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Chemistry - Module 6 > 29 - chromatography and spectroscopy > Flashcards

29 - chromatography and spectroscopy Flashcards

1
Q

overall function of chromatography

A

Used to separate individual components from a mixture of substances

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2
Q

practical uses of chromatography

A

analysis of drugs, plastics, flavourings, air samples and forensic science

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3
Q

stationary phase in TLC

A

solid silica (SiO2) or alumina (Al2O3) held on an inert support material (glass/plastic)

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4
Q

mobile phase of TLC

A

Liquid solvent

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5
Q

advantages of TLC

A

Quick

inexpensive

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6
Q

how does separation occur in TLC

A

via relative adsorption

Components in the mixture adsorb differently to the surface of the stationary phase which result in different rf values

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7
Q

outline the method for TLC

A

Draw a base line approx. 1 cm from the bottom in pencil

Using a capillary tube, spot small sample of solution to be tested onto the baseline

Pour some solvent into the chromatography tank.

Ensure the depth is lower than the pencil line.

Place the TLC plate into the solvent, ensuring the solvent does not cover the spot, and that the edges of the TLC plate do not touch the sides of the tank.

Place a lid on the solvent tank.

Allow the solvent to rise until it is about 1 cm below the top of the plate

Remove the plate, marking the solvent level immediately with pencil

Allow the plate to dry

Circle any visible spots.

may need to hold a UV lamp over the TLC plate or spray the plate with a locating agent to show the position of the spots

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8
Q

compare TLC and paper chromatography

A
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9
Q

rf formula

A

distance moved by component / distance moved by solvent front

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10
Q

limitations of TLC (4)

A

Difficult to measure the exact centre of the component ‘spot’

Similar compounds have similar Rf values – difficult to tell them apart

Difficult to find a solvent which dissolves all components in a sample

may not be a reference chromatogram that exists in the database

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11
Q

overall function of gas chromatography

A

Separate volatile components in a mixture

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12
Q

what types of chemicals can gas chromatography analyse

A

components with low boiling points

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13
Q

stationary phase of gas chromatography

A

high boiling liquid adsorbed onto an inert solid support

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14
Q

mobile phase of gas chromatography

A

inert carrier gas

e.g. helium / neon

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15
Q

how does separation occur in gas chromatography

A

Separation via relative solubility

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16
Q

what is retention time

A

time taken for each component to travel through the column

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17
Q

what are the conditions for adsorption and solubility

A

Adsorption = solid stationary phase = TLC

Solubility = liquid stationary phase = gas chromatography

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18
Q

explain separation by adsorption

A

component molecules bind to surface of stationary phase

stronger the adsorption, more the component molecules are slowed down

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19
Q

explain separation by relative solubility

A

components dissolve in liquid stationary phase

the greater the solubility in the stationary phase, the more the component molecules are slowed down

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20
Q

outline the method for gas chromatography

A

A small amount of volatile mixture is injected into the gas chromatograph

inert carrier gas carries the components in the sample through the capillary column -contains the liquid stationary phase adsorbed onto the solid support.

components slow down as they interact with the liquid stationary phase

4 reach the detector at different times depending on their interactions with the stationary phase in the column

Component that is in the column for the shortest time has the lowest retention time + detected first

21
Q

what information does a gas chromatogram give us

A

each component = detected as a peak

Retention time – used to identify the components present in the sample by comparing these to known components.

Peak integration – can be used to determine the concentrations of components in the sample

(area under the peak)

22
Q

formula for percentage of component in a mixture

A
23
Q

how to plot a calibration curve for concentration of substance against peak area

A

Prepare standard solutions with known concentrations of the compound being investigated.

Run gas chromatograms for each standard solution.

Plot a calibration curve of peak area against concentration

Run a gas chromatogram of sample.

Use the calibration curve to measure the concentration of the compound.

24
Q

limitations of gas chromatography

A

Compounds may have the same retention times

Small amount of one component may hide behind a component with a higher concentration

Unknown components will not have any retention time data to compare it to

25
Q

test + positive observation for alkenes

A

add bromine water drop-wise

should decolourise from orange to yellow

26
Q

test + positive observation for haloalkane

A

add silver nitrate + ethanol and warm to 50°C in water bath

chloroalkane - white precipitate

bromoalkane - cream precipitate

iodoalkane - yellow precipitate

27
Q

test + positive observation for carbonyls

A

add 2,4 DNP

orange precipitate

28
Q

test + positive observation for aldehyde

A

add tollens reagent + warm

silver mirror

29
Q

test + positive observation for primary / secondary alcohol and aldehyde

A

add acidified potassium dichromate + warm in a water bath

colour change from orange to green

30
Q

test + positive observation for carboxylic acids

A

add aqueous sodium carbonate

effervescence

31
Q

property of a nucleus

A

has nuclear spin which is ONLY significant if there is odd number of nucleons
(particles in the nucleus)

E.g. H1 and C13

32
Q

how does NMR work

A

Uses strong magnetic field strength + radio frequency radiation

Causes nuclei of some atoms to absorb radiation + flip between spin states = resonance

Energy absorbed measured + recorded

33
Q

describe the spectrometer used

A

strong superconducting electromagnet

Cooled by liquid helium

34
Q

what is TMS used for

A

standard reference chemical which all chemical shifts are measured against

35
Q

what ppm does TMS have

A

0

36
Q

why do we use TMS

A

Chemically unreactive

1 carbon environment – one peak

Nuclei heavily shielded – rarely peaks below it

Volatile + easily removed from sample after

37
Q

preparation of sample for NMR

A

dissolve in a solvent

38
Q

what is the issue with solvents for NMR + how do we overcome this

A

Most solvents have H1 and C13

would produce a peak

Must use deuterated solvent – all 1H replaced with 2H

Produces no NMR signal in frequency range

39
Q

what is the solvent used for NMR

A

Deuterated solvents

CDCl3

D is an isotope of H with 2 neutrons

40
Q

describe the method of NMR

A

Sample dissolved in deuterated solvents + put into NMR sample tube

Tube inside spectrometer + spun to even out imperfections in magnetic field

Spectrometer zeroed against TMS standard

Given pulse of radiation whilst maintaining constant magnetic field

Absorptions of energy from resonance are detected

41
Q

what information does carbon NMR provide

A

Number of different carbon environments = number of peaks

Types of carbon environment present = chemical shift

42
Q

how to identify carbon environments

A

If two atoms symmetrical – equivalent + have same chemical environment

Carbon environment with higher chemical shift takes priority

43
Q

what information does proton NMR decide

A

Number of different proton environments

Types of proton environments

Relative number of each type of proton

Number of non-equivalent protons adjacent to a given proton

44
Q

what happens if 2+ protons are equivalent

A

absorb at same chemical shift + increase size of peak

45
Q

what are splitting patterns caused by

A

Caused by protons spin interacting with the spin of nearby protons that are in a different environment

46
Q

what is the rule of splitting patterns

A

Number of peaks is 1 more than number of adjacent protons causing the splitting

e.g. 4 peaks (quartet) means 3 protons in adjacent environment

47
Q

how to identify O-H and N-H specifically

A

proton exchange

48
Q

describe the process of proton exchange and the end result

A

Add small volume of deuterium oxide - D2O

Shake mixture

Run NMR as normal

Replaces the OH and NH protons with deuterium atoms

The peak disappears

Chemistry - Module 6 (5 decks)

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