4b: Mass Spec

Question 1

How are OE and EE ions formed?

Answer 1

Odd electrons are formed by an electron knocking an electron from the vapourised molecule.

Even electrons are formed as adduct ions with either H⁺ or Na⁺.

Question 2

How does fragmentation occur and how can it be controlled?

Answer 2

The ionisation energy used to form the ions can also break weak bonds and form fragments. Softer ionisation techniques can be used to reduce fragmentation such as ESI and MALDI.

Question 3

How does chemical ionisation work?

Answer 3

A reagent gas such as methane, ammonia or isobutane is used to give a proton to the analyte to form an adduct ion. The different reagent gases have different proton affinities so the appropriate gas has to be used so the analyte will take up the proton from it.

The difference between the proton affinities makes up the energy avalible for fragmentation.

Question 4

What can you tell about an analyte from the molecular ion cluster?

Answer 4

The mass tells you if the number of nitrogen atoms in the analyte are odd or even.

The A + 1 and A + 2 peaks tell you what other heteroatoms are in the analyte.

The A + 1 peak minus the heteroatoms influence estimates the number of carbons in the structure.

The peak abundances tells you how stable the molecular ion is.

Question 5

How should you account for errors in the calculation for the carbon atoms?

Answer 5

Do the same calculation with ±10% of the original isotope intensity.

Question 6

What is the equation for the degree of unsaturation of an analyte?

Answer 6

du = x - ½y + ½z + 1

x = tetravalent atoms, C

y = monovalent atoms, H

z = trivalent atoms, N

Question 7

What are the three ways a molecular ion can fragment?

Answer 7

M^+• → A⁺ + B^• by breaking one bond and even electron ions forming. This goes via α cleavage or i cleavage.

M^+• → A^+• + B by breaking two bonds and an odd electron ion forming. There must be a rearrangement for this to occur.

Question 8

Draw a mechanism for α-cleavage, the atoms it will likely occur with and give its alternate name.

Answer 8

Homolytic cleavage

Question 9

Draw a mechanism for i-cleavage, the atoms it will likely occur with and give its alternate name.

Answer 9

Heterolytic cleavage.

Question 10

Looking at the m/z vaules of neutral losses, suggest the formula of the loss.

1, 15, 16, 17, 26, 28, 29, 31, 35, 44, 45, 46.

Answer 10

1 H, 15 CH₃, 16 NH₂, 17 OH, 26 CN, 28 CO, 29 CHO, 31 OCH₃, 35 Cl, 44 CO₂, 45 CO₂H, 46 NO₂.

Question 11

Where will the radical be formed on a molecule?

Answer 11

At the orbital with the lowest ionisation energy, Normally this preference is non bonding > π > σ. Identify this by looking at where the charge is best stabilised.

Question 12

Identify the species that has these common ion peaks:

29, 43, 57, 71, 85, 99

Answer 12

Alkane series.

Question 13

Identify the species that has these common ion peaks:

27, 41, 55, 69, 83, 97

31, 45, 59, 73, 87, 101

Answer 13

Alkenes/cycloalkanes

Aliphatic alcohols

Question 14

Identify the species that has these common ion peaks:

38, 39, 50-52, 63-65, 75-78 with clusters common.

Answer 14

Aromatic molecules.

Question 15

Retro-Diels Alder rearrangement affects structures with cyclohexane rings. Draw its fragmentation by α and i cleavage and describe their relative rates.

Answer 15

Question 16

When is α cleavage and i cleavage preffered?

Answer 16

α is preffered for low electronegativity heteroatoms N > S, O, π > X

i cleavage is preffered for highly electronegative heteroatoms X > O, S > N

For many molecules both occur in competition.

Question 17

Why would fragmentation by rearragement occur?

Answer 17

To improve stability of a molecule by transferring a hydrogen to form alkenes for example.

Question 18

How do EE ions formed by α cleavage continue to fragment?

Answer 18

By hydrogen rearrangement

Question 19

What is a common carbonyl fragmentation? How is this used in ketones?

Answer 19

Acylium ions often form. The radical is best stablised by a longer carbon chain so smaller ions from ketones are common but both will be seen.

Question 20

What is the McLafferty rearrangement? Draw the mechanism. How does this react further?

Answer 20

A carbonyl takes a γ hydrogen in a 6 membered ring intermediate, the α-β bond is then cleaved giving an alkene and a protonated carbonyl with the radical and the positive charge.

A double McLafferty can occur if both chains are long enough (not involving the carbonyl).

Question 21

What other fragmentation occurs with a ring of 5-10 atoms and competes with cleavage?

Answer 21

Hydrogen rearrangement can occur to break up a ring. A six-membered ring can break up into two 3 carbon chains, one with an alkene and one with an acylium. This occurs by an α cleavage, then a rH, then an α cleavage.

Question 22

What mass 91 ion is common? What similar mass ion occurs?

Answer 22

The tropylium ion. Benzene with an methyl chain is also common as a McLafferty rearrangement can occur.

Question 23

How do polyaromatic rings appear in a spectra?

Answer 23

They have a very stable π system and the second most common peak may be the +2 peak.

Question 24

Draw how esters and amides can fragment.

Answer 24

By 4-membered transition states.

Question 25

What complications occur for very large molecules?

Answer 25

Mass defect adds up so odd or even masses are irrespective of the nitrogen rule.

Many doubly charged ions will occur.

Question 26

How can structures be modified before MS for additional information?

Answer 26

Labels can be added to double and triple bonds to show where they are in the structure with D₂ by showing where extra mass has been lost (D is heavier than 2H).

Question 27

How can double bonds be used to supress small fragments?

Answer 27

If they are in the middle of the molecule, introducing -SCH₃ will stop small fragments as the radical is stable in that area so fragmentation will occur round there.

Question 28

How can multiple MS machines be used to find more information about an analyte?

Answer 28

A scan can be taken of an unfragmented analyte and then collision with an inert gas causes fragmentation giving a new spectra which can find new information.