Chromatography and spectroscopy

Question 1

What is chromatography?

Answer 1

It is used to separate individual components from a mixture of substances.
The stationary phase does not move and is normally a solid or liquid supported on a solid.
The mobile phase does move, and is usually a gas or liquid.
Used for analysis of drugs, plastics, flavourings, air samples and forensic science.

Question 2

What is Thin Layer Chromatography?

Answer 2

TLC is a quick and inexpensive analytical technique that indicates how many components are in a mixture.
It uses a TLC plate which is usually plastic or glass sheet, coated with a thin layer of a solid adsorbent substance, usually silica.

Question 3

What is the adsorbent in TLC?

Answer 3

It is the stationary phase.
The different componenets in the mixture have different affinities for the absorbent and bind with differing strengths to its surface.
Adsorption is the process by which the solid silica holds the different substances in the mixture to its surface.

Question 4

What is Rf?

Answer 4

TLC are analysed by calculating the value for the Retention Factor for each component.
Each component can be identified by comparing its Rf value with known values.
Rf = distance moved by the component / distance moved by the solvent front.

Question 5

How else can TLC be interpreted?

Answer 5

Run the sample alongside pure samples of compounds that may be present.
It can then be identified visually by seeing which dots appear in the impure sample.

Question 6

What is gas chromatography?

Answer 6

Used to separate and identify volatile organic compounds present in a mixture.
The stationary phase is a high boiling liquid adsored onto an inert solid support.
The mobile phase is an inert carrier gas such as helium or neon.

Question 7

How does gas chromatography work?

Answer 7

The volatile mixture is injected into the gas chromatograph.
The mobile carrier gas carries the components in the sample through the capillary column which contains the liquid stationary phase adsorbed onto the solid support.
The components slow down as they interact with the liquid in the column.
The more soluble the component in the liquid, the more it moves through the capillary column.

Question 8

What is the retention time?

Answer 8

The retention time is the time taken for each component to travel through the column.
The components of the mixture are separated depending on their solubility in the liquid stationary phase.
The compounds in the mixture reach the detector at different times depending on their interactions withe the stationary phase.
The compound retained in the column for the shortest time has the lowest retention time and is detected first.

Question 9

How is a gas chromatogram interpreted?

Answer 9

Retention times identify the components present by comparing to retention times of known components.
Peak integrations determine the concentrations of components in the sample.

Question 10

What is the process of obtaining the concentration of components?

Answer 10

Prepare standard solutions of known concentrations of the compounds.
Obtain gas chromatograms for each standard solution.
Plot a calibration curve of peak area against concentration.
Obtain a gas chromatogram of the compound under the same conditions.
Use the calibration curve to measure the concentration of the compound.

Question 11

The student runs a second chromatogram on the sample using a more polar solvent.
Predict the effect, if any, on the Rf values of the amino acids.

Answer 11

Rf values would be larger.
Amino acids are more soluble in more polar solvent so would travel further up the plate.

Question 12

Explain how the chromatogram can be used to identify amino acids.
The student thinks that there should be three spots on the chromatogram.
Suggest why there are only two spots

Answer 12

Measure distance moved by spot/distance moved by solvent.
Compare Rf value with known values.
Two amino acids have the same Rf value or similar adsorption.

Question 13

What is NMR spectroscopy?

Answer 13

The sample is dissolved in a solvent and placed in the NMR tube, with some TMS.
The sample is spun and produces a chemical shift, dependent on the chemical environment.
After analysis, the sample can be recovered by evaporation of the solvent.

Question 14

What is TMS?

Answer 14

Tetramethylsilane, (CH3)4Si is used as the standard reference chemical against which all the chemical shifts are measured.
TMS has a chemical shift of 0 ppm (parts per million).

Question 15

What are deuterated solvents?

Answer 15

A solvent where the 1H atoms have been replaced by 2H atoms (D).
Deuterium produces no NMR signal in the frequency ranges used in 1H and 13C spectroscopy.
Deuterated trichloromethane CDCl3 is a common solvent in NMR spectroscopy.

Question 16

What is carbon-13 NMR spectroscopy?

Answer 16

It determines: the number of different carbon environments from the number of peaks.
The types of carbon environments, from the chemical shift.

Question 17

How is the chemical environment of a carbon atom determined?

Answer 17

Carbon atoms that are bonded to different atoms or groups of atoms have different environments and will absorb at different chemical shifts.
If two carbon atoms are positioned symmetrically, they have the same chemical shift and contribute to the same peak.

Question 18

How do electron withdrawing groups affect chemical shift?

Answer 18

Carbon atoms nearer to an oxygen atom will be shifted more to the right than other carbon atoms.
E.g. in propanal, the CH2 group is also bonded to a C, like the CH3 group is, but is closer to the O so has a higher ppm.

Question 19

What is proton NMR spectroscopy?

Answer 19

Determines, the number of proton environments and the type of proton environments.
Also the relative numbers of each type of proton - from intergration traces or ratio numbers of relative peak areas.
The number of non-equivalent protons adjacent to a proton, from splitting pattern.

Question 20

What is the intergration trace?

Answer 20

The area under the peak, either an extra line or a number of relative peak areas.
The ratio shows how many hydrogens there are for that peak.

Question 21

What is the n+1 rule?

Answer 21

The number of sub-peaks is one greater than the number of adjacent protons causing the splitting.
n+1, where n is the number of adjacent hydrogens.

Question 22

What are the spin-spin splitting patterns?

Answer 22

n
0 singlet no adjacent H
1 doublet adjacent CH
2 triplet adjacent CH2
3 quartet adjacent CH3
4+ multiplet likely (CH3)2CH

Question 23

What are common spin-spin splitting pairs?

Answer 23

CH2 - CH3 quartet - triplet
CH - CH2 triplet - doublet
CH2 - CH2 triplet - triplet
CH - CH doublet - doublet
CH - CH3 quartet - doublet
CH - (CH3)2 septet - doublet

Question 24

How are hydroxyl and amine groups shown?

Answer 24

OH and NH2 peaks can occur at any chemical shift, and are not involved in spin-spin coupling.

Question 25

What is proton exchange?

Answer 25

The proton NMR is run as normal, then a small volume of D2O (deuterium oxide) is added, and another spectrum run.
Deuterium exchanges the OH and NH2 protons with D atoms.
D doesn’t absorb chemical shift, so the OH or NH2 peak disappears.

Question 26

How do you answer a NMR question?

Answer 26

Draw a table:
Chemical shift | Splitting pattern | intergration | environment
Underline the hydrogens to show which H causes the peak.

Question 27

[1]
15(a). i. State the region of the electromagnetic spectrum used in 1 H NMR spectroscopy.
[1]
ii. Explain why CDCl3 is used as a solvent in 1 H NMR spectroscopy.