TESTING OF ORGANIC COMPOUNDS

Question 1

distillation

Answer 1

• purifies and separates desired products from undesired products
• uses boiling points to separate two or more molecules
  
  • two types - simple distillation and fractional distillation

Question 2

simple distillation

Answer 2

• technique relies on a difference of at least 50 degrees celsius in boiling point between the components to obtain an effective seapration
• equipment used:
  
  • round bottom flask: contains the mixture to be separated and* boiling chips (distribute heat and avoid explosion)*
  • condenser: glassware used to cool hot vapours
    
    • vapours condense back to liquid
  • distillate: the cooled vapour which has condensed back to liquid and collected in the beaker
• liquids w higher boiling points are left in the round-bottom flask

Question 3

fractional distillation

Answer 3

• enables separation of liquids that have boiling points that are close together
• commonly used to separate volatile liquids from a reaction mixture
  
  • a volatile liquid is one that vapourises easily at low temperatures (low boiling point)
• the vapours from the distillation flask contain a higher concentration of more volatile or lower boiling point components than the liquid in the flask
• vapours rise up the fractionating column until they reach a height where the temperature is low enough for condensation to occur
• as the condensed liquid trickled back down the it is reheated by vapours rising from the distillation flask - causes some condensed liquid to evaporate
• vapour now has a higher concentration of the low boiling component
• process of evaporation and condensation repeats multiple times throughout the length of the fractionating column - increasing the concentration of the most volatile component in the vapour
• temp at the top of the fractionating column remains relatively stable
• vapour eventually reaches condenser - it is cooled and the distillated drips into the collection vessel
• the distillate consists of components that condense over a small temp range near the boiling point of the desired organic compound

Question 4

melting point determination

Answer 4

• temp at which melting occurs depends on structure of organic molecule
  
  • pure organic molecule has a sharp melting point
• melting point range: DIFFERENCE between the temp at which the sample begins to melt and the temp at which sample completely melts
• melting point range of 0.5 - 2 degrees celsius is considered small//narrow → indicating pure compound
• in pure organic compounds - all the molecules are the same
  
  • intermolecular attractions are maximised - molecules can pack together in an orderly arrangemet
  • this is why pure organic substances has a sharp melting point
• in a mixture of 2 or more organic compounds (impure) the molecules cannot pack in an orderly array
  
  • intermolecular forces are disrupted - less heat is needed to melt this mixture - melting point is lower
  • impure solid has a broader melting point range because regions of the crystal structure contain different amount of impurity
  • low melting point and a wide melting point range of more than 2 degrees celsius usually indicated an impure substance
  • mixed melting point determination - can be performed to confirm identity of unknown solid

Question 5

testing for carbon-carbon double bonds

Answer 5

• addition reaction between alkene and bromine can be used to test for the presence of a carbon carbon double bond
• red-orange colour of bromine rapidly decolourises - become colourless → colourless dibromo product forms
• bromine doesn’t react w alkanes
  
  • can be used to distinguish between alkanes and alkenes

Question 6

testing for hydroxyl functional group

Answer 6

• can be determined by esterification reaction
• mixture of organic compound and ethanoic acid is gently heated with a few drops of concentrated sulfuric acid
• if there is a hydroxyl gorup - characteristic smell of an ester will be detected
• to determine if the organic compound is primary, secondary or tertiary, acidified potassium dichromate or acidified potassium permanganate can be used
• acidified potassium dichromate:
  
  • primary or secondary - orange colour of the dichromate ion will be reduced to Cr 3+ → green in colour
  • no colour change if tertiary
• acdified potassium permanganate
  
  • primary or secondary - purple coplour of permanganate ions will be reduced to Mn2+ - colourless
  • no colour change if tertiary

Question 7

testing for carboxyl functional group

Answer 7

• sodium hydrogen carbonate can be used to test for the presence of carboxylic acid
• when acids react w metal hydrogen carbonate → carbin diocide, salt and water are formed
• effervescence (fizzing) is observed when the carboxyl group is present
• presence of CO2 gas can be confirmed using lime water test
• when calcium hydroxide reacts with CO2 - calcium carbonate precipitate forms which turns the solution milky or cloudy

Ca(OH)2 (aq) + CO2 (g) → CaCO3 (s) + H2O (l)

Question 8

iodine number

Answer 8

• used to measure degree of unsaturation of fats and oils
• iodine number - mass of iodine in grams that reacts with 100 grams of fat or oil
• the more unsaturated a fat or oil, the higher the iodine number
• oils with higher iodine numbers are grenerally more reactive, less stable, softer and more susceptible to oxidation that fats or oils with lower iodine numbers
• vegetable oils tend to have more unsaturated fats than animal fats - higher iodine values
  iodine decolourises when it reacts with unsaturated hydrocarbons

Question 9

volumetric analysis procedure

Answer 9

• preparation of standard solution
• preparation of sample aliquots
• preparation of burette
• titration (produce 3 concordant titres)
• calculation of the unknown concentration

Question 10

prep standard solution

Answer 10

• standard solution: solution of accurately known conc.
  
  • prepared by dissolving primary standards or diltuin stock solutions
  • ideal primary standard:
    
    • readily obtained in pure form so that the amount in moles can be accurately calculated from their mass
    • known chemical formula
    • easy to store w/o absorbing water vapour or reacting w gases
    • high molar mass (minimise effect of errors in weighing)
    • is inexpensive
• the standard solutions needs to be titrated against a standard solution to determine its actual concentration - becomes standardised solution
• to maximise precision and accuracy, volumetric flask should only be rinsed with water to remove any trace chemicals from the glassware

Question 11

prep aliquot

Answer 11

• known volume of sample transferred to conical flask with pipette
  
  • fixed volume known as aliquot
  • indicator may need to be added
• to maximise precision and accuracy
  
  • conical flask should only be rinsed w deionised water to remove any trace chemicals
    
    • rinsing w other solutions can cause unintended reactions and affect the result
  • pipette should be rinsed with the solution it is to be filled with
    
    • rinsing with water will dilute the aliquot (pipette is hard to dry as it is small)

Question 12

prep burette

Answer 12

• burette contains the standard solution → delivers accurate volume of solution to the aliquot
• to maximise precision and accuracy
  
  • burette should only be rinsed with the solution to be filled with (rinsing w water dilutes the solution → burette is narrow)

Question 13

titration to produce 3 concordant titres

Answer 13

• solutions are mixed until equivalence point is reached
• equivalence point: the point at which the number of moles of both reactants are in the mole ratios of the balanced chemical equation
• end point: permanent colour change of solution in conical flask
• indicator is chosen so that the colur change at the end point occurs near the equicvalence point
• amount dispensed from burette → titre
  
  • calculated by subtracting initial reading from final
  • all readings to 2 decimal places
• to minimise random errors, titration is repeated several times to obtain concordant titres
  
  • reading within 0.10mL of each other
  • average fo 3 concordant titres used in calculation

Question 14

errors in volumetric analysis

Answer 14

• accuracy depends on calibration of equipment used and uncertainties associated w analytical equipment
• parallex error - not reading from bottom of meniscus

Question 15

dilution factor

Answer 15

• dilution factor = volume of the diluted solution/volume of the undiluted solution
• It is often necessary to dilute a solution by adding water to it to reduce its concentration, before carrying out a titration. This is done to obtain concentrations that will result in titres that are within the range of the burette.

Question 16

mass spectrometry

Answer 16

• quantitative technique to detect concentrations in parts per billion to parts per trillion
• mass specturm of an unknown sample can be compared to mass spectra in database to find its identity

Question 17

principles of mass spectrometry

Answer 17

ionisation

• sample is injected to an ionisation chamber
• sample is bombarded by a stream of electrons
  
  • knocks one or more electrons from the particle - resulting in positively charged ions
  • molecule is said to have been ionised

deflection

• positively charged ions are deflected by a magnetic field
• ions are separated in a magnetic field based on there mass (m)/charge (z) ratio (m/z)
• ions with greater m/z will be deflected the least (heavier)

detection

• the ions pass through a detector based on their m/z values
• data is recorded as a mass spectrum

Question 18

fragmentation

Answer 18

• inside the ionisation chamber, high energy electrons from the electron beam ionise the sample by knocking off electrons in the molecule to produce a positively charge moelcular ion
• renmoving electrons weakens the bonds and can cause covalent bonds to break - resulting in the molecule to fragment - break apart
• Any bond in the molecular ion can break, so fragments can be single atoms, small groups of atoms, or large sections of a molecule
• However, as electrons have been lost, these fragments are always positively charged.
• write all fragments with a positive charge

Question 19

features of mass spectrum

Answer 19

• relatively intensity is determined by:
• abundance of peak/abundance of most intense peak x 100
• the base peak is the most intense peak
  
  • produced by the most stable and abundant fragment ion
  • assigned a relative intensity of 100%
• molecular ion peak - parent moelcular ion
  
  • forms when the entire molecule loses one electron and becomes positively charged
  • tells us the relatively molecular mass of the molecule
• other peaks with smaller m/z values than the molecular ion are known as fragment ions
• the relatively intensities o the peaks that form depend on
  
  • energy of the ionising electrons
  • ease with which fragments can be formned
  • stability of the fragment ions that can form

Question 20

empirical formula

Answer 20

simplest whole number ratio of the types of atoms in one molecule

Question 21

spectroscopy

Answer 21

study of interaction between matter and electromagenetic radiation
The absorbed radiation may cause:
* a change in the vibrational state of a molecule: IR
* a change in the ‘spin’ of particles in the nucleus: NMR

Question 22

infrared spectroscopy

Answer 22

• gives qualitive information about the functional groups in the compound
• it can be used to quantitatvely determine the concentration of substances

Question 23

principles of IR spectroscopy

Answer 23

• causes only molecular vibrations of the bonds within the molecules - bending and stretching
• when applied infrared frequency = natural freq of vibration → absorption of IR radiation
• causes a change in the vibrational state of the molecule
• diff functional groups absorb characteristic freq of IR to give a peak value on IR spectrum
• frequency of vibration of the bond is proportional to the energy of the vibration
• the amount of energy of infrared rafiation energy absorbed by the sample is equal to the amount of energy req to move from one vibrational level to the next vibrational energy level

Question 24

frequency of IR radiation

Answer 24

• depends on strength of bond between two identical atoms
  
  • the weaker the bond, the lower the frequency of absorbed radiation
  • C-C is weaker than C=C and absorbs IR radiation of lower frequency
• mass of atoms attached to a bond
  
  • atoms with higher mass absorb lower frequency radiation

Question 25

description of absorption bands

Answer 25

• narrow absorption bands correspond to one specific type of molecular vibration.
• Broader bands may result from a number of related vibrational changes that have similar energies.
• Strong’, ‘medium’ or ‘weak’ describe the amount of radiation absorbed.
• each type of bond absorbs IR radiation over a typical rage of wavenumber
• absorption bands can be compared to ref tables of known functional groups and their associated wavenumbers to identify the compound
• can differentiate between very similar compounds

Question 26

NMR spectroscopy

Answer 26

• uses electromagnetic radiation in the radio frequency range
• energy of radiowaves is low in energy and frequence
• it is too low to cause electronic, vibration or rotational translations
• radio waves cause nuclear spin
• only nuclei with an odd number of protons and/or neutrons such as 1H, 13C, 31P
• the odd number of nuclear particles causes these nuclei to behave like tiny bar magnets
• in the presence of an external magnetic field - magnets can either line up i the same direction as the field or in the opposite direction
  
  • nuclei with spin can also line up in the same direction as field or in opposite direction
  • when magnets or nuclei are lined up against an external magnetic field, they have a higher energy and is cnsidered an unstable arrangement
  • nuclei are normally in lower energy state when aligned w stron magnet
  • radio waves provide energy to flip the nuclei into the higher energy state - nuclei are aligned against magentic field - in an unstable state
  • nuceli will flip back to lower energy state - emitting a pulse of energy
  • this is represented on an NMR specturm

Question 27

types of NMR spectroscopy

Answer 27

• proton NMR
  
  • examines 1H nucleus (1 proton and no neutrons)
• carbon-13 NMR
  
  • examines 13C nucleus (6 protons, 7 neutrons)
• NMR spectroscopy gives info on the number and type of H and C nuclei in an organic compound

Question 28

chemical environments

Answer 28

• atoms in same chemical environment absorb and emit the same energy so they produce the same singal in the NMR specturm
• made up of the atoms and electrons that surround a particular atom
• atoms have the same chemical environment if they are attached the same way to the same atoms

Question 29

NMR spectrum

Answer 29

• difference in energy needed to change spin states in a sample is compared to energy needed to change spin states in TMS
  
  • called chemical shift in parts per million - ppm

Question 30

proton NMR

Answer 30

• no. of signals = no. of different proton environments
  
  • H atoms in the same chemical enevironmnet will have the same chemical shift
• relative peak area = proportional to the no. of hydrogen atoms in the environment it corresponds with
• splitting of signals = info about H atoms in adjacent environmnets
• chemical shifts = idenitfy the chemical environment where protons are located

Question 31

signal splitting - proton NMR

Answer 31

• number of lines = n+1 (n= no. of neighbouring protons)
  
  • protons up to 3 bonds away from a particular H atom
  • attached to neighbouring carbon atoms for eg
    
    • H atoms with no neighbours - one line signal - singlet
• signal produced by H in the OH of ALCOHOLS is not split by H’s attached to neighbouring carbons
• the hydroxyl H also doesn’t cause splitting of other signals of neighbouring H
• signal for OH is usually a singlet

Question 32

interpretation of Carbon 13 spectra

Answer 32

• peak areas are not proportional to no. of carbon atoms in each environmnet
• spectra can only tell us the no. of different C environments

Question 33

chromatography

Answer 33

• the components are swept along the stationary (non-moving) phase by the solvent
• undergo a continual process of adsorption onto the stationary phase an desorption and dissolving into the mobile (moving) phase
  
  • adsorption - molecule adhere to the surface of the phase

Question 34

Rf value

Answer 34

distance the component travels from origin/distance the solvent front travels from origin

Question 35

column chromatography

Answer 35

• solid stationary phase packed into a glass column
• sample mixture is applied to the top of the column
• mobile phase is continuously dripped slowly into the column from above
• a tap at the bottom allows solvent to leave the column
  
  • called the eluent (the solvent that leaves the column)
• components are separated by the time it takes for them to appear in the eluent

Question 36

high performance liquid chromatography

Answer 36

• also known as high pressure liquid chromatography
• same principles as column chromatography
• difference between column and HPLC
  
  • particles are often 10-20x smaller in the HPLC column
  • the higher surface area allows for more frequent adsorption and desorptio resulting in better separation of similar compounds
  • solvent pumped through under high pressure rather than gravity-fed due to high resistance to the flow of mobile phase
• HPLC is sensitive - can detect conc in ppt

Question 37

reverse phase HPLC

Answer 37

• some solids have chemical bonded to the surface to improve separation of particular classes of compounds
• most common form of HPLC is known as reverse phase HPLC
• stationary phase - non-polar
  
  • to make it non-polar - hydrocarbonds have been attahced to the surface of silica
• mobile phase → polar solvent (water or methanol)
• polar molecules move quickly through the column
  - form strong bonds with solvent molecule (dipole-dipole/hydrogen bonding) and are only weakly adsorbed to the non-polar stationary phase
• non-polar molecule move more slowly through the column
  
  • adsorbs strongly to stationary phase by dispersion forces

Question 38

detecting components in HPLC

Answer 38

• detected by passing eluent stream through UV light
• many organic compounds absorb UV light
• when organic compound is passed in front of beam of light - reduced signal is picked up by the detector
• computer produces a chromatogram trace/chart - shows absorbance over time
• time taken for a component to pass through component - retention time
• used to qualitatively identify the components associated with the peaks on a chromatogram
• relative amounts of each components in a micture may be quantitatively determined by comparing the areas under each peak with areas under peaks for standard samples

Question 39

qualitative analysis using HPLC

Answer 39

• the same compounds will give the same retention time if the conditions are kept the same - eg. temperature, mobile phase, stationary phase, flow rate, pressure. etc.
• 2 methods to identify a compound
  
  • running a sample of the known pure compound (reference sample) under exactly the same conditions as the sample and comparing
  • adding the known compound to the sample
    
    • if the peak grows, it confirms that the added compound was the same as the compound to be identified (known as spiking)

Question 40

quantitative analysis using HPLC

Answer 40

• peak area used to detemine concentration of components
• samples are compared to standard solutions of accurately known concentrations
• used to produce a calibration curve
• if a calibration curve needs extrapolation to determine a concentration - samples need to be diluted

Question 41

investigating mixtures and product purity

Answer 41

• analyse sample on MS, IR or NMR and collect spectra
• compare sample’s spectra to a reference sample
• if spectra match with no additional or missing peaks → pure
• if spectra do not match - there are aditional peaks or missing peaks → not pure

Question 42

mass spec and HPLC

Answer 42

• resolution of mass spectrometers can be increased by attaching HPLC to a mass spectrometer
• the HPLC first separates the components of a mixture and the eluent is injected into the mass spectrometer
• the mass spec fragments the molecules and produces mass spectrum that can be compared to databases for identification
• important to ensure retention times and mass spectra match
• calibration curve is prepared using a series of standards and the unknown conc is determined

Question 43

identifying unknown contaminant using IR spectra

Answer 43

• subtract the spectra of the known compounds from the spectra of the unknown mixture
• spectra of the contaminant is then compared with spectra from database

Question 44

why one molecule can produce diff peaks on mass spectrum

Answer 44

• (cations) needed to be formed in order for any peak
  to appear on a mass spectrum.
• fragmentation pattern that occurs because of the initial parent ion being unstable
• molar masses of the different isotopes present in the parent ion.

Question 45

redox titration

Answer 45

• Used to analyse for Vit C, ethanol, SO2 (in wine), iodate in iodised salt
  *

Question 46

effect of column length in HPLC

Answer 46

• Column length affects the retention time and peak separation.
• A longer column means more interaction between the components of the mixture and the stationary phase resulting in higher retention times and greater separation of peaks.
• A shorter column means less interaction between the components of the mixture and the stationary phase leading to lower retention times and smaller separation of peaks.