VCE Chemistry UNIT 4 NOTES PDF

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VCE Chemistry UNIT 4 NOTES...

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Reaction Pathways KDP - the pathways used to synthesise primary haloalkanes, primary alcohols, primary amines, carboxylic acids and esters, including calculations of atom economy and percentage yield of single-step or overall pathway reactions.

Reaction pathways Making haloalkanes, alcohol and carboxylic acids

Making primary alcohols

Making primary amines Making carboxylic acids

Making esters Need to first obtain alcohol and carboxylic acid and then react the two through condensation to form an ester

Theoretical vs actual yields Theoretical yield  amount of yield (in mol) that should be obtained given that all the reactants have turned into products (according to the stoich ratios) O Factors affecting theoretical yield: o Reaction reaches equilibrium and does not continue to proceed forward to completion o Reaction rate – not enough time has elapsed for reaction to have reached completion o Loss of reactant/product – whilst transferring between reaction vessels, decomposition or loss through filtration/evaporation

Percentage yield Compares actual yield to theoretical yield

**ALWAYS calculated using mol, NOT mass

Percentage yields in multistep syntheses O O O

When a reaction proceeds by a number of steps, the overall percentage yield is reduced at each step. The yield for each step has an effect on the overall yield Comparison of the overall percentage yields for different pathways to the same product can be used to determine whether a particular synthetic pathway is the best and most economical way to produce an organic compound.

Atom economy A measure of how many of the atoms in the reactants end up in the desired product O Considers waste products – if the atom economy of a reaction is high, then there are few, if any waste products calculated through:

Analytical Techniques KDP - the principles and applications of mass spectroscopy (excluding features of instrumentation and operation) and interpretation of qualitative and quantitative data, including identification of molecular ion peak, determination of molecular mass and identification of simple fragments KDP - the principles and applications of infrared spectroscopy (IR) (excluding features of instrumentation and operation) and interpretation of qualitative and quantitative data including use of characteristic absorption bands to identify bonds KDP - the principles (including spin energy levels) and applications of proton and carbon-13 nuclear magnetic resonance spectroscopy (NMR) (excluding features of instrumentation and operation); analysis of carbon-13 NMR spectra and use of chemical shifts to determine number and nature of different carbon environments in a simple organic compound; and analysis of high resolution proton NMR spectra to determine the structure of a simple organic compound using chemical shifts, areas under peak and peak splitting patterns (excluding coupling constants) and application of the n+1 rule KDP - determination of the structures of simple organic compounds using a combination of mass spectrometry (MS), infrared spectroscopy (IR) and proton and carbon-13 nuclear magnetic resonance spectroscopy (NMR) (limited to data analysis) KDP - the principles of chromatography including use of high-performance liquid chromatography (HPLC) and construction and use of a calibration curve to determine the concentration of an organic compound in a solution KDP - determination of the concentration of an organic compound by volumetric analysis, including the principles of direct acid-base and redox titrations (excluding back titrations)

Qualitative vs. Quantitative analysis O Quantitative analysis  determining what chemical species are present in the sample O Qualitative analysis  determining how much of each chemical species is present in sample Qualitative techniques usually precede quantitative techniques

The Electromagnetic Spectrum and Spectroscopy Electromagnetic radiation  form of energy that consists of electric and magnetic fields that travel at the speed of light (i.e. visible light, radio waves, infrared waves) O a single unit of electromagnetic radiation = photon o each photon contains a certain amount of energy - the difference between the various types of electromagnetic radiation is the amount of energy found in the photons O Electromagnetic spectrum describes the range of all possible electromagnetic radiation frequencies

Spectroscopy Any type of chemical analysis that involves the use of electromagnetic spectrum (light) – samples are exposed to light of different energy and the amount of light reflected, absorbed or emitted is studies It is used to – O Determine the components of a sample (qualitative) O Determine the concentration of one particular component of a sample (quantitative)

Infrared Spectroscopy Uses infrared light absorbance of organic compounds to determine that carbon bonds and functional groups present O Infrared light is relatively low in energy and therefore does not promote electrons to higher levels – instead it causes bond vibrations: o Stretching (symmetric and asymmetric) o Bending (scissoring, rocking, wagging, twisting) O There are varying frequencies (and energy levels) of infrared light - each different bond in a compound absorbs different wavelengths of infrared light which causes it to either stretch or bend (moves molecule from a ground state vibrational energy level to an excited vibrational energy level) O The amount of energy a bond will absorb is dependent on: o The strength of bonds – the stronger the bond the higher the energy of the radiation absorbed o The mass of the atoms attached to the bond – the higher the mass attached the lower the energy of radiation absorbed O The amount of radiation absorbed is directly proportional to the amount/concentration of sample being analysed (greater conc. = greater radiation absorbed)

Inside the spectrometer: 1. A beam of infrared light or varying wavelengths is passed through the sample 2. Different bonds within the sample will absorb different wavelengths of light 3. Produces an infrared spectrum which can be read to determine the bonds in the compound Reading an infrared spectrum: O Shows transmission of light against wavenumber O Troughs – indicate where particular bonds are absorbing energy of infrared light O Wavenumber indicates position of peak – look at data booklet for values o Each different bond in a compound absorbs different wavelengths of infrared light creating characteristic troughs at particular wave numbers Alcohols– O O-H stretch (large circular trough), O C-H trough O C-O trough O No C=O group – use this to differentiate alcohols from carboxylic acids and esters

Carboxylic acids O O-H stretch, and C-H stretch overlap making a large wide trough of the two o Position and shape of hydroxyl trough is different in carboxylic acid that alcohol o If there is a peak between 3200-3600, it represents O-H group in an alcohol o If there is a peak between 2500-3500 it represents the O-H group in carboxylic acids O C=O stretch (acids)

O

CANNOT SEE C=O stretch that is seen in alcohols – use this to differentiate carboxylic acids from alcohols

Ester O C=O stretch (acids) O C-O stretch (alcohols, esters and ethers) O NO O-H group in ester – use this to differentiate esters from carboxylic acids

Amine O N-H stretch (amine twin peak) o Position of N-H trough is the sane as that for an O-H trough BUT the shape is very different O NO C=O stretch – use this to differentiate amines from

Amide O C=O stretch (amides) O N-H stretch – same as amines but larger in size

Aldehydes and Ketones O C=O (aldehydes) O C=O (ketones) O C-H stretch is wider in aldehydes than ketones

Mass Spectrometry O

Quantitatively analyses samples by detecting mass to charge ratio (m/z) of a molecule in that sample and using it to determine: o molecular mass of the sample’s components o the isotopes of an element o the relative isotopic masses and abundance within an element

Inside the mass spectrometer O

Mass spectrometer  the instrument that ionises, deflects and detects an atom so that its mass to charge ratio can be measured

1. Vaporised sample is fragmented into different ions – a stream of electrons is bombarded at the sample in order to create collisions that will knock electrons off the molecules creating a positive ion with a +1 charge o It is much more difficult to remove further electrons from a positively charged species

2. The positive ions are then accelerated so that they all have the same kinetic energy 3. An electromagnetic field is used to deflect ions, dependent upon their mass and charge o Mass – lighter ions are deflected further than heavier ions o Charge – ions with a +1 charge are deflected further than ions with a +2 charge

4. Ions (only positively charged) are detected as an electric current which can be amplified and recorded - mass/charge ratio is obtained to create a mass spectrum.

Reading a mass spectrum

Base peak also indicates most stable fragment

How isotopes effect mass spectroscopy O

Additional peaks can be seen in mass spectra as a result of isotopes of a particular element o i.e. next to the parent peak is the peak of similar mass – this peak represents an isotope of the atom

Nuclear Magnetic Resonance (NMR) Spectroscopy O

Uses electromagnetic radiation in the radio waves frequency range to determine number of chemical environments and ratio of atoms in an organic compound o Radio waves have too low of an energy to cause electronic, vibrational or rotational transitions

O

Protons neutrons and electrons spin on their axis in either an up/anticlockwise or down/clockwise direction – both directions possess same energy in the absence of external magnetic field o In order for radio waves to interact with nuclei, the atoms nucleus must have a nuclear spin – only atoms with an odd number of nucleons display overall spin (even number does not as spin of protons and neutrons cancel out) and when they do, they act like magnets  When placed in a strong magnetic field these atoms either align themselves with the external field or against the external field  align themselves with the magnetic field = lower energy state  orient themselves against the magnetic field = higher energy state and more unstable

O

1

H and 13C are atoms that have nuclear spin - the nuclear spins of these atoms are used to identify the structures of organic compounds

Inside the NMR spectrometer O O O

The nuclei are in a low energy spin A radio transmitter (device that produces radio waves) is used to provide energy to flip the nuclei into a high energy state Eventually the nuclei flip back to a lower energy state and when they do, they release energy (is this the chemical shift???) – energy released depends on the type of nucleus and the chemical environment surrounding the nucleus o Chemical environment  atoms and electrons that surround an atom o Energy released measured and represented in a spectrum

NMR Spectrum Provides information about the number and type of hydrogen and carbon nuclei in an organic compound by finding their distinct chemical environments O Number of peaks  tells us the number of different hydrogen or carbon environments O O

Areas under the peak  the ratios of hydrogens or carbons in each environment Chemical shift  amount of energy required to change the spin state (is CS of going to a higher level and coming to a lower level the same??) of the particular environment with reference to TMS o TMS = Tetramethylsilane is chemically inert compound that contains protons, carbon atoms and a silicon atom in a symmetrical arrangement (so it has only one chemical environment) – forms only a single peak at 0 and is used as a reference compound to keep analysis consistent  Chemical shift (measured in ppm) of a sample is compared to the chemical shift of TMS (0)  Atoms in the same chemical environment (i.e. attached to the same atom) have the same chemical shift – produce a single signal in the NMR spectrum and are said to be equivalent

  1

Nuclear shielding– electrons also have a spin and a magnetic field that can shield the nucleus from the applied magnetic field effecting chemical shift Neighbouring atoms can also affect the chemical shift (n+1)

H NMR

Observations from spectrum Number of signals (peaks) Chemical shift (ppm)/position of signals Splitting pattern 1 – singlet 2 – doublet 3 – triplet 4 – quartet 5 – pentet 6 – hexet 7 – heptet Relative peak area/area under the curve/ratio of peak area 13

Information Number of hydrogen environments Type of hydrogen environment (refer to data book) Number of neighbouring hydrogen atoms (ONLY occur in High Res) O N represents the neighbouring hydrogen atoms o n+1 = number of peaks o number of peaks – 1 = n Number of hydrogens in each environment

C NMR

Observations form spectrum Number of signals (peaks) Chemical shift (ppm)/position of signals O O O

Information Number of carbon environments Type of hydrogen environment (refer to data booklet Won’t show splitting patterns - each peak represents a different carbon environment Relative peak area also does not represent anything in CNMR CNMR tends to have a larger scale for chemical shift

Chromatography Used to: o identify unknown components of a sample based on the unique retention time of each component (qualitative) o Determine the concentration of substance present by comparing it to a calibration curve constructed using samples of known concentrations All methods of chromatography have a: o Stationary phase – solid that is coated onto a solid surface o Mobile phase – a liquid moves through the stationary phase Chromatography relies on two key ideas: o Component’s adsorption to stationary phase – how well the stationary phase can hold the components through intermolecular forces of attraction o Component’s desorption to mobile phase – how well the mobile phase can dissolve the sample through intermolecular forces of attraction o Components adsorption and desorption depends on the nature of the two phases: o stationary phase is polar – if a component in the sample is also polar it will easily adsorb to it o mobile phase is polar – if a component in the sample is also polar it will easily desorb to it

High Performance Liquid Chromatography (HPLC) Used to determine the quality and purity of compounds 1. A column is set up with a solid stationary phase which consists of fine particles – fine particle size increases surface area and allows for the largest number of interactions between the stationary phase and the components in the sample 2. The smaller the particles, the greater the resistance to flow - therefore the mobile phase has to be pumped in at high pressure

3. When from the end of used to detect its absorption of UV o if an

components emerge the column a recorder is presence of based on its light organic compound is present in the eluent stream, the recorder will a decrease in the UV signal, and this will be recording in the chromatogram

detect light

The effect of a on retention time Change to conditions Longer column Increased temperature Increased flow rate of mobile phase Increased total surface area of stationary phase

change to conditions Effect on retention time Increased Decreased Decreased Increased

Reading a chromatogram (qualitative analysis) O O O

Each peak represents a component in the sample The order in which they come out shows the order of retention time If there are known retention times for particular substances such as glucose (that have been obtained under the same conditions) and they correspond to the retention time of a peak in the chromatogram such as peak C then you can conclude that the sample consists of glucose

Using calibration curves (quantitative analysis) Area under the peak is the concentration of the component Concentration of a component (i.e. glucose) can be determined by: 1. Run a set of standards of known concentrations in the same HPLC column under the exact same conditions. 2. construct a calibration curve (concentration on x-axis and peak area on y-axis) with known concentrations plotted and a ‘line best fit’ drawn connecting first and last points – DO NOT EXTEND beyond these points 3. Using given peak area, use ‘line of best fit’ to match an appropriate concentration

Combining analytical techniques Mass spectroscopy  used to determine molar mass of the entire molecule as well as idenitify fragments to help determine structure IR  Used to identify functional groups NMR  used to identify the types of C and H present in a molecular – help determine bonding environment around C and H atoms 1. Determine empirical formula (if percentages of each compound are given) 1. Turn percentage to molar mass 2. Calculate mol from molar mass 3. Divide mol values obtained from all three atoms by the lowest mol value 4. Obtain ratio and write empirical formula 2. Determine the moleuclar formula of the compound 1) Use the mass specturm to determine the molar mass of the parent ion 2) Divide molar mass of parent ion with the molecular mass of the empirical formula 3) Approximate the number obtained and multiply all atoms in the empirical formula by this number 3. Use CNMR and HNMR to determine the final structure of the molecular

Volumetric analysis Acid-base titrations Important terms O Aliquot – the amount of solution measured by the pipette (e.g. 20.00 mL aliquot of HCl) O Pipette – the glassware used to accurately measure an aliquot O Standard solution – the solution with the accurately known concentration O Volumetric flask – the flask that can be used to prepare the standard solution O Burette – the glassware used to accurately add one solution to another O Titre – the volume of solution added from a burette O Equivalence point - the point at which exactly the right around of one solution has been added to another so that they react stoichiometrically as determined by the balanced equation O End point – the point at which the indicator changes colour

Steps of normal order titrations: 1. An aliquot (accurately measured volume) of the unknown solution is taken using a pipette and placed in a conical flask with the indicator 2. A standard solution with an accurately known concentration is prepared in a volumetric flask 3. The unknown solution is placed under the burette which is filled with the standard solution 4. Standard solution is added to conical flask via burette until end point is reached o Titrations should be repeated until three cooncordant titres are obtained  Concordant titres – titires that differ by a maximum of 0.10 mL from highest to lowest 5. Volume (titre) of standard solution added is recorded 6. Concentration and volume of standard solution is used to calculate the mole of standard solution that was added 7. Mole of standard solution is used to determine the mole of the unknown using balanced chemical equation 8. Concentration of unknown is determined using its mole and volume

Choosing indicators O Indicators need to be chosen so that the end point matches the equivalence point O Indicators need to have a discernable and permanent colour change at the required pH

Redox titrations O O O

O

Involves the titration of an oxidant against a reductant Procedure is identical to acid-base titration but no indicator is required to detect end point as redox titrations are self indicating Redox titrations may be identified if: o There is a spontaneous colour change in the absence of an indicator o The words reduced and/or oxidised appear in the question o One or more of the chemucal species has unde...