1

Advanced Experiments in Physical and Analytical Chemistry

CM3292 (II) Table of contents

1. Constructing a Binary Phase Eutectic Diagram from Cooling

Curves

2. Surface Phenomena : (a) Adsorption at Solid/Liquid Interface

(b) Surface Tension

(c) Micelle Formation

3. Second-Order Reaction Kinetics Determined by Conductometry.

4. (a) Raman and Infrared spectroscopy

(b) Powder X-Ray Diffraction (XRD)

5. Computational Chemistry

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Introduction

This updated and corrected edition has been prepared through the contributions of many people at the Department of Chemistry: members of the teaching staff, technicians, demonstrators and finally you, the undergraduates whose stimulating questions prompted many improvements in the text. The manual contains descriptions of experiments which will help you visualize and understand the theoretical concepts encountered during lectures. It is also a reminder that chemistry is an experimental discipline where all the knowledge is ultimately derived from practical work in the laboratory. Only the most essential details of every experiment are given in the manual. Further reading is therefore strongly recommended. Always use appropriate SI units when reporting your results.

Consult this electronic web page for additional information! (http://www.science.nus.edu.sg/~webchm/lab/phylab.htm) For theoretical background one may consult: 1) P.W. Atkins, Physical Chemistry, 7th ed., Oxford University Press 2002 2) I. N. Levine, Physical Chemistry, 4th ed., McGraw-Hill 1995 For details of a particular experiment refer to: 3) D.P. Shoemaker, C.W. Garland, J.W. Nibler, Experiments in Physical Chemistry, 6th ed., McGraw-Hill 1996 4) R.J. Sime, Physical Chemistry: Methods, Techniques and Experiments, Saunders College Publishing, 1990 5) G.P. Matthews, Experimental Physical Chemistry, Oxford University Press 1985 Useful handbooks: a) Lange's Handbook of Chemistry, 13th ed., McGraw-Hill 1985 b) Handbook of Chemistry and Physics, 77th ed., CRC Press, Boca Raton 1996 c) Quantities, Units and Symbols in Physical Chemistry, prepared by: I.Mills, T.Cvitas, N.Kallay, K.Homann, K.Kuchitsu, Blackwell, Oxford 1993 Sources for numerical data analysis (e.g. linear regression): e) K.Ebert, H.Ederer, T.L.Isenhour, Computer Applications in Chemistry, VCH 1989

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1. Constructing a Binary Phase Eutectic Diagram

from Cooling Curves

Introduction

In this experiment the heterogeneous equilibrium between solid and liquid phases

in a two-component (binary) system consisting of components A (o-nitrophenol) and B

(p-toluidine) will be observed. The cooling curves for systems of different compositions

are obtained by plotting the temperature vs. time. From them a phase diagram for the

system can be deduced. When a homogeneous binary liquid mixture is cooled, a solid

separates but its composition will depend entirely on the nature of the components. If the

molecular (atomic) sizes of the pure compounds (components) are similar or have the

same type of crystal structures the components will form solid solutions. This case often

occurs in metal systems where the solids separating from the liquid phase can have a

wide range of compositions. The two components can also form a solid compound which

has a defined stoichiometry and a crystal structure different from the two components. In

that case, the phase diagram will show a maximum surrounded by two eutectic minima.

In the case of organic compounds there is a high degree of crystal lattice incompatibility

and no solid compound formation the system will show a simple phase diagram with only

one eutectic minimum. In that case each of the two components crystallizes out

separately lowering the freezing point of the other.

Cooling curves often show two types of features: thermal arrest and break. The

former (found in single component systems) is caused by constant temperature achieved

while pure substance crystallizes. All phase changes of pure substances take place at

constant temperatures. The latter represents a change in slope (kink) of the binary system

curve. When crystals are being formed in the liquid the heat thus released in the process

will slow down the cooling of the liquid causing the change in slope. A difficulty often

encountered in this experiment is "super cooling" when the temperature of solution drops

below its freezing point before crystals start to separate. It is caused by a lack of nuclei of

crystallization and manifests itself as a dip in the cooling curve.

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Aim

In this experiment a eutectic diagram for a binary mixture will be constructed from the

cooling curves.

Experiment

The apparatus consists of two boiling tubes and a cork carrying a stirrer and

digital thermometer. Using a top-pan balance weigh the empty inner (smaller) tube first

(standing in a beaker) and then filled with approximately 4 g of component A. The added

sample should cover the probe of the thermometer. The probe is inserted only after the

compound or mixture has melted to liquid and the inner tube is removed from the

hot water bath. Heat the inner tube in hot water bath until A becomes entirely liquid.

Remove the inner tube from the bath and insert it in the outer (larger) tube provided.

Allow the apparatus to cool slowly in a bath with stirring and record the temperature at

1/2 min intervals until all liquid has solidified. The temperature of the cooling bath is to

be adjusted at 10 °C lower than that of the system at all times by the addition of water or

ice (What is the rationale of doing this?). From the plotted cooling curve determine the

freezing point of A.

Weigh approximately 1 g of the second component B, add it to A and obtain the

cooling curve as described above. The new cooling curve will show a distinct change in

slope at the new freezing point of the mixture. The new freezing point will be

considerably lower than the freezing point of pure A. Keep measuring the temperature

until all liquid has solidified.

Add another 1 g of B to the mixture and obtain the cooling curve. Repeat the

procedure (by adding 0.75g, 0.75 g and 0.5g) until a total amount of 4 g of B has been

added. At this stage transfer the mixture into the waste container (in the fumehood),

clean (with acetone and water) and dry the inner tube. Repeat the experiment starting

with 4 g of pure B to which successive 1 g,1 g, 0.75g, 0.75g and 0.5g portions of A are

added.

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Safety precaution: The organic compounds used are toxic and must be handled with due

care, minimizing skin contact and vapour inhalation. Discard all residues into bottle

provided in the fume-cupboard and rinse the glassware with acetone followed by water.

Data analysis

Plot the cooling curves and from them deduce the phase diagram of the binary

system studied. Detect eutectic temperature and eutectic mixture composition. Label

different points and areas on the diagram which represents relationship between freezing

points and composition (mole fraction) of one of the components.

Exercises

1. Why does the temperature remain constant during phase change of pure substance and

what does the rate of cooling depend on?

2. Consider the freezing points of pure water, 0.1M NaCl solution and 0.1M CaCl2

solution. Order the freezing points according to their magnitudes (e.g. 0°C, <0°C, >0°C).

3. If sea water is frozen what is the composition of the resulting solid (refer to the phase

diagram for sea water)?

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Surface Phenomena

(2a) Adsorption at a solid/liquid interface

Introduction

The surface of a solid is in a state of strain similar to that existing in liquids and

has a residual attractive force. All solids consequently tend to adsorb on their surfaces

any gas or liquid with which they come into contact. Many solids are also capable of

adsorbing ions and molecules from solution; this effect is used for separation of

substances by chromatography. The concentration of a solute in solution is often reduced

by addition of a finely divided solid such as charcoal. The solute becomes partially

adsorbed on the surface of the solid. Adsorption from dilute solutions may be represented

by the empirical Freundlich isotherm: x = kcn, where k and n are experimental constants,

c is the equilibrium molar concentration of solute in solution and x is the number of

moles of solute adsorbed per gram of adsorbent. Another important relation describing

adsorption is the Langmuir isotherm: x = ac/(1+bc), where a and b are experimental

constants. Both expressions indicate that x increases with c, but the Langmuir equation

also indicates that at high concentrations, bc 1, ie., the adsorption becomes constant

and independent of further increase in concentration. At that point the surface of the solid

becomes fully covered by a monolayer of adsorbed solute molecules.

Aim

To study the adsorption of a weak acid on charcoal.

Experiment

Prepare 6 solutions in stoppered bottles according to the table:

Flask Number 1 2 3 4 5 6

0.5M acid (cm3) 100 80 60 40 20 10

H2O (cm3) 0 20 40 60 80 90

g of charcoal 2 2 2 2 2 2

aliquot of filtrate

for titration(cm3) 5 5 10 10 25 25

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The charcoal must be weighed accurately and volumes of water and acid added

measured with burettes. Shake the flasks intermittently for about 20 min by which time

equilibrium should be established between adsorbed and dissolved acid. To remove the

charcoal, filter the solution and then titrate the appropriate volume (see Table) of clear

supernatant solution with 0.1 M NaOH using phenolphthalein indicator.

Data analysis

1. Tabulate data and calculate c and x for each flask in mol of the acid per gram of

charcoal.

2. Plot x vs. c and explain its significance.

3. Test the applicability of the Freundlich isotherm by plotting log {x} against log {c}.

If the relationship is linear, evaluate n and k using regression analysis.

4. The Langmuir isotherm can be also be written as: c/x=1/a+bc/a. Test the applicability

of the Langmuir isotherm by plotting c/x vs. c. If you find that the Langmuir isotherm

describes the adsorption process well, calculate a and b using regression analysis (see

Introduction)

5. The a/b ratio gives the value of xmax which in turn allows the calculation of active

surface area of charcoal in m2/g assuming the area of the acid molecule to be 0.21

nm2. Compare the calculated area with the floor area of the laboratory.

Exercises

1. What assumptions are made in deriving the Langmuir adsorption isotherm and discuss

their validity?

2. Which of the two isotherms (Freundlich or Langmuir) describes the adsorption of

a weak acid on charcoal? (use regression analysis to obtain the answer)

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2(b) & 2 (c) Surface tension

Introduction

The surface tension of solutions differs from that of pure solvents. J.W. Gibbs has

shown that for lowering of surface tension the solute concentration is higher at the

surface than in the bulk of the solution (i.e. there is a surface excess of solute) and

conversely the surface tension is larger than that of the pure solvent indicates surface

depletion of solute. The relationship between surface tension and concentration is given

by Gibbs adsorption equation:

= (-c/RT)(d/dc) (1)

where is the surface excess of solute in moles per unit area of surface, c molar

concentration and surface tension. Hence if the surface tension of a liquid is measured

at constant temperature for various concentrations, d/dc may be obtained for any

concentration from the tangents to the curve of against c. Values of may then be

calculated from (1). Assuming that the surface excess is concentrated in a monomolecular

layer, the surface area occupied by 1 molecule (A) can be estimated:

A = 1/( NA) (2)

where NA is Avogadro constant. Assuming further that the molecules in the surface move

freely within the two dimensions of the surface, that their molecular cross-section is small

compared to the area they inhabit (A), and that there is no intermolecular interaction, the

following relationship holds:

A=kT (3)

where is the surface pressure (solvent - solution) and k Boltzmann constant.

Equation (3) resembles the ideal gas equation PV = nkT and is likewise not applicable at

high pressures. At high surface pressures a modified relationship can be used: (A-A0) =

xkT where A0 is the molecular cross section and x coefficient representing intermolecular

attraction.

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Solutions of highly surface active materials exhibit unusual physical properties. In

dilute solution they act as normal electrolytes but at fairly well defined concentrations

abrupt changes in several physical properties such as osmotic pressure, electric

conductance, turbidity and surface tension take place. This behavior can be explained in

terms of organized aggregates or micelles of surfactant ions in which the lipophilic

hydrocarbon chains are oriented towards the interior of the micelle leaving the

hydrophilic groups in contact with aqueous medium. The concentration above which

micelle formation becomes significant is called the critical micelle concentration.

2(b) Aim

To study the surface tension of an alcohol in water

Experiment

Prepare 100 ml of 0.1M t-butyl alcohol in water using volumetric flask and from

that stock solution prepare 0.08M, 0.04M, 0.02M and 0.01M solutions by diluting with

water. (Use 0.08M solution to prepare 0.04M etc.) Measure the surface tensions of water

and the solutions using a Du Nouy tensiometer. Take the mean of 3 readings for each

solution and record the temperature. When measuring the surface tension of different

solutions, start with the least concentrated one first so that the glassware will not be

contaminated in subsequent measurements.

Data analysis

Plot a graph of surface tension γ vs concentration c by drawing a smooth curve

through the data points. Draw tangents to the curve at concentrations 0.01, 0.02, 0.04,

0.06 and 0.08M and determine d/dc at these points. Calculate values from (1) and plot

them against appropriate concentrations. Estimate the limiting value lt from the graph

and calculate 1/lt. Next calculate 1/ for each experimental concentration and thus

obtain A from (2). Finally plot A vs. and from that graph determine A0 and x.

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2(c) Aim

To determine the CMC of a surfactant

Experiment

The second part of the experiment consists of the measurement of critical micelle

concentration (CMC). Prepare 250 ml of 0.02M solution of sodium dodecyl sulfate. By

diluting that solution further, prepare solutions with concentrations: 0.015, 0.01, 0.0075,

0.005, 0.0025, 0.0015 and 0.001M. Determine the surface tension of each of the above

solutions by pouring them into a tensiometer dish and allowing a 5-10 min interval before

starting measurement. This time is necessary for the micelles to form and reach

equilibrium with the bulk of solution.

Data Analysis

Plot the surface tension against the concentration of sodium dodecyl sulfate solution and

determine the CMC from the diagram.

References:

1. E.F. Meyer, G.M. Wyshel, J.Chem. Educ. 63 (1986) 996.

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3. The Rate Constant of a Second-Order Reaction

determined by Conductometry Introduction:

Many hydrolytic reactions are catalyzed by bases as well as acids. The reaction

RCOOR' + OH- = RCOO- + R'OH

can be monitored by titrating the unreacted alkali or measuring the electrical

conductivity. The net result of reaction above is progressive replacement of mobile OH-

by less mobile RCOO- (RCOO- has approx. 5 times lower mobility), which causes the fall

in conductivity.

The rate law for the reaction is

k = (1/at) [x/(a-x)] (1)

where a is the initial concentration of both reactants and x is the number of moles reacted

after time t. (1) can be written in terms of conductivity rather than concentrations to

give

k = (1/at) [(0 - t)/(t - )] (2)

where 0 is initial conductivity of solution, t conductivity after time t and

conductivity after reaction is completed. Rearranging (2) we obtain

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t = (1/ak)[( 0 - t)/t] + (3)

The rate constant k can then be calculated from the known initial concentration of

reactants a and a plot of t vs. (0 - t)/t.

Experiment:

The solutions required are: 500 ml of 0.02M NaOH, 500 ml of 0.02M EtOAc (freshly

prepared) and 100 ml of 0.01M NaOAc made by neutralizing 20 ml of 0.05M NaOH with

20 ml of 0.05M CH3COOH and making up to 100 ml.

All solutions should be prepared with CO2 free water. For that purpose boil 2l of

deionized water, in separate 1l beakers, cover the beaker and let the water cool to room

temperature.

All solutions should be stoppered to avoid evaporation.

Fill a 500 ml volumetric flask 3/4 full with boiled water, pipet the necessary volume of

EtOAc into the flask and make to volume with boiled water (Overall EtOAc

concentration should be 0.02M).

Clean dip-type conductivity cell with deionized water and allow it to dry. Then measure

conductivities at 30° C for the 0.01M NaOH solution (0) and 0.01M CH3COONa

solution ().

Suspend two large boiling tubes in the thermostat at 30° C, one containing 70 ml of

0.02M NaOH and the other 70 ml of 0.02M EtOAc.

Allow the flasks to reach equilibrium temperature (10 min) and in the meantime clean

and dry the conductivity cell. Then mix the two solutions quickly and start the stop-clock.

Dip the conductivity cell in the mixture and take conductivity readings at 2, 3, 4, 5, 6, 7,

9, 11, 13, 15, 18, 20, 25, 30, 35 and 40 min after commencement of the experiment.

Repeat the experiment at 45° C and 60° C remembering to set the temperature switch

on the conductometer accordingly.

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Data analysis:

Obtain linear least-squares fit of t vs. (0 - t)/t and from its coefficients determine

approximate rate constant for the hydrolysis (k) and . Calculate activation energy for

the reaction using Arrhenius equation.

Exercises:

1. Why is it necessary to use freshly prepared EtOAc solution and CO2 free water?

2. Electrodes of the conductivity probe are dipped in water, dried and then dipped in

mercury. In which liquid(s) will the indicator light on the probe light up and why?

3. What is the definition and what are the SI units for conductivity?

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4 (a) Laser Raman Spectroscopy

Features of Raman spectroscopy

In the study of spectroscopy (refer to CM2101), we learn that photons interact with

molecules or matter to induce transitions between energy states. The scattering of photon

by molecules is an example of the type of interaction which include absorption, reflection

and emission. In Rayleigh scattering, the photons are elastically scattered which means

the emitted photon has the same wavelength as the absorbing photon. On the other hand,

the Raman effect is related to the relatively small percentage (about 1 in 107 photons) of

inelastic scattering of photons by molecules, which gives rise to the difference of energies

between the incident and scattered photons.

The energy of the scattered radiation is less than the incident radiation for the Stokes line

but more than the incident radiation for the anti-Stokes line. The energy changes is

related to the vibrational energy spacing in the ground electronic state of the molecule

and therefore the wavenumbers of the Stokes and anti-Stokes lines are a direct measure of

the vibrational energies of the molecule.

In the spectrum, the Stokes and anti-Stokes lines are displaced to opposite sides from the

Rayleigh line. This occurs because in either case, one vibrational quantum of energy is

gained or lost. Anti-Stokes lines are much less intense than the Stokes line because only

molecules that are vibrationally excited prior to irradiation can give rise to the former

line. At room temperature, the percentage of molecules occupying excited vibrational

states is small. Hence, only the more intense Stokes lines are normally measured.

Infrared (IR) and Raman spectroscopy both measure the vibrational energies of

molecules but these method rely on different selection rules. Recall that for a vibrational

motion to be IR-active, the dipole moment of the molecule must change. For example,

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the symmetric stretch in carbon dioxide is not IR-active because there is no change in the

dipole moment while the asymmetric stretch is IR-active due to a change in dipole

moment. For a vibration to be Raman active, there must be a change in the polarizability

of the molecule which means that the symmetric stretch in carbon dioxide is Raman

active because the polarizability of the molecule changes. In general, IR transitions

recorded for a molecule possessing a centre of symmetry will not appear in the Raman

spectrum and vice-versa.

Experimental

In this experiment you will collect infrared and Raman spectra of three samples : (1)

dichloromethane, (2) chloroform and (3) diphenylacetylene in chloroform solution. The

infrared spectra will be collected using a liquid cell and the FTIR spectrometer. Raman

spectra will be collected using the Raman spectrometer with a solid-state laser as the

excitation source. Please consult the demonstrator who will show you the setup of the

Raman spectrometer.

In your laboratory report, you will need to include ;

1. sketch the Raman spectrometer setup and explain how the experiment is carried out.

2. describe the functions of the instruments and apparatus such as the monochromator,

photomultiplier tube, fiber optic cable and other important parts of the Raman

spectrometer.

3. Assign as many as possible transitions recorded in the IR and Raman spectra of the

three samples. It will be necessary for you to use the literature to help assign the peaks to

the particular vibrations in the molecule.

4. discuss the similarities and differences between the IR and Raman spectra of each

sample.

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4(b) Qualitative and Quantitative Analysis of Alkali Halide Mixture by Powder X-Ray Diffraction (XRD) Aims

1. To learn about the use of the X-ray powder diffraction technique for identification of crystal structures

2. To determine the structural parameters for alkali halides 3. To quantitatively analyze the composition of alkali halide mixtures

Introduction This experiment will be conducted in a group of 4 students and they will be assigned beforehand in performing either part A or part B of the experiment. Experiment procedures are similar for both part A and B but in part A alkali halides with different cations (NaCl : KCl) will be analyzed while in part B alkali halides with different anions (KCl : KBr) will be analyzed. This X-ray powder diffraction experiment involves three major procedures: sample preparation, data acquisition and X-ray diffraction patterns interpretation. A complete description of the theory required for this experiment is not provided in this manual. Students are strongly advised to read the relevant textbook on X-ray diffraction before conducting this experiment. Following is a brief introduction and theory about X-ray diffraction: Qualitative X-ray Diffraction Analysis The determination of the chemical structure of molecules is indispensable to chemists in their effort to gain insight into chemical problems. Only a few physical methods are capable of determining chemical structure, and amongst these methods, diffraction methods have been the most successful. Diffraction methods are capable of defining bond lengths, bond angles and other important structural parameters for a particular crystalline solid. Diffraction experiments can be carried out by using electromagnetic wave such as X-ray or tiny particles such as electrons and neutrons. Over the years numerous diffraction methods have been developed in order to determine the structure of a crystallite sample. Among these techniques, single crystal X-ray diffraction (XRD) is a powerful method that is commonly used to determine the structures of new materials. However, the technique is limited by the ability to grow crystals with nearly perfect structures and ideal sizes that are suitable for diffraction analysis. Due to this limitation and the time and cost-intensive nature of the technique, single crystal diffraction is not used for routine structural characterization of known materials. For routine structural characterization of materials, powder XRD is far more common. The samples for powder XRD may be large crystals, or they may be in the form of a powder composed of tiny crystals that are too small to be seen by the human eye. The underlying principles of the experiment are the same in both powder diffraction and single crystal diffraction, although the data analysis is simpler in powder diffraction.

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Figure 1. Illustration of Bragg’s law for X-ray diffraction. Powder diffraction patterns are typically plotted as the intensity of the diffracted X-rays against the angle 2. Peaks will appear in the diffraction pattern at 2values when constructive interference is at a maximum, that is, when Bragg’s Law (eq. 1) is satisfied.

(eq. 1) n = order of the reflection = wavelength dhkl = interplanar spacing of planes with Miller indices h, k and l

= Bragg’s angle From the above equation, it is evident that the structure of the solid (such as the distance between the atoms) can be obtained if a monochromatic X-ray source is used. The constructive interferences, that results at specific values of 2map out the interplanar spacing of the solid. In this experiment the X-ray has a wavelength of 1.5405 Å (Cu K1 line). For a cubic system with simple geometry,

(eq. 2) a = unit cell dimension

n = 2dhkl sin

dhkl2 = a2/(h2 + k2 + l2)

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Combining both equation 1 and 2, a relation between the Miller indices and the Bragg’s angle can be established,

(eq. 3) For the simple cubic (SC) unit cells, all possible combinations of integer Miller indices are possible. However for face-centered cubic (FCC) unit cells, certain combinations of Miller indices are eliminated by destructive interference caused by the planes formed by the atoms in the faces of the unit cell. The result is that for FCC type lattices reflections occur only if h, k, and l are all even or all odd. Some of the possible combinations of Miller indices for cubic unit cells are shown in the following table: Table 1: Reflection planes for cubic unit cell.

Simple Cubic (hkl)

Body-centered Cubic (hkl)

Face-centered Cubic (hkl)

h2 + k2 + l2

100 1

110 110 2

111 111 3

200 200 200 4

210 5

211 211 6

7

220 220 220 8

221, 300 9

310 310 10

311 311 11

222 222 222 12

The systematic extinction of certain planes can then be used to assign the diffraction patterns for a particular crystallite solid with cubic system. Once the Bravais lattice for the specimen has been identified, the cell unit dimension, a, can be obtained by solving equation 3.

h2 + k2 + l2 = (4a2/2)sin2

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Quantitative X-ray Diffraction Analysis Besides qualitative structural analysis of a particular crystallite solid, quantitative analysis is also possible by using powder XRD. Quantitative analysis of diffraction data usually refers to the determination of amounts of different phases in multi-phases sample. While qualitative analysis usually involves the identification of peaks at certain Bragg’s angles, qualitative analysis usually involves the measurement of peak area and intensity. The peak intensities reflect the total scattering from each plane for a particular crystal structure and therefore they are dependent on the distribution of particulars atoms in the structure. The calculation of peak intensities is complicated as a lot of experimental constants are involved. In any circumstances the peak intensities can be approximated by using the following equation:

sαhklI

XK

eK

(eq. 4)

I(hkl) = intensity of reflection of hkl in phase Ke = experimental constant K(hkl) = structural factor for phase X = weight fraction of phase = density for the phase

(/)s = mass absorption coefficient of the specimen (s is the linear absorption coefficient of the specimen while s is the density of the specimen)

If the value of (/)s is known then the amount of the phase in the specimen can be calculated. Unfortunately (/)s is not a constant as its value is dependant on the amounts of the entire constituent phases in the specimen. This problem can be solved by using internal standard method, which is one of the most widely applied techniques for quantitative XRD. Assuming the specimen contains phase (phase which need to be quantified) and phase (standard phase with known amount), the (/)s problem can be eliminated by simply dividing the intensity equation for both phases to yield:

X

Xk

hklI

hklI

(eq. 5)

k = calibration constant derived from the plot of I(hkl) / I(hkl) against X / X

Internal standard method is usually applicable in the determination of the composition of multi-phases specimen. This method can be extended to the quantification of physical mixture consisted of different crystallite solids. In the latter part of the experiment, the students need to determine the composition (in weight percent) of an alkali halide mixture by using the internal standard calibration method.

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Experiment Procedures The major steps in the acquisition of X-ray powder diffraction patterns are:

1. Sample preparation and mounting 2. Selection of suitable instrument parameters

X-ray: kV, mA and Divergence and receiving slits

3. Data collection Scanning range, step size, count time

4. Pattern processing Smoothing Stripping of K2 Peak search Peak correction Store/print the diffractogram

5. Interpretation (NOTE: X-ray radiations are VERY DANGEROUS and all students MUST OBEY the safety regulations listed inside the X-ray instrument room. NEVER attempt to operate the X-ray instrument unless you have been trained properly in handling the X-ray instrument.) Part A. Analysis of NaCl and KCl Sample preparation

1. Prepare samples of NaCl : KCl mixtures with various composition as according to Table 2. (NOTE: Sample A1 and A5 are pure alkali halide)

2. Mix the mixture in the beaker thoroughly by using a metal spatula. 3. Grind the mixtures by using mortar and pestle. 4. Acquire six XRD sample holders and Unknown A from the lab technicians. The

sample holder has a circular well for mounting the sample. 5. Transfer and mount the samples onto the XRD sample holders by using “press

and pull” sample loading method (NOTE: The sample is loaded into the circular well and pressed down with a glass slide. After that the glass slide is carefully pulled to the side. This method produces a mounted sample with an extremely flat and smooth surface).

6. Label the sample holder properly. Acquisition of X-ray powder diffraction patterns

1. Submit the samples to the lab technicians for powder XRD patterns acquisition (NOTE: Please do not attempt to operate the powder XRD instrument).

2. Take note of the important components of the powder XRD instrument. 3. For pure KCl (Sample A1) and NaCl (Sample A5), use the following parameters:

2 range : 20 to 80

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Step size : 0.05 For the rest of the sample (Sample A2, A3, and A4) and Unknown A, use the following parameters: 2 range : 20 to 35 Step size : 0.05

4. After the acquisition of X-ray diffraction patterns, process the patterns by using EVA software. Identify the 2 value and peak intensities for all the peaks.

5. Print out a copy of the XRD diffractograms. Students may duplicate the diffractograms by using the photocopy machine.

Part B. Analysis of KCl and KBr Sample preparation

1. Prepare samples of KCl : KBr mixtures with various composition as according to Table 2. (NOTE: Sample B1 and B5 are pure alkali halide)

2. Mix the mixture in the beaker thoroughly by using a metal spatula. 3. Grind the mixtures by using mortar and pestle. 4. Acquire six XRD sample holders and Unknown B from the lab technicians. The

sample holder has a circular well for mounting the sample. 5. Transfer and mount the samples onto the XRD sample holders by using “press

and pull” sample loading method (NOTE: The sample is loaded into the circular well and pressed down with a glass slide. After that the glass slide is carefully pulled to the side. This method produces a mounted sample with an extremely flat and smooth surface).

6. Label the sample holder properly. Acquisition of X-ray powder diffraction patterns

1. Submit the samples to the lab technicians for powder XRD patterns acquisition (NOTE: Please do not attempt to operate the powder XRD instrument).

2. Take note of the important components of the powder XRD instrument. 3. For pure KBr (Sample B1) and KCl (Sample B5), use the following parameters:

2 range : 20 to 80 Step size : 0.05 For the rest of the sample (Sample B2, B3, and B4) and Unknown B, use the following parameters: 2 range : 20 to 35 Step size : 0.05

4. After the acquisition of X-ray diffraction patterns, process the patterns by using EVA software. Identify the 2 value and peak intensities for all the peaks.

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5. Print out a copy of the XRD diffractograms. Students may duplicate the diffractograms by using the photocopy machine.

Table 2. Composition of alkali halides mixture

Part A. Mixtures of NaCl and KCl

Sample Weight (gram)

Weight %

(200) Peak intensity

(200) Peak intensity %

NaCl KCl NaCl NaCl KCl NaCl KCl

A1 0.0 2.0

A2 0.5 1.5

A3 1.0 1.0

A4 1.5 0.5

A5 2.0 0.0

Unknown A

Part B. Mixtures of KCl and KBr

Sample Weight (gram)

Weight %

(200) Peak intensity

(200) Peak intensity %

KCl KBr KBr KCl KBr KCl KBr

B1 0.0 2.0

B2 0.5 1.5

B3 1.0 1.0

B4 1.5 0.5

B5 2.0 0.0

Unknown B

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Data Analysis Qualitative analysis The structural parameters for a particular crystallite sample can be obtained by interpreting the X-ray diffraction patterns and such data interpretation can be performed conveniently by using the EVA software. However, in this experiment the students should not rely on EVA software in doing the interpretation. Interpretation of X-ray diffraction pattern usually starts from indexing of the significant peaks. Manual indexing of peaks can be performed by following these steps:

1. Identify all the significant peaks from the diffractogram of a pure alkali halide. 2. By using a spreadsheet program such as Excel, enter the positions of each peaks

and label the columns as “2-theta”. 3. Construct a second column that equals sin2. 4. Label the third column as “h2+k2+l2”. Construct this column by dividing the

values of sin2 with sin2 value of the first peak. If the Bravais lattice of the crystal is simple cubic, the result of this operation should give an integer series starting with one (Table 1). If the Bravais lattice is face-centered cubic, then this column need to be multiply by 3 before the integer series can be obtained.

5. Once the Bravais lattice for the alkali halide has been identified, create a fourth column that contains the expected integer values for that Bravais lattice (for example 1,2,3,4,5... for simple cubic). Label this column as “Integer”.

6. Create a plot of sin2 against the integer and obtain the best-fitted line for the plot. The slope of the plot is equals (2/4a2). Calculate the unit cell dimension, a.

7. Calculate the crystallographic density of the alkali halide. Example: The face-centered cubic cell contains four molecules. Therefore the crystallographic density = mass of four molecules / volume of one unit cell.

8. Complete Table 3 and compare the crystallographic density value obtained with the macroscopic density.

9. Sketch out the structure of the alkali halides. It is known that K+ has an ionic radii of 1.34 Å. Find out the ionic radii of the rest of the ions (Na+, Cl- and/or Br-) from the crystal structure, assuming that the cation-anion contacts determine the edge length of the ions.

Table 3. Structural parameters for alkali halides

Salt Molecular

weight (gmol-1)

Density (gcm-3)

Bravais lattice type

Unit cell dimension

(Å)

Crystallographic density (gcm-3)

NaCl 58.443 2.165

KCl 74.551 1.984

KBr 119.002 2.75

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Qualitative analysis

1. Identify all the significant peaks on the diffractograms for sample 2, 3 and 4 and Unknown A (or Unknown B) by making comparison with the diffractogram for pure alkali halides.

2. Complete Table 2 by recording the intensity of the (200) peak on the diffractograms for all the samples.

3. Calculate the peak intensity ratio and plot a calibration curve of peak intensity ratio against weight percent of NaCl (or KBr).

4. Determine the composition of the unknown sample by using the calibration curve. Questions

1. Explain why the (111) peak appears as a weak peak on the diffractograms of NaCl

or KBr, but “disappears” on the diffractogram of KCl. 2. Tungsten, also known as Wolfram, is a chemical element with very high melting

point. It has an atomic weight of 183.84 gmol-1 and a macroscopic density of 19.25 gcm-1. The following diffraction data of Tungsten is obtained with Cu K1 radiation. Determine the Bravais lattice and unit cell dimension of Tungsten by interpreting the diffraction data. It is known that the unit cell of Tungsten is a cubic system.

2 (degree) Relative intensity (%)

40.262 100

58.251 15

73.184 23

86.996 8

100.632 11

114.923 4 References [1] V. K. Pecharsky and P. Y. Zavalij, “Fundamentals of Powder Diffraction and Structural Characterization of Materials”, Kluwer Academic Publishers, 2003 [2] A. Clearfield, J. Reibenspies and N. Bhuvanesh (eds.), “Principles and Applications of Powder Diffraction”, Wiley, 2008 [3] R. Jenkins and R. L. Snyder, “Introduction to X-ray Powder Diffractometry”, John Wiley & Sons, Inc, 1996

1

5. Activated Dienophiles in Diels–Alder Cycloadditions

The most common (and synthetically most useful) Diels–Alder reactions involve electron-rich dienes and electron-deficient dienophiles.

Y

X

Y

X

Y = R, OR

X = CN, CHO, CO2H

The rate of these reactions generally increases with the donor ability of the diene substituent, Y, and with the p acceptor ability of the dienophile substituent, X. This behavior can be rationalized using Frontier Molecular Orbital (FMO) theory by examining the interaction of the diene’s highest-occupied molecular orbital (HOMO) and the dienophile’s lowest-unoccupied molecular orbital (LUMO).

OrbitalEnergy

HOMO

LUMO

diene dienophile

The HOMO–LUMO interaction lowers the reaction barrier by stabilizing the electrons in the diene HOMO, and FMO theory suggests that this effect will be largest when the HOMOdiene − LUMOdienophile energy difference is smallest. In other words, diene reactivity should be positively correlated with HOMO energy, and dienophile reactivity should be negatively correlated with LUMO energy.

Electrostatic interactions between the diene and dienophile may also play a significant role in determining the reaction rate.1 According to this view, rapid reactions occur when

1 S. D. Kahn, C. F. Pau, L. E. Overman and W. J. Hehre, J. Am. Chem. Soc., 108, 7381 (1986).

2

electrostatic interactions between the diene and dienophile are most attractive (least

repulsive). Thus, diene reactivity should increase as the electrostatic potential over the face of the diene becomes more negative, and dienophile reactivity should increase as the

electrostatic potential over the face of the dienophile is made more positive.

The FMO and electrostatic models can also be used to predict the regiochemistry of a cycloaddition reaction. According to the FMO model, the best HOMO–LUMO interaction for a given diene-dienophile pair results when there is good overlap between the interacting orbitals. Thus, the best orientation for two unsymmetrical reactants is usually the one that aligns the HOMO and LUMO so that the atomic orbitals that contribute most to each molecular orbital are able to overlap.2

The electrostatic potential model suggests that the best orientation is the one that brings together the region of most negative potential on the diene with the region of most positive potential on the dienophile.

In this experiment, you will explore to what extent the behaviour of a series of dienophiles (cyanoalkenes) can be correlated with either their frontier molecular orbitals and/or their electrostatic potentials as provided by ab initio calculations at HF/3-21G level.

Procedure

2 I. Fleming, Frontier Orbitals and Organic Chemical Reactions, Wiley, New York, 1976.

Experimental relative rates for Diels–Alder cycloadditions of cyclopentadiene and various dienophiles

3

1) Rates of Diels–Alder cycloadditions:

Build the dienophiles listed in the table below, and optimize their geometries using the HF/3-21G level of theory.

To build the structure:

1) Double click on the GaussView shortcut.

2) Click on “File” and choose “New” and choose “Create MolGroup” and a purple screen to build the molecule will appear (Figure A).

3) From the “Element fragment” , choose a sp2 carbon and build 2 together to form an ethene (Figure B).

4) Substitute a hydrogen with a cyano group from “R-Group Fragments” to form the acrylonitrile (Figure C)

5) Minimize the molecule using the “Clean” icon .

6) Undo any errors with Ctrl + Z. Save the molecule on desktop as .gjf file.

3 J. Sauer, H. Wiest and A. Mielert, Chem. Ber., 97, 3164 (1964).

dienophile 3relative rate (krel)

acrylnitrile 1

trans-1,2-dicyanoethylene 78

cis-1,2-dicyanoethylene 88

1,1-dicyanoethylene 4.4 × 104

tricyanoethylene 4.6 × 105

tetracyanoethylene 4.1 × 107

4

7) When acrylonitrile has been optimized, add cyano groups to the optimized structure to build the other dienophiles. (Optimized structure can be found by opening the log file in Guassview in the same location as the gjf file)

Mouse Control

Left click: Press and hold on to left click and drag to rotate

Centre click/wheel: Press and hold to translate/move the molecule

Right click: Press and hold; dragging up and down Zooms in and out of the molecule.

Dragging left and right rotates the molecule in the plane of the screen.

Figure A

5

Figure B

Figure C

6

To optimize a structure:

1) Go the “Calculate” drop down menu and choose the Gaussian option.

2) Under the “Job Type” drop-down menu choose “Optimization” (Figure D).

3) Under method, choose “Hartree-Fock” and use the 3-21G basis set. (Figure E).

4) Under “Link 0”, set the RAM as 1000Mb. ** Remember to have a different .chk file name for each dienophile as you will need the LUMO map for the later exercises.

5) Input the additional keywords “pop=npa” to calculate the charge.

6) Repeat steps 1 to 5 for all the 6 dienophiles

Figure D.

7

Figure E.

Record the HF/3-21G LUMO energy (in eV) of each dienophile, and plot these energies against the log of the experimental krel values. You can assess the LUMO for HOMO–LUMO band gap by looking at the log file for the first virtual eigenvalue (Alpha virt. eigenvalues).

Using MS word to open the log file, using the “Find” function (Ctrl + F), change the options from “More” to “up” instead of “down” and search “Alpha” from the last page to search for the last optimization eigenvalues (Figure F). Convert the values from hartrees to eV.

From log file in MS Word:

Alpha occ. eigenvalues -- -15.65673 -15.65503 -15.65385 -11.40083 -11.38117 Alpha occ. eigenvalues -- -11.34508 -11.34262 -11.34195 -1.31718 -1.30679 Alpha occ. eigenvalues -- -1.30193 -1.20515 -1.01572 -0.87084 -0.80044 Alpha occ. eigenvalues -- -0.71553 -0.62530 -0.60821 -0.60578 -0.59781 Alpha occ. eigenvalues -- -0.55117 -0.53845 -0.53102 -0.52840 -0.51820 Alpha occ. eigenvalues -- -0.42897 HOMO Alpha virt. eigenvalues -- -0.01183 LUMO 0.14692 0.17737 0.19951 Alpha virt. eigenvalues -- 0.22303 0.23290 0.28462 0.30118 0.33623 Alpha virt. eigenvalues -- 0.40943 0.45500 0.53610 0.56665 0.59386

Figure F

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The FMO model actually depends on two factors: the HOMO − LUMO energy difference and the degree of the HOMO–LUMO overlap. Is there a relationship between the value of the LUMO and the dienophile reactivity? If there is, is this relationship consistent with the FMO model?

Next, consider the electrostatic argument.

1) Open the log file (Figure G) in Gaussview by clicking on “File” and picking the “Open” option in the drop down menu. Choose “File of Type” as “Gaussian Output Files” and look for the log file in the same folder as the input file (.gjf file).

2) Go to “Results” and under the drop down window, choose “Charges” and “Show charge number” (Figure H).

3) Record the most positive charge found on the sp2 bond forming carbon. Plot this quantity against log(krel).

What does the charge describe about the electronic properties of each atom? Are the two correlated? If so, is the direction of the correlation consistent with the electrostatic character of the dienophiles?

Figure G

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Figure H

2) Regioselectivity in Diels-Alder cycloadditions:

You will use the reaction of 1-methylcyclopentadiene (MCP) with various unsymmetrical dienophiles (acrylonitrile, 1,1-dicyanoethylene, and tricyanoethylene) to examine FMO and electrostatic potential-based analyses of regioselectivity.

** (Regioselectivity is not equivalent to endo/exo selectivity)

10

Build 1-methylcyclopentadiene, optimize its HF/3-21G geometry.

1) In the “Ring Fragments” , choose cyclopentadiene and build. (Figure I)

2) Add a sp3 carbon to the H you want to substitute as the methyl using the “Element

Fragment”

3) Repeat optimization step 1 to 5 from Exercise 1 to optimize the structure.

Figure I

Display the HOMO of the diene and LUMO of the dienophiles using the checkpoint file.

1) Under “File”, choose “Open” in the drop down menu and set “File of Type” as “Gaussian Checkpoint Files”. The .chk files are located in the directory C:\G03W\Scratch.

2) Go to the “Results” menus again and choose “Surfaces”.

3) Under the “Cube Actions…” drop-down menus choose “New Cube”. Yet another dialog appears called the “Generate Cubes” dialog.

4) Under the “Orbitals:” drop-down menu, choose the “HOMO” or “LUMO” accordingly. (Figure J)

5) Go to the “Cubes Available:” drop down menu and choose the “HOMO” or “LUMO” accordingly.

6) Go to the “Surface Actions” and in the drop down menu and choose “New Surface”.

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Figure J

Which end of the diene is best suited for overlap with an unsymmetrical dienophile? Next, examine LUMO maps of the three unsymmetrical dienophiles. Which alkene carbon is best suited for overlap? What product regioisomers are predicted by the overlap? Do you arrive at the same predictions using resonance arguments? (Your answer should include drawings of the specific resonance structures that your predictions are based on.)

Next examine the NBO charges of the diene and the three dienophiles. Which sp2 carbon involved in the bond formation of the diene has the most negative potential? Which end of each dienophile has the most positive potential? What product regioisomers are predicted by the electrostatic model? Are the Molecular Orbitals and electrostatic predictions the same?

3) Lewis Acidity effects on Rate

Lewis acids such as BF3 generally increase both the rate and regioselectivity of Diels–Alder reactions. These effects might be due to enhanced and more selective frontier orbital interactions, or they might be due to changes in the nature of key electrostatic interactions. Examine the effect of a simple “Lewis” acid, H+, on the reaction of acrylonitrile and 1-methylcyclopentadiene. Lewis acids preferentially bind to the strongest Lewis base present, the nitrogen lone pair of acrylonitrile.

12

Therefore, build an N-protonated acrylonitrile cation, optimize its geometry using the HF/3-21G level of theory, and use the results to evaluate the reactivity and regioselectivity of this dienophile.

1) Use the “Add Valence” icon to add a hydrogen atom to the N of acrylonitrile.

2) Repeat optimization step 1 to 5 from Exercise 1 but CHANGE THE CHARGE TO 1.

Which factors appear to contribute to this dienophile's enhanced reactivity and selectivity?

4) Transition state and kinetics

The most direct way to evaluate Diels–Alder selectivity is to calculate energy barriers for reactions involving the dienophiles. Build 4 transition states for the 4 possible products of the reaction between 1-methylcyclopentadiene and acrylonitrile, using the “template” method. That is, add cyano groups to a simpler transition state “template” of cyclopentadiene and ethene:

Create the geometry of the transition state for constrained optimization

a. Open the optimized 1-methylcyclopentadiene.log and build an ethene inside the same screen as the MCP (Figure K)

b. Orientate the ethene towards the two bond forming carbons in the optimized MCP (Figure L). Press and hold on “Alt”, use the centre click to shift the molecules individually to the desired position of the pseudo C-C bond formation in the Diels Alder reaction.

c. Add single bonds to the bond forming carbons using the “Modify Bond” option and minimize the molecule using the “Clean” function (Figure M).

d. Set both the bond lengths to 2.19 Å (Figure M) using the Modify Bond option .

13

Figure K

Figure L

Figure M

14

e) Add the constraints to the two “forming” C-C bonds.

I. Click on the “Redundant Coordinate Editor” icon ( Figure N)

II. Click on the “Add new constraints” icon (Figure O).

III. Set the parameter (Bond) that you want to constrain under “Coordinates” and choose “Freeze coordinate” (Figure O). Choose the two atoms which constitute the bond you want to constrain.

IV. Go to “Set value” and set it as 2.19 A.

V. Repeat step II to IV for the other bond.

f) Optimize the constrained geometry by repeating optimization step 1 to 4 from Ex 1, EXCLUDING the step of inputting the additional keyword.

Constrained optimization

Figure N

Click on “R” icon to input constraints

15

Figure O

Use the optimized constraint structure to search for the HF/3-21G transition state.

1. Under Job type choose “Opt+Freq”. (Figure P)

2. Under Optimize choose “TS(Berny)”,

3. Under Calculate Force Constants, choose “Once” and under Compute Raman, choose “No”

4. Under method, input the usual Hartree-Fock /3-21g, 0 charge, singlet. (Figure Q)

5. Insert keyword “ opt=noeigentest “ and run the calculation

Add the cyano substituent at different position of the ethene in the obtained transition state to generate the inputs for the different stereoisomers. Repeat the above steps 1-5. . (DO NOT RUN CONSTRAINED OPTIMIZATION STEPS a to f AGAIN )

Operation specified Parameters chosen

Parameters to be altered Value to be constrained Add new

constraint

16

Figure P

Figure Q

17

Verify that these are transition states by their vibrational frequencies (a transition state has only one imaginary frequency). Use the HF/3-21G energy of each transition state along with those of the reactants to obtain the energy barriers for each cycloaddition. Compare the activation barriers and use Boltzmann distribution to identify the composition of the products (assuming the degeneracy is the same for all the stereoisomers at standard state, 298K/1 atm). Is the composition correlated with the regioselectivity predicted? What are the reasons for the different in stabilities of the transition states? EA = ETS – EReactants = ETS – EMCP – Edienophile

To extract the energy of the operation, open the .log file in MS words and go to the end of the file and find the line “\HF=”. The energy is in expressed in hartrees.

Sometimes even though the TS search did not terminate properly, both the vibration calculation and TS search are completed. Open the log file in Guassview and click on “Results” and choose “Vibration”. If the vibration frequencies are present means the job is completed. To extract the energy from the improperly terminated files, open the log file in MS words, go to the bottom of the file and search “#” upwards. It will bring you to the start of the frequency calculation. Above it is the termination of the TS search and you can look for the /HF there.