fragmentation pattern in mass

FRAGMENTATION PATTERN AND ITS RULES AND INTERPRETATION OF MASS SPECTROSCOPY: 

FRAGMENTATION PATTERN AND ITS RULES AND INTERPRETATION OF MASS SPECTROSCOPY Guided by:- Mr. NISHIT S. PATEL M.Pharm (Q.A.) Assistant Professor Department of Pharmaceuticcal chemistry Dharmaj degree Pharmacy College Presented by:- BHUMIN PATEL Roll No.:-2 1 st semister (2010-2011) M.Pharma (Pharmaceutics) Dharmaj degree Pharmacy College

Fragmentation Pattern and Rules: 

Fragmentation Pattern and Rules Definition:- Molecular ions(Parent ions ):-If the electron beam energy is excess than ionization potential electrons may be ejected from a lower lying molecular orbital leading to the formation of ions called “Molecular ions”. M + e M + + e - Fragment ions :- if the electron beam is further increased to the apparent potential of a molecules, then the excited molecular ions undergo decomposition to give rise to varieties of fragment ions which is having smaller masses than parent ions. CH 3 -CH 2 -Cl + e CH 3 -CH 2 -Cl + + e - parent ion CH 3 -CH 2 + + Cl - CH 3 -CH 2 Cl CH 2 -CH 2 + + HCl Fragment ion

Slide 3: 

Fragmentation is initiated by electron impact Driving force for fragmentation is energy transferred as result of the impact. Fragmentation of the odd-electron molecular ion ( radical- cation , M . + )-----occur by homolytic or heterolytic cleavage of the single bond. For homolytic cleavage:- For heterolytic cleavage:-

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Further Fragmentation of an even-electron cation –--result in another even electron cation and even-neutral molecule or fragment. Cleavage of particular bond ----related to the --bond strength --possibility of low energy transition --stability of the fragments --both charge and uncharged. Ease of fragmentation to form ions increases in the order:- CH 3 + <RCH 2 + <R 2 CH + <R 3 C + <CH 2 =CH-CH 2 + <C 6 H 5 - CH 2 + DIFFICULT EASY

Rules for predicting the prominent peaks in EI spectra:-: 

R ules for predicting the prominent peaks in EI spectra:- Relative height of the molecular ion peak ---greatest for the straight-chain compound and decreases as the degree of branching increases. The relative height of the molecular ion peak ---decreases with increasing molecular weight in a homologous series. ( fatty ester---exception.)

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Cleavage---favored at alkyl-substituted C-atoms—more substituted, the more likely is cleavage because of the stability of a tertiary carbocation . Double bonds, cyclic structure and aromatics(or heteromatics )rings stabilize the molecular ions---increase the probability of its appearance. CH 3 + <RCH 2 + <R 2 CH + <R 3 C +

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Double bond favor allylic cleavage and give the resonance stabilize allylic carbocation . Saturated rings tends to loose alkyl side chain at  bond. Unsaturated ring can undergo a retro- Diel‘s -Alder reaction.

Slide 8: 

in alkyl substituted aromatic compound, cleavage ----very probable at the β to the ring, giving rise to the resonance stabilized benzyl ring ion or more likely, tropylium ion. The C-c bonds next to the hetero-atoms---frequent cleaved, leaving the charge on the fragment containing the hetero-atom whose bonding electrons provide resonance stabilization.

Slide 10: 

Cleavage----often associated with elimination of small, stable, neutral molecules . Eg. CO, H2O, olefines , ammonium, hydrogen sulfide, hydrogen cyanide, ketone or aldehyde ---often with the rearrangement.

McLafferty Rearrangement: 

McLafferty Rearrangement Involving the intra-molecular rearrangement. A γ -H atom---transferred through 6 membered transition state to an electron deficient centre ----followed by the cleavage of β -bond. Result in the elimination of a neutral molecule. To undergo a McLafferty rearrangement, a molecule must posses:- A side chain containing at least 3-C atoms—last bearing γ -H atom. An appropriately located heteroatom ( eg . O) and  -system, usually a double bond which could be a carboxyl group or olefinic double bond or aromatic system.

Mass spectrum of some chemical classes: -: 

Mass spectrum of some chemical classes: - Hydrocarbons:- Saturated hydrocarbons : The fragmentation pattern---- characterised by cluster of peaks----corresponding peak of each cluster are 14(CH 2 ) mass unit apart. Largest peak----C n H 2n+1 -----occur at m/z=4n+1 ---accompanied by C n H 2n-1 and C n H 2n fragments. EG.cyclohexane

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Alkenes ( olefines ): In cyclic (specially polycyclic) alkenes, location of the double bond----result of a strong tendency for allyllic cleavage without much double bond migration(rule 5). Conjugation with a carbonyl group ---fixes the position of the double bond. Eg. Β - myrcene .

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Aromatic and Aralkyl Hydrocarbon : An aromatic ring in a molecule stabilizes the molecular ion peak.(rule 4) Branching at the α -C leads to masses higher than 91by increment of 14( rule 3), largest the substitution being eliminated most readily. The ion of mass 91 ---- tropylium ion rather than benzylic cation Eg. Xylenes

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Peak at m/z=65---results from elimination of acetylene molecules from the tropylium ion Hydrogen migration with elimination of neutral alkenes molecule----peak at m/z=92----observed when alkyl group is longer than C 2. Alkylated polyphenyls and alkylated polycyclic hydrocarbon exhibit the same β -cleavage .

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Hydroxyl Compounds:- Alcohols :- Cleavage of the C-C bond next to the oxygen atom(Rule 8). Thus primary alcohol show prominent peak result of CH 2 =OH R R C=OH (m/z=45, 59,73,etc.) C=OH H R Where R” > R’ or R when R and/or R’=H, M-1 peak –usually seen. In the long chain (>C 6 ) alcohol, fragment becomes dominated by the hydrocarbon pattern

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A distinct and sometimes prominent peak are found due loss of water (M-18).

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Benzyl alcohol ---fragments ---gives M-1, C 6 H 7 + ions by loss of CO and C 6 H 5 + ion by loss of benzene. Loss of water to give a distinct M-18 peak----common in o-substituted in benzyl alcohol.

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Phenol:- M-1---small peak Eg. (1) In cresol , M-1 peak ---large than the molecular ion as a result of facile benzylic C-H cleavage. The rearrangement peak at m/z=77 and peaks resulting from loss of CO (M-28) and CHO (M-29)----usually found in phenol.

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(2) 0-Ethyl phenol :-spectrum shows that a methyl group---lost much more readily than an α -H.

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Ethers:- Aliphatic ethers (and acetal ):- Fragmentation occurs in 2 principles:------- Cleavage of the C-C bond next to O- atom ( α , β -bond, rule 8).

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C-O bond cleavage with the charge remaining on the alkyl fragment. The spectrum of the long –chain ethers becomes dominated by the hydrocarbon pattern.

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Aromatic ethers:- The molecular ion peak –prominent. Primary cleavage occur at the β bond to the ring. Eg. Anisol :----m/z=93 and 65 The characteristic aromatic peak at m/z=78 and 77---raise from anisol .

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When the alkyl portion of an aromatic ether --- C 2 or larger—cleavage β to the ring ---accompanied by hydrogen migration .

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Ketones :- Aliphatic ketones :- Cleavage at C-C bond adjacent to the O-atom---charge remaining with resonance-stabilized acylium ion . Cleavage ---rise to a peak at m/z=43 or 57 or 71…… The base peak---result of loss of large alkyl group.

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When alkyl chains attached to C=O group is C 3 or longer----cleavage of C-C bond---removed( α , β -bond) from C=O group occurs with hydrogen migration to give a major peak (McLafferty rearrangement).

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Aromatic Ketones:- Cleavage of aryl alkyl ketone occurs at β -bond to the ring, leaving a characteristic ArC =O fragment ----base peak. When the alkyl chain is C 3 or longer, cleavage of the C-C bond once removed from the C=O group occurs with hydrogen migration.

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Aldehydes:- Aliphatic Aldehyde :- Cleavage of the C-H and C-C bonds next to the O-atom----result in an M-1 and M-R peak (m/z=29 CHO + ). The M-1 ---good diagnostic peak even for the longer aldehyde. In C4 or higher aldehyde---McLafferty cleavage of the α , β -C-C bond occur to give a major peak at m/z=44, 58,or 72……..depending on the α - substitution. This resonance-stabilized ion formed through the cyclic transition state.

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In the straight-chain aldehyde, the other unique, diagnostic peak is ---- M-18----loss of water, M-28---loss of ethylene, M-43----loss of CH 2 =CH-O . , M-44—loss of CH 2 =CH-OH

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Aromatic aldehyde :- M-1 ---always larger than molecular ion peak. M-1 ion C 6 H 5 ≡O + eliminate CO to give the phenyl ion (m/z=77), which in turn eliminates HC ≡CH to give the C 4 H 3 + ion.(m/z=51).

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Carboxylic acids :- Aliphatic Acids :- Molecular ion peak of straight-chain mono-carboxylic acid---weak. Branching at the α -C enhances cleavage.

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In short chain acid----peak at M-OH and M-CO2H----prominent-----represent cleavage of bond next to C=0. In long-chain acids---the spectrum consist of 2 series of peak resulting from cleavage at each C-C bond with retention of charge either on the oxygen containing fragment.(m/z=45, 59,73, 87,….) o on alkyl fragment(m/z=29, 43, 57, 71, 85,….)

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Aromatic acids :- Molecular ion peak----larger. Other prominent peaks----formed by loss of (M-17) and of (M-45). Loss of water(M-18)---noted---if hydrogen bearing o-group---available.

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Amines :- Aliphatic Amines :- Molecular ion peak of aliphatic monoamine----odd number----but usually ---weak. In long chain or highly branched amines----undetectable. Base peak----result from C-C cleavage next ( α , β ) to the atom (rule 8). Loss of largest branch from the α -C atom----preferred----

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Cyclic fragments---occurs during the fragmentation of longer chain amines. Cleavage of amino acid ester occurs at both C-C bonds next to the nitrogen atom,---loss of the carbalkoxy group.

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Aromatic amines :- Molecular ion peak----odd number—intense. Loss of the amino H-atom of aniline gives moderately intense peak at M-1. Loss of a neutral molecules of HCN—followed by loss of H-atom ----gives rise to prominent peak at m/z=66 and 65 respectively.

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Halogens:- If more than one chlorine atom or bromine atom is present in a molecule, distinctive “isotope cluster patterns” are seen in the mass spectrum. If we have two Cl atoms present in an ion, the distribution of masses in each ion may be 35 Cl 35 Cl, 37 Cl 35 Cl, 35 Cl 37 Cl, or 37 Cl 37 Cl. 35 C1: 37 Cl abundance is 100:33 Bromine has two isotopes, 79Br and 81Br, and they are approximately equal in abundance. A compound RBr containing 1 Br atom will have masses of R + 79 and R + 81, and they will be about equal in abundance.

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If 2 Br atoms are present, as in RBr2 , there will be masses of R +158, R + 160, and R + 162 from R 79 Br 79 Br, R 79 Br 81 Br, R 81 Br 79 Br, R 81 Br 81 Br,& their relative abundances will be 51:100:49, based purely on probability.

Interpretation of Mass Spectra:---: 

I nterpretation of Mass Spectra:--- A mass spectrum is a plot or table of the mass-to-charge ratio, m/z, of detected ions vs their relative abundance (relative concentration). A typical mass spectral plot for a small organic molecule, benzene.

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The most abundant peak in the spectrum is called the base peak. The base peak is assigned an abundance of 100% and the other peak heights are plotted as percentages of that base peak. There are two ways to interpret such spectra:-- i ) to compare the spectrum you have with those in a searchable computerized mass spectral database. ii) to evaluate the spectrum using the interpretation procedure described subsequently.

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The procedure for interpreting a mass spectrum consists of the following steps: Identify the molecular ion if present Apply the “nitrogen rule” Evaluate for “M + 2” elements Calculate “M + 1” and “M” elements Look for reasonable loss peaks from the molecular ion Look for characteristic low mass fragments Postulate a possible formula Calculate “rings β -double bonds” Postulate a reasonable structure

General Rules for interpretation of Mass Spectrum:-: 

General Rules for interpretation of Mass Spectrum:- The Exact Molecular weight :- Mass spectrometer----used to determine the molecular weight of the pure compound from identification of the parent peak. From molecular weight ----determine molecular formula. The Isotopic Effect:- Mass spectrometer---used to determine the distribution of naturally occurring isotopes. Heavier isotopes gives peak at M+1, M+2, M+$, etc from the height of such peaks molecular weight ---determine.

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Nitrogen Rule :- State that:--- If m/z=even for the parent ions----than it contains even number(including zero) of nitrogen atom. If m/z=odd than contains odd number of nitrogen atoms. Fragmentation at the single bond gives an odd numbered fragment ions from even numbered molecular ion and even numbered ion fragment from the odd numbered molecular ion.

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Applied to all compound which are covalent bonds and contain any combination of C, H, N ,O, S, Si, As, P, halogen and alkaline earth metals. In this rule---molecular weight to be used is sum of the abundant isotopes .

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Ring Rule :- Used to calculate number of unsaturated centre ( R ). R= no. of rings + no. of double bonds +twice the no. of triple bonds. The ring rule for molecule C w H x N y O z R = w + 1 + y - x 2

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Eg. Benzene (C 6 H 6 ) w=6, x=6 , y=0, z=0 Therefore, R= 6+1-3=4 Thus, it contain 1 rings and 3 double bonds.

REFERENCE:-: 

REFERENCE:- Silverstein Robert M., Webster Francis X. ; Spectrometric Identification of Organic Compounds; Sixth Edition; Wiley Student Edition; page no.12-35. Pavia Donald L., Lampman Gary M., Kriz George S.; Introduction to Spectroscopy ,a Guide for Students of Organic Chemistry ; third Edition ; Brooks/Cole Thomson Learning;;404-41. Robinson James W., Skelly Frame Eileen M., Frame George M. II; Undergraduate Instrumental analysis ; Sixed Edition; Marcel Dekker; Page no. 652-57. Dr. Kasture A. V., Dr. Mahadik K. R.,Dr . Wadodkar S. G., DR.More H.N. ; Pharmaceutical Analysis Volume-II, Instrumental Method ; Nirali Prakashan ; Page no. 248-49. Chatwal Gurdeep R., Anand Sham K.; Instrumental Methods of Chemical Analysis ; Himalaya Publishing House ; Page no.287-88.

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Thank You!!!