10. Other Spectroscopies 1.IR 2.NMR. Set equal VIBRATIONAL CHROMOPHORES Any bond can act as a spring which can be described as the balance Between the

10. Other Spectroscopies1.IR2.NMR

F ma

F ky

Set equal

VIBRATIONAL CHROMOPHORES

Any bond can act as a spring which can be described as the balanceBetween the force due to acceleration and to the restoring force of the spring

ky m a ad y

d t

2

2

ky m

d y

d t

2

2

Need a function which reinvents itself

d a t

d ta t

d a t

d t

co ssin

d a t

d ta a t

co ssin

d a t

d t

d a a t

d ta a t

d a t

d t

2

2

co s sinco s

d a t

d ta a t

2

22

co sco s

ky m

d y

d t

2

2

d a t

d ta a t

2

22

co sco s d y

d t

ky

m

2

2

y a t co s ak

m2

a is a frequency

a m 2

2mk

m

m

k

m

1

2

The frequency of the oscillation isRelated to the bond constant, k,And m, which is a mass term

Natural mechanical frequency

mm m

m m

1 2

1 2

m

k

m

1

2

Bond k(Newton/m)Single 500Double 1000Triple 1500

N

m

kg m

s m

kg

s

2 2

Example: Calculate double bond band forR-C=O

mg

mole

m ole

x a tom s

kg

gx kgc

1 2 1

6 1 0 1 02 1 0

2 3 32 6

mg

mole

m ole

x a tom s

kg

gx kgO

1 6 1

6 1 0 1 02 7 1 0

2 3 32 6.

m

x kg x kg

x kg x kgx kg

2 1 0 2 7 1 0

2 1 0 2 7 1 01 1 1 0

2 6 2 6

2 6 2 6

2 6.

..

m

kg

sx kg

x

sx

s

1

2

1 0 0 0

1 1 1 0

9 0 9 1 0

6 2 84 8 1 0

12

2 6

2 81 3

.

.

..

m

kg

sx kg

x

sx

s

1

2

1 0 0 0

1 1 1 0

9 0 9 1 0

6 2 84 8 1 0

12

2 6

2 81 3

.

.

..

For IR usually reported as reciprocal cm

c

c

1

14 8 1 0

1

3 1 01 6 0 0

1 3

1 0

1

cm

xs

xcm

s

cm .

,

The average reported value is 1690-1760

1600 cm-1 What is the predicted frequency of C-O?

m

k

m

1

2


Example: Calculate single bond band forR-C-O

m

kg

sx kg

xs

1

2

5 0 0

1 1 1 03 3 9 3 1 0

12

2 61 3

..

c

c

1

13 3 9 3 1 0

1

3 1 01 1 3 1

1 3

1 0

1

cm

xs

xcm

s

cm .

,

1600 cm-1 1100 cm-1

What wouldBe this frequency?

m

k

m

1

2


Example: Calculate single bond band forR-N-Pb

m

kg

sx kg

xs

1

2

5 0 0

2 1 9 1 02 4 0 4 1 0

12

2 61 3

..

c

c

1

12 4 0 4 1 0

1

3 1 08 0 1 6

1 3

1 0

1

cm

xs

xcm

s

cm .

.

mg

mole

m ole

x a tom s

kg

gx kgN

1 4 1

6 1 0 1 02 3 3 1 0

2 3 32 6.

mg

mole

m ole

x a tom s

kg

gx kgPb

2 0 7 2 1

6 1 0 1 03 4 5 1 0

2 3 32 5.

.

m

x kg x kg

x kg x kgx kg

2 3 3 1 0 3 4 5 1 0

2 3 3 1 0 3 4 5 1 02 1 9 1 0

2 6 2 5

2 6 2 5

2 6. .

. ..

1600 cm-1 1100 cm-1

800 cm-1

Bonds cm-1

R-N-Pb 800R-C-O 1100R-C=O 1600

R-CO=O ?

Our calculations

Bonds cm-1

R-N-Pb 800R-C-O 1100R-C=O 1600

R-CO=O ?

0

10

20

30

40

50

60

70

80

90

100

05001000150020002500300035004000

cm-1

no

rmal

ized

%T

Lead EDTA spectra 2009 data

Pb-N

Them Us 2009strength strength Their assignments

cm-1 breadth cm-1 breadth350 m,b Pb-O400 m,b 409 vw,sh Pb-N stretch450 w,sh 432

509 m,sh555 m,sh

595 w,b640 m,sh

710 w,b 721 m,sh800 w,b 806 m,sh845 w,b918 vs 914 m,sh C-C acetate stretch975 w,b 957 w

1020 w 1040 m1085 w 1090 w C-N stretch1140 m,b 1160 m 1200 w 1230 m CH2 wag in CH2COOH1265 m,b1280 vw,sh1335 s 1320 w1406 vs 1380 s,sh carbonyl1435 m,sh carbonyl1450 m,sh 1490 s,sh carbonyl1600 m,b 1550 s,b carbonyl

23602841 2850 vs,sh C-H in HCOO-2936 s 2930 vs,b C-H in CH2,COO-

We once tried looking for the Pb-N Changes but this region is too far into the“fingerprint region” to get good data

420

430

440

450

460

470

480

0 0.5 1 1.5 2 2.5 3

q/r

cm-1

(M

-N b

on

d)

Pb(1.32)

Cu(0.71)

Higher charge density = higher frequency

0

20

40

60

80

100

120

050010001500200025003000350040004500

cm-1

%T

2009 Data, 0, 0.25, 0.5, 0.75, and 1 fraction Pb EDTA

Pb-N bond

CO=O, 1750cm-1

Pure EDTA, Na

What would influence the degree of the frequency shift of CO=O?

Hint – who would have a stronger attraction for the electrons on O?

424, cm-1

Bonds cm-1

R-N-Pb 800R-C-O 1100R-C=O 1600

R-CO=O ?

More energy

Greater electrostatic attraction

Shift in

C=O fr

equen

cy

Covalent type bonding (H) at COLeaves double bond character atCO=O

Ionic bonding at C-O;Pulls electrons fromC=O, creating bond1.5 order

1750cm-1

lead

Higherenergy

m

k

m

1

2


mm m

m m

1 2

1 2

Lower reduced mass has higher frequency

Cm-1 Bonds

3600-3200 -O-H

3500-3300 -N-H

3100-3010 R=CH-H R=C-H

3100-2970 -C-H

2280-2210 -C=N

2260-2100 -C=C

1760-1690 -C=O

1680-1500 -C=C

1360-1180 -C-N

1300-1050 -C-O

GroupFrequencyRegion

What do you observe?

m

k

m

1

2


mm m

m m

1 2

1 2

Cm-1 Bond

3600-2700 X-H

2700-1900 X=Y

1900-1500 X=Y

1500-500 X-Y

“fingerprint” region

CH3

CH3CH3

CH3

OH

CH3CH3

CH3H

CH3CH3

CH3

Which is which?Can you identify all of them?

http://orgchem.colorado.edu/hndbksupport/specttutor/irchart.html

Where is the expected vibrational absorption band forR-C=O?

We predicted a band at 1600 cm-1 for R-C=O

Is this the only band we should observe?

IR bands Observed

1. Degrees of freedom set total possible bands2. Requires a change in dipole moment3. Requires a high molar absorptivity4. Requires good instrumental “window”5. Complicated by passing of overtones

Degrees of freedom

Bands observed = total DF – (translation + rotational)3N - ( 3 + (2 or 3))

Possible BandsLinear molecule: 3N-5Non-linear: 3N-6

EXAMPLEHow many bands should we observe for O=C=O?

Linear molecule: 3N-5 = 9-5=4Each possible band

must have a change in dipole momentshould not be degenerate

1. O=C=O symmetric stretch

2. O=C=O asymmetric stretch

3. O=C=O scissoring

4. O=C=O scissoring

Which have a change in dipoleMoment?

Which are (if any) are degenerate?

EXAMPLEHow many bands should we observe for O=C=O?

Linear molecule: 3N-5 = 9-5=4Each possible band


1. O=C=O symmetric stretch

2. O=C=O asymmetric stretch

3. O=C=O scissoring

4. O=C=O scissoring

No dipole moment change

Degenerate – only one is observed

Total of two bands!

http://www.succeedingwithscience.com/labmouse/chemistry_a2/2906.php?LabmouseOnline=191f6382b179f1a28fb4da8c34523817

Animation associated with spectra:

EXAMPLEHow many bands should we observe for H-O-H?

Non-Linear molecule: 3N-6 = 9-6=3Each possible band


H

O

H

H

O

H

H

O

H

Symmetric stretch

Asymmetric stretch

scissoring

http://www.lsbu.ac.uk/water/vibrat.html

Animation of water vibrations

http://www.lsbu.ac.uk/water/vibrat.html

Fused silica Si-O

http://www.laseroptik.de/?Substrates:Transmission_Curves:%26%23150%3B_IR-FS

Typical region of IR interest!

Can not use the quartz materials we were using in UV-Vis

Quartz vibrational bands overlap C-H bonds

Some times we observe more bands than predicted.Why?

1st order 2nd order 3RD ORDER 4th orderSource light cm-1 cm-1 cm-1 cm-1

100 200 300 400200 400 600 800300 600 900 1200400 800 1200 1600800 1600 2400 3200

Our calculationsR-N-Pb 800R-C-O 1100R-C=O 1600

fn nector

m irrorm irrord et

1 2 1 12

FT instrument

Light that passes at 100 cm-1 will also pass at 200, 300, and 400 cm-1 That means that it could excite absorption at 400 cm-1!

Our calculated band at 800 cm-1 could be observed at 200 4th order; 400 2nd order and 800 first order!

This is one reason that the “fingerprint” region is difficult to interpret

INSTRUMENTATIONGenericSource not = UV-VisCell not = UV-VisMonochromatorDetector not = UV-VisSpatial Arrangment Components not = UV-VisReadoutSolvents not = UV-Vis

IR SOURCES:Nernst Glower (Blackbody)Tungsten-filament bulb (household)Hg arcTunable CO2 laser

IR DETECTORSWhat is the main problem in using a UV-Vis detector here?

IR DETECTORS

1. ThermalWhat kind of problem can youAnticipate here?

a. Thermocouple

6 to 8 mV/uW

b. Bolometer or thermister

Change in resistance on T change

Bi/Sb/BiVoltage difference

2. Pyroelectric1. Move charge with temperature

= capacitor= capacitor current with T

NH CH COO3 2

NH CH COO3 2

3. Photoconductivity low bandgap semiconductor

LUMO

HOMO+

e

DESIGN OF INSTRUMENT

1. Double Beam vs Single Beam?2. To Chop or not to Chop?3. Whither the sample?4. To FT or not to FT?

An old (not FT instrument): Describe what you see

UV-Vis:Source - Monochrom – Sample – Detector

Place the sample after the monochromator to decrease theTotal power on the sample. Energy of individual frequenciesIs large enough to remove an electron (break bonds) andDecompose the sample

IRSource – Sample – Monochrom – Detector

Sample not decomposed. Placing the monochromator After the sample can help prevent scattered radiationFrom entering the detector.

To FT or not to FT?

fve locity

cectorm irrow

d e t

2

Mirror velocity in a typical instrument is 0.01 to 10 cm/s

Do we FT the instrument or scan through each individual wavelength?Why?

http://www.wooster.edu/chemistry/analytical/ftir/ftir.swf

Shows how the FT is generated by the movable mirror

Describe this instrument in terms ofVarious functional parts

1. Optical resolution2. Controls mirror3. Eliminates Phase Shift

Why should eliminating phaseShift be so important?

Instrument Range Resolution Cost TimeSingle Beam 1.3 to 29 4 16k 1sFT 0.4 to 1000 8 to 0.01 150k >1min

SAMPLE HANDLING

1. GasesEffusion of volatile liquid through a pin hole

2. Solution: need a cell that minimizes IR activity of background

Background? solvents (H2O) Air (H2O, CO2) windows (Si-O band in quartz)

One way to minimize solvent effects is to minimize the total amount of solvent

Use a very thin cell (small value of b)

How could we then determine the value of b?

-2.5

-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

2.5

0 0.5 1 1.5 2 2.5

Film Thickness

Am

plit

ud

e

-2.5

-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

2.5

0 0.5 1 1.5 2 2.5

Film Thickness

Am

plit

ud

e

-2.5

-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

2.5

0 0.5 1 1.5 2 2.5

Film Thickness

Am

plit

ud

e

-2.5

-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

2.5

0 0.5 1 1.5 2 2.5

Film Thickness

Am

plit

ud

e

-2.5

-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

2.5

0 0.5 1 1.5 2 2.5

Film Thickness

Am

plit

ud

eInterference Fringes

2b N film

2b N film

N N N b1 21 2

21 1

1

2 film

N

b

1

21

1

film

N

b

1

22

2

film

N

b

Example

50070090011001300150017001900

cm-1

Peak at N1900 11400 2900 3

Count the peaks between two known frequencies

3 1 2 1 9 0 01

9 0 01

b

cm cmb cm

1

1 0 0 0

50070090011001300150017001900

cm-1

N N N b1 2 1 22 Try this one

SOLID SAMPLES

Crush sample into a solid matrix (could also use as the “window”

What kind of solid material would you chose and why?

HINT

Bond force constantSingle 500 N/mDouble 1000N/mTriple 1500N/mIonic ????

Use an ionic solid to avoid backgroundIn the region of interesting covalent bonds

KBr One difficulty here is the factThat the windows are nowWater soluble

Any other considerations important?

Hint consider the scattered light equation

m

k

m

1

2


Example: Calculate single bond band forR-C-F;

m

kg

sx kg

xs

1

2

5 0 0

1 2 2 1 03 2 2 1 0

12

2 61 3

..

c

c

1

13 3 9 3 1 0

1

3 1 01 0 8 0

1 3

1 0

1

cm

xs

xcm

s

cm .

mg

mole

m ole

x a tom s

kg

gx kgc

1 2 1

6 1 0 1 02 1 0

2 3 32 6

mmole

m ole

x a tom s

kg

gx kgF

1 9 1

6 1 0 1 03 1 1 0

2 3 32 6.

m

x kg x kg

x kg x kgx kg

2 1 0 3 1 1 0

2 1 0 3 1 1 01 2 2 1 0

2 6 2 6

2 6 2 6

2 6.

..

The C-F bond is shifted out of theRegion of the C-H bond.

3100-3010 cm-1

In your lab you used PTFE for your IR “plate”

http://www.internationalcrystal.net/polycard.htm

Polytetrafluorethylene

0

10

20

30

40

50

60

70

80

90

100

200700120017002200270032003700

cm-1

no

rmal

ized

%T

"blank PTFE"

1210 1150

11001030

3100 (C-H)

2009 student data

1080 predicted C-F

0

20

40

60

80

100

120

400600800100012001400160018002000

cm-1

%T

Pretty hard to pull Understandable changesOut of here

Why is the “far” IR hard instrumentally?

Three primary reasons1. Overlapping orders of light create a “forest” of bands2. Intensity of the light source is low so we are trying to measure changes in small

signals3. The beamsplitters used in the FT instrument do not transmit both forms of

polarized light equivalently4. The detectors must be sensitive to very low amounts of light – implies that any

environmental noise will be a problem.

http://infrared.phy.bnl.gov/pdf/homes/homes_01ins.pdf

Here is an article detailing problems and solutions to creating Far IR instruments

Mov

ing

mirr

or

IR sourceBeamsplitter

Fixed mirror

B

CA

detector

Constructive interference occurs when

AC BC n 1

2

fcnector

m irror ligh td et

2

Frequency of light

From “really basic optics”

AC BC n 1

2

n/2 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5overtonesin cm-1

cm-1 um

100 100 200 100 66.66667 50 40 33.33333 28.57143 25 22.22222 20125 80 250 125 83.33333 62.5 50 41.66667 35.71429 31.25 27.77778 25150 66.66667 300 150 100 75 60 50 42.85714 37.5 33.33333 30175 57.14286 350 175 116.6667 87.5 70 58.33333 50 43.75 38.88889 35200 50 400 200 133.3333 100 80 66.66667 57.14286 50 44.44444 40225 44.44444 450 225 150 112.5 90 75 64.28571 56.25 50 45250 40 500 250 166.6667 125 100 83.33333 71.42857 62.5 55.55556 50275 36.36364 550 275 183.3333 137.5 110 91.66667 78.57143 68.75 61.11111 55300 33.33333 600 300 200 150 120 100 85.71429 75 66.66667 60325 30.76923 650 325 216.6667 162.5 130 108.3333 92.85714 81.25 72.22222 65350 28.57143 700 350 233.3333 175 140 116.6667 100 87.5 77.77778 70375 26.66667 750 375 250 187.5 150 125 107.1429 93.75 83.33333 75400 25 800 400 266.6667 200 160 133.3333 114.2857 100 88.88889 80425 23.52941 850 425 283.3333 212.5 170 141.6667 121.4286 106.25 94.44444 85450 22.22222 900 450 300 225 180 150 128.5714 112.5 100 90475 21.05263 950 475 316.6667 237.5 190 158.3333 135.7143 118.75 105.5556 95500 20 1000 500 333.3333 250 200 166.6667 142.8571 125 111.1111 100525 19.04762 1050 525 350 262.5 210 175 150 131.25 116.6667 105550 18.18182 1100 550 366.6667 275 220 183.3333 157.1429 137.5 122.2222 110575 17.3913 1150 575 383.3333 287.5 230 191.6667 164.2857 143.75 127.7778 115600 16.66667 1200 600 400 300 240 200 171.4286 150 133.3333 120625 16 1250 625 416.6667 312.5 250 208.3333 178.5714 156.25 138.8889 125650 15.38462 1300 650 433.3333 325 260 216.6667 185.7143 162.5 144.4444 130675 14.81481 1350 675 450 337.5 270 225 192.8571 168.75 150 135700 14.28571 1400 700 466.6667 350 280 233.3333 200 175 155.5556 140725 13.7931 1450 725 483.3333 362.5 290 241.6667 207.1429 181.25 161.1111 145750 13.33333 1500 750 500 375 300 250 214.2857 187.5 166.6667 150775 12.90323 1550 775 516.6667 387.5 310 258.3333 221.4286 193.75 172.2222 155800 12.5 1600 800 533.3333 400 320 266.6667 228.5714 200 177.7778 160825 12.12121 1650 825 550 412.5 330 275 235.7143 206.25 183.3333 165850 11.76471 1700 850 566.6667 425 340 283.3333 242.8571 212.5 188.8889 170875 11.42857 1750 875 583.3333 437.5 350 291.6667 250 218.75 194.4444 175900 11.11111 1800 900 600 450 360 300 257.1429 225 200 180925 10.81081 1850 925 616.6667 462.5 370 308.3333 264.2857 231.25 205.5556 185950 10.52632 1900 950 633.3333 475 380 316.6667 271.4286 237.5 211.1111 190975 10.25641 1950 975 650 487.5 390 325 278.5714 243.75 216.6667 195

1000 10 2000 1000 666.6667 500 400 333.3333 285.7143 250 222.2222 2001025 9.756098 2050 1025 683.3333 512.5 410 341.6667 292.8571 256.25 227.7778 2051050 9.52381 2100 1050 700 525 420 350 300 262.5 233.3333 2101075 9.302326 2150 1075 716.6667 537.5 430 358.3333 307.1429 268.75 238.8889 2151100 9.090909 2200 1100 733.3333 550 440 366.6667 314.2857 275 244.4444 2201125 8.888889 2250 1125 750 562.5 450 375 321.4286 281.25 250 2251150 8.695652 2300 1150 766.6667 575 460 383.3333 328.5714 287.5 255.5556 2301175 8.510638 2350 1175 783.3333 587.5 470 391.6667 335.7143 293.75 261.1111 2351200 8.333333 2400 1200 800 600 480 400 342.8571 300 266.6667 2401225 8.163265 2450 1225 816.6667 612.5 490 408.3333 350 306.25 272.2222 2451250 8 2500 1250 833.3333 625 500 416.6667 357.1429 312.5 277.7778 2501275 7.843137 2550 1275 850 637.5 510 425 364.2857 318.75 283.3333 2551300 7.692308 2600 1300 866.6667 650 520 433.3333 371.4286 325 288.8889 2601325 7.54717 2650 1325 883.3333 662.5 530 441.6667 378.5714 331.25 294.4444 2651350 7.407407 2700 1350 900 675 540 450 385.7143 337.5 300 2701375 7.272727 2750 1375 916.6667 687.5 550 458.3333 392.8571 343.75 305.5556 2751400 7.142857 2800 1400 933.3333 700 560 466.6667 400 350 311.1111 2801425 7.017544 2850 1425 950 712.5 570 475 407.1429 356.25 316.6667 2851450 6.896552 2900 1450 966.6667 725 580 483.3333 414.2857 362.5 322.2222 2901475 6.779661 2950 1475 983.3333 737.5 590 491.6667 421.4286 368.75 327.7778 2951500 6.666667 3000 1500 1000 750 600 500 428.5714 375 333.3333 3001525 6.557377 3050 1525 1016.667 762.5 610 508.3333 435.7143 381.25 338.8889 3051550 6.451613 3100 1550 1033.333 775 620 516.6667 442.8571 387.5 344.4444 3101575 6.349206 3150 1575 1050 787.5 630 525 450 393.75 350 3151600 6.25 3200 1600 1066.667 800 640 533.3333 457.1429 400 355.5556 3201625 6.153846 3250 1625 1083.333 812.5 650 541.6667 464.2857 406.25 361.1111 3251650 6.060606 3300 1650 1100 825 660 550 471.4286 412.5 366.6667 3301675 5.970149 3350 1675 1116.667 837.5 670 558.3333 478.5714 418.75 372.2222 3351700 5.882353 3400 1700 1133.333 850 680 566.6667 485.7143 425 377.7778 3401725 5.797101 3450 1725 1150 862.5 690 575 492.8571 431.25 383.3333 3451750 5.714286 3500 1750 1166.667 875 700 583.3333 500 437.5 388.8889 3501775 5.633803 3550 1775 1183.333 887.5 710 591.6667 507.1429 443.75 394.4444 3551800 5.555556 3600 1800 1200 900 720 600 514.2857 450 400 3601825 5.479452 3650 1825 1216.667 912.5 730 608.3333 521.4286 456.25 405.5556 3651850 5.405405 3700 1850 1233.333 925 740 616.6667 528.5714 462.5 411.1111 370

Light at these frequencies Will appear at 400 cm-1

cm-1

200400800140018002000


Three primary reasons1. Overlapping orders of light create a “forest” of bands2. Intensity of the light source is low so we are trying to measure changes in

small signals3. The beamsplitters used in the FT instrument do not transmit both forms of



0 500 1000 1500 2000 2500 3000 3500

nm

Inte

nsi

ty

I T 4

Wien’s Law

m ax b

T

Stefan-Boltzmann Law

8

1

3

3

h

ch

kTex p

= Energy density of radiationh= Planck’s constantC= speed of lightk= Boltzmann constantT=Temperature in Kelvin= frequency

1 8

13

hhc

kTex p

1. As ↓(until effect of exp takes over)2. As T,exp↓,

Planck’s Blackbody Law

From: “Really Basic Optics”






T

R

Angle of transmittenceIs controlled byThe density of Polarizable electronsIn the media asDescribed by Snell’s Law

The intensity of light (including it’s component polarization) reflected as compared to transmitted (refracted) can be described by the Fresnel Equations

parallel

perpendicular

Perpendicular will transmitted to a greater degree than parallel at the interface since it oscillates into the second medium

interface

Medium 1

Medium 2

s

p TE

TM


T ti i

i i t t

2

22

co s

co s co s

T ti i

i t t i/ / / /

co s

co s co s

2

22

R ri i t t

i i t t

2

2

co s co s

co s co s

R rt i i t

i i t t/ / / /

co s co s

co s co s

2

2

The amount of light reflected depends upon the Refractive indices and the angle of incidence.

We can get Rid of the angle of transmittence using Snell’s Law

sin

sin

i

t

t

i

Since the total amount of light needs to remain constant we also know that

R T

R T/ / / /

1

1Therefore, given the two refractiveIndices and the angle of incidence canCalculate everythingFrom “really basic optics”

Consider and air/glass interface

Here the transmitted parallel light isZero! – this is how we can selectFor polarized light!

This is referred to as the polarizationangle

i

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0 10 20 30 40 50

Angle of incidence

Tra

nsm

itta

nce

Parallel

Perpendicular

“Take-home message” can selectFor different polarization of light byControlling the angel of incidence


Notice that as you changeThe angle of the beam splitterYou can have very largeAsymmetry in the light Reflected which leads to “false absorbance”

http://infrared.phy.bnl.gov/pdf/homes/homes_01ins.pdf




polarized light equivalently4. The detectors must be sensitive to very low amounts of light – implies

that any environmental noise will be a problem.

www.rsc.org/ej/PC/2000/b001200i/

Where does the noise come from?

Pyroelectric detector frequency Response is

S igna l V H z

V op tica l ve locitycm

s

cmrad ia tion

0 3 1 6

1

.

S igna lcm

sH z

0 3 1 6.

400cm-110 cm-1 Range for Far IR

S igna lcm

s cmH z H z

0 3 1 6 4 0 01

1 2 5.`

~

S igna lcm

s cmH z H z

0 3 1 6 1 01

3.`

~

Room noise from the electric lines is 60 Hz which smack dab in the middleSo we will get a lot of room noise also

8. Other Spectroscopies1.IR2.NMR

http://www.cem.msu.edu/~reusch/VirtualText/Spectrpy/nmr/nmr1.htm

http://www-keeler.ch.cam.ac.uk/lectures/understanding/chapter_5.pdf

Excellent sites for information on NMR can be found within the Analytical SciencesDigital Library: www.ASDLib.orgBelow are some great ones found there



http://www.asdlib.org/

+spin

Causes aLocal magnet

External Magnetic Field, Bo, causes the energy associated with the spin to split

Nuclear Magnetic Resonance

http://www.cit.gu.edu.au/~s55086/qucomp/qubit.html

Orange= magnetic moment

Green equals externalMagnetic field

NMR active species

magnetic moment in magnetons (5.05078x10-27 J/T)1H 2.792719F 2.623331P 1.130513C 0.7022

I = spin quantum number ½ is active

Odd mass nuclei have fractional spins1H, 13C, 19F I = ½17O I = 5/2

Even mass nuclei with odd numbers of protons and neutrons have integral spins2H, 14N I = 1

Even mass nuclei with even number protons and neutrons have zero spin12C, 16O I = 0

NM

R a

ctiv

e

What are the spin numbers of the four isotopes ofLead shown below?

Mass Average % Relative Abundance I?204 1.36206 25.4207 22.7208 52.1

Calculate the nuclear magnetic moment for Pb207

% spin 5.05078e-27 J/T 10e7 rad/(Ts)Natl Abund I nuclear Magnetons magnetogyric ratio

1H 99.9844 0.50 2.7927 26.7532H 0.0156 1.00 0.8574 4,10711B 81.17 1.50 2.688 --13C 1.108 0.50 0.7022 6,72817O 0.037 2.50 -1.893 -3,62819F 100 0.50 2.6273 25,17929Si 4.7 0.50 -0.5555 -5,31931P 100 0.50 1.1305 10,840207Pb 22.7 6

What didWe get FromPrecedingSlide?

Example 1H

magnetonsx

J

Tm agnetons

rad

cycle

x J s Ix

rad

T s I

5 0 5 0 7 8 1 02

6 6 2 6 1 04 8 1 0

2 7

3 4

7

.

..

2

hI

4 8 1 0

2 7 9 2 7

0 52 6 7 1 0 2 6 7

1 07 87

..

.. .x

rad

T s

m agnetonsx

rad

T s

rad

T s

Bo

Eh

B o

2

1

2B o

c c

B o

2

E h

v c

All describe the splittingOf nuclear states equally Well BUT…. One is moreconvenient

c

xm

s

frequencys

3 1 0

1

8

Larmorfrequency

m=-1/2β m=+1/2

m=-1/2β

m=+1/2

1

2B o

c c

B o

2

Splitting of Orbitals as a function of Applied magnetic field

Because there is a linear relationship between the applied magneticField and the splitting frequency is the preferred unit

0

10

20

30

40

50

60

70

80

0 1 2 3 4 5 6 7 8 9 10

Bo, Tesla

1H W

avel

eng

th m

0.00E+00

5.00E+07

1.00E+08

1.50E+08

2.00E+08

2.50E+08

3.00E+08

3.50E+08

4.00E+08

4.50E+08

1H F

req

uen

cy H

z

FMradio

tv

-250

-200

-150

-100

-50

0

50

100

150

200

250

0 1 2 3 4 5 6 7 8 9 10

Bo, Tesla

Fre

qu

ency

, MH

z

1H

1H

13C

13C

Hold the Magnetic field constant and scan the frequency while monitoring Absorption …….Where will we find the various resonances?

300 MHz

300 MHz

Signal

m=+1/2

m=-1/2β

1

2B o

The designation of an instrument as 300, 200, or 100 MHz NMR is Referring to the natural resonance of the proton under the applicationOf magnets varying between 7.05 to 2.35 Tesla

As an exercise in preparation for an exam you should fill in the table.

Where will we find 207Pb resonances?

Change table

Tesla7.05 4.7 2.35

% spin 5.05078e-27 J/T 10e7 rad/(Ts) resonant proton frequencyNatl Abund I nuclear Magnetons magnetogyric ratio 300 MHz 200 MHz 100 MHz

1H 99.9844 0.50 2.7927 26.753 3002H 0.0156 1.00 0.8574 4,107 46.111B 81.17 1.50 2.688 --13C 1.108 0.50 0.7022 6,728 75.517O 0.037 2.50 -1.893 -3,62819F 100 0.50 2.6273 25,179 28329Si 4.7 0.50 -0.5555 -5,31931P 100 0.50 1.1305 10,840 122207Pb 22.7 6

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

0 50 100 150 200 250 300 350

Frequency, (in MHz)

Rel

ativ

e S

ign

al

300 MHzDifferent atoms have different magnetogyricRatios and therefore absorb at different frequencies

1H

31P

19F

Why are the sensitivities so low?

100 MHz

Frequency Relative Sensitivity

1H 300 12H 46.1 1.45E-06

13C 75.5 1.76E-0419F 283 8.30E-0131P 122 6.63E-02

MHz

0

0.0001

0.0002

0.0003

0.0004

0.0005

0.0006

0.0007

0.0008

0.0009

0.001

0 50 100 150 200 250 300 350

Frequency (in MHz)

Rel

ativ

e S

ign

al

13C

Argument on the size of signals that follows is from Atkins, Phys. Chem. p. 459, 6th Ed

Photons can stimulateEmission just as muchAs they can stimulateAbsorption(idea behind LASERsStimulated Emission)

*

o

Stimulated Emission

The rate of stimulated event is described by :

w B N *

Is the energy density of radiation already present at the frequency of the transition

B= empirical constant known as the Einstein coefficient for stimulated absorption or emission

N* and Noare the populations of upper state and lower states

Where w =rate of stimulated emission or absorption

w B N o

8

1

3

3

h

ch

kTex p

can be described by the Planck equation for black body radiation at some T

If the populations of * and o are the same the net absorption is zero as a photon isAbsorbed and one is emitted

In order to measure absorption it is required that the Rate of stimulated absorption is greater than the Rate of stimulated emission

w B N w B Nabsorp tion o enussion *

N No *

Scale the difference to the total population to get relative signal one can expectTo get:

S igna lN N

N

N N

N No

to ta l

o

o

* *

*

Boltzman distribution can help us describe this

N

N o

E

kt

h

kT*

ex p ex p

S igna lN N

N N

o o

h

kT

o o

h

kT

h

kT

h

kT

ex p

ex p

ex p

ex p

1

1 v c

S igna l

hc

kT

hc

kT

1

1

ex p

ex p

In a UV-Vis experiment, 400 nm, at room temperature (298 K)

h=6.626x10-34 Jsk= 1.381x10-23 J/Kc=3x108m/s

N

N

x J s xm

s

xJ

KK x m*

ex p

.

.

0

6 6 2 6 1 0 3 1 0

1 3 8 1 1 0 2 9 8 4 0 0 1 0

3 4 8

2 3 9

N

N o

*ex p . 1 2 0 7

S igna l

hc

kT

hc

kT

1

1

ex p

ex p

S igna l

1

11

1 2 0 7

1 2 0 7

ex p

ex p

.

.

For an NMR experiment the exponential term is larger

Signal relative to the concentrationIs 1

N

N o

E

kt

h

kT

h B

kT

o

*ex p ex p ex p

2

N

N o

x J s xT s

x KJK

*ex p ex p

. . .

. .

6 6 2 6 1 0 2 6 7 1 01

4 7

1 3 8 1 1 0 2 9 8 0 0 0 0 2 0 2

3 4 8

2 3

For a proton at 4.7 Tesla, magnetogyric ratio of 2.67x108rad/(Ts), at Room Temp, 298 K

h=6.626x10-34 Jsk= 1.381x10-23 J/Kc=3x108m/s

For an NMR experiment

S igna l x

1

1

1 0 9 9 9 8 0 0 2

1 0 9 9 9 8 0 0 26 6 6 1 0

0 0 0 2

0 0 0 25ex p

ex p

.

..

.

.

NMR signal relative to the concentration is 6.66x10-5 Relatively WEAK!

In order to get a signal that is measureable a relaxation experiment is performed

heat

SmallEnergyGapMeansHighN*population

Monitordecay

Consider, for example, a temperature jump

As T (hc/kT)↓ so exp so N* soSignal based on decay

N

N o

E

kt

h

kT*

ex p ex p

-250

-200

-150

-100

-50

0

50

100

150

200

250

0 1 2 3 4 5 6 7 8 9 10

Bo, Tesla

Fre

qu

ency

, MH

z

1H

1H

13C

13C

We can accomplish a “temperature jump” by altering the appliedExternal field

Magnetic Pulse populatesA range of energy states (frequencies)

Signal will be the sum of all the various protons that respond to those differentfrequencies

which changes the energy difference between the Ground and excited states and alters the populations of the two

E A B sin ( ) s in

Consider two frequencies summed but not too far apart – a “beat” frequencyWill result

-2.5

-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

2.5

0 100 200 300 400 500 600 700 800

time

amp

litu

de

E A B A B 21

2

1

2sin ( co s( )

-2.5

-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

2.5

0 100 200 300 400 500 600 700 800

time

amp

litu

de

This is our signal up toThis point

But~remember it will decay!

In many types of experiments this is done by performing a temperature jumpWhich creates a greater population of the excited state: consider the reaction

M Mk

k

f T

bT

,

, *1

1

At Temp1

d M

dtk M k MfT bT 1 1 *

At equilibrium

d M

dt 0

Therefore

k M k MfT equ ilib rium bT equ ilib rium1 1 *

Consider the decay involved in a temp jump experiment

Increase the temperature instantaneously, the temp 1 equilibrium concentrationsMust change to the new temp 2 equilibrium values with a rate driven by the newRate constants

M x M

and

M x M

equ ilib rium T equ ilib rium T

equ ilib rium T equ ilib rium T

, ,

, ,* *

1 2

1 2

The old equilibrium values deviate from the new equilibrium by some valuex

k M k MfT equ ilib rium T bT equ ilib rium T2 2 2 2, ,*

The concentration of M therefore changes as:

d M

dtk x M k x MT equ ilib rium T bT equ ilib rium T 2 2 2 2, ,*

k M k MfT equ ilib rium bT equ ilib rium1 1 *

recall

d M

dtk x k M k x k MT T equ ilib rium T bT bT equ ilib rium T 2 2 2 2 2 2, ,*

so

d M

dtk x k x x k kT bT fT bT 2 2 2 2

First order reaction with solution of

x x o

t

ex p

1

k kf b

Represents the decayIn the signal afterA jump

http://www.cit.gu.edu.au/~s55086/qucomp/qubit.html

Our “decay” represents a flip in the magnetic moment.

How can we easily represent a flip?

Describe as the sum of two vectors which cancelIn the xy plane and sum value in the z plane

Define an xy plane

Flip of the magnetic moment is described asThe motion of two vectors

Longitudinal Relaxation, T1

Related to motion of moleculesWhich usually is not very interesting to us

M Mz t eq

t

T( ) ex p

1

Motion of molecules

Transverse Relaxation, T2

In addition there is the “fanning outOf the spins” in the xy plane

M y t

t

T( ) ex p

2

T2 < = T1

Note that this imageSuggests out of phaseSignals implying phaseManipulation of the acquired data

What is this component of the signal due to?

Due to the fact that the local magneticField is not entirely in the z direction dueTo contributions from neighboring nuclei

+spin


Local magneticFields associated withElectrons alter the magnitudeOf the external field experienced

1

2B o

Spin/spinCouplingLifetime ofT2

+spin


CouplingDescribed byConstant, J, Which is a measureOf the energyOf the interaction

What we have “learned” so far:

1. Pushed sample into an excited state which will emit a radiofrequencyresponse on return to equilibrium

signa l S in A ( )

Radiofrequency

2. Spin spin coupling can also alter the energy levels as described by a coupling constant, J.

ex p

t

T2

4. The net signal in the time domain is described by:

S in At

T( ) ex p

2

3. Spin spin coupling results in fixed lifetime of the response, leading to anexponential decay of the signal

-1.5

-1

-0.5

0

0.5

1

1.5

0 50 100 150 200 250 300 350 400

t (s)

Am

plit

ud

e

Free Induction Decay (FID) due to T2

S in At

T( ) ex p

2

Sample frequency

Decay of signal

We learned (a long time ago) that a square wave can be modeled as The sum of a series of sine waves which can be “uncovered” by an FT

F f f t d tjft

ex p 2

FID= Free Induction Decay

The longer the signal lastsThe larger T2

Digitally filter the high frequency

subtractFT

-15000

-10000

-5000

0

5000

10000

15000

0 0.2 0.4 0.6 0.8 1 1.2

Time (s)

Am

plit

ud

e

Anti-FT

An exponential decay can also be described by the sum of a series of sineWaves.:

0

20

40

60

80

100

120

1 4 7 10 13 16 19 22 25 28 31 34 37 40 43 46 49 52 55 58 61 64 67

frequencies

Am

pli

tud

e

0

0.2

0.4

0.6

0.8

1

1.2

0 200 400 600 800 1000 1200

Time

Am

pli

tud

e

ex pt

T2

-1.5

-1

-0.5

0

0.5

1

1.5

0 50 100 150 200 250 300 350 400

t (s)

Am

plit

ud

e

Free Induction Decay (FID) due to T2

S in At

T( ) ex p

2

Sample frequency

Decay of signal

0

10

20

30

40

50

60

1 4 7 10 13 16 19 22 25 28 31 34 37 40 43 46 49 52 55 58 61 64 67 70 73

Frequences

Am

pli

tud

eFT of S in A

t

T( ) ex p

2

This distribution is fit by a Lorentzian population

Why did we “all of a sudden” get some frequenciesOn the left side of the peak?

FT

NMR peak shapes are NOT described Gaussian, but Lorentzian distribution

x

b

tan x

b

tan 1 x

b

dx

b

dx

b

1

12

2

dbdx

b x

2 2

Probability ofLength x as a functionOf b and theta

The cumulative probability over all angles is

P

b

b x mx( )

12 2

Where m is the peak location and bIs the half width at half height

y

a

c x b

2

Another form is

In this form a is the halfPeak width and b is theLocation of the peak

For those interested: how a Lorentzian population is derived mathematically

y

a

c x b

2

This can be converted to a quadratic

1 22 2

y

x

a

bx

ac

b

a

In this form a is the half Peak width b is the Location of the peak

Lorentzian Peak shape The signal width at half height is related to this lifetime

aT

1 2

2

1/

Calc. the transverse relaxation time if a peak has a half height width of 10 Hz

Where c is a/H and h is the height of the peak

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

-5 -4 -3 -2 -1 0 1 2 3 4 5

Gaussian

Lorentzian

PG

x b

1

2

2

12

ex p

P

a

x ba

L

1 2

22

2

Lorentzian is sharper

What have we learned so far that affects instrumentation?

1. External magnetic field2. Signal depends on exact magnetic field so it will need to be uniform3. Need to pulse the magnetic field (width of pulse is related to number of frequencies)4. Signal will involve several frequencies (beats) due to chemical information5. Need to use FT to find those frequencies.6. Signal will decay (causes a Lorentzian shape)7. Decay will have a phase component so need to worry about phasing of the signal8. Want to be able to capture the peak at half height because it gives chemical

Information.

Typical Experiment:0. Insert sample and lock1. Shim (tune) the magnetic field to be constant within the sample cell Starting population ratio set by starting applied magnetic field; temperature2. Jump the population (Pulse applied magnetic field) by applying a square

wave through the coil. 3. Watch the magnetization decay: cause small electric currents in a coil

surrounding the sample.4. Currents are time based and are the sum of all the frequencies

Set a) acquisition time and b) sampling frequency 5. Repeat as desired to increase the S/N, allowing a delay time to equilibrate6. Convert from time to frequency by FTing the measured electric signal7. Display the signals as a function of frequency.8. Adjust for the phase of the T2 signal9. Scale the frequencies relative to some standard compound.

http://images.google.com/imgres?imgurl=http://nmr.chem.ualberta.ca/nmr_news/figures/283_8360.JPG&imgrefurl=http://nmr.chem.ualberta.ca/AOVNMR_course/chapter_5.htm&h=1338&w=460&sz=132&hl=en&start=42&tbnid=-8mLZIhy1rkvZM:&tbnh=150&tbnw=52&prev=/images%3Fq%3Dvarian%2BNMR%2Bscreen%2Bimage%26start%3D40%26gbv%3D2%26ndsp%3D20%26hl%3Den%26sa%3D

Insert SampleMonitor 2H signal as a way of checking for magnetic field driftHere you attempt to get a nice large signal for 2H (46.1 MHz)


Locking

Monitor thisSignal as a checkOn the stabilityOf the homogeneityOf the magnetic field

Typical Experiment:0. Insert sample and lock1. Shim (tune) the magnetic field to be constant within the sample cell Starting population ratio set by starting applied magnetic field; temperature2. Jump the population (Pulse applied magnetic field) by applying a square




http://bouman.chem.georgetown.edu/nmr/nuts/shims.htm

Shimming involves tuning the magnetic field across the sample so that it is Homogeneous – in effect you are “tuning” the x axis of the experiment

Shimming

ManipulatesThe magneticField inMultiple dimensionsTo achieveconsistency


One way to ensure that your sample has as homogeneous a magneticField as you can get is to:

rmn.iqfr.csic.es/guide/man/bsms/chap5.2.htm

Spin the sample

Link worked in 2008, source of material, but no longer works

Typical Experiment:1. Shim (tune) the magnetic field to be constant within the sample cell Starting population ratio set by starting applied magnetic field; temperature2. Jump the population (Pulse applied magnetic field) by applying a square




Things YOU control:a. Shimmingb. Pulse width

-15000

-10000

-5000

0

5000

10000

15000

0 0.2 0.4 0.6 0.8 1 1.2

Time (s)

Am

plit

ud

e

The pulse width will control the range of frequenciesThat can be sampled.It will have to change with respect to different1proton resonance frequencies

-250

-200

-150

-100

-50

0

50

100

150

200

250

0 1 2 3 4 5 6 7 8 9 10

Bo, Tesla

Fre

qu

ency

, MH

z

1H

1H

13C

13C

Example of Pulse Width

Assume 500 MHz NMR. The proton frequency range at this field strength is 5000 to 7000 Hz. Will a pulse width of 8 uz sample that range?

8s pulse

-15000

-10000

-5000

0

5000

10000

15000

0 0.2 0.4 0.6 0.8 1 1.2

Time (s)

Am

pli

tud

e

y t ft ft ftsquare ( ) s in sin sin . . . . .

42

1

36

1

51 0

TYPICALLY ASSUME MINIMUM 5 FREQUENCIES

fcyc les

s pu lsebase

1

2

At a minimum we should get the 5th frequency above this soThat the bandwidth would be

fpu lse

5

2

For our example this is:f

x x sH z

5

2 8 1 03 1 2 5 0 0

6

Number of points is important for 2 reasonsaliasing (Nyquist frequency minimum)and 2^x for FT processing





Aliasing

-1.5

-1

-0.5

0

0.5

1

1.5

0 200 400 600 800 1000 1200

Sampling of a sine wave at too low number of points results in somePsychedelic types of patterns, need at least three points to describe a sine wave

-1.5

-1

-0.5

0

0.5

1

1.5

0 200 400 600 800 1000 1200

-1.5

-1

-0.5

0

0.5

1

1.5

0 20 40 60 80 100 120 140 160 180 200

D ig ita l so lu tionacqu isition tim e a t

xsw eep w id th sw

number o fda ta po s npR e

in t

1 2

number of data points (np): 4KFourier number (fn): 4Kacquisition time: 0.35 seczero-filling: none



D ig ita l so lu tionacqu isition tim e a t

xsw eep w id th sw

number o fda ta po s npR e

in t

1 2


number of data points (np): 24KFourier number (fn): 32Kacquisition time: 2.00 seczero-filling: 1.33 x

number of data points (np): 24KFourier number (fn): 64Kacquisition time: 2.00 seczero-filling: 2.66 x

http://www-keeler.ch.cam.ac.uk/lectures/understanding/chapter_5.pdfThis link is still good, 2009

This chapter describes the mathematics of the instrumentIn terms of “getting” the signal.

delay

pulseacquire





http://www.chemistry.nmsu.edu/Instrumentation/NMSU_NMR300_J.html


This chapter describes the mathematics of the instrumentIn terms of “getting” the signal.

delay

pulseacquire





Free Induction Decay of an NMR signal in a 1D experiment with a 50 msec expansion showing the digitization of the signal.


A 90 degree pulse followed by turning on the detector leads to a Free Induction Decay (FID)

1.91.81.71.61.51.41.31.21.11.00.90.80.70.60.50.40.30.20.1 sec

13C NMR FID of Methyl -D-Arabinofuranoside in CD3CN. Collected at 11.7 T by Jim Rocca in AMRIS.

(J) and altered frequency

The signal decays

Different decay rates can be seen

http://edison.mbi.ufl.edu/bch6745/lecture3.pdf

0.200.180.160.140.120.100.080.060.040.02 sec

Expansion of previous FID

Now we can see complicated fine structure

Most of the signal in the fast relaxing peak is gone by around 100 ms.

Peak to peak spacing is just under 50 ms, contains J spin spin energy coupling


The longer T2* the greater the resolution(narrower the peaks)

Fourier Transform of previous FID

120 110 100 90 80 70 60 50 40 30 20 10 ppm

The FID had several different frequencies. These are called chemical shifts.

Some peaks are bigger than others (CD3CN)

This peak has other stuff going on.


0.200.180.160.140.120.100.080.060.040.02 sec

120 110 100 90 80 70 60 50 40 30 20 10 ppm

Expansion of big peaks

The value for the J coupling is the inverse of the spacing between beats on the FID!

The width of this peak is about 10 Hz. This is R2* (or 1/T2*). The * indicates that it is the natural linewidth + experimental sources of inhomogeneities.

Scalar (J) coupling. This J is just over 20 Hz.

The width of these peaks is around 1 Hz. They come from the slowly relaxing part of the FID.


What is effect of phase shifting on FT?

Typically measure only oneOf the phases ORConvert both to a single phase





-30

-20

-10

0

10

20

30

40

50

1 5 9 13 17 21 25 29 33 37 41 45 49 53 57 61 65 69 73 77 81 85 89 93

Frequency

Am

pli

tud

e-30

-20

-10

0

10

20

30

40

50

1 5 9 13 17 21 25 29 33 37 41 45 49 53 57 61 65 69 73 77 81 85 89 93

Frequency

Am

pli

tud

e

-1

-0.8

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

1.2

0 50 100 150 200 250 300 350 400

Time

Sig

nal

(si

n)

-30

-20

-10

0

10

20

30

40

50

1 5 9 13 17 21 25 29 33 37 41 45 49 53 57 61 65 69 73 77 81 85 89 93

Frequency

Am

pli

tud

e

StartingFT dataPoint 90o

Starting FT data point (0 o)

Starting FT data point (180o)





+spin


External Magnetic Field, Bo, causes the energy associated with the spin to split

Nuclear Magnetic Resonance

Local magneticFields associated withElectrons alter the magnitudeOf the external field experienced

1

2B o

1

21B o

ShieldingEffect fromLocal electrons

-400

-300

-200

-100

0

100

200

300

400

0 2 4 6 8 10 12 14 16

Bo, Tesla

Fre

qu

ency

(M

Hz)

0

0

0 3.

0 3.

0 8.

0 8.

1

21B o

The change in the absorption frequencyGives information about the chemical environment

The change is compared toA standard (TMS)

CH3

Si CH3CH3

CH3

1

21

1

21

1

2

B B

B

o TM S o Sam ple

o Sam ple TM S \

-400

-300

-200

-100

0

100

200

300

400

0 2 4 6 8 10 12 14 16

Bo, Tesla

Fre

qu

ency

(M

Hz)

0

0

0 3.

0 3.

0 8.

0 8.

The magnitude of the shielding effectDepends directly on the External AppliedMagnetic Field

To be able to compare betweenInstruments the data is normalizedBy the external magnetic field resonantFrequency for a non shielded electron

1

2B o Sam ple TM S \

1

2

1

2

B

B

o Sam ple TM S

o

Sam ple TM S

\


These normalized differencesAre typically 10-6 in magnitudeSo the numbers are multiplied by106 to avoid working with small numbers

Sam ple TM S or o ther re ference 1 0 6

Chemical shift

C-H O-H

Less electronDensityLess shielding

More shielding

Aryl protons experienceLarger local magneticField rather than lowerSo it is seen as lessshielding

Chemical Shifts

Chemical shifts are influenced by the electronic environment. Therefore, they are diagnostic for particular types of molecular structures. The following figure indicates average ranges of chemical shifts for different types of molecules.

Table from: http://www.cem.msu.edu/~reusch/OrgPage/nmr.htm


http://www.chem.uic.edu/web1/ocol/spec/HTable.htm

Here is nice correlation chart from the web

Link still works 2009

Try Problem 1:

http://www.chem.uic.edu/web1/ocol/spec/NMR.htm


http://www.chem.uic.edu/web1/ocol/spec/NMR3.htm

For more problems see

CH3-CH2-OH

# of Peaks Observed in Proton NMR

2 1N I peaks #Proton I= 1/2

N peaks 1 #

# of atoms

2+1 Split TheseProtonsInto 3 peaks

3+1

Split theseProtons into4 peaks

Split these protonsInto 2

1+1

2x4=8

Expected peak heights (if not split) 3: 2:1

3 chemical environmentsProportion of protons (3:2:1) is correct

Splitting pattern observed:3: 4: 1

Expected splitting (by order of peakSize) was expected to be 3:(4x2):3

4 instead of 8 tells us what?

Keep thinking – we will comeBack to this

CH3-CH2-OH

Try Problems 2&3Correction to 3:C2H4O


N

N

O

O-

O O-

O

O-

O

O-

2x2=4

4x2=8

How many peaks for EDTA?

pH

11.3

10.3

9.3

8.3

7,3

6.3

5.3

Starting at pH 11.3 and moving more acidic ,protonation of N canBe seen in the shift of peak location, pKa

Values for NH3 around 9.2

What happened to the peaks outlinedIn red here?

How many peaks for Pure EDTA?

2 with relative intensity of 8:4 or2:1

N

N

O

O-

O O-

O

O-

O

O-

pH

11.3

10.3

9.3

8.3

7,3

6.3

5.3

aT

1 2

2

1/

The broader peaks at ~ pH 9-10 meanThat the lifetime is SHORTER.

What might make the lifetime shorter atThat pH?

N+

N

O

O-

O O-

O

O-

O

O-

HN

+

N

O

O-

O O-

O

O-

O

O-

H

N+

N

O

O-

O O-

O-

O

HO

-

ON

+

N

O

O-

O O-

O

O-

O-

O

H

N

N

O

O-

O O-

O

O-

O

O-

pH

11.3

10.3

9.3

8.3

7,3

6.3

5.3

aT

1 2

2

1/

N+

N

O

O-

O O-

O

O-

O

O-

H N+

N

O

O-

O O-

O

O-

O

O-

H

N+

N

O

O-

O O-

O-

O

HO

-

O

N+

N

O

O-

O O-

O

O-

O-

O

H

N

N

O

O-

O O-

O

O-

O

O-

Why do we choose this model insteadOf?

1. pH at which we observe the effects2. Both types of protons are affected3. Would expect splitting of carboxylic arm protons due to different spatial

orientation of protons

Would not Expect muchEffect here

aa’

N

NO

O-

O

O-

OH

O

O

O H

pH

11.3

10.3

9.3

8.3

7,3

6.3

5.3

aT

1 2

2

1/

N+

N

O

O-

O O-

O

O-

O

O-

H N+

N

O

O-

O O-

O

O-

O

O-

H

N+

N

O

O-

O O-

O-

O

HO

-

O

N+

N

O

O-

O O-

O

O-

O-

O

H

N

N

O

O-

O O-

O

O-

O

O-

pKa EDTA 1.99; 2.67; 6.16; 10.2

We predicted 3 peaks for this OH group

And remember we just didThis one

As our EDTA example suggestedWe are looking at conformationalchanges

We predicted 8 peaks (4x2) for the CH2 group what happened?

Lines can be broader if the local chemical environment is shiftingDue to the dynamics of equilibrium

Peter Atkins 6th Ed. Physical Chemistry, Freeman

Fast = single line, mean of bothIntermediate = very broadSlow = two lines

Predicted Number of Peaks in addition to the main proton peaksFor the Pb-EDTA compound from the 207 Pb

2 1N I peaks* #

I Pb 2 0 7 1 2/

2 11

21 22 0 7Pb peaks*

N

N

O

O-

O O-

O

O-

O

O-

2x2=44x2=8

Expect to that EDTA bound to 207 should split into two peaksFor each type of proton

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8

-1 X ppm scaled to standard =0

sig

nal

sca

led

to

lar

ges

t P

bE

DT

A p

eak

J – coupling of207 Pb to H

1.66 1.72

300 MHz instrument

1

2

1

2

B

B

o Sam ple TM S

o

Sam ple TM S

\

What is the coupling, J, in Hz?

What is the lifetime?

Lifetime = 1/((3.14)*3(1/s))=106ms

Using the area related to spin couplingOf Pb207 what is the % 207 in the sample?

Peak width at half height =(1.69-1.68)x10-6)*300x106)= 3 Hz

(1.72-1.66)x10-6)*300x106)=18 Hz

aT

1 2

2

1/

1.68 1.69

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

-5 -4 -3 -2 -1 0 1 2 3 4 5

Gaussian

Lorentzian

PG

x b

1

2

2

12

ex p

P

a

x ba

L

1 2

22

2

Lorentzian is sharper

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

1.4 1.45 1.5 1.55 1.6 1.65 1.7 1.75 1.8


sig

nal

sca

led

to

larg

est

Pb

ED

TA

pea

k

Deconvolution of NMR peaks – assumed Gaussian

Calculated relative areas 10.3830.289

% *. .

. .Pb

Pb

Pb other2 0 7 1 0 0

2 0 7

2 0 7

0 3 8 3 0 2 8 9

1 0 3 8 3 0 2 8 91 0 0 4 0 %

Expected value = 21%

-0.1

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

1.5 1.55 1.6 1.65 1.7 1.75 1.8

ppm (reversed)

Sig

nal

I cheated to account for the poor baseline = due to T2* which is an instrumentalInability to get perfect fields and shimming

Relative areas by Lorentzian:Main peak: 1Side 1 0.0959Side 2 0.0966

% *. .

. ..Pb

Pb

Pb other2 0 7 1 0 0

2 0 7

2 0 7

0 0 9 5 9 0 0 9 6 6

1 0 0 9 5 9 0 0 9 6 61 0 0 1 6 1 4 %

Lorentzian Decovolution for the areas

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8


sig

nal

sca

led

to

lar

ges

t P

bE

DT

A p

eak

1.66 1.72

300 MHz instrument

What about these features?

Suggests that the protons on the carboxylic armsAre not-equivalent

Spinning side bands?Could be due to poor Shimming of x and y

Could be due to poor shimmingOn z

But! Would expect these featuresAt both peaks!

ON+

N+

O

O

OO

O-

O

OH

Pb

Non-equivalentProtons due to binding

aa’

aa’

Increased splitting patternCan be due to differencesOf orientation of the protonsThis supposition is increasedBy the very nearly observableSplitting in the central peak

Peak Width

Lifetime depends upona) chemistry (for example a deprotonation reaction

b) rotation of the solution lattice of molecules (T1)c) spin/spin exchange (T2)

HA H A

A A A A* *

PbEDTA L PbEDTA LC N C N6 4 22. . . .

PbEDTA H Pb HEDTAC N62 3

. .

These reactions can reduceThe lifetime of Pb207 whichIn the complex. Since Pb207 isCausing the splitting a reductionIn it’s lifetime can eliminate thesignal

PbNN

O

OOOExcess Pb2+

PbNN

O

OO

OPb

PbNN

O

OOO

Excess EDTA

Pb

N

NO

NOO

O

O

N

O

O

O

Pure EDTA1:1 Pb-EDTA

1:2 Pb EDTA

Excess EDTA

Main peak splitsDue different orientationOf protons around Pb

Adding more EDTARemoves the mainPeak split (protonsAround Pb now lookThe same). The splitting

Due to Pb207 also canBe lost

(It might also be thatLead is not complexing at thisHigh pH)

Application of lead isotope analysis in shooting incident investigations. Zeichner, Arie; Ehrlich, Sarah; Shoshani, Ezra; Halicz, Ludwik. Division of Identification and Forensic Science, Israel Police National Headquarters, Jerusalem, Israel. Forensic Science International (2006), 158(1), 52-64. Publisher: Elsevier Ltd., CODEN: FSINDR ISSN: 0379-0738. Journal written in English. CAN 145:57143 AN 2006:202948 CAPLUS

Abstract

A study was conducted to examine the potential of the considerable variability of the lead isotope compns. in bullets (projectiles) and primers in shooting incident investigations. Multiple-collector inductively coupled plasma mass spectrometry (MC-ICP/MS) was used to analyze lead isotopic compns. in projectiles, cartridge cases, firearms discharge residues (FDR) in barrels of firearms and in the gunshot entries. .22 caliber plain lead and plated ammunition and 9 mm Luger full metal jacket (FMJ) ammunition were employed in shooting expts. using semiautomatic pistols. Cotton cloth served as the target material and two firing distances were tested; 1 cm (near contact) and 2 m distances. It was obsd. that various mech. or chem. means of cleaning do not completely remove lead deposits ("lead memory") from barrels of firearms. Nonetheless, it was shown that anal. of lead isotopic compn. may provide valuable evidence in investigating specific scenarios of shooting incidents. For instance in a shoot-out where several firearms and ammunition brands are involved, it may be feasible to point out which ammunition and/or firearm caused a particular gunshot entry if the ammunition brands involved (bullets and primers) differ considerably in their lead isotopic compn.

Origin assignment of unidentified corpses by use of stable isotope ratios of light (bio-) and heavy (geo-) elements - A case report. Rauch, Elisabeth; Rummel, Susanne; Lehn, Christine; Buettner, Andreas. Institute of Forensic Medicine, Ludwig-Maximilians-University Munich, Munich, Germany. Forensic Science International (2007), 168(2-3), 215-218. Publisher: Elsevier Ltd., CODEN: FSINDR ISSN: 0379-0738. Journal written in English. CAN 147:316041 AN 2007:474376 CAPLUS

Abstract

An unknown male body was found near an expressway in Germany. As different criminalistic and forensic methods (e.g. tooth status, fingerprint or DNA-anal.) could not help to identify the person, multielement stable isotope investigations were applied. The combined anal. of stable isotope ratios of light (H, C, N) and heavy elements (Pb, Sr) on the man's body tissues supported to assign him to Romania. The case report demonstrates an application of multielement-isotope anal. in the forensic fields and its potential.

END HERE

“Coalesance of the two lines occurs when the lifetime of a conformation, tao,Gives rise to a linewidth that is comparable to the difference of resonance frequencies,”

2

Example, when the chemical shifts differ by 100Hz what is the maximumLifetime of a single conformation?

The rate of interconversion is the inverse of the lifetime.

What is the rate of interconversion?

Peter Atkins 6th Ed. Physical Chemistry, Freeman

http://www.physics.sjsu.edu/becker/physics51/mag_field.htm

Notice that the magneticMoment can be decomposedInto two components alongThe y and x directions.

http://www.urmc.rochester.edu/smd/Rad/MRIweb/mri_html/Animation_spin01.gif

http://www.papimi.gr/Ilika%20kimata.jpg

http://home.tiscali.nl/physis/deHaasPapers/DiracEPR/DiracEPR.html

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