L9--mass

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Problem

• The exact mass of a compound determined byHRMS is 212.0833. What is the molecularformula of the compound?

• . , . , . , .• Nitrogen rule? N yes or No?

• Rule of 13 ? Subtract heteroatom equivalents

• C16H20 ?• C15H16O ?

• C14H12O2 ?

• Index of H deficiency? = 14 – 12/2 + 1 = 9

• 212 (12%), 167 (4%), 105 (100%), 91 (45%), 77(30%), 63 (10%), 51 (12%). Structure?

PhCOOCH2Ph

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Fragmentation• Molecular ion = a radical cation

• Homolytic cleavage or heterolytic cleavage

1. Bond strength

2. Possibilit of low ener transition

3. Stability of the fragments

• Our knowledge of pyrolysis can be used• These are mostly unimolecular decompositions

• Organic reactions are initiated by reagents, heat or light

suffered by an organic molecule at a vapor pressure of

10-6 mm Hg struck by an ionizing beam of electrons

• There are a number of rules to predict fragmentation

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Fragmentation Rules (for EI only)

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Remember, the rules are applicable to EI mass spectrometry only.

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McLafferty Rearrangement

• Rearrangement involving migration of a γ- hydrogenatom to the heteroatom leading to elimination of alkene

• low energy transitions and increased product stability

• A molecule must possess a heteroatom, a double bond,.

Prominent

Peaks in MS

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Simple Cleavage Rearrangement

Even numberedmolecular ion gives

Even numberedmolecular ion gives

odd numberedfragment and even numberedfragmentvice a versa

Cleava e of a sin le Two cleava es at abond – Nitrogen rule time

Involves muchcom lex mechanism

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of

Some Chemical Classes

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Hydrocarbons• Linear hydrocarbons always show M+ peak• M+ is of low intensity for long chain compounds•

clusters of peaks, with peaks of each cluster 14 massunit apart (CH2)• ar est ea in each c uster re resents a C H +1

peak and thus occur at m/z = 14n+1•

This is accompanied by CnH2n and CnH2n-1 fragments.• ,the abundance decrease in a smooth curve to [M-C2H5]; [M-CH3] peak is almost missing or very weak.

• • Characterization is based on M+ peak.• For branched ones, the curve is not continuous

n 2n

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169

CnH2n

[M‐15]Loss of H

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• Groups of

ions

correspond

to

loss

of

one

‐, two

‐,

three‐, and four‐carbon fragments.

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Branched & cyclic hydrocarbons

• Branched: More or less similar to linear ones but the smooth curve isbroken by preferred fragmentation at each branch.

• E. . in 5-meth l entadecane, a M+ ion, a C12 fra ment, absence of C11unit, peak at M-15, peaks for fragments on both sides of the branchconfirm the structure.

• Cyclic compounds give more intense M+, as frag. need two C-C

• Fragmentation of cyclic compound is characterized by loss of two Catoms as C2H4 and C2H5

• Also C-C cleavage is accompanied by loss of hydrogen to get CnH2n-1Loss of

C2H4

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• The most stable carbocation fragments form in greateramounts.

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Alkenes

• M+ is usually distinct

• Location of double bonds in acyclic olefins is difficult due to itsac e m gra on

• In cyclic olefins the double bond won’t migrate much and thereis a strong tendency for allylic cleavage. Distinct M+ is seen.

• Conjugation with carbonyl also fixes the position

• Clusters of peaks at intervals of 14 units

• C H - , C H are more intense than C H -

• Limonene shows adistinct mode of

c eavage. 2 isoprenes.• Can you tell the name

reaction?

It’s the Retro Diels-Alder reaction

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Allylic cleavagesDouble bond isomerization(increased conjugation)

Bi-allylic bondn 2n-1

41, 55, 69

n = 4

n = 3

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ass S ectra o Alkenes Resonance-stabilized cations favored.

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Alcohols

• Small M+ peak for pri./sec.; very weak for tert.Alcohols.

-

(rule 8). Thus Pri. Alcohols show a prominent peak at31 due to +CH2OH (diagnostic of primary alcohols if

Sec. alcohols at +RCH-OH (45, 59, 73 etc.).

Tert. alcohols at +RR’C-OH (59, 73, 87 etc.) • T e argest su stituent is expe e most rea i y

• Occasionally C-H bond next to Oxygen is cleaved -

• Occasionally [M-2] and [M-3] peaks are also seen• In addition to this, primary alcohols show pattern

e a enes w pea s o omo ogues n ecreas ng

intensity resulting from cleavage of C-C bond

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Alcohols

• Distinct peak at M-18 by loss of H2O.

• Peak for M-(H2O + alkene) by thermal

decomposition at higher temperatures

• M-33 for M-(H2O+CH3) in branched ones • yc c a co o ow comp ex ragmen a on

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Benzyl Alcohols

• Benzyl alcohols are unique inshowing prominent M+,

•Moderate M-OH and weak M-2,M-3 may be seen.

• n enzy a co o ea s a

107 (M-1), 79 (-CO), 77 (-H2)

• Loss of H2O is common in ortho

in phenols

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Phen-ols

• Distinct molecular ion peak

• In phenol M+ is base peak (94). 77 (M-17, loss of OH),

M-18 (-H2O), M-28 (-CO) and M-29 (-CHO) are common.In cresols M-1 (loss of benzylic H) > M+

• n o -et y p eno - 3 s ntense > +.

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Ethers

• + Acetals show,• Strong peaks at 31, 45, 59, 73

for RO+ & ROCH2+ fragments – 2 ways

weak M+,Prominent M-R &/or M-OR& weak M-H

• C eavage o C-C on next to O

• Cleavage of C-O bond; charge remaining on Carbon

All facile, mediated by O

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Ketones• Distinct M+

• Cleavage of one of the R group (α,β C-C bond)

• For Alkyl chain longer than C3, hydrogen migration

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Aromatic ketones• Prominent M+,

• In Aryl alkyl ketones ArCOR cleavage of a C-C bond-

then loss of CO to Ar+• If R is lon er than C3 the McLaffert Rn is seen

Diar l etone

P-chlorobenzophenoneM+2 is isotopic

=

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Aldehydes • Usua y visi e M+

• Cleavage of C-H and C-C bond next to Oxygen resultsin M-H and M-R m/z 29 CHO+ or C H + M-18 H O

• For Alkyl chain longer than C3, hydrogen migrationMcLafferty Rearrangement is seen – stabilized ion

and M-1 peak which could be even larger.further fragments to give Ar+ (77 for phenyl) which

4 3

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Acids• Usually visible M+

• Major (base) peak at m/z 60 comes

• Apart from this, a series of clusterat the interval of 14 is seen – y rocar on. n eac c uster a

major peak is of CnH2n-1O2.

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• Loss o O -17 COO -45

• H2O (-18) is major if ortho gr. Contg. H

• Ortho Effect – six memb. TS, eliminate, ,

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Esters• Distinct M+ peak

• Major (base) peak at m/z 74me y es ers rom o

comes from McLaffertyRearran e ent

• Apart from this, a series of

cluster at the interval of 14.Cleavage of on both sides ofCO.

Aromatic esters

Loss of neutral ketene

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• + -

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Amines

• M+ of aliphatic monoamine is an odd number and givesa weak peak

• Base peak results from α,β cleavage (rule 8)

• m/z = 30 (CH2NH2) for pri. and sec/tert without α- su s uen .

• If no α-substituent – [M-1] is frequent.

• • a series of cluster at the interval of 14 (30,44,58,72).

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Amines

• During fragmentation of long chain amines, cyclicfragments occur 6 or 5 member rings formed

, -

• For cyclic amines M+ peak is strong,

• Cleavage next to N leads to loss of α-H to give M-1 or

cleavage of ring followed by elimination of ethylene

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Aromatic Amines (Anilines)

• M+ of aromatic monoamine is an odd number and givesan intense peak

• Loss of one of the amino α-H give M-1

• Loss of neutral HCN followed by H give strong peaksa an

• In alkyl aryl amines, the cleavage is controlled by N

Leadin to cleava e of al ha C-C bond

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Amides

• M+ of amide is usually seen

• In lon chain amines a base eak at 59 for.CH

2

(=OH)NH2

results from McLafferty Ren

• Primary amide gives a strong peak at 44 fromc eavage o - 2 on = - 2

• A moderate peak at 86 results from gamma-delta C-C

bond cleava e accom anied b c clization.

• romatic ami es – + an t en oss o to give

PhCO+ and then loss of C=O to give Ph+.

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Example

N l

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Nitriles

• M+ of aliphatic nitriles is very weak (exceptacetonitrile and ro ionitrile , M-1 is seen

• A Strong peak at 41 (CH2=C=N+

-H) due to McLaffertybut lacks diagnostic value as C3H5 (41) is common in y rocar ons

• Intense peak at 97 is characteristic for C8 and more

• series of peaks at the interval of 14 due to loss of

CH2.

Ni d

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Nitro compounds

• M+ of aliphatic nitro compounds (odd

lower homologs) • a n pea s are a ser es o pea s at t e

interval of 14 upto M-NO2.• Presence of NO2 group is indicated byappreciable peak at m/z = 30 (NO+) and

a smaller peak at 46 (NO2+)

A ti Nit d

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Aromatic Nitro compounds

• M+ of aliphatic nitro compounds (odd number)

• Presence of NO2 group is indicated by-

peak in nitrobenzene) and a peak at M-30 i.e.

loss of neutral NO molecule withrearrangement to form phenoxy cation

• Loss of acetylene from M-46 gives M-72

(strong)• Loss of CO from M-30 gives M-58

• Diagnostic peak at m/z 30 for NO+ ion

Nit ili i

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Nitro-anilinines• O,m,p-nitroanilines – strong M+ peak (even

numbered)

• Two sequences –

• one loss of NO2 (M-46) = m/z 92 which loses HCNo ro uce z

• Second, loss of NO (M-30) = m/z 108 which loses

• M and p comps. give a small peak at 122 by loss ofO whereas ortho com ound loses OH to ive 121

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• Sulfur has three isotopes: 32S (95%), 33S (0.8%),. .

• The M+ peak of ethyl methyl sulfide has an M+2eak that is lar er than usual about 4% of M+ .

• If more than oneThe number of sulfur atoms can

be found from the intensity of M+2.

S lf d

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Sulfur compounds

• Thiols show M+ & M+2,

CH2=SH+ (47), cleavage of beta gamma-

in intensity as to 47, cleavege of χ,δ

and cleavage of δ,ε bond gives a mpeak

presumably due to cyclization

S lfid s

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Sulfides• Cleavage modes resemble ethers. Cleavage of larger

subst. favored

• A strong peak at 61 in straight chain sulfides is due to

• Cleavage of C-S bonds leaves a

c arge on u ur

leading to cyclization

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• Isoto es are resent in their usual

abundance.• Carbon has a 13C isoto e resent in 1.1%abundance. The spectrum will show the

normal M+

and small M+1 peak.• Bromine has two isotopes: 79Br (50.5%)and 81Br (49.5%). Since the abundances

are a most equa , t ere w e an+

peaand and M+2 peak of equal height.

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• 79 . .81Br. The molecular ion peak M+ has 79Br be astall as the M+2 peak that has 81Br.

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• 35 .

24.5% 37Cl. The molecular ion peak M+ is 3times hi her than the M+2 eak.

53

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• Co ounds containin two chlorines or two

bromines or a chlorine and a bromine showdistinct M+4 peak in addition to M+2 dueto presence o two atoms o eavy isotope.

• Thus three atoms give peaks at M+2, M+4an + – e re a ve a un ance can epredicted

compounds

•

Beynon 1968

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Isotopic Peak Pattern

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Isotopic Peak Pattern

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i• Intense M+

•Cleavage of the bond beta to the ring• Localize the charge on the heteroatom

•

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For example

HRMS

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For example

, 2, 2 2 4M. Wt. = 28The sum of formulamasses of the mostabundant isotope gives

HRMSFor CO x . + x . =

27.9959For N2

14.0031 x 2 = 28.0062For CH2N1x12.0 + 2x1.00783 +

1x14.0031 = 28.0187 2 4

2x12.0+4x1.00783 =28.0312.Atomic weight is however

e average o a eisotopes

Index of Hydrogen Deficiency/

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es egrees o nsa ura on

• .removed from the corresponding saturated formula to get

the compound of interest.• , ,

of double bonds and twice the number of triple bonds.(benzene =4, nitro = 1)

• n m x y z .Index = (n) – (m/2) – (x/2) + (y/2) + 1

• Calculate for C7H7NO – . + . + =

• The index give some idea about the structure of molecularion. Even-electron ions give an odd multiple of 0.5, whileo e ec ron ragmen ons g ve n eger va ues o eindex.

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•

looking at the answers first. After you

with the model answer.


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f y g f y

es egrees o nsa ura on

• .removed from the corresponding saturated formula to get

the compound of interest.• , ,

of double bonds and twice the number of triple bonds.(benzene =4, nitro = 1)

• n m x y z .Index = (n) – (m/2) – (x/2) + (y/2) + 1

• Calculate for C7H7NO – . + . + =

• The index give some idea about the structure of molecularion. Even-electron ions give an odd multiple of 0.5, whileo e ec ron ragmen ons g ve n eger va ues o eindex.

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Sinapinic acid Nicotinic acid

Niacin

Vitamin B3

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L9--mass

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