Module 6 (chapter 29) - chromatography and spectroscopy

Question 1

method for basic chromatography

Answer 1

• cut the chromatography to fit your beaker. it needs to sit on the edges of the beaker (1cm from the sides and at least 2cm from the bottom)
• take the chromatography paper out of the beaker and draw a pencil line on the paper. about 2cm from the bottom
• put a small spot of ink from each of your 4 pens on your pencil line. spread them out
• label the colour underneath each spot with a pencil
• put some water into the glass and Lowe the chromatograph into the beaker so that the bottom of the paper is in the water but the water level is below the spots
• leave the paper in the beaker until the water reaches near the top of the paper
• take the paper out and immediately use a pencil to mark the location of the solvent front before it evaporates and leave it to dry

Question 2

Rf Value

Answer 2

distance travelled by the substance/ distance travelled by solvent
-this tells you the retention factor

Question 3

stationary phase and mobile phase

Answer 3

does not move and is normally a liquid supported on a solid

the mobile phase does more and is normal a liquid or gas

Question 4

what is chromatography

Answer 4

a technique used to separate individual components from a mixture of substances. it can be used in drug analysis, plastic flavourings and forensic science, airport security

Question 5

thin layer chromatography

Answer 5

• quick and inexpensive technique that indicates how many components are in a mixture.
• uses a TLC plate which is usually a plastic sheet or glass coated with a thin layer of a solid absorbent substance (usually silica)
• the absorbent is the stationary phase. different components in the mixture have different affinities fir the absorbent and bind with different strengths to its surface

Question 6

adsorption

Answer 6

the process by which the solid silica holds the different substances in the mixture of its surface
-separation is achieved by the relative adsorptions of substances with a stationary phase.

Question 7

method for carrying out a TLC

Answer 7

• take a TLC plate. Using a pencil, draw a line across the plate about 1cm from the end of the plate. This is the base line
• using a capillary tube, spot a small amount of a solution of the sample not the base line
• prepare a chromatography tank for the TLC plate. (a small beaker with a watch glass placed on top) and pour some solvent into the beaker
• place the prepared TLC plate I the beaker, making sure that the solvent does not cover the spot. cover the beaker with the watch glass and leave it undisturbed allowing the solvent to she up the plate
• allow it to ties until it is about 1cm from the top. remove the plate from the beaker and immediately mark the solvent front with a pencil
• if there are any visible spots, circle them with a pencil. or do the same holding UV light over them. Sometimes the plate is sprayed with a chemical/ locating agent such as iodine to show the position of the spots invisible to the naked eye

Question 8

gas chromatography

Answer 8

• useful for separating and identifying relative organic compounds present in a mixture
• small amount injected into the apparatus called a gas chromatograph. the mobile carrier gas carries the components in the sample through the capillary column which contains the liquid stationary phase absorbed onto the solid support
• compounds slow down as they interact with the liquid stationary phase inside the column. the most soluble a component is in a liquid stationary phase, the slower it moves through the capillary column
• components separate depending on their solubility in the liquid stationary phase. the compounds reach the detector at different times depending on their interaction with the stationary phase in the column

Question 9

results of gas chromatography

Answer 9

• compounds retained in the column for the shortest times have the lowest retention time and are detected first
• retention time is time taken to travel through the column
• the temperature is controlled as by the flow rate of the gas

Question 10

interpretation of gas chromatography

Answer 10

• each component is detected as a peak
• retention times can be used to identify the components present in the sample by comparing these to retention times for unknown components
• peak integrations (areas under each peak - basexheight/2) can be used to determine the concentrations of components in the sample
• can calculate concentrations of components by comparing peak integration with values obtained from standard solutions of the components

Question 11

how do you work out concentrations by comparing peak integrations?

Answer 11

• prepare standard solutions of known concentrations of the compound being investigated
• obtain gas chromatograms for each standard solution
• plot a calibration curve of peak integration against concentration. this is called external calibration and offers a method for converting a peak area into a concentration
• obtain a gas chromatogram of the compound being investigated under the same conditions
• use the calibration curve to measure the concentration of the compound

Question 12

what can retention times be used for organically?

Answer 12

• to identify the functional groups present in organic compounds
• add 2,4-dinitrophenylhydrazine to a carbonyl and an orange precipitate will form
• add Tollens’ reagent and warm an aldehyde to from a silver mirror

Question 13

what is NMR spectroscopy?

Answer 13

• nuclear magnetic resonance (invented 70 years ago) is used to analyse organic compounds
• uses a combination of a very strong magnetic field and radio frequency radiation
• with the right combination, the nuclei of some atoms absorb this radiation and the energy for the absorption can be recorded as an NMR spectrum

Question 14

what is nuclear spin

Answer 14

• nuclei can spin and this is significant if there is an odd number of of nucleons
• almost all organic molecules contain carbon and hydrogen (mostly H-1 and C-12, with a small proportion 1.1% of the C-13 isotope)
• this means the H-1 and C-13 are the isotopes with odd numbers of nucleons
• NMR spectroscopy can be used to detect isotopes of other elements with odd numbers of nucleons

Question 15

resonance

Answer 15

• an electron has two different spin states
• the nucleus also has two different spin states and there have different energies
• with the right combination of strong magnetic field and radio frequency radiation, the nucleus can absorb energy and rapidly flips between the two spin states. this is called resonance
• as the strength of the magnetic field increases, the larger the energy gap between different spin states
• the nucleus will flip from the less stable to the more stable form
• energy required to do this depends on the strength of the external magnetic field used (frequencies 25-100MHz)

Question 16

the NMR spectrometer

Answer 16

• the frequency required for resonance is proportional to the magnetic field strength and it is only in strong and uniform magnetic fields that this small quantity of energy can be detected
• typically a very strong support conducting electromagnetic is used, cooled to 4K by liquid helium
• most routine spectrometers operate at radio frequencies of up to 400MHz but they aren’t viable for schools
• MRI body scanners in hospitals use similar techniques

Question 17

chemical shifts

Answer 17

• all atoms have electrons surrounding the nucleus, which shifts the energy and radio frequency needed for nuclear magnetic resonance to take place. it is measured on a scale called chemical shift in units of parts per million (ppm)
• electrons modify the magnetic field experienced by the nucleus
• the external magnetic field needed to bring the carbon into resonance will be smaller if attached to a more electronegative element because the C-13 nucleus feels more of the field

Question 18

TMS

Answer 18

• tetramethylsilane
• used as a standard reference chemical against which all chemical shifts are measure
• given a chemical shift value of 0ppm
• it is used as it has 12 protons all in the same chemical environment giving single sharp peak that is easy to identify
• it is chemically unreactive

Question 19

how is the spectrum run?

Answer 19

• chemicals need to be in solution with a small amount of TMS otherwise they are harder to absorb and flip
• tube is placed inside the NMR where it is spun to even out say imperfections in the magnetic field
• it is zeroed against TMS and a single pulse of radiation is given containing a range of radio frequencies and whilst maintaining a constant magnetic field
• any absorptions are detected and displayed
• after analysis the sample can be recovered by evaporation from the solvent.

Question 20

why are deuterated solvents used un NMR?

Answer 20

• if the solution contains hydrogen or cabin then a signal will be produced
• a deuterated solvent is used meaning that the H-1 atoms are replaced by the H-2 atoms and this produces no signal in the frequencies used in H-1 and C-13 spectroscopy
• it produces no signal as there is an even number of protons and neutrons
• deuterated trichloromethane CDCl3 is commonly used as a solvent but this will still produce a peak in a carbon-13 NMR spectrum (usually filtered out before displayed)

Question 21

what does carbon-NMR spectroscopy tell us?

Answer 21

• the number of different carbon environments from the number of peaks
• the types of carbon environments present from the chemical shifts

Question 22

what determines chemical shifts

Answer 22

• determined by the type of environment.
• the environment of a carbon atom is determined by the position of the atom within the molecule
• 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 within a molecule then they will absorb radiation at the same chemical shift and contribute to the same peak
• they may be outside of the typical ranges depending on solvent, concentration and substituents

Question 23

paper chromatography summary

Answer 23

• mobile phase is the solvent
• stationary phase is the solvent attached to the paper (trapped by cellulose fibres)
• partition

Question 24

thin layer chromatography summary

Answer 24

• mobile phase is the liquid solvent that moves over thin layer
• stationary phase is the thin later if either SiO2 or Al2O3 coated onto the plastic plate
• adsorption

Question 25

gas liquid chromatography summary

Answer 25

• mobile phase is the inert carrier gas (e.g. helium)
• stationary phase is the high boiling point liquid adsorbed onto an inert solid support. a non reactive solvent such as octadecane coated onto the surface of SiO2 packing in the column
• partition

Question 26

partition

Answer 26

separation of components between two liquid phases using a column

• distribution of solid parties between two solvents
• an equilibrium is always set up (concentration of solvent in mobile phase/ concentration of solvent in stationary phase = constant)

Question 27

proton-NMR

Answer 27

• more commonly used than C-13 NMR as99.9% of all hydrogen gas one proton so it will have magnetic spin
• proton NMR are interpreted similarly to C-13 but the peaks can tell us the properties of the protons in each environment
• spin spin coupling tells us about adjacent protons

Question 28

what can proton-NMR tell us?

Answer 28

• the number of differed proton environments (number of peaks)
• the types of proton environments (from the chemical shift)
• the relative numbers of each type of proton (integration traces)
• the number of non-equivalent protons adjacent to a a given proton from the spin-spin splitting pattern

Question 29

solvents for proton-NMR

Answer 29

• samples need to be in solution, no hydrogen or carbon present as they will produce a signal
• therefore, use deuterated solvent (H2) as it has an even number of nucleons hence no signal is produced
• CDCl3 is commonly used as no hydrogen is present and carbon can be removed from the spectrum

Question 30

proton-NMR signal ranges

Answer 30

• much narrower than C-13
• can lead to overlapping of signals and so care needs to be taken as actual chemical shifts will differ depending on the environment of the protons
• if two or more protons are equivalent, they will absorb at the same chemical shift, increasing the size of the peak

Question 31

proton-nor integration traces

Answer 31

• tell us the relative number of H in an environment
• the area under the peak is in proportion to the number of protons that caused that absorption
• the area under the peak can be measured and info displayed as the integration trace
• this shows some steps increasing in height as they pass each peak, the ratio of step heights is equal to the ratio of each proton

Question 32

high resolution NMR

Answer 32

• gives more complex signals (doublets, triplets, quartets, multiplets)
• these are caused by the proton’s spin interacting with the spin states of nearby protons that are in different environments
• the signal produced indicates the number of protons on adjacent carbon atoms
• the splitting pattern depends on the number if hydrogen atoms on adjacent atoms
• for a proton with n protons attached to an adjacent carbon atom, the number of sub-peaks in a splitting pattern = n+1
• multiplets are shown

Question 33

spin-spin coupling in pairs

Answer 33

• if you see one splitting pattern there must be another as it occurs in pairs. each proton splits the signal of the other
• this makes it easy to spot structural features

Question 34

steps for proton-NMR

Answer 34

• get the formula of the compound
• draw out the structure
• go to each atom and as the census questions (peak, chemical shift, type of H, relative no of H on the environment, splitting, splitting pattern, adjacent protons)
• work out what the spectrum would look like . signals due to H’s near electronegative elements are shifted downfield.

Question 35

explaining OH peaks

Answer 35

• OH peaks can have a range of shift positions due to the complex interaction of the OH groups
• concentrations of the sample, the solvent used and any water present will affect the position of the single peak

Question 36

heavy water

Answer 36

• H2 (neutrons are heavier)
• not magnetic as not magnetic nuclei
• add a few drops to your sample, shake it and let the sample settle.
• re-run the NMR and the OH peak disappears. This is because the positive and negative ions formed got lost in the normal equilibrium of water

Question 37

what order would combined techniques take place in?

Answer 37

• elemental analysis (e.g. empirical formula)
• mass spectra (molecular formula)
• infrared spectra (use of absorption peaks to identify bonds present and functional groups)
• NMR spectra (determine the number of types of carbon and hydrogen atoms and the order of atoms using splitting patterns)