In this lesson you will learn: How to fragment molecules and what component weights signify what parts. Fragmenting straight chain alkanes Fragmenting functional groups

USES of the Mass Spectrograph It is mainly used in labs but it is also used in: Archeologists use them for radioactive carbon dating to find the age of artifacts. Space exploration – to find out what the soil on Mars is made from and the atmosphere. Mass spectrometers were aboard many of the probes that go to other planets. Toxins in water supplies Industry – such as crude oil refineries to find out the parts in the oil and the impurities.

FRAGMENTATION PATTERNS

Mass Spectra Fragmentation : Molecular ions are unstable, and will break up into smaller pieces. The simplest case is that a molecular ion breaks into two parts - one of which is another positive ion, and the other is a free radical. Neutral fragments cannot be deflected and never reach the detector ! fragment ions

Fragmentation Patterns The collision of a high energy electron with a molecule not only causes the loss of a valence electron and cause the molecule to fragment. Fragmentation causes most stable cations with higher relative abundances CH3 – CH3 (30) CH3 = 15

Fragmentation Patterns Alkanes Very predictable – apply the lessons of the stability of carbocations (or radicals) to predict or explain the observation of the fragments. Example: iso-butane Example: iso-butane C3H7+ = 43 (58) (CH3)3 CH (15) Question: Why, is the second route not viable? Answer: because 20 carbocations are more stable than 10 (+CH3 )

Fragmentation Patterns Fragment Ions : n-alkanes…Memorize these m/z 15 29 43 57 71 85 etc. CH3+ C2H5+ C3H7+ C4H9+ C5H11+ C6H13+ 71

ncarbons = [M+1] / ((0.011)M+) = 5 / (38)(0.011)) = 12 carbons. Which Alkane is this? A student is told this straight chain alkane MS readout. Find the #of carbon atoms first ncarbons = [M+1] / ((0.011)M+) = 5 / (38)(0.011)) = 12 carbons. mass is 12x12= 144.The M= peak has a mass of 170 , so 170-144 = 26. We have dodecane C12H26 , can you ascertain the other peaks? SEE next slide for ALL dodecane fragments 57 71 85 43 99 113 38% M+ at 170 29 127 5% M+1

Dodecane [C4H9]+ [C3H7]+ [C5H11]+ [C6H13]+ 99 [C7H15]+ 127, [C9H19]+

Fragmentation Film 8 min fragmentation clip: Advance and start at 2:30 https://www.youtube.com/watch?v=stIwRio9FeM Or 8 min Royal Chem Society https://www.youtube.com/watch?v=J-wao0O0_qM

Fragmentation Patterns Alcohols– Fragment Ions Dehydration (M - 18) is a common mode of fragmentation Elimination – occurs from hot surface of ionization chamber There is a mass loss for water of 18 (H2O = 1.01x 2 +16.00) = 18 (May also give 17 for [OH]+)

Abundance % m/z IDENTIFYING THE COMPOUND Other important frags: Peaks appear due to characteristic fragments: 17 for [OH]+ and H2O = 18 28 for [CO]+ , [HCO+] =29 for ketones and aldehydes 10 20 30 40 50 60 70 80 90 100 110 120 130 140 m/z Abundance % 20 40 60 80 100 28 105 106 77 57 43 51

USE IR and MS to figure this out Next page for answer

IR tells us that this is an alcohol. The MS says it weights 60 IR and MS IR tells us that this is an alcohol. The MS says it weights 60 Answer: 60 -17 (mass of [OH]+)= 43. Thus the carbons and hydrogen's need to add to 43. For 43 [C3H7]+ , looks like propanol . You could just try 3x12 = 36 and 7 more hydrogen's (there was a hydrogen in the OH), so C3H7OH..so why is the 31 base peak so large , where does it come from? DRAW IT OUT, first, it makes it much easier 31 is [CH2OH]+ Alcohols dehydrate (loss water or 18). 18 did not show up on the MS as it is an uncharged molecule, but there is a line at 42 and 60-18=42. [CH3CH2CH2OH]+  .CH3CH2 + CH2OH+ (29) (31)

Acetic Acid MS CH3COOH [COOH]+ [CH3COOH] + 60 15 [CH3]+ [HC=O]+ 29 The carbonyl of the Carboxylic Acid (C=O) mass 28 which is also here, with an additional hydrogen for [HCO]+ . The real point is to see the [COOH]+ at 45, nice to remember this number. Mass 60-45 = 15 Which is a methyl, so CH3 COOH

What is this? Answer: Note the strong 45 line, which is [COOH]+ , the mass total is 74. Thus 74-45 =29 29 is [C2H5]+ thus C2H5 COOH (propanoic acid)

FRAGMENTATION PATTERNS ALDEHYDES AND KETONES Cleavage of bonds next to the carbonyl group (C=O) is a characteristic fragmentation of aldehydes and ketones. A common fragment is carbon monoxide (CO)+ has a mass drop of 28

FRAGMENTATION PATTERNS Aldehydes and ketones The position of the carbonyl group influences the fragmentation pattern because the molecular ion fragments either side of the carbonyl group. O CH3 C C4H9 MOLECULAR ION has m/z = 100 • +

FRAGMENTATION PATTERNS Aldehydes and ketones The position of the carbonyl group influences the fragmentation pattern because the molecular ion fragments either side of the carbonyl group. O CH3 C C4H9 Example; MOLECULAR ION has m/z = 100 • + O C4H9 C+ O CH3 C+ m/z = 28 CH3• C4H9• m/z = 85 m/z = 43 O C4H9 C• O CH3 C• CH3+ C4H9+ m/z = 15 m/z = 57 A further peak occurs at m/z = 72 (100-28) due to loss of CO+

Who might this be?

Hmmmm? The IR shows a C=O at 1700, thus it is possibly an Aldehyde of Ketone? It has a mass of 58 and you may think a 43 is [C3H7]+, but wait that is too large it leaves 58-43=15 And the main part of Ketones (aldehydes) is [CO]+ is 28….. DRAW it out. There are 15 fragments, or CH3 and a small [CO]+ fragment. 15+15 for 2 methyl's AND 28 for [C=O]+ (carbonyl)=58 If this is propanone, then why the 43 peak ? [CH3CO]+ = 43 again , draw it out

COMMON FRAGMENTS Mass of [ion]+ Formula of [ion]+ Name of ion Source 15 [CH3]+ methyl Cleavage Alkane 17 [OH]+ hydroxyl Cleavage Alcohol 18 [H2O]+  water 28 [CO]+  carbonyl Cleavage Ketone/Aldehyde 29 [C2H5]+ and [HCO]+  ethyl Hydrogen carbonyl Cleavage Alkane, Cleavage Ketone or Aldehyde or Carboxylic Acid 43 [C3H7]+  propyl 45 [COOH]+ carboxylic acid Cleavage Carboxylic Acid 57 [C4H9]+  butyl 71 [C5H11]+  pentyl 78 [C6H6]+ Aromatic Cleavage aromatics

The mass spectrum of chlorine For molecules with Cl . The M+2 isotope is prevalent. Two isotopes, 35Cl and 37Cl, in the approximate ratio of 75% 35Cl, 25% 37Cl.

Isotopes of Br are lovely, they are and very easy: Mass Spectrometry Isotopes of Br are lovely, they are and very easy: 79Br is 50.52% and 81Br is 49.48% 1:1 ratio If a molecule contains a single bromine atom, the molecular ion would appear: M+ M+2 The M+2 peak would be about the size of the M+ if one Br is present relative abundance m/e

FRAGMENTATION PATTERNS for HALOGENOALKANES See 2 peaks for the molecular ion of [R-Br]+, one for the molecule with isotope 79Br and the other 81Br Because the two isotopes are of similar abundance, the peaks are of similar height. Who might this molecule be? Answer: If we use 81Br and subtract M+ (110) we get 110-81 = 29 [C2H5]+ Thus C2H5Br , check by adding up masses. 29+ 81 = 110. We could have used 79Br at 108 -79 = 29, same 10 20 30 40 50 60 70 80 90 100 110 120 130 140 m/z 20 40 60 80 100 Abundance % molecular ion contains...79Br 81Br

The mass spectrum of chlorine For molecules with Cl . The M+2 isotope is prevalent. Two isotopes, 35Cl and 37Cl, in the approximate ratio of 75% 35Cl, 25% 37Cl.

What is this HALOALKANE???? Is it bromine or chlorine? Answer: Although it is hard to see, it is Cl (3:1) ratio, 35 Cl larger at 78, so let’s use this one. 78-35 = 43 as we see the base peak, most abundant at 43, which is[C3H7]+, thus we have chloropropane. You could use 81 then take away 37Cl still equals 43

Mass Spectrometry: For a straight chain (n-aliphatic) Have a Guess at this one. If you are told there are no functional groups, just a straight chain alkane. Answer: If we have 100, then we could try 8 carbons (12x8)=96, not enough room for hydrogen's. So try 7. 7X12 =84 THEN 100-84 = 16 for hydrogen's, thus heptane (C-C-C-C-C-C-C) C3H7+ 29 C2H5+ C4H9+ M+

IDENTIFY THE COMPOUNDS

Abundance % m/z IDENTIFY THE COMPOUND 43 29 122 124 79 81 122 124 29 79 81 43 20 40 60 80 100 Abundance % m/z 10 20 30 40 50 60 70 80 90 100 110 120 130 140

Abundance % m/z IDENTIFY THE COMPOUND 43 29 122 124 79 81 Answer: 122 124 29 79 81 43 20 40 60 80 100 Abundance % m/z 10 20 30 40 50 60 70 80 90 100 110 120 130 140 Answer: This is a Br as there is and M+ and M+1 peak that is 50/50. Remove the bromine from the total Mr. or 122-79 (or 124-81) = 43. This is the fragment for [C3H7]+ Thus we have C3H7Br

Abundance % m/z IDENTIFY THE COMPOUND 56 57 113 43 71 142 10 20 30 40 50 60 70 80 90 100 110 120 130 140 m/z Abundance % 20 40 60 80 100 142 113 71 56 57 43

Abundance % m/z IDENTIFY THE COMPOUND 56 57 113 43 71 142 10 20 30 40 50 60 70 80 90 100 110 120 130 140 m/z Abundance % 20 40 60 80 100 142 113 71 56 57 43 Answer: There are no F. Groups here, we have 71 [C5H11]+ . If you continue adding 14 in the homologous series 71+14 is [C6H13]+ = 85 and another 14 to 99 then 14 more to 113 we have [C8H17]+ = 113 113 is 29 from 142 or 2 more carbons for 10 carbons or 10x12 = 120. The M+ ion is 142-120 = is 22 hydrogen's C10H22