Option A

The following notes were written for the previous IB syllabus (2009). The new IB syllabus for first examinations 2016 can be accessed by clicking the link below.

IB syllabus for first examinations 2016

Modern Analytical Chemistry


This option builds on some of the key ideas in both physical and organic chemistry that were introduced in the core. Many of the concepts appear in the higher component of the chemistry course and where possible links have been made to these notes. (Refer to the syllabus statements for relevance)

A.1 Analytical techniques (1h)

A.1.1:- State that the structure of a compound can be determined by using information from a variety of spectroscopic and chemical techniques. Students should realise that information from only one technique is usually insufficient to determine or confirm a structure.

A.1.2: Describe and explain how information from an infrared spectrum can be used to identify functional groups in a compound. Restrict this to using infrared spectra to show the presence of the functional groups:

                 

and to match the fingerprint region to a known spectrum.

A.1.3     Describe and explain how information from a mass spectrum can be used to determine the structure of a compound. Restrict this to using mass spectra to determine the relative molecular mass of a compound and to identify simple
fragments, for example :

  •                 (Mr - 15)- loss of CH3
  •                 (Mr - 29)- loss of C2H5 or CHO
  •                 (Mr - 31)= loss of CH3O
  •                 (Mr - 45)+ loss of COOH


A.1.4: Describe and explain how information from a 1H NMR spectrum can be used to determine the structure of a compound. Restrict this to using NMR spectra to determine the number of different environment in which hydrogen is found the number of hydrogen atoms in each environment. Splitting patters are not required.

A.1.5: Describe and explain the structure of benzene using chemical and physical evidence.

Consider the special stability of the ring system (heat of combustion or hydrogenation of C6H6 in comparison to that of cyclohexene, cyclohexadiene and cyclohexatriene), as well as benzene's tendency to undergo substitution rather than addition reactions.


 

A.2: Principles of spectroscopy (2h)

A.2.1: Define the terms rate constant and order of reaction.

A.2.2: Derive the rate expression for a reaction from data.

Rate = k[A]m[B]n

where k = rate constant, [A] = concentration of A in mol dm-3 etc.
m and n = integers, m + n = overall order of the reaction.

A.2.3: Draw and analyse graphical representations for zero-, first- and second- order reactions.

A.2.4: Define the term half-life and calculate the half-life for first-order reactions only. The half-life should be calculated from graphs and by using the integrated form of the rate equation. The integrated rate equation for second-order reactions is not required.


 

A.3: Infrared spectroscopy (IR) (3h)

A.3.1: Define the terms rate-determining step, molecularity and activated complex.

A.3.2: Describe the relationship between mechanism, order, rate-determining step and activated complex. Limit examples to one- or two-step reactions where the mechanism is known. Students should understand what an activated complex (transition state) is and how the order of a reaction relates to the mechanism.


 

A.4: Mass spectroscopy (2h)

A.4.1: Distinguish between primary, secondary and tertiary halogenoalkanes.

A.4.2: Describe and explain the SN1 and SN2 mechanisms in nucleophilic substitution. Students must be able to draw a stepwise mechanism. Examples of nucleophiles should include -CN, -OH and NH3 for each reaction type.

A.4.3:Describe and explain the molecularity for the SN1 and SN2 mechanisms. The predominant mechanism for tertiary halogenoalkanes is SN1 and for primary halogenoalkanes it is SN2. Both mechanisms occur for secondary halogenoalkanes.

A.4.4: Describe how the rate of nucleophilic substitution in halogenoalkanes depends on both the identity of the halogen and
whether the halogenoalkane is primary, secondary or tertiary.


 

A.5: Nuclear magnetic resonance (NMR) spectroscopy (2h)

A.5.1: State the expression for the ionic product constant of water (Kw).

Kw = [H+ (aq)][OH-(aq)] = 1.0 x 1.0-14 mol2 dm-6 at 298 K, but this varies with temperature.

A.5.2: Deduce [H+(aq)] and [OH-(aq)] for water at different temperatures given Kw values.

A.5.3: Define pH, pOH and pKw.

A.5.4: Calculate [H+(aq)], [OH-(aq)], pH and pOH from specified concentrations. The values of [H+(aq)] or [OH-(aq)] are directly related to the concentration of the acid or base.

A.5.5: State the equation for the reaction of any weak acid or weak base with water and hence derive the ionization constant expression.

In general HA(aq) === H+(aq) + A-(aq)

B(aq) + H2O(l) === BH+(aq) + OH-(aq) (base hydrolysis)

Then Ka and Kb

Examples used should involve the transfer of only one proton.

A.5.6: State and explain the relationship between Ka and pKa.

A.5.7: Determine the relative strengths of acids or their conjugate bases from Ka or pKa values.

A.5.8: Apply Ka or pKa in calculations. Calculations can be performed using various forms of the acid ionization constant expression (see A.5.5). Students should state when approximations are used in equilibrium calculations. Use of the quadratic expression is not required.

A.5.9: Calculate the pH of a specified buffer system. Calculations will involve the transfer of only one proton. Cross reference with 9.4.

A6 : Atomic absorption (AA) spectroscopy

 

A7: Chromatography

 


HIGHER LEVEL

A8: Visible and ultraviolet (UV-Vis) spectroscopy - (3h)

 

A9: Nuclear magnetic resonance (NMR) spectroscopy - (2h)

 

A10: Chromatography - (2h)

 

The structure of a chemical compound can usually not be determine accurately with information from only one source. This stems firstly from the great number of possible organic compounds, many of which have very similar chemical properties but different physical properties, similar physical properties but different chemical properties or very similar properties in both categories. There are a variety of possible techniques which go beyond both chemical and physical properties, and by combining the information from all these sources it is generally possible to deduce the structure of a compound.

Most of the techniques involve either the absorption of electromagnetic energy of specific frequency or the emission of electromagnetic radiation of specific frequencies.

Summary

Technique Nature Effect
IR Spectrometry Absorption of radiation in the IR region of the electromagnetic spectrum Changes the vibrational states of the bonds in the molecules
Mass Spectrometry This is the exception - it does not involve either absorption or emission of radiation. The molecule is ionised and broken apart. The fragment masses (also ions) are measured.
NMR Spectrometry Absorption of radiation Changes the spin state energy of the nuclei of the hydrogen atoms
UV Spectrometry Absorption of radiation in the UV region of the electromagnetic spectrum Affects the molecular energy levels of the electrons in atomic and molecular orbitals
Visible Spectrometry Absorption of radiation in the UV region of the electromagnetic spectrum Affects the energy levels of the electrons in atomic and molecular orbitals


Infra-red spectra

Covalent bonds are not of a static length, but rather they vibrate rapidly. This vibration may take the form of stretching or bending and the amount by which they bend or stretch is determined by the quantity of energy they contain. his energy can be absorbed in discrete packets (quanta) in the infrared region of the spectrum. Consequently if a range of wavelengths (in the IR region of the spectrum) is passed through a sample there will be absorption corresponding to the specific energies that can be absorbed by the covalent bonds. As the number of bonds is usually large and the number of different ways that these bonds can stretch and bend is also large then the spectrum obtained by absorption of energy is complex. The spectrum can be divided into two regions.

  • A fairly simple region with few absorptions
  • A complex region with many absorptions (the "fingerprint" region)

Absorptions occuring in the first region can sometimes be identified by consulting the data book. Typical strong absorptions in this region include C=O (the carbonyl group) stretches at around 1600 - 1800cm-1 (see example below)

The fingerprint region can be matched to other previously taken spectra to see if an unknown sample corresponds to one previously scanned

IR Spectra, however, is usually insufficient as it does not offer enough information about the relative placement of the bonds, or their quantity. The spectrum information is given in the data book and can be matched to any given data.

Example:

IR Spectrum of propanone CH3COCH3


The embedded FTIR of acetone was recorded as a thin film using a PE1605 FTIR

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Mass spectra

See Atomic Theory section 12.1

There will usually be a peak in the spectrum at the mass of the unbroken molecule - the molecular ion - corresponding to the whole molecule which has lost just one electron (and consequently still has the same mass). Other peaks appear as the fragments of the main molecule are broken off in the bombardment process. These fragment masses will correspond to the most obvious parts of the molecule.

Example: Propan-1-ol

If Propan-1-ol is used for the sample, fragments will appear at

  • m/e: 15 corresponding to loss of a [CH3]+ fragment

  • m/e: 17 corresponding to an [OH]+ fragment

  • m/e: 29 corresponding to a [CH3CH2]+ fragment

  • m/e: 43 corresponding to a [CH3CH2CH2]+ fragment

 

Example: Pentan-2-ol

mass spectrum 1

The molecular ion is designated "m+" , the next fragment m with one mass unit removed etc.

mass spectrum 2

More examples here


Nuclear Magnetic Resonance NMR

The nuclei of atoms with an odd number of nucleons (protons and neutrons) can absorb energy when in the presence of an external magnetic field and change their spin states. Hydrogen has only one nucleon and its nucleus can have spin states of +1/2 and -1/2.

When the hydrogen nuclei are in a magnetic field this can be detected as absorption of radiation at specific frequencies depending on the actual environment (position in the molecule) in which the hydrogen atoms find themselves.

If a hydrogen atom is attached to an electronegative atom (such as oxygen) the electrons are drawn away from the hydrogen leaving it more exposed to the externally applied magnetic field. It is in a specific environment and will absorb radiation of a specific frequency.

The amount of absorption (area under the curve) is related to the number of hydrogens within the molecule having the same environment and so comparison of the peaks (called integration) gives the relative numbers of hydrogens in each part of the molecule.

The spectrum is taken in conjuction with an internally added standard compound called TetraMethylSilane (TMS) which sets the zero on the spectrum.

Example:

Ethanol CH3CH2OH has three different environments for hydrogen

CH3 -Three hydrogens in a methyl group not attached to an electronegative atom

CH2 -Two hydrogens in a group not attached to an electronegative atom but much closer to an oxygen

OH - A hydrogen attached to an electronegative

Consequently there will be peaks corresponding to three hydrogens at 1,2 (compared to TMS=0), two hydrogens at 3,4 and one hydrogen at 3,7. See examples of the spectrum here

 

 



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