Qualitative Analysis of Unknown Compounds 1. Infrared

Qualitative Analysis of Unknown Compounds

1. Infrared Spectroscopy • Identification of functional groups in the unknown • Will take in lab today 2. Elemental Analysis • Determination of the Empirical Formula of the Unknown • Smallest whole number ratio of elements in the formula (mole ratio of elements) • Given on form with Assignment ID number 3. Mass Spectral Analysis • Determination of the Molecular Formula of the Unknown • Actual number of atoms in the formula • Molecular Weight of the Unknown • Halide identification • Hand-out with both graph and table 4. Proton NMR • Symmetry: Number of chemically different protons • Chemical Shift: Chemical environment of protons • Integration: Ratio of protons • Splitting Patterns: arrangement of neighboring protons • Hand-out with spectrum (x-axis only)

Elemental Analysis –the empirical formula Remember, in order to calculate the empirical formula, you need to take the percentages from the elemental analysis and assume these are gram amounts which need to be converted to moles (because formulas are mole ratios).

Ex. Calculate the empirical formula for a compound whose elemental analysis is the following: %C = 68.31 %H = 11.47 %O = 20.22

What next? %C = 68.31/12.011 = 5.687 %H = 11.47/1.008 = 11.380 %O = 20.22/16.00 = 1.264

What next? %C = 68.31/12.011 = 5.687/1.264 = 4.499 ≈ 4.5 %H = 11.47/1.008 = 11.380/1.264 = 9.003 ≈ 9 %O = 20.22/16.00 = 1.264/1.264 = 1

What do you do when the value comes up at a “half”, as in 4.5 or 2.5? Multiply all of your values by by 2…

Empirical Formula: C9H18O2 (MW of 158 g/mol)

What do you do if you have a third (as in 3.3333) or two thirds (4.66667)? Multiply all by 3!

There are only three options when you do the math: (a) it should result in perfect or almost perfect whole numbers, (b) possible half values (multiply by 2) or (c) or possibly thirds (one-third or two thirds values – multiply by 3). Anything else is a sure sign of a mathematical error!

Formulas are helpful when determining structures. Not so much the empirical formula - being only the smallest whole number ratio, it doesn’t contain all of the atoms in the molecule you are studying so it is better to know the molecular formula before determining a structure. When you know the molecular formula, you can have some added guidance by being able to determine a possible structure.

The Unsaturation Number (or Degrees of Unsaturation) can be calculated from the molecular formula. Remember that every degree of unsaturation is equivalent to a loss of 2 hydrogen atoms from the maximum allowed in an alkane with the same number of carbon atoms. This occurs every time there is a ring structure or a pi bond (double bond has one pi bond, triple bond has two pi bonds).

How many unsaturations in: O

NH 2

Two rings (2), 7 pi bonds = 9 unsaturations.

Each of the following has two degrees of unsaturations, and fit the formula C6H10O: O O O

H O

To calculate the Unsaturation Number (degree of unsaturations) from a formula: #C – ½(#H + #X) + ½(#N) + 1

Calculate the number of unsaturations in: C12H16INO2

UN = 12 – ½(17) + ½(1) + 1 = 12 – 8.5 + 0.5 + 1 = 5

Always helpful to keep in mind: if your molecular formula results in an Unsaturation Number of 4 or greater, there’s a very strong chance that you have an aromatic ring in your structure (ring structure plus three pi bonds).

It is also always helpful to calculate this information as soon as you have a formula as it may prevent you from making incorrect assumptions. Consider C8H18O2 – with two oxygen atoms, you may automatically jump to the conclusion that you have a carboxylic acid or ester functional group but when you calculate the UN (= 0), you realize that neither of these functional groups are possible, as they both have one unsaturation in their structure. O O

OH OR 1 unsaturation

Perhaps you have two alcohols or an alcohol and an ether:

O no unsaturations

To determine the molecular formula, we will need to discuss mass spectroscopy.

Mass Spectroscopy: Purpose: Determination of a compound’s molecular weight (as well as structural information)

Here’s the basic process: A small sample to be studied is vaporized then ionized by being bombarded by high energy electrons (like the molecule was zapped with a lightning bolt). The energy of the electrons in that lightning bolt correspond to >1000 kcal/mol of energy. Under these conditions, the organic molecule is destroyed, breaking into a number of fragments and we can use the fragmentation to help us determine structural information.

When the high-energy electron (e-) impacts an organic molecule (M), one of the molecule’s valence electrons (e-) is ejected, resulting in the formation of an ion call the Molecular Ion (M+•): 3 M + e- → M+• + e- + e-

This new ion is a “radical cation”. A radical is a species that contains an unpaired electron (but is neutral in charge) and a cation is a species missing an electron, resulting in an overall positive charge.

One this happens, the molecular ion can then begin to fragment to give a number of other cations and neutral, radical species. These fragments are then passed through a magnetic field where they are separated and detected according to mass. Only the charged species (cations) can be detected by mass spectroscopy since neutral species (radicals) cannot be affected by the magnetic field. Technically, we are actually measuring the “mass to charge ratio” (m/z, where m = mass and z = charge) for each ion but since the charge is virtually almost always +1, this really is simply the mass of the fragment.

The Molecular Ion: The molecular ion is still the entire molecule, minus only one electron, the weight of which is negligible, thus the weight of the molecular ion equates to the molecular weight of the entire molecule. Shown below is the mass spectrum for

Butane. Knowing that carbon weighs 12 and hydrogen atoms weigh 1 each, butane (C4H10) has a molecular weight of 58. Find the molecular ion (M+•) peak that corresponds to the MW.

The mass spectrum graph is a plot of Mass-to-Charge Ratio, m/z, (X axis) versus Abundance (Y axis).

4 • Mass-to-Charge Ratio can be simplified to “mass of fragment” since the charge on each is +1. • Abundance is the relative number of times that each fragment passed through the detector. The more often the fragment formed and passed by, the larger the abundance for the fragment (the larger the “peak”).

What can you get from knowing the molecular weight? You can figure out a molecular formula.

1. Let’s assume you know you have a compound that only has carbons and hydrogens and whose molecular weight (a.k.a. molecular ion) is 58 a.m.u. What can you determine? Start by determining how many carbon atoms can fit into a weight of 58. • How many carbon atoms fit into 58? § 4 carbons weigh 48 § 58 – 48 = 10. • That leaves you 10 left over for hydrogens.

• Molecular formula: C4H10.

2. What if the compound also has oxygen in its structure (with a molecular ion of 58)? What do you have to save room for in this case? • Start by subtracting 16 (from the oxygen) from the total of 58. § 58-16 = 42. • How many carbons fit into 42? § 3 carbons weigh 36. § 42 – 36 = 6 • That leaves 6 left over for hydrogens

• Molecular Formula: C3H6O.

Fragmentation occurs when the molecular ion is produced with sufficient internal energy to cause spontaneous dissociation. This is similar to the spontaneous dissociation of water in the dehydration of an alcohol to form a stable carbocation.

Fragmentation occurs in a “logical” fashion to form stable intermediates, both cations and radicals. Tertiary systems are more stable than secondary and these are more stable than primary. The fragments that we see on the spectrum are called “daughter peaks”.

The largest peak on the spectrum is called the Base Peak, whose relative intensity is set to 100 (“999” on the table on your unknown’s spectrum). The base peak MAY be the same as the molecular ion but more often than not, they are different peaks. Notice

5 the large number of peaks found to the left of the molecular ion – those are from the fragment ions that formed.

Note the tiny peak at m/z = 59? This is called an M+1 peak (one amu higher than the molecular ion). How did THAT get in there? What weighs more than the entire molecule? To understand this, we need a little reminder on isotopes…

Isotopes are atoms of a particular element that differ in their number of neutrons. The atomic masses we see on the period table are actually the weighted average of all isotopes based on their percentages (natural abundance) and masses (variation caused by differing numbers of neutrons in the nucleus), for any given element. Here is a table summarizing mass and natural abundance for several isotopes relevant to organic molecules and mass spectroscopy:

6 Isotope Atomic Mass Natural Abundance (%)

1H 1.0079 99.985 2H (D) 2.010 0.015 12C 12 98.9 13C 13.003 1.1 14N 14.003 99.63 15N 15 0.37 16O 15.995 99.76 17O 16.999 0.038 18O 17.999 0.2 19F 19.998 100 35Cl 34.969 75.77 37Cl 36.966 24.23

79Br 78.918 50.69

81Br 80.916 49.31 127 I 126.905 100

The most important atoms in organic molecules include carbon (12C), hydrogen (1H), oxygen (16O) and nitrogen (14N). These each have one major isotope. These isotopes are the only isotopes found in the molecular ion. However, carbon-13 is abundant enough (1.1%) that it can be observed on a mass spectrum. Thus the M+1 peak in the mass spectrum of butane is due to the relatively small number of butane molecules where one of the four carbon atoms is a carbon-13 isotope, instead of carbon-12.

CH3CH2CH2CH3 MW = 58 13 CH3CH2CH2CH3 MW = 59 (1 C isotope indicated)

The chance of most molecules having two 13C isotopes is extremely small so you’ll never see an M+2 caused by two 13C isotopes. The instrument could never detect that microscopic percentage. Thus you won’t see a peak for M+2 caused by carbon-13 isotopes. HOWEVER, there is something called an M+2 that occurs and its caused by chlorine and bromine atoms. These show significant M+2 peaks in their mass spectra as they have alternative isotopes that are relatively abundant. As a result, we can use mass spectroscopy to detect and determine the presence and identity of chlorine and bromine in organic molecules.

Consider a molecule like bromomethane, CH3Br. The natural abundance for bromine shows the two isotopes (79Br and 81Br) are present in almost a 1:1 ratio (50.7 : 49.3). This means you should have two peaks (M+. and M+2) of almost equal intensity.

7 M+. is the molecular ion that contains all of the atoms (standard isotopes) and includes the 79Br isotope. MW = 94 M+2 is the version that contains all of the atoms (standard isotopes) and includes the 81Br isotope. MW = 96 Roughly 1:1 intensity

+ H M 79 H C Br + (M+2) H

H H C 81Br H + CH3

Notice that the base peak is, in this case, the same as the molecular ion. Notice also, when bromomethane (M+ = 94 or M+2 = 96) fragments off the bromine atom (weight 79 amu), it leaves a fragment whose m/z = 15, which corresponds to the remaining CH3 piece.

The molecular weight of bromomethane was 94 or 96 (those are WHOLE numbers). If you use the values from the periodic table, you probably would have calculated: (12.011) + 3(1.0079) + 79.904 = 94.939 and perhaps rounded that to 95. If you look at the mass spectrum, you will see that the fragment indicated by the peak at 95 is much smaller than the peak at 94. Remember that the atomic masses given on the period table are a weighted average of the masses of each isotope based on their natural abundance. Since 79Br and 81Br are present in about a 1:1 ratio, this averages out to an atomic mass of ~80, but there is no 80Br. You must always calculate molecular masses for mass spectroscopy using an individual isotope and never use the average masses from the periodic table.

Consider chloromethane. The two isotopes of chlorine (35Cl and 37Cl) are present in a 3:1 12 1 ratio of abundance. Using the C and H isotopes, the molecular weight of CH3Cl is 50. The molecular ion peaks for the two chlorine isotopes are observed at m/z = 50 (M+, 100%) and m/z = 52 (M+2, 33%).

8 + M H + 35 CH3 H C Cl H H + (M+2) H C 37Cl H Roughly 3:1 intensity

The Nitrogen Rule: Just a quick heads up for the future If the mass of the molecular ion peak is an odd weight (such as 93), then the molecule contains an odd number of nitrogen atoms, as in aniline, C6H7N.

NH 2 M+

C6H7N

M+1

The odd value, 93, sort of screams “I have a Nitrogen Atom” so perhaps you may be looking at an amine, amide, nitrile or nitro-containing compound. Even values don’t give you any clues to the elements inside the molecule.

Fragmentation Information: 9 Mass spectroscopy can also be used to identify structural pieces of molecules, if you know what to look for. Remember that this process is all about taking the original molecule and forming fragments, which are pieces of your molecule. Remember that these fragments are always cation fragments, as the detector in the instrument cannot detect any radicals.

The table below shows masses of a variety of fragments that you might see during the fragmentation process, depending on your structure. It can definitely be a little overwhelming, trying to track all of the pieces.

Mass Fragment Structure Fragment Name Possible Inference(s)

14 CH2 methylene alkyl chain 15 CH3 methyl methyl group 16 NH2 amino amine, primary amide 17 OH hydroxy alcohol, carboxylic acid 18 H2O water alcohol 19 F fluoro organofluorine compound 26 CN cyano nitrile 28 CO (or C(O)) carbon monoxide carbonyl compound 29 CHO, CH2CH3 formyl, ethyl aldehyde, ethyl group 31 OCH3 methoxy methyl ester or methyl ether 35, 37 Cl chloro organochlorine compound 41 C3H5 (CH2=CHCH2-) allyl allyl group propyl or isopropyl group, C H , CH CO (or CH C(O)-) propyl, isopropyl, acetyl 43 3 7 3 3 methyl ketone carboxylic acid, ethyl ester CO H, OCH CH carboxyl, ethoxy 45 2 2 3 or ethyl ether 46 NO2 nitro nitro group

57 C4H9 butyl-, tert-butyl butyl or tert-butyl group 77 C6H5 phenyl phenyl group (aromatic ring) 79, 81 Br bromo organobromine compound 91 C6H5CH2 benzyl benzyl group 105 C6H5CO (or C6H5C(O)-) benzoyl phenyl ketone

Keep in mind the previous discussions in lecture on carbocations and their stability. Hyperconjugation, inductive effect and even resonance can contribute to the stability of a carbocation, so we can logically approach the fragmentation of a molecule remembering what cations might be formed (3º > 2º > 1º, for stability). Sometimes its easier to determine the fragmentation from a “What was lost?” approach, focusing on mass differences.

+ Consider butane, CH3CH2CH2CH3. Its total molecular weight is 58 amu and its M appears at 58.

What is causing the other peaks on the spectrum?

The smaller peaks seen below 58 are those “daughter peaks”, caused by the spontaneous dissociation of the molecular ion into smaller fragments. The “fragmentation patterns” can provide some information to help characterize the structure of the molecule.

loss of 15

(CH3)

loss of 29

(CH2CH3) M+

loss of 43

(CH2CH2CH3)

11 An aromatic ring example: Propyl Benzene

MW = 120

+ CH3 (loss of 15)

MW = 105

+ CH2CH3 (loss of 29)

MW = 91

(loss of 43)) + CH2CH2CH3

MW = 77

C9H12

M+=120

77 105 121

+ M =120 105 91 77

Why is the peak at 91 so very large? The base peak is the most stable cation that can form. This cation is in the Benzylic position, and is resonance stabilized. 91 12 Warning: Sometimes fragmentation is so extensive and so quick that the molecular ion itself is not observed. Neopentane, or 2,2-dimethylpropane, is one such compound whose molecular ion at 72 a.m.u. is not visible. Fragmentation occurred too quickly and so extensively to form a tertiary carbocation that the molecular ion itself never survived to reach the detector.

CH3

H3C C CH3

CH3

Note the large peak at 57. It corresponds to the tertiary carbocation easily formed by loss of a methyl fragment: + CH 3 CH3 H C C CH 3 3 H3C C CH3 CH3 CH 3

This is problematic if your goal is to prove your molecular weight using mass spectroscopy…

Remember, every compound should have a molecular ion (M+) and M+1.

Functional Group Information you need to remember: Every unknown in the Qual Lab has at least one of the following: 1. Halide Chlorine: You should be able to find your molecular ion (M+) as well as the M+2 in a ratio of 3:1.

Bromine: You should be able to find your molecular ion (M+) as well as the M+2 in a ratio of 3:1.

14 2. Alcohol or Phenol: • You will also find a fragment caused by the loss of the acidic proton (M-1) from an alcohol or phenol. • Molecules that contain alcohol functional groups are generally recognized by the fragmentation resulting from dehydration of the molecular ion (M-18) but only if the hydroxyl is attached to an sp3 hybridized carbon. Those hydroxyl groups attached to aromatic rings (phenols) cannot dehydrate. • These peaks (M-1, M-18) may not be large, significant peaks but you still need to look for them if you have the alcohol/phenol functional group.

3. Aromatic Rings:

Aromatic compounds contain an aromatic ring and often have a peak for phenyl, C6H5, at

77 or benzyl, C6H5CH2 at 91. Identify these if they are pertinent to your molecule. Similar to benzyl, you may find yourself with a peak at 91 if your aromatic ring is disubstituted and one group is a CH3:

or or

More on the Molecular Ion… The most important peak on the mass spectrum is the molecular ion, because it comprises the entire molecule and tells you the molecular weight.

15 How does one determine which peak is the molecular ion? • Since it involves the molecular weight of the entire compound, it will be one of the highest weight peaks, found on the far right of the spectrum.

Why is it not ALWAYS the furthest one on the right? • Isotopes! Remember the M+1 peak!

So - which peak is the molecular ion? Again, as a rule, the molecular ion always contains only the following isotopes: Carbon-12, Hydrogen-1, Oxygen-16, Nitrogen-14, Chlorine-35 and Bromine-79.

1. Calculate your empirical formula using the Elemental Analysis. 2. Then calculate the weight of the empirical formula, using the above mentioned isotopes. Remember: Never, never, never use the values from the periodic table when determining anything to do with mass spectroscopy. Those values are averages of elemental weights, not actual atom weights and can mislead you in your search for the molecular ion. 3. Then go look for your Molecular Ion…

Now: Tie Elemental Analysis and Mass Spectroscopy Together…

Example 1: Unknown Compound X has an empirical formula of C3H6O and has a molecular weight of 58 a.m.u.

Identify the molecular ion. Remember that the molecular ion has to match or be a multiple of the empirical formula weight. Which one of the peaks above is the molecular ion?

16 58. If the molecular ion matches the weight of the empirical formula, then your empirical formula is the same as your molecular formula. The molecular formula for

Unknown Compound X would be C3H6O.

Example 2: Unknown Compound X has an empirical formula of C3H6O (58 a.m.u.).

Is 58 the mass for the molecular ion? Where is 58 on this spectrum? Right hand side or middle? Middle.

If the mass spectrum for Compound X shows several peaks, many of which have masses that are higher than 58, then the empirical formula cannot be a match to the molecular formula. It must be a multiple of the empirical formula. Try multiplying your empirical formula by 2 or 3 or 4. On this spectrum, 58 x 2 = 116, which appears on the far right hand side. So - if the empirical formula is C3H6O, what is the molecular formula for Compound X?

Since the molecular ion mass is TWICE the mass from the empirical formula, the molecular formula must be twice the empirical formula, or C6H12O2.

End Qual Lecture One. What should you be working on? 1. Calculate the empirical formula for your unknown compound. You’ve been given the percentages for the elements in your molecule. If there is a blank next to the %O, %N or %X, that means you do not have any of that atom type in your molecule. “X” = Cl or Br. Determine the identity of X from your mass spectrum.

17 2. Calculate the weight of your empirical formula using the values discussed for mass spectroscopy (carbon is 12, hydrogen is 1, etc).

3. Find your Molecular ion. It either has to MATCH the weight of your empirical formula or be a multiple of.

4. Calculate the number of unsaturations in your molecule.

5. Consider other possible fragmentation information, like the presence of a phenyl group or benzyl group. When you determine your other functional groups (such as perhaps an alcohol), find the other required peaks for the mass spectrum (halides have M+2, alcohols have an M-1 and M-18).

6. Begin filling out the answer sheet.