Mass spectrum for the ionization of acetone MS of Acetone + Relative Abundance CH 3 H 3 C O + M 15 (loss of methyl) + O H 3 C CH 3 43 58 0 10 20 30 40 50 60 70 m/z It is difficult to identify the ions a priori, but we can take the parent ion (M or P) and look for the mass of lost fragments. Loss of 15 mass units, for example, corresponds to loss of a methyl group from the parent.
Common Fragmentations: M? Fragmentation of the parent can give structural clues P 18 OH loss of water M-18 = alcohol P 15 loss of methyl M-15 = parent had a methyl group O P 29!-cleavage m/z 29 α-cleavage and M-29 = an ethyl group next to a functional group or heteroatom O
Actual Mass Spectra
Isotopic Peaks and Which Daughter Ion? We can use exact mass for M-daughter ion fragments. 4 M-15 = methyl, M-17 = NH 3, M-18 = H 2 O, M-2 - ethyl or CHO, usually ethyl M-44 = CO 2, M-57 = butyl, etc. O + Mass Fragment Mass Fragment 15.0235 CH 3 16.0187 NH 2 29.0391 CH 3 CH 2 31.0184 CH 3 O H 3 C 1 CH 3 17.0265 NH 3 18.0106 H 2 O 27.0235 CH 2 =CH 41.0391 CH 2 =CHCH 2 43.0547 (CH 3 ) 2 CH 43.9898 CO 2 28.0061 N 2 28.0313 CH 2 =CH 2 29.0027 CHO 57.0704 C 4 H 9 77.391 C 6 H 5 91.0547 C 7 H 7
Determining Molecular Weight of Parent: M, M+1, M+2 The mass of the parent ion should be the highest mass M = molecular ion (also known as P = parent ion) It is not! Ratio of M+1/M and M+2/M M There is a peak with one mass unit higher than M the M+1 Relative Abundance 57 2.7% M 59 54 56 58 60 0.35% M How? There is a peak with two mass units higher than M the M+2 m/z
Isotope Relative Abundance (%) 12 C 100 13 C 1.11 1 H 100 2 H 0.016 14 N 100 15 N 0.38 16 O 100 17 O 0.04 18 O 0.20 32 S 100 33 S 0.78 34 S 4.40 35 Cl 100 37 Cl 32.5 79 Br 100 81 Br 98.0 Common Isotopes fluorine ( 19 F), phosphorus ( 31 P), and iodine ( 127 I) do not have M+1 or M+2 isotopes that can be used here.
#C, #N and #O: Formulas from M+1 & M+2 always use the isotope of highest natural abundance to calculate the molecular weight. Since we know the isotopic ratio, measurement of the M+1 peak relative to M should allow us to calculate the number of carbons. This is the basic premise of two formulas that can be used to determine the empirical formula for a molecule. Since the parent ion also provides the molecular weight, if the charge z = +1, we can determine the molecular formula for a sample. Based on the isotope table, the two formulas are: M + 1 = (# C) (1.11) + (# N) (0.38) 13 C 15 N M + 2 = (# C) (1.11) 2 + (# O) (0.20) 200 14 C 18 O
Even M = Even MW (0 N), Odd M = Odd MW (1 N): An Assumption Therefore: If M = even mass, assume 0 nitrogen atoms and M+1 = 1.11(#C) If M = odd mass, assume 1 nitrogen atom, and M+1 = 1.11(#C) + 0.38 (1 N) Note: For M+2 to work with oxygen, with only a 0.2 % isotope, Assume no Cl, S or Br. For the time being - but not forever, a molecule will contain S, Cl, or Br, but only one of each and, if the molecule contains S, Cl, or Br, it will not contain O.
m/z m/z m/z P (162) (100) P+1 (163) (12.21) P+2 (164) (0.95) % % % Example 1 C 11 H 14 O even mass, so 0 nitrogen, so P+1 = 12.21 = 1.11(#C) #C = 12.21/1.11 = 11 C 11 P+2 = 0.95 = [(1.11)(11)] 2 /200 + 0.20 #O 0.95 = 0.74 5 + 0.2 #O #O = 0.95 0.75 / 0.2 = 1 O 1 Cannot see H in mass spec, therefore obtain H by difference from the parent ion (162) 11 C x 12 = 132 + 1 x 16 = 148 162 148 = 14, so 14 H????? Remember the alkane formula C n H 2n+2 = maximum number of H for a molecule For C11, C 11 H 24, so there cannot be more than 24 H, but there can be less.
P (73) (100) P+1 (74) (4.80) P+2 (75) (0.10) Example 2 C 4 H 11 N odd mass, so 1 nitrogen, so P+1 = 4.80 = 1.11(#C) + 0.38 (1 N) #C = 4.80 0.38 = 4.42/1.11 = 3.98 = 4 C 4 N P+2 = 0.1 = [(1.11)(4)] 2 /200 + 0.20 #O 0.1 = 0.098 + 0.2 #O #O = 0.1 0.1 / 0.2 = 0 no O Cannot see H in mass spec, therefore obtain H by difference from the parent ion (73) 4 C x 12 = 48 + 1 N x 14 = 62 73 62= 11, so 11 H????? Remember the alkane formula C n H 2n+2 = maximum number of H for a molecule For C4, C 4 H 10, should be more than 10H, but this has N, with an odd valence therefore, use C n H 2n+3 so 11 is OK
Example 3 P (162) (100) P+1 (163) (6.66) P+2 (98) C 6 H 11 Br even mass, so 0 nitrogen, so P+1 = 6.66 = 1.11(#C) + 0.38 (0 N) #C = 6.66 /1.11 = 6 C 6 P+2 =98% of P = Br assume no oxygen, so C 6 Br Cannot see H in mass spec, therefore obtain H by difference from the parent ion (162) 6 C x 12 = 72 + 1 Br x 79 = 151 (note use 79 isotope for Br parent) 162 151 = 11, so 11 H????? Remember the alkane formula C n H 2n+2 = maximum number of H for a molecule replace H with Br - use alkane formula For C6, C 6 H 14, so there cannot be more than 14H so 11 is OK
Which is the Parent? In the mass spectrum, the parent will be the highest Mass (not necessarily B), but there are M+1 and M+2. Look for the highest mass units. The largest (most abundant of M, M+1, M+2) will be the parent M. The much smaller peaks to the right of M will be M+1 and M+2. Remember that M+1 is typically <10% of P in abundance, or perhaps 5-15%. M+2 is usually <1%, or 0.2-2%. M should be easy to spot. Read section on exact mass
We Know the Formula. How Can We Determine Functional Groups? With an unknown, mass spectrometry gives us the molecular weight and the empirical formula Before we can identify the molecule, we must know the functional group We use infrared spectroscopy In mass spectrometry, so much energy is applied to the molecule that it fragments. It is also possible to apply only enough energy to make the molecule vibrate but not break bonds. This is what happens in infra-red spectroscopy, where we give the molecule the equivalent of a good, swift kick and then observe what happens. In general, functional groups vibrate at unique frequencies and we use these absorptions to identify these groups. Knowledge of functional groups in a molecule will obviously provide a critical structural component.
Molecules Absorb Infrared Light The infrared light used to analyze organic molecules is usually limited to 4000 cm -1 to 400 cm -1 and we are particularly interested in the interaction of infra-red light with molecules. With a simple diatomic molecule, we must further categorize that molecule as symmetrical or asymmetric. A symmetrical molecule has two identical atoms such as H 2, O 2, or N 2, whereas an unsymmetrical molecule will have two different atoms. If we look at the dipole of diatomic molecules, unsymmetrical molecules will have a dipole and the greater the disparity of electronegativities between the two atoms, the greater the dipole and the stronger the IR signal. A symmetric molecule will no have a dipole and, hence, no dipole moment. They give very weak signals in the IR.
Infrared (IR) Spectrophotometer http://homepage.mac.com/stray/ib/chem/options/analytical.html A double beam Infrared Spectrometer: A spectrometer has a source of infrared light (hot coil of nichrome wire) that emits radiation over the whole of the frequency range of the detector. The beam is then split into two beams of equal intensity. One beam is passed through the sample, while the other is a reference beam Via an rotating disk, the beams are passing alternately. The intensities of the two beams are then compared and the wavelength over which the comparison is made is calibrated and dispersed on the detector via prism
Salt Plates and Water Note: IR spectra are typically recorded by placing the sample on a clear plate of pressed NaCl or KBr. orgchem.colorado.edu/ procedures/ir/irprep.html Both dissolve in water Do not wash plates with water, or solvents that contain water - such as acetone or ethanol. Your fingers have moisture. The correct way to hold plates is