Download Lecture 7 - UIC Department of Chemistry

12/07/2012
Physical Chemistry Lab
Chem343
Lecture 7 (12/07/12)
 Class Schedule/Grading
 Final Review
Final Review
Final Exam & Grading Schedule
• Final Exam Schedule
Dec 13 (Thr) From 1 PM (2hours) at 130 SES; ~60 % is multiple choice questions ■ Your
Your TA will send the grade (Final & Overall) by the end of Dec 13 TA will send the grade (Final & Overall) by the end of Dec 13
by e‐mail. If you need any corrections, please claim it before Dec 14 at 3 pm. ■ We will return your graded notebook on the day of the Final. ■ We will not return the Final or Quiz 4, but let you review them in case that your grade is very close to the border line. ■ All other grades are final.
All th
d
fi l
■ Quiz 4 (Today): All the 6 experiments + lecture
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Final Exam
• 6 Experiments + Lecture materials
• 60‐70 multiple choice questions (60‐70%) • 6 calculation or long questions 2 hours
NMR • Review the class materials on NMR from the lectures (Lectures 2, 3)
• Review the NMR presentation • Principle of NMR (Spin magnetic moment, Transition energy, Frequency)
• 13C signal assignments, 1H signal assignments
• DEPT
• HETCOR & COSY
• The principle of molecular diffusion • Long questions are probably on signal assignments for amino acid(s) or small compounds. id( )
ll
d
• For Long questions, I may ask you to draw a possible NMR spectrum for a compound. Check out the structures of amino acids and possible chemical shifts. 2
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Zeeman Interaction Spin in a Magnetic Field
• When a spin is placed in a magnetic field, the energy is given by
Em = ‐ μZB0 = ‐γmħB
ħ 0
1
Figure 1. Zeeman energy levels for spins of I =1/2 and 1.
Lower energy level attracts more spin population.
 It induces spin polarization (bulk magnetic moment)
4C ~ H
NMR Frequency
• The transition energy between the two states is
|E| = γħB0.
• NMR
NMR frequency is easily obtained as
f
i
il bt i d
0 = E/h = B[Q1]
γ/2π.
0
1
4C ~ H
The frequency is typically in a range of 1‐900 MHz  RF (Radio Frequency)
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12/07/2012
Induction Effects in 1H shifts
1H chemical shifts of CH Cl (ppm) n 4‐n
CH4
0 23
0.23 CH3 Cl 3.05
CH2Cl2
5.3
CHCl3 7.27
1H
e-
CH3 Cl
Less e-,
Less Shielded
shifts of CH3X (ppm)
F
OH
NH2
H
Me3Si
4.26
4.26
3.38
2.47
0.23
0.0
TMS
Example 13C and 1H NMR spectrum spectra of diethyl ether (CH3CH2)2O
(CH
(CH33CH
CH22))22OO
Q. Which is C1H2?
Quartet
Triplet
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Miller plane & index
Q. What is the Miller index below?
b
b
a
a
(1,2)
(2,1)
Gas Effusion
Theory Questions
• What is an ideal gas?
– Negligible volume compared to gas line
– Elastic Collisions = Kinetic Energy remains constant
• What are the differences between effusion and diffusion?
– Pressure
– Mean free path
• What is the main factor which affects the mean free path?
– Pressure Pressure
• Factors that influence average speed / probability density(M‐B Distribution)
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Experimental Questions
• Why is the diffusion pump important?
– Mean free path
Mean free path
– Back Effusion
• What type of gas can be monitored by gas effusion?
– Non toxic and non flammable gases (No cold trap on the pump)
p p)
– Molecule size smaller than pinhole
• Constants
– Area, Molar Mass, Gas Constant, and Volume
Calculation Questions
• Calculate the increase/decrease in speed if the t
temperature or molar mass t
l
increases/decreases
• Any calculations from the lab
– Appropiate Equations will be given
Appropiate Equations will be given
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CHEM 343_F12_FTIR questions for final_examples
Q1: Consider the following analytes and answer which would absorb in IR frequency range
(Y/N):
A: Phenol
B: Water
C: CH2Cl2
D: CO
E: O2
a.C and E
b.Only E
c.A, B and D
d.None of the above
Q2: Consider the following analytes and answer which would show a FINE ROTATIONAL
STRUCTURE in its FTIR absorption spectra (assume analyzed molecules are polar):
A: Solution of Cholesterol in HCCl3/CH2Cl2
B: Pulverized solid PVC paletted with KBr
C: Air sample containing gaseous contaminants
D: Aqueous peptide solution
E: gaseous pyrolisis products
a.A and B
b.Only D
c.C and E
d.All of the above
CHEM 343_F12_FTIR questions for final_examples
Q3: Consider following statements related to FTIR spectroscopy and choose an answer that applies:
A: Heavier the atoms in a diatomic molecule are, smaller the vibrational wavenumber is
B: Energy of rotational levels is usually ~ 103 times smaller than that of the corresponding vibrational
levels
C: Value of rotational constant is dependent on the vibrational energy level
D: According to anharmonic oscillator potential energy curve it takes much more energy to move
atoms closer together that it takes to move them apart
a) A is the only one correct
b) All are correct but C
c) All are correct
Q4: Considering a model of a rigid rotor choose an answer that best describes given symbol or term:
A)
DJ 2 ( J  1) 2
B) Term for moment of Inertia
a) Term for transition energies in P-branch
P branch
b) Term for vibrational-rotational interaction constant
c) Term expressing contribution of centrifugal distortion
T  T ( v, J ) final  T ( v, J ) initial
a)
b) I=µr2
c) F ( J )  BJ ( J  1)
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12/07/2012
CHEM 343_F12_FTIR questions for final_examples
Q5: FTIR absorption spectrum for DCl with labeled R and P branches with initial state J values is shown
below. Which statement(s) is/are correct:
A
a) In case A: Jinitial is equal 1
b) In case B: Jfinal equals 0
c) In case B peak due to D37Cl is observed
d) In case A peak due to D35Cl is observed
e) a and b
f) c and d
g) none of the above
B
p
of 12C16O shows vibrational absorption
p
band centered at 2170cm-1. At what
Q6: The IR spectrum
wavenumber would the corresponding peak for 14C16O occur ?
Use: Ar(12C)=12; Ar(14C)=14.0032; Ar(16O)=15.995

M1  M 2
M1  M 2
~0 ,12 CO  

~0 ,14 CO  
1/ 2



14
CO 
12
CO
Final Exam Preparation On Nano Diffraction
1. [Theory]: Explain the Bragg equation and deduce the question based on the used model for 2D and 3D dimensions.
2. [Sample]: Lattice system of the crystal in the experiment and parameter comparison of each sample(d or D). 3. [Experiment set up]:Wavelength of the laser; dos and don’ts for better data. 4. [Diffraction pattern]: Difference and reasons for multiple domain and single domain. 5. [Calculation]: How to calculate the d spacing based on the known factors. Establish relationship between D and d based on each structure
Establish relationship between D and d based on each structure. 6. [Miller indices]: Simple models. 8
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Examples:
Photophysics and Molecular Spectroscopy Review
Brian K. Yoo
December 7, 2012
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Jablonski Diagram
http://elchem.kaist.ac.kr/vt/chem‐ed/quantum/graphics/jablon.gif
• What is the basis for the observed Stoke’s shift?
• Why do we expect to see a mirror images for the Wh d
tt
i
i
f th
excitation and emission profiles?
• Know numerical calculations, basis and meaning of variables, i.e. Beer‐Lambert Law, Henderson‐
Hasselbalch…
• What are the differences between fluorescence and UV, vis‐spectroscopy (absorbance)?
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Error Propagation Example
V RT

, T  0.00  0.08 C, P  1.000  0.005 atm
P
n

T(C)
T(K)  
+ 273.15K

 C
0.00

T(K)  
+ 273.15K
 C

T(K)  273.15
273 15  0.08
0 08 K
Vm 
L  atm
 273.15 K
K  mol
Vm 
1.000 atm
Vm  22.41 L  mol-1
0.08205746
Error Propagation cont.
2
2
2
2
2
V 
V 
V 
V 
V 
 V2   m   R2   m   T2   m   V2   m   T2   m   V2
 R T ,V
 T R,V
V R ,T
 T V
V T
m
2
R 
V 
2
 RT  2
  V
 V 2 
 V2     T2  
m
 V2
m
2

L  atm 
0.08205746 K  mol 
2
 
 0.08 K 
1.000
atm





2
L  atm

273.15
K
0.08205746


2
K  mol

 0.005 atm
2
1.000 atm




 V  0.1 L  mol-1
m
Vm  22.4  0.1 L  mol-1
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12/07/2012
Flash Photolysis of
Benzophenone
final presentation
Chemistry 343
Fall 2012
THEORY
Reactive
electrones
 The total electronic spin
angular momentum is 2S+1
S for S0 is 0 (1/2-1/2) = singlet
 Flouroscence X
 Phosphoroscence: Liquid N2 T
RT
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12/07/2012
REACTION
- Isopropanol (take H)
- Protonated ketyl radicals
Benzoph.
Benzoph
+
Isoprop.
- Basic
- Deprotonated ketyl radicals
- Dimerization
Benzopinacol
NaOH
N
OH
+
H2O
Fixed wavelength
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12/07/2012
CALCULATION
• Kobs =slope*el (e:5000 /M.cm, l: 5 cm)
• Kobs = k7
k x [H] / k6
• [H] = 10-ph
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