--%>
Assignment Help - homework help

  Structure Elucidation of Organic Molecules, Chemistry tutorial

Introduction:

The information that might be obtained from ultraviolet-visible, infrared, proton and carbon-13 nuclear magnetic resonance and mass spectra is complementary, and it is much simpler to recognize the structure of a compound if all the spectra are considered. Each and every spectrometric method gives characteristic data to assist in the final recognition of the sample.  These have been considered in the individual topics; however necessitate to be combined to extract the maximum information.

Whenever an unknown material is presented for analysis, it must first be found out whether the sample is a single substance, or a mixture. The purity of substances like pharmaceuticals is extremely significant. Separation via an appropriate method must reveal the number of components in the sample.

Information from each Spectrum:

In order to study and recognize any unknown analytical sample by using spectroscopic methods, the analytical chemist should first get good quality spectra and then use these to choose the information from each method that is of most value. It is as well significant to recognize that other analytical observations must be taken into account. For illustration, if the sample is a volatile liquid and the spectral information recommends that it is an involatile solid, clearly, there is conflicting proof.

The use of computerized library databases can help in the matching of spectra to recorded illustrations. If difficulties are found in differentiating between the two possibilities for the sample identity, then it might be essential to consult reference texts or additional computer databases in such a way that an exact match is found. Several databases give information that assists when working by samples which are new compounds or whose spectra are not present in the database. For illustration, the presence of a strong peak in an IR spectrum close to 1700 cm-1 must recommend a high probability that the sample might be a carbonyl compound.

Spectroscopic Identification:

The conditions under which each and every spectrum has been obtained should be taken into consideration. For illustration, if the UV, IR and NMR spectra were run in solution, what was the solvent? The instrumental parameters as well require to be considered. In MS, the kind of ionization employed will influence the spectrum obtained.

Whenever the source of the analytical sample is known, this can be a great help in describing the identity of the material. It is a valuable exercise to follow the similar general scheme and note down the information that is expressed from the study of each spectrum. One recommended scheme is given below; however the value of 'feedback' in checking the deductions should not be overlooked.

1) Empirical Formula:

Occasionally, if the sample has been analyzed to determine the percentage of carbon, hydrogen, nitrogen, sulphur and other elements and to express the percentage of oxygen by difference, this can be a helpful primary step. If this information is not available, it might be found from the MS if an accurate relative molecular mass has been measured.  

Example: A solid sample contained C 75.5%, H 7.5% and N 8.1% by weight. Determine the empirical formula of the sample?

Dividing by the relative atomic masses provides the ratio of numbers of atoms, noting that there should be (100-75.5-7.5-8.10) = 8.9% oxygen.

C = 75.5/12 = 6.22

H = 7.5/1 = 7.5

N = 8.1/14 = 0.578

O = 8.9/16 = 0.556

This corresponds (approximately) to C11H13NO, having an RMM of 175, which might give a molecular ion in the mass spectrum.

2) Double Bond Equivalents:

The presence of instauration in a structure must be considered. As a saturated hydrocarbon has the formula CnH2n+2, and since a single-bonded oxygen can be thought of as equivalent to -CH2-, and a single-bonded nitrogen as -CH<, the number of double bonds or rings, known as the double bond equivalents (DBE) for the compound is represented by:

DBE = (2n4 + 2+ n3-n1)/2

Here, 

n4 is the number of tetravalent atoms (example: carbon)

n3 is the number of trivalent atoms (example: nitrogen), 

1 is the number of monovalent atoms (example: hydrogen or halogen).

Thus, for benzene, C6H6, the DBE is (14-6)/2, that is, three double bonds and one ring.

For illustration in (1) above, C11H13NO, the DBE is (24 + 1-13)/2 = 6, that would correspond to one benzene ring (4) plus one -C=C- plus one >C=O. Note that other spectra should be employed to differentiate between the ring and a double bond or between a -C=C- and a >C=O.

3)  The IR spectrum provides proof regarding the presence of functional groups. The illustration in (2) above would be resolved if the infrared spectrum showed no carbonyl to be present. The presence of aliphatic groups, or unsaturated or aromatic structures might be inferred from the position of the -C-H stretching bands around 3000 cm-1, and confirmed via the presence of other bands. A helpful application of Raman spectrometry is the detection of groups that encompass very weak absorbance in the infrared region, like substituted alkynes.

4) The UV spectrum does give a number of structural information, even when there is little or no absorbance, which would recommend the absence of any aromatic, conjugated or ketonic structures. If there are double bonds or unsaturated rings present, the UV spectrum must give further information.

5)  Much helpful information might be derived from the mass spectrum. A short summary of what you must look for should comprise:

a) The m/z of the molecular ion. This corresponds to the molecular formula that might be a multiple of the empirical formula. An odd value for the m/z of the molecular ion needs that an odd number of nitrogen atoms are present, as in the illustration above. Prominent isotope peaks point out the presence of Cl, Br or S.

b) The exact value of m/z of the molecular ion. For illustration, the nominal RMM of the illustration is 175, and the formulae might have been expressed if the exact mass was found out as 175.0998, as, excluding some impossible formulae, a few others are:

C8H5N3O2     175.0382

C7H13NO4     175.0845

C11H13NO     175.0998

C10H13N3      175.1111

c) The fragments present and the fragments lost.

6)  Both the 1H-NMR and the 13C-NMR provide necessary information regarding the kinds of protons and carbons present, their environment and their connections to neighboring atoms.

 7)  Before the final report is provided, it is for all time a good idea to retrace the steps above the check whether the data is self-consistent. For illustration, if there is no proof for the aromatic structures in the IR spectrum, is this consistent by the NMR spectrum?  If an isomer should be recognized, do the positions of the peaks in the IR and NMR spectra correspond, and does the fragmentation in the mass spectrum give proof?

Tutorsglobe: A way to secure high grade in your curriculum (Online Tutoring)

Expand your confidence, grow study skills and improve your grades.

Since 2009, Tutorsglobe has proactively helped millions of students to get better grades in school, college or university and score well in competitive tests with live, one-on-one online tutoring.

Using an advanced developed tutoring system providing little or no wait time, the students are connected on-demand with a tutor at www.tutorsglobe.com. Students work one-on-one, in real-time with a tutor, communicating and studying using a virtual whiteboard technology.  Scientific and mathematical notation, symbols, geometric figures, graphing and freehand drawing can be rendered quickly and easily in the advanced whiteboard.

Free to know our price and packages for online Chemistry tutoring. Chat with us or submit request at [email protected]