6.3 Organic chem and analysis 29.1-29.6

Question 1

chromatography

Answer 1

used to separate individual components from a mixture of substances
all forms of chromatography have a stationary phase and a mobile phase

Question 2

stationary phase

Answer 2

doesn’t move and is normally a solid or a liquid supported on a solid

Question 3

mobile phase

Answer 3

does move and is normally a liquid or a gas

Question 4

thin layer chromatography

Answer 4

indicates how many components are in a mixture. the technique uses a TLC plate which is usually a plastic sheet or glass, coated with a thin layer of a solid adsorbent surface- usually silica
the adsorbent is in the stationary phase
the different components 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. separation is achieved by the relative adsorptions of substances with the stationary phase

Question 5

carrying out TLC

Answer 5

1. draw a line in pencil across the TLC plate about 1cm from one end of the plate. this is the base line.
2. using capillary tube, spot a small amount of a solution of the sample onto the base line of the plate
3. pour some solvent into a beaker to a depth of 0.5cm
4. place the prepared TLC plate in the beaker, making sure that the solvent doesn’t cover the spot
5. cover the beaker with clingfilm and leave it undisturbed. solvent will rise up plate
6. when solvent is 1cm below plate top, remove it from the beaker and immediately mark the solvent front with a pencil. allow plate to dry

Question 6

interpretation of TLC plate

Answer 6

calculate the value for for the retention factor Rf for each component
each component can be identified by comparing its Rf value with known values recorded using the same solvent system and absorbent

Question 7

calculate Rf

Answer 7

Rf= distance moved by component/ distance moved by solvent front

Question 8

gas chromatography

Answer 8

useful for separating and identifying volatile organic compounds present in a mixture
the stationary phase is a high boiling liquid adsorbed onto an inert solid support.
the mobile phase is an inert carrier gas such as helium or neon

Question 9

gas chromatography steps

Answer 9

a small amount of the volatile mixture is 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
the components slow down as they interact with the liquid stationary phase inside the column. the more soluble the component is in the liquid stationary phase, the slower it moves through the capillary 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 with the stationary phase in the column. the compound retained in the column for the shortest time has the lowest retention time and is detected first

Question 10

retention time

Answer 10

time taken for each component to travel through the column

Question 11

interpretation of a gas chromatogram

Answer 11

each component is detected as a peak on the gas chromatogram. two pieces of info can be obtained:

• retention times can be used to identify the components present in the sample by comparing these to retention times for known components
• peak integrations (areas under each peak) can be used to determine the concs of components in the sample

Question 12

retention times interpretation from a gas chromatogram

Answer 12

peaks at different retention times show how many components are present
you can see from the relative sizes of the peaks that there are different proportions of each component present

Question 13

conc of components interpretation from a gas chromatogram

Answer 13

the conc of a component in a sample is determined by comparing its peak integration (peak area) with values obtained from standard solutions of the component

Question 14

test for alkenes

Answer 14

add bromine water drop-wise

bromine water decolourised from orange to colourless

Question 15

test for haloalkanes

Answer 15

add silver nitrate and ethanol and warm to 50 decrees C in a water bath
chloroalkane- whit precipitate
bromoalkane- cream ppt
iodoalkane- yellow ppt

Question 16

test for carbonyls

Answer 16

add 2,4-dinitrophenylhydrazine

orange precipitate

Question 17

test for aldehydes

Answer 17

add tollens reagent and warm

silver mirror

Question 18

test for primary and secondary alcohols, and aldehydes

Answer 18

add acidified potassium dichromate (VI) and warm in a water bath
colour change from orange to green

Question 19

test for carboxylic acids

Answer 19

add aqueous sodium carbonate

effervescence

Question 20

test for phenols

Answer 20

phenols are acidic compounds and can be tested using pH indicator paper
not as acidic as carboxylic acids and do not react with sodium carbonate

Question 21

nmr spectroscopy

Answer 21

nuclear magnetic resonance spectroscopy
uses a combination of a very strong magnetic field and radio frequency radiation
with he right combination of these, the nuclei of some atoms absorb this radiation
the energy for the absorption can be measured and recorded as an nmr spectrum

Question 22

nuclear spin

Answer 22

the nucleus has nuclear spin, this is significant if there is an odd number of nucleons
almost all organic molecules contain carbon and hydrogen
nmr is relevant for 1H and 13C, the isotopes with an odd number of nucleons
nmr spectroscopy can be used to detect isotopes of other elements with odd numbers of nucleons

Question 23

nucleons

Answer 23

protons and neutrons

Question 24

resonance

Answer 24

the nucleus has two different spin states and these have different energies
with the right combination of a strong magnetic field and radio frequency radiation, the nucleus can absorb energy and rapidly flips between the two spin states
this is called resonance and the whole process gives the name nucelar magnetic resonance

Question 25

the nmr spectrometer

Answer 25

radio freq radiation has much less energy than the infrared radiation used in IR spectroscopy. the freq required for resonance is prop to the magnetic field strength and it is only in strong and uniform magnetic fields that this small quantity of energy can be detected.

Question 26

chemical shift

Answer 26

in an organic molecule, every carbon and hydrogen atom is bonded to other atoms. all atoms have electrons surrounding the nucleus, which shifts the energy and radio freq needed for nmr to take place.
the freq shift is measured on a scale called chemical shift, in units of parts per million

Question 27

TMS

Answer 27

Tetramethylsilane
used as the standard reference chemical against which all chemical shifts are measured
TMS is given a chemical shift value of 0ppm

Question 28

standard reference chemical for chemical shifts

Answer 28

TMS tetramethylsilane

Question 29

running the spectrum- nmr

Answer 29

in a nmr specrometer, the sample is dissolved in a solvent and placed in a narrow nmr sample tube, together with a small amount of TMS
the tube is placed inside the nmr spectrometer, where it is spun to even out an imperfections in the magnetic field within the sample
the spectrometer is zeroed against the TMS standard and the sample is given a pulse of radiation containing a range of radio frequencies, whilst maintaining a constant magnetic field. any absorptions of energy resulting from resonance are detected and displayed on a computer screen. after analysis, the sample can be recovered by evaporation of the solvent

Question 30

deuterated solvents- nmr

Answer 30

molecules of most common solvents contain carbon and hydrogen atoms, which will produce a signal in both 13C and 1H nmr spectra.
a deuterated solvent is usually used in which the 1H atoms have been replaced by 2H atoms (deuterium, D). D produces no nmr signal in the freq ranges used in 1H and 13C nmr spectroscopy.
deuterated trichloromethane, CDCl3 is commonly used as a solvent in nmr spectroscopy, but this will still produce a peak in a carbon-13 nmr spectrum.
the computer usually filters out this peak before displaying the spectrum

Question 31

CDCl3

Answer 31

deuterated solvent

Question 32

what two pieces of info does a carbon-13 nmr spectrum provide about a molecule?

Answer 32

• the number of diff carbon environments- from number of peaks
• the types of carbon environment present- from the chemical shift

Question 33

environemtns

Answer 33

the chemical environment of a C atom is determined by the position of the atom within the molecule

• carbon atoms that are bonded to diff atoms or groups of atoms have diff environments and will absorb at diff chemical shifts
• if two carbon atoms are positioned symmetrically within a molecule, then they are equivalent and have the same chemical environment. they will then absorb radiation at the same chemical shift and contribute to the same peak

Question 34

what does the number of peaks on a carbon-13 nmr spectrum indicate?

Answer 34

number of environments

Question 35

what four pieces of info does a proton nmr spectrum provide about a molecule?

Answer 35

• the number of different proton environments- from the number of peaks
• the types of proton environments present-from the chemical shift
• the relative numbers of each type of proton- from integration traces or ratio numbers of the relative peak areas
• the number of non equivalent protons adjacent to a given proton- from the spin-spin splitting pattern

Question 36

equivalent and non equivalent protons- proton nmr

Answer 36

• if two or more protons are equivalent they will absorb at the same chemical shift, increasing the size of the peak
• protons of different types have different chemical environments and are non-equivalent- they absorb at different chemical shifts

Question 37

relative numbers of each type of proton- proton nmr

Answer 37

in proton nmr the ratio of the relative areas under each peak gives the ratio of the number of protons responsible for each peak
the nmr spectrometer measures the area under each peak as an integration trace
the integration trace is shown either as an extra line on the spectrum or as a printed number of the relative peak areas

Question 38

spin-spin coupling- proton nmr

Answer 38

a proton nmr peak can be split into sub-peaks or splitting patterns
these are caused by the proton’s spin interacting with the spin states of nearby protons that are in different environments
this provides info about the number of protons bonded to adjacent carbon atoms
n+1 rule

Question 39

n+1 rule- proton nmr

Answer 39

the splitting of a main peak into sub-peaks is called spin-spin coupling or spin-spin splitting, and the number of sub-peaks is one greater than the number of adjacent protons causing the splitting

Question 40

splitting pattern names

Answer 40

singlet 
doublet
triplet
quartet 
multiplet

Question 41

calibration curve for conc of component in gas chromatography

Answer 41

1. prep standard solutions of known concs of the compound being investigated
2. obtain gas chromatograms for each standard solution
3. plot a calibration curve of the peak area against conc. this is called external calibration and offers a method for converting a peak area into a conc.
4. obtain a gas chromatogram of the compound being investigated under the same conditions.
5. use the calibration curve to measure the conc of the compound