Derek Wong

Chemistry 273 Lab

Unknown Report

Introduction

 

In this lab, three unknowns were given and were determined through several methods, including nuclear magnetic resonance (NMR), infrared spectroscopy (IR), and mass spectroscopy.  The first step was to do preliminary observations which included what the physical state of the unknowns were, and any other physical characteristic such as smell and appearance.  The second step would be to examine the NMR, IR and mass spectrometer data to help determine the structure of the unknown.  Finally, a chemical test was performed to confirm the structure and analysis.

           

The degrees of unsaturation indicates how many double bonds, triple bonds, and/or rings are present.  The formula is: ((2(n) + 2) – (number of hydrogens)-(number of additional elements)) n = number of carbons. If there are any Cl or N atoms in the unknown, for each of these atom, an additional one must be subtracted.

 

Nuclear magnetic resonance, or NMR, uses the magnetic property, called spin, of a nucleus in an atom. When a sample is set in a strong magnetic field, it is possible to transfer energy into the spin system in the form of radiofrequency pulses and change the state of the system. After the pulse, the system relaxes back to its state of equilibrium, sending a weak signal that can be recorded. Because every nuclear spin in a molecule senses also the small magnetic fields of its nearest neighbors, it is possible to separate the signals coming from different atomic surroundings. A map of carbon-hydrogen framework and the structure of the molecule itself can be determined from these individual signals.

           

When an external magnetic field is applied to the molecule, the moving electrons create a local magnetic field which acts in opposition to the applied field so that the effective field actually felt by the nucleus is smaller than the applied (i.e. B_effective = B_applied – B_local, where B is the magnetic field applied).  This effect is called the shielding effect and can help explain why certain signals are more downfield (left hand side of the spectrum) then others due to the fact that each specific nucleus in a molecule is in a slightly different electronic environment.   If de-shielded, the signal will appear more downfield and require a lower field strength for resonance, and if more shielded, more up-field and will require a higher field strength for resonance. 

 

            Infrared Radiation can be utilized in determining the organic structure by making use of the fact that IR is absorbed by interatomic bonds in organic compounds. Chemical bonds in different environments will absorb varying intensities and at varying frequencies, so IR spectroscopy involves collecting absorption information and analyzing it. The frequencies at which there are absorptions for the IR are given in "peaks" or "signals" can be correlated directly to bonds within the compound in question.  Infrared spectroscopy is used to determine the presence of functional groups such as hydroxyl, nitrous, and amides.

            In mass spectrometry, a substance is bombarded with an electron beam fragmenting the molecule. The fragments produced are accelerated in a vacuum through a magnetic field and are sorted on the basis of mass-to-charge ratio. This can then be use to identify the presence of certain functional groups and the molecular size of the molecule.

           

Once the preliminary observations and the interpretation of the NMR, IR, and mass spectrometer are complete, a sketch of possible structures for the unknown can be made and will be confirmed through experimentation.

 

 

Unknown No. 1 – 013

 

Preliminary observation of 013 was a fine white powder with no odor.  From the given data (NMR, IR, and mass spectrophotometer) the molecular weight of the unknown was 136 grams, with 9 carbons, 12 hydrogen atoms, and one oxygen atom.  The molecular formula is C₉H₁₂O.  The degree of unsaturation was calculated by multiplying 9 by two to get 18, and then add 2 to get 20. Subtract 12 (number of hydrogen atoms) from 20 to get 8 which is then divided by 4 to get 4 degrees of unsaturation. 

 

Formula:   ((2(9) + 2) – 12 = 8/2 = 4 degrees unsaturation

 

This means that there are either 4 double bonds, rings, or a benzene ring present (benzene ring being 4 degrees of unsaturation). 

 

 

NMR (Nuclear Magnetic Resonance)

 

According to the NMR analysis, there were five NMR signals at:

1.25 PPM (Doublet) indicating presence of a saturated alkane (ex. CH₃-CH₂-CH₃)

2.75 PPM (Septet) indicating presence of alkynyl

4.75 PPM (Exchangeable) indicating presence of an OH group

6.75 PPM (Doublet) and 7.15 PPM (Doublet) indicating presence of rings or aromatics.

 

In NMR, the number of signals indicates how many hydrogens are located on the adjacent carbon.  The formula is to add one to the number of hydrogens present to give the number of signals.  For example a doublet means that there is only one hydrogen adjacent, and for a triplet there are two hydrogens adjacent.  For non-symmetrical sides, the signals are multiplied (no such case in this unknown).  If there are no adjacent peaks, then there would still be a signal, but only a single peak (i.e. singlet).

 

 For example at 1.25 PPM, there is a doublet meaning there is one H near another H on the adjacent carbon.  For example, H₁-C=C-H₁.  Here each of the H₁’s will register two (doublet) peaks since there is only one hydrogen adjacent.  At 2.75 PPM there is a septet showing that there is an H with 6 adjacent hydrogens.   I.e. CH₃-CH₁-CH_3. The H₁ in the center is surrounded by 6 hydrogens.  With six hydrogens present, the NMR will register seven peaks (septet).  The 4.75 PPM is labeled as exchangeable and most likely indicates the presence of OH group.  The signal at 6.75 PPM and 7.15 indicate the presence of an aromatic ring.

 

 

 

IR (Infrared spectroscopy)

 

According to the IR analysis, there is a stretch at 3600 cm^-1 indicating presence of OH group (confirming that the exchangeable signal in the NMR at 4.75 PPM is an OH group), 1200 cm^-1 indicating presence of C-O bond, 1600-1450cm^-1 indicating presence of an aromatic ring, and at 3100 cm^-1 indicating the presence of an aromatic C-H bond.

 

 

Mass Spectrometer

 

According to the mass spectrometer for the unknown 013, there is a benzene ring at 91(D), and an alcohol at 107 (D). Not much other data could be extrapolated.

 

Structure of molecule:

 

From the NMR, mass spectrometer and IR information, the following structure

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

OH

           

 

Both of the CH₃ groups seen at the top of the molecule represent the saturated alkane seen at 1.25 PPM. Both of the CH₃ groups has one adjacent H (H₁) represented by the doublet seen in the NMR.  The H₁ represents the septet seen.  H₁ has 6 hydrogen atoms located adjacent to it, accounting for the 7 peaks seen at 2.75 PPM.  There is an OH group (making a phenol) indicated by the C-O bond on the IR (1600-1450cm^-1) located at the bottom of the molecule.  The signal seen at 6.75 PPM  and 7.15 PPM is shown in H₃ and H₄ where each of the hydrogens are located next to an adjacent hydrogen. The difference in the NMR is that  H₃ is adjacent to an alkane giving it a signal of 6.75 PPM while H₄ is located to 7.15 because it is near oxygen which de-shields the hydrogen causing it to be more downfield.. This one hydrogen creates the two peaks seen.  The signal seen around this range also indicates an aromatic ring, here a phenol group.

 

The phenol group accounts for the four degree of saturation.  Each double bond is one (seen has three double bonds), and each ring is one, here the phenol group consists of one ring with three double bonds. Checking for each element, there are indeed nine hydrogens, one oxygen, and nine carbons accounted for.

 

Test Used to Verify Results: Phenol Identification Test

 

In order to ensure that structure and identity of the unknown is correct, the Ferric Chloride test was used to test for the presence of the phenol.  Procedure 2 (water-insoluble Phenols) was used (page 518 in the “Organic Laboratory Techniques”). 

 

The procedure is to dissolve 20mg of the unknown in 1 ml of chloroform.  One drop of pyridine and 3-5 drops of 1% ferric chloride solution in chloroform was added   The results of the test ran was blue, indicating a positive result – a phenol group is present.

 

 

 

Unknown No. 2 – 014

 

Preliminary observation was a clear liquid with a distinct order of “sweetness,” similar to an ester. From the given data (NMR, IR, and Mass spectrophotometer) the molecular weight of the unknown was 174 grams, with 8 carbons, 14 hydrogen atoms, and four oxygen atoms.  The molecular formula is C₈H₁₄O₄.  The degree of unsaturation was calculated by multiplying 8 by two to get 16, and then add 2 to get 18. Subtract 14 (number of hydrogen atoms) from 18 to get 4 which is then divided by 2 to get 2 degrees of calculations. 

  ((2(8) + 2) – (4))/2 = 2

Indicating at least one double bond.

 

 

NMR (Nuclear Magnetic Resonance)

 

 

According to the NMR analysis, there were three NMR signals at:

1.3 PPM (Triplet) indicating presence of a saturated alkane

2.6 PPM (Singlet) indicating that H atom(s) is adjacent near zero hydrogens, and is located at least two carbon atoms away from a carbon with hydrogens

4.1 PPM (Quartet) indicating presence of an ester group

 

As stated before in NMR, the number of signals indicates how many hydrogens are located on the adjacent carbon. 

 

Here the 1.3 PPM indicates there is a triplet meaning there is two H near another H on the adjacent carbon.  For example, CH₃-CH₂-, the CH₃ will register 3 signals (triplet) since it is adjacent to 2 hydrogens.  At  2.6 PPM (Singlet) there is only one signal meaning that  there should be no hydrogen adjacent to the hydrogen.  For example CH₃-CO-C.  The CH₃ will register as a singlet because there are no hydrogens located on the adjacent carbon (the adjacent carbon being connected to oxygen and another carbon). Using the reference handout given, at 4.1 PPM an ester group is indicated.  The formula for an ester group is R-COR’ with R and R’ being any alkyl (or functional) group and the carbon being double bonded to the oxygen.

 

 

IR (Infrared spectroscopy)

 

 

According to the IR analysis, there is a stretch at 1700 cm^-1 indicating presence of ester group (confirming that the signal in the NMR at 4.1 PPM is indeed an ester group).  The ester-carbonyl group (C=O) peak is usually a strong absorption.  The C=O stretch normally appears approximately 1735 cm^-1.  A C-O stretch can give two or more absorptions in the region of 12850-1050 cm^-1.  However, in this unknown, no such stretch was seen in the IR taken.

 

Mass Spectrometer

 

According to the mass spectrometer for the unknown 014, there is a C₂H₅ group located at 73 (D). The last line at 174 (D) indicates the molecular mass of the unknown. No other data was able to be extrapolated from the mass spectrometer read out.

 

Structure of molecule:

 

For this molecule, it was determined that it was symmetrical.  This was done by taking the ratio of each absorption.  It was found that it did not add to the number of hydrogens present in the molecule and therefore indicated symmetry was to be found in the molecule. 

 

Absorption                   Ratio

Triplet                          2 H’s

Singlet                          2 H’s

Quartet                         3 H’s

There is a total of seven H’s which do not equate to the 14 H’s present, so you need to multiply by two to get 14 (14 being the amount of hydrogens actually present in the molecule).

 

 

 

 

From the NMR, mass spectrometer and IR information, the following structure of 014 was made:

  

        H    H           O      H      H     O           H    H

        |       |             ||       |        |      ||             |      |

H – C₁ – C₂ – O- C₃ – C₄ –  C₅ – C₆ - O – C₇ – C₈ - H

        |       |                      |       |                      |      |

       H     H                    H     H                    H    H

 

The blue line indicates the symmetry of the molecule between the two C₄O₂H₇ groups.  Recall that the formula for an ester is R-COR’ with R and R’ being any alkyl (or functional) group and the carbon being double bonded to the oxygen. For each side there is an ester being composed of an ethyl group (R) joined to a carboxyl group (CO) by an oxygen atom bounded to the other half of the molecule (R’). The NMR supports this arrangement (structure being two esters formed together).

 

All of the hydrogens attached to C₁ and C₈ have the signal of 1.3PPM on the NMR and have 2 adjacent hydrogens, giving a triplet splitting pattern.     The hydrogens attached to C₂ and C₇ have a signal 4.1 PPM on the NMR and have three adjacent hydrogrens, giving it a quartet splitting pattern.  A signal at 4.1PPM on the NMR is also indicative of an ester group present.  This unknown has a structure of two esters “merged” together, giving it a symmetrical geometry.  C3 and C6 are carbonyls and are shown in the IR at 1700cm^-1.  The hydrogens attached to C₄ and C₅ give a signal at 2.6 PPM and although they are adjacent, since the molecule is symmetrical, it is taken as if there is no other hydrogen adjacent (i.e. giving it a single splitting pattern).   The signal at 2.6 PPM indicates that the hydrogen is located on the alpha carbon of the carbonyl group (i.e. on the carbon adjacent to the carbonyl group).

 

In the preliminary calculations, it was found that there were two degrees of unsaturation, indicating either two double bonds or one triple bond.  In the molecule drawn, there are two double bonds accounting for the two degrees of unsaturation.  Each carbonyl group represents one degree of unsaturation and since there are two carbonyl groups, there are two degrees of unsaturation.  Upon counting each atom, all eight carbons, fourteen hydrogens, and four oxygen have all been accounted for.

 

 

Test Used to Verify Results: Ester Identification Test

 

In order to ensure that structure and identity of the unknown is correct, the Ferric hydroxamote test was used to test for the presence of the ester. 

 

There are two parts to this test.  The first will test for enolic character.  This is done by dissolving one or two drops in ethanol, hydrochloric acid and ferric chloride.  If a definite color, except yellow, appears the ferric hydroxamate test cannot be used.  If no enolic character was found, then the second part of the test consisted of adding liquid ester, droxoylamine hydrochloride, and sodium hydroxide to the sample.  The sample was then heated and ferric chloride solution was added and the color was observed.  A positive test should give a deep burgundy or magenta color.

 

When heated with hydroxylamine, esters are converted to the corresponding hydroxamic acids:

     O                                   O

     ||                                      ||

R-C-OR’ + H₂NOH à  R-C-NH-OH + R’-OH

                 (Hydroxylamine)                             

The hydroxamic acids form strong, colored complexes with the ferric iron, prouducing the observed burgundy red color.

 

      O

       ||

3 R-C-NH-OH + FeCl₃ à 3HCl + (R-C=O)  - - Fe

                                                              |

                                                             NHO             

 

 

The sample did turn yellow, indicating no enolic character.  From there, the second part of the procedure was preformed.  The result was the sample turning to dark red burgundy, indicating the presence of the ester groups as predicted.

 

 

Unknown No. 3 – 015

 

Preliminary observation was a fine brown powder with no odor. From the given data (NMR, IR, and Mass spectrophotometer) the molecular weight of the unknown was 201 grams, with seven carbons, four hydrogen atoms, one nitrogen atom, one chloride atom, and four oxygen atoms.  The molecular formula is C₇H₄ NClO₄.  The degree of unsaturation was calculated by multiplying 7 by two to get 14, and then add 2 to get 16. Subtract 4 (number of hydrogen atoms) from 16 to get 12 and then subtract one for the chloride atom and subtract one for the nitrogen atom, which equals to10.  10is then divided by 2 to get 2 degrees of calculations. 

  ((2(7) + 2) – (4)-(1)-(1))/2 = 10

Indicating at either double bonds (maximum being 5), a benzene ring, and possible triple bonds

 

 

NMR (Nuclear Magnetic Resonance)

 

According to the NMR analysis, there were four NMR signals at:

7.7 PPM (Doublet) indicating presence of an aromatic

8.2 PPM (Doublet) indicating presence of an aromatic

8.6 PPM (Singlet) indicating presence of an aromatic

~10-12 PPM (Exchangeable) indicating presence of a carboxylic acid

 

As stated before in NMR, the number of signals indicates how many hydrogens are located on the adjacent carbon. 

 

Using the reference handout, it was determined that 7.7 PPM, 8.2 PPM, and 8.6 PPM  are all aromatics.  Carboxylic acid (COOH) is the only functional group that will register has an exchangeable signal at ~10-12 PPM. 

 

IR (Infrared spectroscopy)

 

In the IR, a carboxylic group (COOH) should have two signals.  One at the C=O stretch which is very strong and board between 1725cm^-1 and 1690cm^-1, but this was not seen. An O-H bond was seen in the region of 3300 cm^-1 to 2700 cm^-1 as a broad stretch. 

The Nitro group should have been seen in the IR has two strong bands near 1560 cm^-1 and 1350 cm^-1 however from the IR taken only the signal at 1560 cm^-1 shows.  Halides (Chlorine) is often difficult to detect especially since when doing the IR, the unknown was dissolved in CCL₄ solution. the C-Cl band should show as a stretch from 850-500 cm^-1, however in this sample, no such stretch was seen.

 

Mass Spectrometer

 

According to the mass spectrometer for the unknown 015 there is a Nitro group (NO₂) group at 46 (D) and a carboxylic acid (COOH) group at 105 (D). No other data was able to be extrapolated from the given mass spectrometer printout.

 

 

Structure of molecule:

 

Originally, there were six different structures, or isomers, that could be the unknown.  After doing a melting point, the melting range was found to be 152-156^oC.  Upon researching the melting point for each of the isomers, three of the unknowns could be eliminated since their melting points were either extremely higher or lower than the melting point taken.

 

However there were three different isomers that could be unknown 015.  In order to determine which structure were unknown 015 two methods could be used.   The first involved using the NMR given to compare against the standard NMR for each of the isomers.  By comparing each of the NMR for each isomer, it is possible to determine which the true structure is.  Although all will have the same types of signal, the nuclear shifts for each isomer will be distinct.

 

The second way is to do a mix melting point analysis.  For a pure substance, the melting point is very exact and the boiling point is within a few degrees of temperature.  If a sample of the unknown and a sample of another isomer is mixed together and then melted as a compound, the melting point would decrease and melt over a larger range of temperature.  Since the compound is one of the isomers, the compound that has a precise melting point and melts at the given melting point for that isomer must be the unknown.

 

 

 

 

 

 

From the NMR, mass spectrometer and IR information, the following structure of 015 was made (4-Chloro-3-Nitrobenzoic acid):

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Cl

           

 

 

All of the atoms were accounted for, 7 carbons, 4 hydrogens, 1 chloride, 1 nitrogen, 4 oxygens.  The degrees of unsaturation have been all accounted for.  Four degrees of unstaturation were found in the benzene ring formed.  The fifth degree of unsaturation was found in the carboxlic acid (COOH).  In COOH, there is a double bond of C=O.  The COOH also appeared in the NMR has an exchangeable (as a fold over) at ~10-12 PPM. The structure of COOH is:  O

                                             ||

                                       R - C – OH  (R being another group attached to the COOH)

According to the NMR, there is a singlet at 8.582 PPM. This can been seen in the structure at H₁.  H1 appears more downfield because of the deshielding effect (similar to the hydrogens in unknown 013) because H₁ is in between two very strong electron withdrawing groups, NO₂ and COOH.  Two signals were seen at 7.7 PPM and 8.2 PPM, both as doublets.  As stated before, doublets indicate the hydrogens being next to another adjacent hydrogen.  H₂ and H₃ will each give a doublet signal.  H₂ is next to COOH, which is a stronger electron withdrawing group than Cl, therefore it will appear more downfield. Most likely H₂ will have the signal of 8.2 PPM and H3 will have the signal   7.7 PPM which is more up field.

 

 

Test Used to Verify Results: Ferrous Hydroxide Test

 

In order to ensure that structure and identity of the unknown is correct, the Ferrous Hydroxide test was used to test for the presence of the nitro group. 

 

This test involves placing aqueous ferrous ammonium sulfate with the unknown.  The solution is then mixed with sulfuric acid and potassium hydroxide.  The solution is shaken vigorously and the detection of a red-brown precipitate is indicates the presence of a nitro group.

Here, the nitro group oxidized the ferrous hydroxide to ferric hydroxide, which is the red-brown solid.  The equation is given below:

 

R-NO₂ + 4H₂0 + 6 Fe(OH)₂ à R-NH₂ + 6 Fe(OH)₃

 

When the experiment was preformed, in less than 30 seconds, a reddish-brown precipitate appeared at the bottom of the solution confirming the unknown had a nitro group has predicated. 

 

 

Conclusion

 

In this lab we learned many types of functional groups and learned how to identify them through a battery of tests and through the use of instruments such as the NMR, IR and mass spectrometer.  From there, we learned how to apply this knowledge to determine the identity of the three unknowns given.

 

In conclusion three unknowns were given (013, 014, and 015) and were determined through a variety of methods, mainly by NMR, IR and mass spectrometer.  The identity was then confirmed through identification tests such as the Ferric Chloride test, the Ferric hydroxamote and the Ferrious hydroxide test.