Mwt.
2.
Calculate the
degree of unsaturation
from the (
MF
).
3.
Determine
structural
and
residual
units (using NMR, UV and IR).
–
Examine the IR spectrum to determine the presence of suspected
functional groups.
–
Examine the NMR spectrum and note the number of peak groups and their
integrals, chemical shifts and attempt to determine the proximities of
nearby protons to the functional group.
–
Examine the NMR spectrum for peak multiplicity in order to determine the
number of protons on neighbouring groups. This will give the proton
sequence that determines which isomer is the correct structure.
4.
Combine simple units into complex unit.
5.
Arrange simple and complex units in molecular structure Confirm
by
MS
fragments.
•
When constructing structural formulae from
individual units polyvalent units are combined
first in all possible ways then monovalent groups
or units are attached in all possible ways.
•
Finally check the various possible structures to
see if they are consistent with all the known
information about the unknown compound.
Problem
Problem
•
Propose all possible structures
Practice Example
Practice Example
Questions !
Understanding elemental analysis and structure elucidation is crucial in chemistry. Learn the process, calculations, and methods involved, such as mass spectrometry, infrared spectroscopy, and nuclear magnetic spectroscopy. Discover how to determine empirical and molecular formulas using examples like histidine, an amino acid. Explore the importance of molecular structure determination.
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Presentation Transcript
Elemental analysis & Elemental analysis & Structure Elucidation Structure Elucidation Alkhair Adam Khalil, B. Pharm., M. Pharm. Department of Pharmaceutical Chemistry College of Pharmacy Karary University
There are four data required for structure elucidation: Elementalanalysis Mass spectrometry(MS) Infrared spectroscopy(IR) Nuclear magnetic spectroscopy(NMR)
Elemental analysis Calculation of the Empirical formula and molecular formula The empirical formula of a compound is the simple ratioof atoms present in such compound. The molecular formula is the actual number ofatoms present in a compound. Suppose we have a new compound whoseelemental analysis gave the following: Carbon 67.95% Hydrogen 5.69% Nitrogen 26.2% Total 99.84 Since the total of these percentage is close to 100% the small difference is experimental error that is there isno other element like oxygen.
To calculate the empirical formula: 1. Convert the % mass of each element to moles by dividing the % mass with atomic weight of the particular element C = 67.95/ 12 = 5.66 mole H = 5.69/1 = 5.69 mole N = 26.2/14 = 1.87 mole 2. Divide the mole ratios with the smallest number and then round ratio to the nearest whole number C: 5.66/1.87 = 3.03 which 3 H: 5.69/1.87 = 3.02 which 3 N: 1.87/1.87 = 1 Thus the empirical formula of the compound is C3H3N To find the molecular formula it is necessary to know the molecular weight.
Example: Histidine an amino acid isolated from protein gave the following elemental analysis: C, 46.38%, H, 5.90%, N, 27.01%. Total is 79.29 in this case the difference is oxygen i.e. 100- 79.29 = 20.71% To get the ratios as moles divide by atomic weight: C: 46.38/12 = 3.86, H:5.9/1.0 = 5.90, N: 27.29/14 = 1.94, O: 20.71/16 = 1.29
Now divide with the smallest number which is 1.29 If the ratios numbers are too far multiply with factor (2 or 4). In the above example we need to multiply by 2 and the empirical formula is C6H9N3O2
Mass spectrometry showed that histidine has the molecular weight 155 The empirical formula weight 155, therefore molecular formula of histidine is also C6H9N3O2 The way in which these atoms are joined together to give histidine is called the molecular structure of histidine this is determined by both physical and chemical methods.
Problem 1 Aspartame (artificial sweetener agent) shows an elemental analysis as follows: 57.14% C, 6.16% H, 9.52 N calculate the empirical formula and find the molecular formula (M.wt = 294.3 g/mol).
57.14+6.16+9.52 = 72.82% The remaining is oxygen = 27.18 57.14% C, = 57.14/12 = 4.762 moles 6.16% H, = 6.16/1 = 6.16 moles 9.52% N, = 9.52/14 = 0.68 mole 27.18% O = 27.18/16 = 1.7 moles Divide by 0.68 C7H9NO2.5 Multiply by 2 C14H18N2O5 Molecular mass = 294 Therefore the empirical formula is the molecular formula
Aspartame C14H18N2O5 294.3 gram/mole Degree of unsaturation = 7 ( (aromatic ring = 4) + 3 C=O)
Problem 2 Citral is a chemical responsible for the odour of lemon. Combustion analysis shows that citral to contain 78.9% C, 10.6% H. Assuming that the reminder is oxygen what is the empirical formula of citral. If citral has a molecular wt of 152 what is its molecular formula
Steps to be followed for structure elucidation Determine the possible molecular formula MF from data provided by elemental analysis and Mwt. Calculate the degree of unsaturation from the (MF). Determine structural and residual units (using NMR, UV and IR). Examine the IR spectrum to determine the presence of suspected functional groups. Examine the NMR spectrum and note the number of peak groups and their integrals, chemical shifts and attempt to determine the proximities of nearby protons to the functional group. Examine the NMR spectrum for peak multiplicity in order to determine the number of protons on neighbouring groups. This will give the proton sequence that determines which isomer is the correct structure. Combine simple units into complex unit. Arrange simple and complex units in molecular structure Confirm by MS fragments. 1. 2. 3. 4. 5.
When constructing structural formulae from individual units polyvalent units are combined first in all possible ways then monovalent groups or units are attached in all possible ways. Finally check the various possible structures to see if they are consistent with all the known information about the unknown compound.
Problem Propose all possible structures
