WM ij: Modern analytical techniques: mass spectrometry; IR spectroscopy

Question 1

Describe the premise of mass spectrometry.

Answer 1

• Molecules are cationised, forming M⁺ ions
• These are accelerated towards detector
• Some M⁺ ions fragment, losing uncharged species + resulting in charged fragments
• Produces characteristic fragmentation pattern

Question 2

On a mass spectrum, % intensity is plotted against m/z. What does m/z give values for?

Answer 2

The molecular mass of ions

M/z means mass to charge ratio, and is equal to charge, assuming that all ions have charge +1

Question 3

What is the heaviest ion on a mass spectrum known as?

Answer 3

The molecular / M⁺ / parent ion

Question 4

The most abundant ion on a mass spectrum is set to intensity 100%. What is it known as?

Answer 4

The base peak

Question 5

What is the cause of the M+1 peak, and why is it often small?

Answer 5

• Presence of ¹³C in molecules
• Due to low abundance of ¹³C

Question 6

This is the mass spectrum of 2-ethoxybutane.

• Draw 2-ethoxybutane
• Explain, using skeletal formulae, the presence of each peak

Answer 6

Question 7

How do the mass spectrums of structural isomers compare?

Answer 7

• Same M⁺ peak
• Different fragmentation patterns

Question 8

This is the mass spectrum of a molecule which is a benzene ring with an -OH group and a -COOH group.

• Draw all the possible isomers
• Use the mass spectrum to decide which isomer it is

Answer 8

Peaks at 39, 63, 64, 92, 120, 138

M_r = 138

Ignore -COOH and -OH M_rs since all isomers can fragment this way

Peak at 120 indicates loss of H₂O with Mr = 18. This forms due to hydrogen bonding between carboxyl + phenol groups. This can only occur for 2-hydroxybenzoic acid, so this is the compound.

Question 9

2-methylbutane and pentane are isomers.

Explain why the peak at m/z = 57 is much larger on the mass spectrum of 2-methylbutane.

Answer 9

• CH₃ species fragment off both molecules, resulting in fragments of M_r 57
• Pentane can fragment to release CH₃ in 2 ways, + 2-methylbutane in 3 ways
• So 2-methylbutane has higher frequency of m/z = 57 fragments

Question 10

Explain the premise of infrared spectroscopy.

Answer 10

• Molecules absorb specific frequencies of IR radiation, since energy is quantised
• Specific bonds vibrate more
• Stronger bonds absorb higher frequencies
• Characteristic absoption frequencies recorded

Question 11

Suggest why there is only one peak on the IR spectrum for HCl, whereas there are several for ethanol.

Answer 11

• HCl contains only one bond, so can only stretch, so there is one vibrational IR absorption
• Ethanol contains more bonds, so more bond deformations are possible, so there are several absorptions

Question 12

Suggest why bonds which are more polar have more intense peaks on an IR spectrum.

Answer 12

• Polar bonds undergo large changes in polarity during vibration
• This gives stronger absorptions + more intense peaks

Question 13

Explain why the peak for the O-H bond in ethanol is broad in the liquid phase and sharp in the gas phase.

Answer 13

• In liquid phase, hydrogen bonding between OH groups alters vibration of O-H bond, so greater range of absorption frequencies
• In gas phase, little to no hydrogen bonding gives smaller range

Question 14

1. Wavenumber is plotted instead of frequency on infrared spectrums. Show how wavenumber is calculated and give its unit.
2. Suggest why it is preferable.

Answer 14

1. c = λf, so f = c / λ

c is a constant, so f ∝ 1/λ (∝ bond enthalpy)

1/λ = wavenumber, in cm^-1

2. Gives cleaner numbers.

Question 15

What wavelength, in µm, corresponds to wavenumber 2360 cm^-1?

Answer 15

1/λ = 2360 cm^-1

λ = 1/2360 cm

x 10⁴ = 4.24 µm

Question 16

What wavenumber corresponds to frequency 2.5 x 10¹³ Hz?

Answer 16

c = λf, so λ = c/f

λ = (3 x 10⁸) / (2.5 x 10¹³)

= 1.2 x 10^-5 m = 1.2 x 10^-3 cm

ans^-1 = 833 cm^-1

Question 17

1. Where is the fingerprint region on an IR spectrum?
2. How is it typically used in analysis?

Answer 17

1. Below 1500 cm^-1
2. Used for comparison, rather than for identifying functional groups (since it’s messy)

Question 18

Use the absorptions below to divide the IR spectrum into 4 general regions.

Answer 18

Question 19

Work out which functional groups are present in the molecule which produced this IR spectrum.

Suggest what type of molecule this is, and give an example.

Answer 19

• ~3600, sharp → O-H in alcohol / phenol (although sharp)
• ~3100, sharp → C-H in alkene / arene. Position suggests aromatic
• Aromaticity corroborated by “several peaks in range 1450-1560” → aromatic C=C
• ~1750 → C=O in ester / acyl chloride / acid anhydride

Overall, suggests aromatic carboxylic acid, e.g. benzoic acid

Question 20

The spectra below represent compounds C, D and E. One is an ester, one a carboxylic acid and one an alcohol.

1. Identify the key peaks and corresponding bonds in each spectrum.
2. Decide which spectrum represents each type of compound.

Answer 20

C

• ~1700 → C=O
• ~3000, broad → OH in carb acid

D

• ~1650 → C=O
• ~3000, sharp → C-H
• ~3400, broad → OH in alcohol

E

• ~1250 → C-O
• ~1750 → C=O
• ~3000, sharp → C-H

C is carboxylic acid, D is alcohol, E is ester.