IMIN200 Lecture 5: 263, F14, C10 NMR (1)

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amaranthotter401

28 May 2018

School

University of Alberta

Department

Immunology and Infection

Course

IMIN200

Professor

Edan Foley

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Chapter 10 1H and 13C NMR (Nuclear Magnetic Resonance):

Tools for Structural Determination

Chemistry 263, Fall 2014, Brzezowski

I. Introduction

Spectroscopy: results when energy in the electromagnetic spectrum is imparted to

a molecule and causes some temporary or permanent change to the molecule

- the change in the molecule or the change in the energy that was used is monitored

- the energy can be absorbed, emitted, transmitted or cause a chemical change

Spectrophotometer: an instrument which imparts energy to the molecule and records

the change to the molecule or the change of energy

Electromagnetic Spectrum

- the electromagnetic spectrum is shown below, along with the type of spectroscopy associated

with some of the regions

- energy decreases from left to right

- the more energetic a wave is, the higher its frequency and the shorter its wavelength

according to the following formulas

- for instance, x-rays are relatively high energy waves that break bonds and can cause

permanent changes to molecules (and damage to biological systems)

E = hν

E = energy in J

h = Planck’s constant; 6.63 X 10-34 Js

ν = frequency in Hertz (Hz)

E = hc ν α 1

λ λ

c = speed of light,

λ = wavelength

x-rays vacuum

ultraviolet

(UV)

near

ultraviolet

(uv)

visible

(vis)

near

infrared

(IR)

infrared

(IR)

Increasing ν and E

Increasing λ

uv- vis spectroscopy IR spectroscopy NMR

spectroscopy

microwave radiowave

- here is a summary of the several types of spectroscopy important in organic chemistry:

IR Spectroscopy (Chapter 2, Chemistry 261 PLEASE REVIEW)

- energy in the infrared (IR) region causes stretching and bending of bonds in the molecule; the

molecule is only temporarily changed

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- the energy absorbed is monitored and is very characteristic of the types of bonds (e.g. double

bonds vs. single bonds, carbonyls vs. carbon-carbon double bonds) that a molecule has

- this is an excellent way to identify the types of functional groups in a molecule

- this will be used in tandem with NMR and molecular formula to find the structures of

molecules

NMR Spectroscopy

- energy in the radiofrequency range is applied to a molecule in the presence of a strong

magnetic field

- this causes the nuclei of certain atoms (i.e. 1H and 13C) to absorb energy

- the wavelength of energy absorbed is then monitored and gives important information about

the connectivity of atoms in a molecule

- the medical application of this is called MRI (magnetic resonance imaging)

Mass Spectrometry (will not be discussed in this course)

- high energy particles such as electrons are used to bombard a molecule and make it fall apart

into ionic fragments

- the type of fragments produced are characteristic of the structure of the molecule

- for example ketones fall apart in certain ways that give information about the groups attached

to the carbonyl

Ultraviolet-Visible Spectroscopy (Chapter 11)

- molecules with several double bonds in a row absorb light in the UV-Vis range

- this absorption gives information about how conjugated a molecule is (i.e. how many

double bonds in a row it has) and will help determine if a molecule is aromatic

- this type of spectroscopy is used routinely in biology labs because many biologically

interesting molecules have many double bonds in a row and they can be monitored most

easily with this technique

II. The Theoretical Basis of NMR Spectroscopy

NOTE: YOU WILL NOT BE ASKED TO REPRODUCE ANY OF THE THEORY HOWEVER A

GENERAL UNDERSTANDING OF IT WILL HELP YOU SOLVE THE SPECTROSCOPY

PROBLEMS

A. Atomic Nuclei as Bar Magnets

- note that this is an extremely rudimentary (an not entirely accurate) explanation but it is still very

useful in helping to interpret NMR spectra

- certain atomic nuclei such as the regular isotope of hydrogen (1H) and the 13C isotope of carbon

have a nuclear spin

- note that the regular 12C isotope of carbon or the 16O isotope of oxygen do not have spin (and are

consequently invisible to NMR)

- when nuclei which possess spin are put into a strong magnetic field they behave like bar magnets

and align with or against the magnetic field (slightly more with the field)

randomly alligned nuclei (which possess nuclear spin)

with no external magnetic field

direction of

external

magnetic

field

nuclei alligned with strong

magnetic field

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- when the nuclei in the strong magnetic field are subjected to exactly the right amount of

electromagnetic energy they can flip from being aligned with the magnetic field to being aligned

against the magnetic field

- what makes this technique useful is that different 1H nuclei will need slightly different energy to

flip depending on the environment they are in i.e. different protons in the same molecule flip with

slightly different amounts of energy

1H in

environement a

1H in

environement b

E1E2

direction of

magnetic field

- the NMR machine generates a strong magnetic field and has the ability to subject the molecule to

different wavelengths of electromagnetic radiation

- the absorption of certain wavelengths of electromagnetic radiation is monitored; this information

allows one to “see” the different types of protons in the molecule and infer the connectivity of the

molecule

- different NMR machines can have magnets of different field strength; stronger magnets interact

with the nuclei more strongly and therefore more energetic electromagnetic radiation is required to

flip the nuclei between states

- very powerful magnets are used to decode the structure of more complex molecules; they allow

the different nuclei to be seen better because peaks are much less likely to overlap (i.e. they are

better resolved)

- a very low field NMR has a magnet of about 1.4 T (tesla) and uses energy in the 60 MHz range to

flip the proton nuclei

- routine NMR’s today are in the 8.6 T range and use energy frequencies of about 360 MHz to flip

the proton nuclei

- note that NMR machine strengths are identified not by the magnet strength but by the approximate

energy required to flip the proton nuclei

direction

magnetic

field

energy

required to

flip

nucleus

1.4T 4.7 T 8.6 T

60 MHz 200 MHz 360 MHz

- in primitive NMR machines (called continuous wave or CW) the magnetic field was kept constant

and the radio-frequencies were slowly scanned; this was very time-consuming

- modern machines are called FT (Fourier transform) NMR

- a pulse of all appropriate radiofrequencies is applied to the sample all at once and the data is

recorded as a sine wave

- a Fourier transform operation then separates the individual absorptions and allows them to be

recorded as an easy-to-read NMR spectrum

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Document Summary

Chapter 10 1h and 13c nmr (nuclear magnetic resonance): Spectrophotometer: an instrument which imparts energy to the molecule and records the change to the molecule or the change of energy. E = energy in j h = planck"s constant; 6. 63 x 10-34 js. E = hc 1 c = speed of light, Increasing and e x-rays vacuum ultraviolet (uv) near ultraviolet (uv) visible (vis) near infrared (ir) infrared (ir) microwave radiowave. Nmr spectroscopy here is a summary of the several types of spectroscopy important in organic chemistry: Ir spectroscopy (chapter 2, chemistry 261 please review) energy in the infrared (ir) region causes stretching and bending of bonds in the molecule; the molecule is only temporarily changed. Note: you will not be asked to reproduce any of the theory however a. General understanding of it will help you solve the spectroscopy.

Related Questions

PLEASE HELP ME ANSWER THE POST LAB QUESTIONS. I did THE PRE LAB. I ALSO FILLED in the tables.

Determination of Cobalt (II) Chloride

by UV/VIS Spectroscopy

Introduction

The absorption of specific quantities of electromagnetic (light) radiation by an element or compound allows electrons to move from lower energy level to a higher energy level. We say that the energy levels are quantized. Electrons, returning to lower energy levels, will release energy within the electromagnetic spectrum. The wavelengths may be in the ultraviolet, visible, or infra red regions of the spectrum. For cobalt (II) chloride, the compound of study, the visible region, of the spectrum will be addressed.

By knowing the wavelength (l) or frequency (n) of radiation, one may determine the energy of the transition. Relationship between energy, wavelength and frequency are:

c= ln and E= hn or E= hc/l

The light radiation absorbed in a solution of this salt is proportional to its concentration through a relationship provided by the Beer-Lambert Law:

A = kC

where A= absorbance, k is Beerâs Law constant, and C is the molar concentration in solution.

The absorbance (A), a unit-less number is dependant on the quantity of light energy absorbed and transmitted by the solution through the following equation:

A= -log T or A = 2-log %T

where T= transmittance= I/I⁰ where I is the absorbed light and I⁰ is the impinging light source.

Purpose

The purpose of this laboratory experiment is to practice, learn and carefully prepare molar solutions to investigate the relationship of concentration and spectrophotometer response. A 0.150 M solution of cobalt (II) chloride will be provided. You will prepare diluted solutions through the serial dilution method, measure the transmittance and calculate the absorbance values. You will prepare a Beerâs law graph of data; molar concentration vs. absorbance. Then determine Beerâs law constant using Microsoft Excel program.

You will also be given a concentrated sample of cobalt (II) chloride of unknown concentration and you must determine the appropriate dilution to prepare and thereafter determine the unknown concentration.

Procedure

In order to carry out this analysis procedure, it will be necessary to determine the wavelength of transition. Recall, that one specific energy maximum exists for each transition. We will study the transition in the range of 400- 600 nanometers (nm). This is the visible region of the spectrum.

Step 1. Prepare the following solution concentrations using the 0.150 M cobalt (II) chloride stock solution, 18 mm x 150 mm test tubes and Mohr pipets.

TABLE 1

Solution No.	0.150 M CoCl₂ (mL)	Volume water (mL)
1	5.0	0.0
2	4.0	1.0
3	3.5	1.5
4	3.0	2.0
5	2.5	2.5
6	2.0	3.0
7	1.0	4.0

Step 2. Carefully place a cork or rubber stopper over each solution and mix the contents.

Step 3. Measure the transmittance of a 0.150M solution between 400 nm and 600 nm at 25 nm intervals.

To accomplish this, follow these steps:

Caution: instruments are delicate. If you are uncertain about instrument operation, ask your instructor for assistance. (Instructions on instrument operation- see next page.)

SPECTRONIC is a registered trademark of Spectronic Instruments, Inc.

Instrument operation- edited version of product ^*Operatorâs Manual

Transmittance and Absorbance

1. Turn on the instrument by turning the Power Switch (10) clockwise. Allow the spectro-photometer to warm up for at least 15 minutes to stabilize.

2. After the warm-up period, set the desired wavelength with the Wavelength Control Knob.

3. Set the filter lever to the appropriate position for the selected wavelength (not required for SPECTRONICÂ® 20D).

4. Adjust the display to 0%T with the Zero Control (10). Make sure that the sample compartment is empty and the cover is closed.

5. Set the display mode to TRANSMITTANCE by pressing the MODE control key until the appropriate LED is lit.

6. Fill a clean cell with water and wipe the cell with a tissue (Kimwipe^TM) to remove liquid droplets, dust and fingerprints.

7. Place the cell in the sample compartment and align the guide mark on the cell with the guide mark at the front of the sample compartment. Press the cell firmly into the sample compartment and close the lid.

8. Carefully adjust the display to 100%T with the Transmittance/Absorbance Control (9). Move the knob slowly as you approach 100%T.

9. Remove the cell from the sample compartment and empty the water.

10. Rinse the cell twice with small volumes of the solution to be measured and fill it with the solution.

11. Wipe the cell with a tissue and insert the cell into the sample compartment. Align the guide marks and close the lid.

12. Read the appropriate value (%T) from the display.

13. Remove the cell from the sample compartment and repeat steps 10 through 12 for any remaining sample solutions.

14. When all measurements are completed, turn off the spectrophotometer by turning the Power Switchcounterclockwise until it clicks.

Step 4. Using solution no. 1, follow the directions for instrument operation and record (%) transmittance values for each wavelength in Table 2.

Remember to reset the instrument to 100% for each wavelength investigated.

DO NOT DISCARD THIS SOLUTION,

Step 5. The wavelength that gives the largest absorbance value will provide the greatest sensitivity to concentration change. Generate an Excel graph of wavelength (x) vs. absorbance (y) using these values. At zero absorbance, there isnât any CoCl₂ so donât forget to include the (0,0) data point. Connect your data points with a smooth curve. This is the absorption spectrum of aqueous CoCl₂. Submit this with your lab report.

Step 6. Reset your spectrometer using this wavelength (step 5), and measure this solution and the other six solutions listed in Table 1, and record your % transmittance values in Table 3.

TABLE 2

Wavelength (nm)	% Transmittance	Absorbance
400	88.3	0.055
425	73.7	0.132
450	43.6	0.362
475	23.8	0.623
500	15.7	0.804
525	16.0	0.797
550	38.9	0.410
575	73.0	0.37
600	85.3	0.069

MAXIMUM ABSORBANCE IS LOCATED AT __________500______nm.

TABLE 3 -Results measured at _500_____nm.

Test tube No.	% Transmittance	Absorbance	Molarity(M)
1	15.8	0.802
2	22.8	0.643
3	27.4	0.562
4	33.1	0.480
5	40.2	0.396
6	47.6	0.323
7	69.3	0.159

Step 7. Plot a graph of Molarity (M) vs. Absorbance for all seven solutions using Excel plotting techniques. This is called a calibration curve, whereby one establishes a measured response to a known solution concentration. Determine the slope of the line and include the linear equation of fit, and the correlation coefficient.

Recall the relationship A = kC or y = mx + b; where b is the (0,0) data point

Then: k = A/C

Step 8. The unknown solution to be analyzed is more concentrated than any of the standard solution concentrations that were measured to prepare the calibration curve.

Points to consider:

You must achieve a dilution such that the absorbance is within the absorbance range of the standards that were previously measured.

By using serial dilutions, will multiply many errors; thus it is best to obtain a dilution is one-step if possible.

Consider reducing the concentration by Â¾, or by Â½, and by Â¼ or to 0.10 of the original concentration and test the solution to see if the absorbance response is within range. You may also consider using the dilutions in Table 1. Once this is determined, proceed to the next step. Complete the dilution table for the diluted concentrations you prepared:

Trial	mL of unknown	mL of water	(%T)	A
1			51.9	0.284
2			70.7	0.151
3			79.3	0.101

Step 9. Using the desired dilutions, determined in step 8, measure two of the diluted solutions and record the measured transmittances in Table 4.

UNKNOWN NUMBER _____B_____

TABLE 4

Diluted concentration (M)	Transmittance (%T)	Absorbance (A)
Pure unknown B	8.7	1.062
1 mL of unknown in 1 mL of water	29.5	0.531

Calculation of absorbance, for diluted Unknown solution:

Step 10. Determine the concentration of the diluted solution using the linear equation of fit, knowing the absorbance and k, the slope of the line.

C = A/ k

Step 11: Determine the concentration in your original Unknown solution. Show your

back-calculation below to support your final answer to this experiment.

Calculations:

If needed I did a graph slope and got y= 0.1896x

R^2=0.9951

Unknown Solution No. ________ Concentration: ___________M

Prelab Questions

1. Consider a solution of 0.400M Co(NO₃)₂ provided by your lab instructor.

You are required to make the following dilute concentrations:

0.160M and 0.240M solutions. You are provided the following glassware: 5.0 mL Mohr pipet and 18 mm x 150 mm test tubes. Describe how you would prepare each solution, and support your description with your calculations. (Recall (M_c x V_c = M_d x V_d) where M_c is the molarity of the concentrated solution, V_c is the volume of the concentrated solution, M_d is the molarity of the dilute solution, and Vd is the volume of the diluted solution.)

Given 0.400 M Co(NO3)2 solution.

Lets say we pipette out 5 mL of this solution and dilute it to get 0.160 M solution. Then

Using, Mc x Vc = Md x Vd

0.400 x 5 = 0.160 x Vd or Vd= (0.400 x 5)/0.160 = 12.5 mL

Thus we need to add 7.5mL distilled water to 5 mL of cobalt nitrate. Since we are only provided with a 5 mL pipette, we can only measure out volumes in multiple of 5.

thus pipette out (5 x 2) 10 mL of cobalt nitrate solution in the test tube and dilute it by addition of (7.5 x2) 15 mL of distilled water.

Again for preparing 0.240 M solution

Lets say we pipette out 5 mL of this solution and dilute it to get 0.160 M solution. Then

Using, Mc x Vc = Md x Vd

0.400 x 5 = 0.240 x Vd or Vd= (0.400 x 5)/0.240 = 8.33 mL

Thus we need to add 3.33mL distilled water to 5 mL of cobalt nitrate. Again taking volumes only in multiples of 5, pipette out (5 x 3) 15 mL of cobalt nitrate solution in the test tube and dilute it by addition of (3.33 x3) 10 mL of distilled water.

2. In your own words, explain, in detail, what you will be learning during this lab. Address the chemical principles and, laboratory skills, in your answer.

Post - Lab Questions:

a) Letâs say, for a given concentration of salt solution, the maximum response at a wavelength maximum of 485nm, is 0.750. If an unknown solution is analyzed at a longer or shorter wavelength would the investigator obtain the same concentration for the Unknown solution? Explain why.

b) List all possible sources of errors that may result in this experiment.

THE ONE IN ITALICS AND BOLD IS THE PRELAB QUESTION PLEASE ANSWER THAT. PLEASE HELP ME ANSWER THE POST LAB QUESTIONS. I did THE PRE LAB. I ALSO FILLED in the tables.

Determination of Cobalt (II) Chloride

by UV/VIS Spectroscopy

Introduction

By knowing the wavelength (l) or frequency (n) of radiation, one may determine the energy of the transition. Relationship between energy, wavelength and frequency are:

c= ln and E= hn or E= hc/l

The light radiation absorbed in a solution of this salt is proportional to its concentration through a relationship provided by the Beer-Lambert Law:

A = kC

where A= absorbance, k is Beerâs Law constant, and C is the molar concentration in solution.

The absorbance (A), a unit-less number is dependant on the quantity of light energy absorbed and transmitted by the solution through the following equation:

A= -log T or A = 2-log %T

where T= transmittance= I/I⁰ where I is the absorbed light and I⁰ is the impinging light source.

Purpose

Procedure

Step 1. Prepare the following solution concentrations using the 0.150 M cobalt (II) chloride stock solution, 18 mm x 150 mm test tubes and Mohr pipets.

TABLE 1

Solution No.	0.150 M CoCl₂ (mL)	Volume water (mL)
1	5.0	0.0
2	4.0	1.0
3	3.5	1.5
4	3.0	2.0
5	2.5	2.5
6	2.0	3.0
7	1.0	4.0