chm 3402 experiment 8

8/7/2019 CHM 3402 Experiment 8

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LABORATORY REPORT OF SPECTROSCOPY( CHM 3402 )

Name : Lai Sok Fuen

Matric No : 110484

Partners Name : Tan May Thing

Matric No : 110123

Programme : Bacelor Sains ( Kepujian ) Chemistry

Expt.Tittle : Nuclear Magnetic Resonance Spectrometer

No.Expt : 8

Date : 13/1/2004

Demo : Masnizaayu Abd.Manaf, Chen Lai Min

Lab : MP4


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Experiment 8: Nuclear Magnetic Resonance Spectrometer

Objectives:

1. To understand the basic concept of NMR spectroscopy

2. To know the basic components/part of NMR spectrophotometer and their functions.3. Confirm the possible structures for the sample given from chemical shifts and the

number of protons represent by each peak.4. To determine completely the structure of an unknown molecule.

Introduction:

Nuclear magnetic resonance (NMR) is a spectroscopic method that is even moreimportant to the organic chemist than infrared spectroscopy. Many nuclei may be studied

by NMR techniques, but hydrogen and carbon are most commonly available. Whereas

infrared spectroscopy reveals the types of functional groups that present in a molecule, NMR gives information about the number of magnetically distinct atoms of the type being studied. When hydrogen nuclei (protons) are studied, for instance, one candetermine the number of each of the distinct types of hydrogen nuclei as well as obtaininformation regarding the nature of the immediate environment of each type. Similar information can be determined for the carbon nuclei. The combination of IR and NMR data is often sufficient to determine completely the structure of and unknown molecule.

The chemical shift and shielding

Nuclear magnetic resonance (NMR) has great utility because not all protons in amolecule have resonance at the same frequency. This variability is due to the fact that the

protons in a molecule are surrounded by electrons and exits in slightly different electronicenvironment from one another, the valence-shell electron densities vary from one protony to another. The protons are shielded by the electrons that surround them. In an appliedmagnetic field, the valence electrons of the protons are caused to circulate. This circulate,called a local diamagnetic current, generates a counter magnetic field, which opposes theapplied magnetic field. Figure 8.1 illustrates this effect, which is called diamagneticshielding or diamagnetic anisotropy.

Circulation of the electrons around a nuclear can be viewed as being similar to theflow of an electric current in an electric wire. From physics, we know that the flow of acurrent through a wire induces a magnetic field. In an atom, the local diamagnetic currentgenerates a secondary, induced magnetic field, which has a direction opposite that of theapplied magnetic field.


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Figure 8.1: Diamagnetic anisotropy the diamagnetic shielding of a nuclear caused by the circulation of valence electrons.

As a result of diamagnetic anisotropy, each proton in a molecule is shielded fromthe applied magnetic field to an extend that depends on the electron density surrounding

it. The greater the electron density around a nucleus. The greater the induced counter field that opposes the applied field. The counter field that shields a nucleus diminishesthe net applied magnetic field that the nucleus experiences. As a result, the nucleus is at alower frequency. This means that it also absorbs radiofrequency radiation at this lower frequency. Each proton in a molecule is in a slightly different chemical environment andconsequently has a slightly different amount of electronic shielding, which results in aslightly different resonance frequency.

The continuous-wave(CW) instrument

Figure 8.2: Schematically illustrates the basic elements of a classical 60-MHz NMR spectrometer. The sample is dissolved in a solvent containing no interfering protons(usually CCl 4 or CDCl 3),and a small amount of TMS is added to serve as aninternal reference. The sample cell is a small cylindrical glass tube that is suspended inthe gap between the faces of the pole pieces of the magnet. The sample is spun around itsaxis to ensure that all parts of the solution experience a relatively uniform magnetic field.

Figure 8.2: The basic elements of the classical nuclear magnetic resonancespectrometer.


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Also in the magnet gap is a coil attached to a 60-MHz radio frequency (RF)generator. This coil supplies the electromagnetic energy used to change the spinorientations of the protons. Perpendicular to the RF oscillator coil is a detector coil. Whenno absorption of the energy taking place, the detector coil picks up none of the energy off the RF ascillator coil. When the sample absorbs energy, however, the reorientation of the

nuclear spins induces a radiofrequency signal in the plane of the detector coil and theinstrument responds by recording this as a resonance signal or peak.

At constant field strength, the distinct types of the protons in a molecule precessat slightly different rates. Rather than changing the frequency of the RF oscillator toallow each of the protons in a molecule to come into resonance, the typical NMR spectrometer uses a constant-frequency RF signal and varies the magnetic field strength.As the magnetic field strength is increased, the precessional frequencies of all the protonsincrease. When the precessional frequency of a given type or proton reaches 60-MHz,ithas resonance. The magnet that is varied is actually a two-part device. There is a mainmagnet, with strength of about 1.41 Tesla, which is capped by electromagnet pole pieces.

By varying the current through the pole pieces, the worker can increase the main fieldstrength by as much as 20 parts per million(ppm).Changing the field in this waysystematically brings all of the different types of protons in the sample into resonance.

Chemical environment and chemical shift

If the resonance frequencies of all protons in a molecule were the same, NMR would be of little use to the organic chemist. Not only do different types of protons havedifferent chemical shifts, but each also has a characteristic value of chemical shifts. Everytype of protons has only a limited range of values over which it gives a clue as to thetype of proton originating the signal, just as an infrared frequency gives a clue as to the

type of bond or functional group. It is important to learn the ranges of chemical shiftsover which the most common types of protons have resonance. Figure 8.3 is a correlationchart contains the most essential and frequency encountered types of proton.

Figure 8.3: A simplified correlation chart for proton chemical shift values.


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Methodology:

The given sample is weighted and put inside the NMR tube.

0.5ml solution and tetrametilsilane-TMS 1% are put into the NMR tube to dissolve thesample. Make sure that the volumes of sample come to minimum volume that is 0.5ml or

35mm.

Mop the NMR tube to make sure it is in dry condition.

The spectrometer has to standard with 5%TMS and 35% chloroform before usedit.

The NMR tube with the sample inside it is put into the probe that has been prepared.

The spectrum of NMR sample has been obtained.

The integration of each peak has been measure by measuring the height of the peak.

The integration relative for the peaks have been compare, and expected the sum of the protons for every peak.

The chemical shift for every peak is marked.

According to the data that we get, the structure of the sample can be obtained.

Result and the analysis data:


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A) The spectrum of the sample A/COCl 3

/ppm The data that we get0.000 TMS7.336

7.302

Doublet peak. There is 2-atom hydrogen, which is

neighbour to each other at one-C aromatic ring.7.021 This is the peak for the solvent.6.974 Singlet peak. There has no neighbour to the atom-H

at C-aromatic ring.3.8613.7752.606

Singlet peak.

The formula molecule for sample A is:C 10H1203While, the value of the reference, (TMS) = 0.00ppm

The proposed structure for the sample A is as below:

There are 2 readings for doublet peak in the spectrum for the part (a) and (b). Theyare at 7.336ppm and 7.302ppm. According to the (n+1) rule, n+1=1+1=2.This mean that

part (a) and (b) are neighbour for each other. By this, there occurred a doublet peak at thisrange. The H proton bonded to the benzene ring give the reading of 6.974ppm. It has noneighbour with hydrogen atom, thus according to the (n+1) rule, n+1=0+1=1, so there isonly a peak in the spectrum, it is a singlet peak foe the part(c).

As for the part (d), (e) and (f), they have no neighbour which contain proton H. So,they give only a singlet peak. The electronegative effect from the carbonyl group, C=O

produced deshielded effect and this will cause the chemical shift move to the lower fieldwhich is to the left side. The reading for the part (d), (e) and (f) is 3.861ppm, 3.775ppmand 2.6061ppm respectively. Actually, it is found that each carbon atom bonds three

protons. But, because of the chemical vibration, it gives only one peak, which is high as a

group. The peak for the chemical shift 2.325ppm, shown clear here, but we cantdetermine it. This maybe cause by the contamination of the solution.

B) The spectrum for sample B/COCl 3


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/ppm The data that we get0.000 TMS1.1381.2571.377

Triplet peak. There is a proton from CH 3 which possesses 2 neighbours.

2.039 The methyl group has no proton nearby singlet peak.3.9614.0784.1974.317

Quartet peak. The proton from CH 2 which is equivalentgives one peak that gives 3-proton neighbour.

The formula molecule for sample B is: C 4H802While, the value of the reference, (TMS) = 0.00ppm

The proposed structure for the sample B is as below:

The TMS absorption for this sample B is shown at = 0.00ppm.There are 3 types of proton which shown with 3 peaks in the spectrum.

The peak consist of a set of triplet peak at = 1.138,1.257 and 1.377ppm. As for thesinglet peak, that is at = 2.039ppm and for the quartet peak are at = 3.961, 4.078,

4.197 and 4.317ppm.

The part (a) is the triplet peak. We found that the spin coupling occurred and the proton of CH 3 give a peak in the spectrum. The proton for CH 3 has two neighbour, so byusing the (n+1) rule, (n+1) = 2+1=3, there were 3 peaks we can see. A chemical shiftoccurred for this triplet peak.

There is a methyl group that has no H proton for the singlet peak. According to the(n+1) rule, (n+1)=0+1=1, means that is only one peak (singlet) there. The chemical shiftfor the methyl group in part(c) is greater than the methyl group in part (a). This is becausethe methyl group in part (c) is bonded to the carbonyl group, C=O which is a group has a

strong electronegative effect that can cause the CH 3(c) move to the lower field.

For quartet peak, there is a peak produced from the CH 2 group of (b). The 2 protonsfrom CH 2 are equivalent and there are 3-proton neighbour nearby which are the hydrogenfrom CH 3 group (a). So, by using the (n+1) rule, (n+1) = 3+1= 4,therefore the quartet

peak is produced.

Discussion:


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Nuclear Magnetic Resonance (NMR) Spectroscopy is the studied of the molecular structure through the measurement of the interaction of radio-frequency electromagneticradiation with a collection of nuclei immersed in a strong magnetic field. NMR is the firstmethod of structure determination organic chemists turn to provide a map of carbon-

hydrogen framework of an organic molecule. This involved measuring the absorption of radiofrequency radiation by a material placed in a strong magnetic field. Many nucleimay be studied, but H and C are most commonly available.

The spectrum NMR that we used is spectrum NMR proton ( 1H). Spectrum NMR will show out the series of peak. The wide of each peak is linear with the number of

proton where it is represent by the integrator. The wide of the peak is measured by the pengamir electronic, which will show a series, which is related to the certain absorption,or we called it kamiran curve. Ratio, height step will give different ratio nucleus proton inthe compound.

Not every nucleus1

H or in the molecule will absorb the energy radio frequency, rf atthe same frequency. The entire nucleus is surrounding by the electrons. When they cometo the outside magnet field, they will form their own magnet field. This magnet field iscontrary with the outside magnet field, so that this field is effective for the nucleus to

become smaller. By this, we can say that nucleus is shielded.

Every nucleus in the compound is in the different electronic surroundings, soshielding for every nucleus is different, and this mean that effective magnet field is notthe same for every nucleus. This differentiate can be observe through the NMR spectroscopy, and we can see this signal for every C or H in the compound.

The spectrum NMR is given in the chart to show the strength field is increased fromleft to the right. So, the left side of the chart is down field and the right side is high field.To define position of absorption, reference point used, TMS for both proton and C-13

NMR-a singlet peak. Chemical shift is referring to the exact place on the chart at which anucleus absorbs. The chemical shift of NMR absorption given in ppm or units isconstant, regardless of the operating frequency of the spectrometer.

The chemical shift for the reference (TMS) is 0ppm. The scale for the NMR chartis 1 = 1ppm frequency spectrometer. So, if a spectrum 1H is getting from the instrumentoperated at 60MHz, 1 is 1ppm from 60,000,00 or 60Hz. If we use the instrument300MHz, so 1 = 300Hz.

= chemical shift (Hz from the TMS) x 10 6

Frequency spectrometer (Hz)

Chemical shift can reflect the structure of the molecule, so it can use to analysisthe structure of the unknown compound. Hydrogen is most abundant particle in manyorganic compounds, so this method is widely used as we mention before. The strength for the line in the spectrum is linear to the sum of the proton inside it. We can know the


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distribution of the atom hydrogen in the organic molecule and get the relativedistribution. The chemical shift and the integral can give the information about thenumber and types of hydrogen contained in a molecule. A third type of information to befound in the NMR spectrum is that derived from the spin-spin splitting phenomenon. This

phenomenon can be explained by the n+1 rule.

The formula molecule for sample A is C 10H1203. The proposed structure for thesample A is as below:

There are 2 readings for doublet peak, they are 7.336ppm and 7.302ppm. The H proton bonded to the benzene ring give the reading of 6.974ppm. The reading for thesinglet peak is 3.861ppm, 3.775ppm and 2.6061ppm respectively.

The formula molecule for sample B is C 4H802. The proposed structure for thesample B is as below:

The peak consist of a set of triplet peak at = 1.138,1.257 and 1.377ppm. As for thesinglet peak, that is at = 2.039ppm and for the quartet peak are at = 3.961, 4.078,4.197 and 4.317ppm.

To get the more accurate result, we should take into consideration on the followingprecaution steps :

Use distilled water to rinse all the relevant apparatus. To get a more homogen solution, the sample should be rinse with the solvent. The NMR tube must be in the dry and clean condition. The NMR tube must handle with carefully because it is very expensive and

fragile.

Make sure that the tube is placed vertically into the prob. Make sure that the height of the sample is ~35mm, if too little it cannot be

detected, and the result will become inaccurate if the sample is too much in thetube.

Use the suitable solvent to dissolve the sample completely.

Conclusion:


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1) The structure of sample A is

\

2) The structure of sample B is

3) The formula molecule for sample A is C 10H1203.4) The formula molecule for sample B is C 4H802.5) The spectrum of the sample A.

/ppm The data that we get0.000 TMS7.3367.302

Doublet peak. There is 2-atom hydrogen, which areneighbors to each other at one-C aromatic ring.

7.021 This is the peak for the solvent.6.974 Singlet peak. There have no neighbors to the atom-H

at C-aromatic ring.3.8613.7752.606

Singlet peak.

6) The spectrum of the sample B.

/ppm The data that we get0.000 TMS1.1381.2571.377

Triplet peak. There is a proton from CH 3, which possesses 2 neighbors.

2.039 The methyl group has no proton nearby singlet peak.3.9614.0784.197

4.317

Quartet peak. The proton from CH 2, which isequivalent, gives one peak that gives 3-protonneighbors.

References:


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1) Introduction to Spectroscopy; 3 rd edition, Donald L. Pavia, Gary M. Lapman,George S. Kriz.

2) Laboratory Manual CHM3402 Spectroscopy, Chemistry Department Faculty Of Science & Environmental Studies.

3) Analytical Chemistry, 5 th edition,Gary D Christion.

4) Analytical Chemistry : An Introduction, 7th

editionD. A. Skoog, D. M. West, F. J. Holler, and S. R. Crouch

chm 3402 experiment 8

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