Major Chemistry B.Sc.III ( Semester V)

Name of paper- Spectral analysis

CHE/DSC/T/301

Mass Spectroscopy Part-II

Isotopic peaks. Nitrogen rule, rule of 13 for determination of empirical formula and molecular formula. Fragmentation pattern of compounds up to 5 carbons – aldehydes, ketones, alkyl halides and alcohols. McLafferty rearrangement, Retro-Diels–Alder reaction.

Isotopic peaks:

Molecules in nature do not occur as isotopically pure species. Virtually all atoms have heavier isotopes that occur in characteristic natural abundances. With the possible exception of fluorine and a few other elements, most elements have a certain percentage of naturally occurring heavier isotopes. Peaks caused by ions bearing those heavier isotopes also appear in mass spectra. The relative abundances of such isotopic peaks are proportional to the abundances of the isotopes in nature. Most often, the isotopes occur one or two mass units above the mass of the ―normal‖ atom. Therefore, besides looking for the molecular ion (M+) peak, one would also attempt to locate M + 1 and M + 2 peaks. Therefore, it should be noted that in the mass spectrum of organic compounds the molecular ion peak is followed by two peaks namely the one of m/z one mass unit greater than the molecular ion and a second peak of m/z two mass units greater than the molecular ion. These peaks are called the molecular ion + 1 (M+1) and the molecular ion + 2 (M+2) peaks. The intensity of the molecular ion, molecular ion + 1, and molecular ion + 2 peaks depends the relative abundance of the respective isotopes i.e., the ratio of peak pattern is the result of natural isotope abundance of the atoms in the molecule. The appearance of this type of peak arrangement at the high end of the m/z scale on the mass spectrum is characteristic of the molecular ion.

Table 2: Isotopes and their relative abundance

The isotopic distributions of several elements commonly found in organic compounds are shown in Table above. From the isotopic distributions it is seen that the contributions to the M+1 peak by isotopes of hydrogen, oxygen, and the halogens are very small or nonexistent. This means the mass spectrum would have only M+ ion peak and the M+1 isotopic peak may be very small or absent. Mass spectra can show M+2 peaks as a result of a contribution from 18O or from having two heavy isotopes in the same molecule (say, 13C and 2H or two 13C). Most of the time, the peak is very small. The presence of a large M+2 peak isevidence of a compound containing either chlorine or bromine, because each of these elements has a high percentage of a naturally occurring isotope that is two units heavier than the most abundant isotope. From the natural abundance of the isotopes of chlorine and bromine in Table above, one can conclude that if the M+2 peak is one third the height of the molecular ion peak, then the compound contains one chlorine atom because the natural abundance of 37Cl is one-third that of 35Cl (see spectrum below).

If the M+ and M+2 peaks are about the same height, then the compound contains one bromine atom because the natural abundances of 79Br and 81Br are about the same. Since their percentage of natural abundance is 50.5% for 79Br and 49.5% for 81Br (see spectrum below).

If a compound contains sulphur atom then its M+2 peak would be larger than M+1 peak. This is because the abundance of 34S (M+2) is 34 % whereas 33S (M+1) is only 0.8% (see spectrum below).

The spectrum becomes much more complicated when one considers the relative abundances of ions containing several polyisotopic elements; the presence of two bromine atoms in an ion gives rise to three peaks at M, M + 2 and M + 4, the relative intensities being 1:2:1, while for three bromine atoms the peaks arise at M, M + 2, M + 4, M + 6, with relative intensities 1:3:3: 1. These figures ignore any contribution from 13C that may be present (see image below).

Molecular weight from isotopic peaks:

One of the ways to determine the molecular formulas is to examine the relative intensities of the peaks due to the molecular ion and related ions that bear one or more heavy isotopes (the molecular ion cluster). Use of the molecular ion cluster can be useful, though, for a relatively quick determination of the molecular formula that does not require the much more expensive high-resolution instrument. Accurate calculation of the relative intensities of isotope peaks in a molecular ion cluster for compounds containing several elements with isotopes is time consuming to do by hand as it requires polynomial expansions. For compounds containing only C, H, N, O, F, Si, P, and S, the relative intensities of M + 1 and M + 2 peaks can be estimated quickly using simplified calculations. The formula to calculate the M + 1 peak intensity (relative to M1 = 100) for a given formula is given below:

[M+1] = (number of C × 1.1) + (number of H × 0.015) + (number of N × 0.37) + (number of O × 0.04) + (number of S × 0.8) + (number of Si × 5.1)

Similarly, the intensity of an M + 2 peak intensity (relative to M1 = 100) may be found by using the below equation:

[M+2] = (number of C × 1.1)2 /200 + (number of O × 0.2) + (number of S × 4.4) + (number of Si × 3.4)

This method would not be commonly used by researchers who have a high-resolution mass spectrometer at their disposal or are able to submit their samples to a service laboratory for exact mass analysis. This method is useless, when the molecular ion peak is very weak or does not appear. Sometimes the isotopic peaks surrounding the molecular ion are difficult to locate in the mass spectrum, and the results obtained by this method may at times be rendered ambiguous.

Nitrogen rule:

Nitrogen rule states that a molecule of even-numbered molecular mass must contain no nitrogen or an even number of nitrogen atoms. An odd-numbered molecular mass requires an odd number of nitrogen atoms.

This rule holds for all compounds containing C, H, N, O, S, and halogens, as well as less usual atoms like P, B, Si, As, etc.

An important corollary of this rule states that the fragmentation at a single bond gives an odd-numbered ion fragment from an even-numbered molecular ion. Similarly, an even-numbered ion fragment results from an odd-numbered molecular ion. However, the fragment ion must not contain all the nitrogen atoms of the molecular ion.

Example: Nitrobenzene (C₆H₅NO₂)

The molecular ion appears at m/z 123, an odd-numbered molecular mass, since the compound contains one (odd number) nitrogen atom.

Two important fragment ions formed in the mass spectrum are:

NO₂⁺ at m/z 46

NO⁺ at m/z 30

Both fragment ions appear at even mass numbers.

Now consider 2,4-dinitrophenol. This compound contains two (even number) nitrogen atoms. Its molecular ion (M⁺) appears at m/z 184.

The fragment ions appearing at (M⁺ − H) occur at m/z 183, and (M⁺ − (M − CO)) at m/z 155. Thus, the fragment ions containing both nitrogen atoms appear at odd mass numbers. This proves the validity of the nitrogen rule.

The inclusion of any common stable isotope except ¹⁸O alters the use of the nitrogen rule.

Rule of 13:

Rule of thirteen is a useful method for determining the possible molecular formula of a compound from its Molecular Mass. In the Rule of Thirteen first, a base formula is generated which consists of only hydrogen and carbon atoms. This base formula is calculated by dividing the molecular mass by 13 (C + H: 12+ 1 =13). When a molecular mass, M+ , is known, a base formula can be generated from the following equation:

The base formula will be: CnHn+r The index of hydrogen deficiency (IHD) or Double Bond Equivalent DBE will be: DBE = (n – r + 2)/2

Determination of Molecular weight

The mass spectrum is a plot representing the m/e values of the parent as well as the fragment ions against their corresponding relative abundances. The peak on the extreme right (i.e., particle of highest mass) corresponds to the molecular mass of the original molecule. In case of straight chain hydrocarbons, the abundance of the parent ion peak is fair. It also gives (M⁺ + 1) peak which is of 9.9% abundance compared to the parent peak. Consider that a compound forms peaks at m/e values of 100, 85, 71, 57, 43 (100%) etc. Evidently, it is a straight chain alkane because fragment peaks are formed 14 units apart. A base peak due to C₃H₇⁺ is most abundant in straight chain hydrocarbons. Thus, a molecular formula of the compound can be obtained. In case of chloro and bromo compounds, the pair of peaks are in the intensity ratio of 1 : 3 and 1 : 1 respectively. In some cases, a McLafferty rearrangement ion peak gives an important clue in the determination of molecular formula of the compound. For example, all straight chain aldehydes containing a γ-H atom form a base peak at m/e 44. Also in aldehydes a fairly abundant M⁺ parent peak follows a less abundant (M⁺ − 1) peak.

Fragmentation pattern of Ethane

Mass spectrum of Ethane:

Fragmentation pattern:

Fragmentation pattern of Propane

Mass spectrum of propane

Fragmentation pattern

Fragmentation pattern of Butane

Mass spectrum of Butane (C4H10)

Fragmentation pattern

Fragmentation pattern of pentane

Mass spectrum of pentane (C5H12)

Fragmentation pattern

Aliphatic Aldehydes and ketones:

Important features of their mass spectra are:

The intensity of the parent peak decreases as the alkyl chain length increases.

The main fragmentation processes are α- and β-cleavage. In α-cleavage, the bigger group on either side of the carbonyl group (ketone) is preferably lost.

In lower aldehydes, α-cleavage is prominent with retention of charge on oxygen.

In aldehydes and ketones containing γ-H atom, McLafferty rearrangement ion is most significant. In an aldehyde which is not α-substituted, a base peak due to this is formed at m/e 44.

The McLafferty rearrangement ion in methyl ketones which are not α-substituted appears at m/e 58.

In C₄ and higher aldehydes, cleavage of the C—C bond once removed from the C=O group occurs with hydrogen rearrangement to give a major peak at m/e 44, 58 or 72 etc., depending upon the α-substituents.

In straight chain aldehydes, other diagnostic peaks are at m/e 18 (loss of water), m/e 28 (loss of C₂H₄), m/e 43 (loss of CH₂=CH—O•) and m/e 44 (loss of CH₂—CH—OH).

In aldehydes, methyl or alkyl radical is preferably lost compared to hydrogen radical. Consider fragmentation pattern of butanal.

In aliphatic ketones, the cleavage of the C—C bonds adjacent to oxygen atom gives rise to a peak at m/e 43, 57 or 71. The base peak results from loss of the larger alkyl group. When one of the alkyl chains attached to the C=O group is C₃ or larger, cleavage of the C=C bond (α, β bond) from the C=O group occurs with hydrogen rearrangement to give a major peak at m/e 58, 72 or 86 etc.

In cyclic ketones, the base peak occurs at m/e 55. The mechanism involves hydrogen rearrangement from a primary radical to a conjugated secondary radical followed by formation of the resonance stable ion.

Aromatic aldehydes and Ketone

1. The parent ion peak is intense in these compounds.

2. The ion Ar—C≡O⁺ eliminates CO to give C₆H₅⁺ (m/e 77) which in turn eliminates HC≡CH to give the C₄H₃⁺ ion (m/e 51). Consider benzaldehyde in which M−1, M−28 are formed by the removal of CO.

Reaction sequence:

3. In ketones, the loss of larger group is preferred by α-cleavage. For example, fragmentation of alkyl phenyl ketone occurs as follows:

Reaction sequence:

4. When the alkyl chain is C₃ or longer, cleavage of the C—C bond once removed from the CO group occurs with hydrogen rearrangement. Unsymmetrical diaryl ketones cleave to give ArC≡O⁺ ions.

Alkyl halides:

Important Features of Mass Spectra of Halogen Compounds

The molecular ion abundance of a particular alkyl halide increases as the electronegativity of the halogen substituent decreases.

The relative abundance of the molecular ion decreases with increase in chain length and branching.

Compounds containing chlorine and bromine exhibit isotopic peaks. A compound having one chlorine atom shows an M + 2 peak which is 1/3 in intensity of the parent peak.

In the parent ion, the charge resides on the halogen atom.

Important fragmentation mode is α-cleavage with charge retention by the halogen-containing fragment. Another mode leads to the loss of halide radical.

Fragmentation pathways:

Straight-chain chlorides longer than C6 give C3H6Cl⁺, C4H8Cl⁺ (most intense), and C5H10Cl⁺ ions.

Aliphatic iodides give the strongest parent peak of the aliphatic halides. Iodides cleave much as do chlorides and bromides, but the C4H8I⁺ ion is not so evident.

The base peak is at m/e 69 due to CF₃⁺ in all perfluorocarbons. The stable ions C₃F₅⁺ and C₄F₇⁺ give large peaks at m/e 131 and 181, respectively.

Benzyl Halides: The benzyl or tropylium ion formed by the loss of the halide is favoured even over β-bond cleavage of an alkyl substituent. The α-bond cleavage is prominent when the ring is polysubstituted.

Aromatic Halides: The molecular ion peak is apparent for all compounds in which the halogen atom is attached directly to the aromatic ring.

Alcohols.

The parent peak of a primary and secondary alcohol is usually small. It is not detectable in tertiary alcohols.

The parent ion peak is formed by the removal of one electron from the lone pair on the oxygen atom of primary and secondary alcohols.

Fragmentation mode depends on the nature of the alcohol, whether it is primary, secondary or tertiary.

The fragmentation of the C–C bond adjacent to the oxygen atom (α-cleavage) is of general occurrence.

Aliphatic primary alcohols show a prominent signal at m/e 31. This signal corresponds to the formation of the oxonium ion (CH₂=OH⁺) and is formed by the cleavage of the C–H bond.

Primary alcohols show M⁺ – 18 peaks corresponding to the loss of water (H₂O).

Higher alcohols show a peak corresponding to:

The olefinic ion then decomposes by successive elimination of ethylene.

Long-chain alcohols may show peaks corresponding to the successive loss of hydrogen radicals at: M – 1, M – 2 , M – 3

Alcohols containing branched methyl groups (e.g., terpenes) show a fairly strong peak at m/e 33 resulting from the loss of CH₃ and H₂O.

A peak at m/e 31 is quite diagnostic for a primary alcohol, provided it is more intense than the peaks at m/e 45, 59, 73, etc. However, the first-formed ion of a secondary alcohol decomposes further to give a moderately intense m/e 31 ion.

In addition to α-cleavage, primary alcohols also undergo β-, γ-, and δ-cleavages to form peaks at: m/e 45,m/e 59 , m/e 73 etc.

Secondary alcohols cleave to give prominent peaks due to R—CH=OH⁺ at m/e 45, 59, 73, etc.Tertiary alcohols fragment to give prominent peaks due to RR′C=OH⁺ at m/e 59, 73, 87, etc.

Cyclic Alcohols: The fragmentation patterns are quite complicated. In the case of cyclohexanol, the molecular ion peak appears at m/e 100. It loses a hydrogen radical to form the M⁺ – 1 peak at m/e 99. This, in turn, loses a molecule of water to form a signal at M⁺ – H – H₂O at m/e 81.

Allyl Alcohols: Here the (M – H)⁺ ion is formed due to its high stability.