IR Spectroscopy Problem Set 3

7/29/2019 IR Spectroscopy Problem Set 3

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Name SOLUTIONS Ch 1b, Problem Set ThreeSection Side 1 of 16 Due Friday, Jan. 27, 2012Rec TA at 4 PM in the Drop Box

1. Vibrational Motion, Infrared (In)active Modes & Degeneracy (10 points)In lecture you saw different types of vibrational motions for a linear triatomic moleculeand a bent molecule. (To get a better feel for these molecular motions, you may want tolook at the animated .gifs for the six different vibrational modes for the -CH2 group at:

http://en.wikipedia.org/wiki/Infrared_spectroscopy#Theory.) When the stretching andbending motions were presented in class, we briefly discussed which vibrational motionswere infrared active and which were infrared inactive for each molecule.

Consider the normal modes for the vibrational motions of a planar molecule, GaI3.

(a) In the Figure above: are there too many, too few or just enough diagrams drawn tocapture all modes ofvibrational motion for this molecule? Briefly explain youranswer. (4 points)

(b)Identify which modes are infrared active and which are infrared inactive for GaI3.Briefly explain your answer. (6 points)

Solution for Problem 1aThere are just enough diagrams since the molecule has six vibrational degrees of freedom(4 points). Student does not need to show work.

Solution for Problem 1b

Only the first mode, 1, is infrared inactive; all other modes are infrared active. Themolecule has no permanent dipole moment and the symmetric stretch in 1 does notcause a dipole moment. All other motions, 2 - 6, produce a dipole moment.
http://en.wikipedia.org/wiki/Infrared_spectroscopy#Theoryhttp://en.wikipedia.org/wiki/Infrared_spectroscopy#Theoryhttp://en.wikipedia.org/wiki/Infrared_spectroscopy#Theory


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2. Molecular Electronic Spectroscopy (40 points)You perform experiments and measure the following spectroscopic constants for theground and first excited states of carbon monoxide, CO.Parameter (Units) Ground state (X

1+) First Excited State (A1)re () 1.1281 1.2351e (cm

-1) 2170.21 1515.61

exe (cm-1) 13.46 17.25

De (cm-1) 90 230 25 160

Data from: A. Ellis, M. Feher & T. Wright, Electronic and Photoelectron Spectroscopy:Fundamentals and Case Studies. Cambridge University Press (2005), p. 30.

For this problem refer to the following potential energy figure.

You should notice that the curves for potential energy are no longer parabolic, as theywould be for an ideal harmonic oscillator. To include anharmonicity we often model thespring (oscillator) potential using the Morse potential, which leads to this equation forvibrational energy levels,

( ) ( )22

1

2

1

,++= vxhvhE

eeevvib ,

or as a term value,

( ) ( ) ( )22121 ++= vxvvG eee ,with e expressed in wavenumbers. The symbols are defined as:

e: the oscillation frequency (the same as in OGC/OGN ore);xe: a dimensionless quantity, called the anharmonicity constant, which accounts fordeviations from an ideal harmonic oscillator;Other symbols commonly used:

re: the equilibrium internuclear distance (average bond length, same as reqin figure).De: the dissociation energy measured from the bottom of the potential well. Thedissociation energy is the energy required to dissociate a molecule into atoms with zerovelocity.


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2. Molecular Electronic Spectroscopy (continued)

(a)On a single plot, sketch potential energy versus internuclear distance for both theground and excited states. (You may use wavenumbers instead of energy for the

y-axis.) Label important features (re, De) with their values for CO. In addition,you should sketch the first six vibrational energy levels for each state. (10 points)

(b)In the electronic ground state at room temperature, which vibrational state is mostlikely to be occupied? Justify your answer. (5 points)

(c) In the electronic ground state, what is the spacing between lines in rotationalspectroscopy? Express your answer in wavenumbers. (5 points)

(d)Calculate the spacing between lines in rotational spectroscopy for the first excitedstate. Express your answer in wavenumbers. (5 points)

(e)Calculate the zero-point energy for the molecule in its excited state. Express youanswer in Joules. (5 points)

(f) By what percentage does the force constant increase when the moleculefluoresces from the first excited state back to its ground state? Report your answerwith three significant figures. (5 points)

(g)When we analyzed vibrational motion in class, we assumed that the force constant which is a measure of bond stiffness was proportional to the bond energy sowe could make approximations when we didnt have exact values available. Forthis problem, use the data in the Table to estimate the value of the dissociationenergy for the first excited state. How does it compare to the measured value?(5 points)


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Solution for Problem 2a

10 points total:1 point for general shape of curves (not just parabolas)1 point for labeling curves with words or symbol to denote the state

2 points for minimum of excited state slightly to right of ground state minimum

2 points for labeling equilibrium radii (1 point each)

2 points for labeling energy levels in some way1 point for vibrational energy levels about equally spaced in ground state

1 point for vibrational energy levels getting smaller in excited state

It is not important if the student uses energy or wavenumber for the y-axis. It is notimportant if the student puts the zero energy mark at the bottom of the well for theground state or at the dissociation level.


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Solution for Problem 2b

The student can either perform the complete calculation or use the approximations wediscussed in class. Ill only present the approximation here.

The Boltzmann distribution is,

=

Tk

E

g

g

n

n

B

exp . (2 points)

We know that kBT200 cm-1 at room temperature (1 point; from lecture notes) and that a

typical vibrational frequency is on the order of 1000 cm-1. (1 point) Since the ratio ofE/ kBTis large, the Boltzmann distribution predicts that the population ratio between theexcited state and the ground state is very small; so mostly just the ground state ispopulated at room temperatures. (1 point for final answer)

Give full credit to a student who refers to page 8 of Lecture #2 (or other reliable source not wikipedia) and quotes the text Vibrational energy levels: mostly just the ground stateis populated at room T.

Give no credit if the student just writes the statement/answer without justification/source.


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Solution for Problem 2c

The lines in rotational spectroscopy are separated by 2B,

I

hB 242 = ,

and the moment of inertia is 2RI = and )/(2121

mmmm +=

(2 points using the three equations above in some way)

2

21

21

2

1

42

Rmm

mmhB

+=

Since 2B is a frequency and we want to express this result in wavenumbers, divide bothsides by the speed of light,

2

21

21

2

1

4

~2

Rmm

mm

c

hB +=

.

(1 point for dividing by speed of light somewhere in problem to convert to wavenumbers)

( ) ( )( ) Jsmkg

cm

m

m

A

kgx

u

Auu

u

smx

sJxB

22

22

220

27282

34/

10

10

106605387.1)1281.1(

1

9949.150000.12

9949.27

/109979925.24

106260688.6~2

=

8640.3~

2 =B cm-1 (2 points total: 1.5 point answer; 0.5 point sig figs)(intermolecular distance had five sig figs and all constants should have more than that in

the above calculation)

Solution for Problem 2d

The internuclear distance changes but the reduced mass does not, which means you onlyhave to take a ratio of the internuclear distances to find the new value.(3 points for taking a ratio OR for substituting the new value into the equation from theprevious part OR any other logical way of getting an answer)

2235.3~

2 =B cm-1 (2 points total: 1.5 point answer; 0.5 point sig figs)

Side note: Why are there so many sig figs for an equilibrium internuclear distance? Ifyou remember, we usually estimate the internuclear distance from the rotational constants(not the other way around as we did in this problem). If you look at tables of rotationalconstants for molecules, youll even see numbers with more sig figs sometimes evennine or ten. No measurement can be made as precisely as frequency, so when utmostaccuracy is important, one should strive to measure the phenomenon of interest in termsof a frequency. (H.A. McGee, Jr., Molecular Engineering)


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Solution for Problem 2e

Starting with the equation for the anharmonic oscillator,

( ) ( )22

1

2

1

,++= vxhvhE

eeevvib .

The zero-point energy can be found from the v=0 state and the equation for vibrationalenergy levels,

eeezeropoxhhE

2

int2

1

2

1

= . (2 points)

In the Table the frequency is in wavenumbers; so we must multiply by the speed of light,

( )eeeeeezeropoxhcxhchcE 2

1

2

int

2

1

2

1

2

1=

=

(1 point for realizing that you have to multiply by the speed of light)

Solving for the zero-point energy,

( )( )m

cmsmxsJxE

100cm25.17

2

11515.61cm/109979925.2106260688.6

2

1 1-1-834

=

2010497.1 = xE J

(2 points total: 1.5 points answer; 0.5 points sig figs)NOTE: the value forexe only has four sig figs.


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Solution for Problem 2f

Start with the relationship between the frequency and the force constant,

k

21= . (2 points)

Solving for the force constant,224 =k .

We want to find the percent increase in the force constant after flourescence and use thefact that the reduced mass does not change during this transition,

1001004

44100

2

*

2

*

2

2

*

2

2

*

222

*

*xxx

k

kkgroundgroundground

=

=

(1 point: definition of percent increase)(1 point: recognizing reduced mass doesnt change)

( )( )

10061.1515

61.1515)21.2170(100

21

2121

*

*x

cm

cmcmx

k

kkground

=

%105100*

*=

x

k

kkground

The force constant increases by about 105%. (1 point: answer)


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Solution for Problem 2g

Well assume that the dissociation energy,De, is proportional to the force constant,

kDe~ .

Start again with the relationship between the frequency and the force constant,

k

2

1= .

Again, well solve for the force constant,224 =k .

We want to solve for the dissociation energy of the first excited state in terms of thedissociation energy for the ground state and the two frequencies. Again, the reduced

mass does not change during the transition.

22

2

*

2

,

,*

4

4

groundgrounde

e

D

D

= ,

2

*

,2

2

*

,,*

==

ground

grounde

ground

groundeeDDD

,

(2 points for getting to this equation or following similar steps)

( )

2

1

1

1

,* 21.2170

61.1515

90230

=

cm

cm

cmDe

1

,* 44007

= cmDe

(2 points answer: order of magnitude and first two sig figs are important)Turns out that were not even close. In class you saw that we were able to makeestimates of properties to within 10% in many cases, following similar solution schemes.This process tends to breakdown when the two atoms get very close (like in carbonmonoxide). The bond length for CO is about 1.13 , which is significantly smaller thanthe bond length for a C-C triple bond (1.20 ).(1 point: any comparison that demonstrates the student realizes the estimate isnt even

close)

3. 1H NMR: Number of Peak Groupings and Relative Areas (18 points)


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Use different letters of the Greek alphabet (, , , , ) to clearly mark groups ofchemically equivalent hydrogen atoms. From your analysis of hydrogen equivalence,estimate the relative peak areas for NMR spectra. Use the two molecules in the top rowas examples. (3 points each)

Molecule Relative Areas Molecule Relative Areas

() ()CH3OH

:3:1

() () () ()CH3CH2OCH2CH3

:6:4

CH3CH2CH2BrCH3a a

CH3CHCH2Cl

CH3CH2 CH2CH2CH3


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(Graders: Give full credit if students reduce the ratios. For instance, if they report 1:2

instead of 2:4 then dont subtract anything.)


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4. 1H NMR: Interpreting Chemical Shifts (4 points)

Assume the reference (=0.00 ppm) is comprised of hydrogen atoms on the methylgroups in the reference compound, tetramethylsilane (TMS). Explain why the chemicalshift increases ( values become larger) for this series of molecules.

Molecule (ppm)CH3I 2.10

CH3Br 2.70CH3Cl 3.05CH3F 4.30

Solution

Hydrogen atoms on carbon atoms with electronegative elements (e.g, oxygen, halogens)have less electron density around them and deshield the nucleus. As the degree ofshielding decreases (i.e., as the element has a greater deshielding effect), the chemicalshift increases.

The Table shows CH3X, where X is a series of halogens. Electronegativity increases asyou go up the column of halogens in the periodic table. (Remember that flourine has thehighest electronegativity, 4.0.) The more electronegative an atom is, the stronger itsaffinity for electrons and the stronger its deshielding effect on the hydrogen nucleus.

In summary, as the electronegativity of X increases in the CH3X series, X causes greaterdeshielding and a larger chemical shift.

1 point for recognizing halogen series (or that all atoms are in a single column of theperiodic table)1 point for recognizing effect of electronegativity on deshielding2 points for discussing deshielding and effect on chemical shift

5. 1H NMR: Interpreting Spectra (16 points)a. You are told that the proton NMR spectrum of an unknown material shows two groupsof peaks. The first group (a doublet) has six times the area of the second group (a septet).You are told that the molecular formula for the unknown material is C6H14O. Draw yourproposed structure of the unknown material. Briefly explain your answer. (7 points)

b. You do elemental analysis and find that the molecular formula of your unknownmaterial is C4H8O. The proton NMR spectrum is shown below. Draw your proposedstructure and briefly explain your answer. (9 points)


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spectrum source: Aldrich chemical.TMS is tetramethylsilane.

Red line is the total integrated area.

Solution for 5a

Since there are two groups of peaks, there are only two types of hydrogen atoms. (1point) Ill call them types and .Since the peak area is 6:1 and there are 14 total hydrogen atoms in the molecule, theremust be 12 hydrogen atoms of type and two hydrogen atoms of type . (2 points)

The first group (type ) is a doublet, which means that there is only one hydrogen atomattached to adjacent atom(s) that exhibit spin-spin coupling. The second group (type ) isa septet, which means that six hydrogen atoms are attached to the adjacent molecule(s)that exhibit spin-spin coupling. (2 points; NOTE: students dont need to use the termspin-spin coupling they may use splitting or discuss neighboring atoms.)

Diisopropyl ether satisfies these three criteria. (2 points for structure)


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Solution for 5b

Since there are three groups of peaks, there are three types of hydrogen atoms. Ill callthem types , , . (1 point)

From the red line you deduce that the relative peak areas are 2:3:3 for::, moving leftto right along the chemical shift axis (x-axis). (2 points)

Since there are a total of eight hydrogen atoms in the molecule, you can deduce from thepeak areas that there are two hydrogen atoms of type , three hydrogen atoms of type ,and three hydrogen atoms of type . (2 points)

The first group (type ) is a quartet, which means that three hydrogen atoms are attachedto adjacent molecule(s) that exhibit spin-spin coupling. The second group (type ) is asinglet, which means that no hydrogen atoms are attached to the adjacent molecule(s) thatexhibit spin-spin coupling. The third group (type ) is a triplet, which means that twohydrogen atoms are attached to adjacent molecule(s) that exhibit spin-spin coupling. (2points; NOTE: students dont need to use the term spin-spin coupling they may usesplitting or discuss neighboring atoms.)

Methyl ethyl ketone (aka, MEK, 2-butanone) satisifies these criteria.

(2 points for structure)


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6. IR/NMR Combination Analysis (12 points)This is a suite of combined IR/NMR spectroscopy problems. From the spectra andinformation below, derive the molecular structure. Explain your reasoning.a. The chemical formula of this molecule is C6H12O2, and in the NMR spectrum below

the peaks near $\delta$=3.9 and 0.8 are triplets, that near 1.9 is a singlet. The relativeintegrated areas from high to low chemical shift are 2, 3, 2, 2, 3. (6 points)

Solution to 6a: Butyl Acetate

H3C

O

O CH3 From the IR we can confirm CH3, C=O, and C-O stretches. The NMR shows a singlet at1.9 ppm from the carbonyl methyl group, a triplet at (ppm):3.9 due to the CH2 next tothe ester oxygen atom, and multiplets at decreasing values depending on the distancefrom the ester oxygen.

(3 points for structure and 3 for explanation)


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b. The two NMR peaks below are the only ones seen in this molecule with chemicalformula C5H8O. The integrated areas of the two NMR features are equal. (6 points)

Solution to 6b: CyclopentanoneO

From the IR we can identify the CH and C=O stretches as well as CH2 bends. Mostnotably, the IR shows only 2 mutiplets, indicating that there are only 2 sets of chemicallyequivalent protons. Given the formula, it should be easy to come up with the abovestructure.

(3 points for structure and 3 for explanation)

This is the end of Problem Set Three. Remember this set is due in the Ch1homework box near the mailboxes by Lloyd Hovse. If you are unsure where this

is located, ask an upperclassman to show you.

These sets will be collected five minutes after the due date and time. Be

prompt in turning in your set! As with all sets, PS3 is due at 4PM on

Friday.

IR Spectroscopy Problem Set 3

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