experiment 11: nuclear magnetic resonance spectroscopyfaculty.ccbcmd.edu/~cyau/201 11 nmr...

Experiment 11: NUCLEAR MAGNETIC RESONANCE SPECTROSCOPY

87

Purpose: This is an exercise to introduce the use of nuclear magnetic resonance spectroscopy, in conjunction with infrared spectroscopy, to determine the structures of organic compounds. Theory: A detailed discussion of the theory behind nuclear magnetic resonance (NMR) is provided by any standard organic chemistry text (such as Organic Chemistry by Brown, Foote & Iverson, 4th Ed, Chapter 13, in particular, Sections 13.4-13.10). In this exercise we will concentrate on the applications of 1H-NMR (henceforth referred to as simply NMR) towards the structure determination of organic compounds. Just keep in mind that IR spectra reflect the absorption of energy due to molecular vibrations (bond stretching and bond bending). NMR spectra reflect the absorption of energy due to nuclear spin transitions. The proton (H) is equivalent to a minute magnet, spinning (precessing) at a particular frequency. When the frequency of an external rotating magnetic field and the frequency of the precessing proton become equal, they are said to be in resonance, and absorption of energy by the nucleus can occur. The surrounding electrons shield the proton, thus each proton has its characteristic chemical shift (given the symbol δ, in units of ppm, along the x-axis of the spectrum), depending on the environment created by the surrounding electrons. Aside from the location of the peak, we need to also consider the splitting pattern of the peak (singlet, doublet, triplet, quartet, quintuplet, or multiplet) which gives us information on the number of protons on the adjacent carbon(s). The y-axis reflects the peak height, but it is the area under each set of peaks that is of more concern to us. The ratios of peak areas are equal to the ratios of nonequivalent protons. Considering that this is the first time you are being exposed to NMR spectroscopy, we will keep this simple and tell you the number of protons associated with each set of NMR peaks, rather than having you calculate the ratios and deduce the number of protons from the ratios.

All this probably sounds very complicated, but the complexity is reduced by looking at examples and actively working out practice problems. Do not sit back passively and merely watch your instructor work out problems for you. The interpretation of NMR and IR spectra will not be taught in the lecture portion of the course at CCBC-Catonsville. In Organic Chemistry II, both lecture and lab (Chem 202 and 203), you will be expected to remember what you have learned in this exercise and a significant portion of that course will depend on your understanding of this topic.

88 EXPERIMENT 11: NUCLEAR MAGNETIC RESONANCE SPECTROSCOPY

ANALYSIS OF THE NMR SPECTRUM About Chemical Shifts: (Appendix 3 provides a summary of the chemical shifts discussed here. Know how to make use of this appendix. A copy will be provided at your lab final.) • The location of the peaks in the NMR spectrum (called chemical shift, recorded in units

of ppm) tells us what kind of protons we are looking at (aliphatic H? vinyl H? allylic H? aryl H? H bonded to carbonyl C? etc.)

• The reference point is 0 ppm at the far right. Usually a peak shows up at 0 ppm, due to the tetramethylsilane (TMS) added to the sample as a reference point. This peak is not part of what you are analyzing. This is like “zeroing” a balance.

• Aliphatic H’s (as in alkanes) are “shielded” by electrons and lie close to the right of the spectrum (near 0 ppm). You can consider these as the “normal H.” They are said to be shielded and are upfield.

• When the H is bonded to an electronegative element (such as O, N, Cl, Br etc.) the electron cloud around H is pulled away from the H and it is said to be deshielded. The H shows up further to the left and is said to be shifted downfield.

• Protons are deshielded by other factors such as C=C, C=O, C≡C, and aromatic ring, to

different extents. The chart in Appendix 3 tells you how far downfield these H are shifted and gives you the approximate range for each type of H. Although you will be provided with this chart at exams, it would be wise to commit to memory roughly where you would expect to find these protons. Here is a suggestion on how to learn these shifts. Start at the far right (upfield).

1 − 2 ppm: “normal” unshielded aliphatic H in a hydrocarbon. 2 − 3 ppm: H-C-C=O (H bonded to C that is joined to carbonyl group) H-C≡C (acetylenic H) 3 − 4 ppm: H-C-O or H-C-N or H-C-Cl or H-C-Br 4.5−7 ppm: H-C=C (vinyl H) 7 − 8 ppm: aromatic H (H bonded to benzene ring) 9 −10 ppm: H-C=O (aldehydic H) 10-12 ppm: H-O−C=O (carboxylic H)

• H of –OH in an alcohol has a wide range (anywhere between 1 − 5 ppm) and should be considered last. Generally, it would show as one single peak (not split) and would be assigned to only 1 H.

• H of –NH2 (as in 1° amine) or −NHR (as in 2° amine) also has a wide range (between 1.5−3 ppm). These should also be analyzed last.

• H of −CONH2 (as in amides) are between 5 − 8 ppm.

(upfield, more shielding) (downfield, more deshielding)

3 2 1 0 ppm

1.6 ppm: H-C-C=C (allylic H) 2.2 ppm: (benzylic H) CH

EXPERIMENT 11: NUCLEAR MAGNETIC RESONANCE SPECTROSCOPY 89

About Equivalent Protons • If you want to see whether an NMR spectrum fits a particular structure, the first thing to

do is look for symmetry in the structure. If all the protons in the molecule are equivalent, there would be only one signal. Each of the following molecules are expected to show one signal:

CH3 O CH3 C CCH3

CH3 CH3

CH3

Nonequivalent protons would show different signals:

CH3 CH2 O H CH3 CH2 O CH2 CH3 C CH

CH3 CH3

HO CCH3O

OCH3

(3 signals) (2 signals) (2 signals) (3 signals) (2 signals) About the Splitting Patterns: • By splitting pattern, we are referring to whether a signal is showing up as a singlet,

doublet, triplet, quadruplet, quintuplet…. or multiplet. • The splitting pattern (or multiplicity) reflects the number of protons on the adjacent

atom. The number of peaks equals the number of protons on the adjacent atom plus one. This is commonly referred to as the "n + 1" rule. o Singlet means the proton is bonded to a C where the neighboring atom has no

protons. o Doublet means the proton is bonded to a C where the neighboring atom has 1 proton.

The two peaks is approximately the same height. o Triplet means the proton is bonded to a C where the neighboring atom has 2 protons.

Peak heights are approximately 1:2:1. o Quadruplet means the proton is bonded to a C where the neighboring atom has 3

protons. Peak heights are approximately 1:3:3:1.

singlet doublet triplet quartet • Equivalent protons would not split each other. • If the proton is on a C joined to an atom with 1 proton on one side, and joined to an atom

with 2 protons on the other side, then the pattern is a triplet of a doublet. (It is split into a doublet due to the 1 proton on one side, and then each peak of the doublet is split into triplets.) You may clearly see two sets of triplets, but sometimes these peaks overlap and all you will see is a multiplet.


• A spin-spin coupling constant refers to spacing between adjacent peaks within a set of signals. It depends on what type of protons are causing the splitting and give you a lot of information, but we will keep to simple structures and will not need to deal with coupling constants in this course.

• A common exception to the splitting pattern described above is the H of O−H in acids and alcohols and N−H in amines. It is often undergoing a rapid chemical exchange and is not bonded to the molecule long enough for its signal to split with the signal of its neighbor:

About Peak Areas We had talked about relative peak heights. It is actually the area under the peak rather than peak height that we should be considering. The area under each set of signals corresponds to the number of protons that generated the set of signal. We will not deal with the measurement of peak area this semester. You will be given the ratio of protons in all NMR spectra that we give you. Just remember you are given a ratio and not necessarily the exact number of protons. For example, a ratio of 2H:3H might mean that you are looking at 4H and 6H. Now, let’s try to predict what the NMR of ethyl alcohol might look like.

By examining the structure you should see that there are 3 types of protons and should expect 3 signals. CH3−CH2−O−H X Y Z • The X hydrogens (in the methyl group CH3− ) should be furthest upfield, being

furthest away from the electron-drawing oxygen atom (around 1 or 2 ppm). • The Y hydrogens (in the methylene group –CH2−) should be downfield from X

hydrogens because it is deshielded by the adjoining O. • The Z hydrogen (in −O−H) could be anywhere between 1 and 5 ppm.

What splitting pattern would you expect of these three sets of protons? • The X hydrogens should be split into a triplet by the two hydrogens “next door”. • The Y hydrogens should be split into a quartet by the three hydrogens “next door”. • The Z hydrogen (alcoholic H) probably would not be split because of the rapid

chemical exchange. So, one might expect something like this….

Note that the sets of signals are not entirely symmetrical. For example, the peak heights in the triplet is not exactly 1:2:1. You will find that the peak heights are slightly skewed towards the signal of the proton(s) that caused the split. This is sometimes helpful to know.

R-O-H R−O− + H+

0 ppm

about 3-4 ppm about 1-2 ppm

2H 1H 3H


Practice Questions Your instructor will spend the entire lab period going through a series of examples (Compounds A thru L) to show you how to make use of IR and NMR spectroscopy to determine the structures of organic compounds. You will then solve 10 other problems (Compounds #1 thru 10) and turn in your work the following week. You will want to review the experiment on IR before coming to the lab. There is no actual lab work involved. You may work with a partner to discuss your thoughts on the matter, but as usual you will have to turn in your own work. The lab final will include problems of this sort and you will not have a partner to do your work for you.


Practice Compound A: C2H6O

2H

1H

3H

(liquid film)


Practice Compound B: C4H10O

(liquid film)

2H 3H


Practice Compound C: C3H6O2

(liquid film)

1H 2H 3H


Practice Compound D: C3H9N

(liquid film)

1H 2H 6H


Practice Compound E: C8H10

2H 3H


Practice Compounds F, G, H: C6H4Cl2

1H 3H

1H 1H

Compound F

Compound G

Compound H


Practice Compound I: C2H4O

(liquid film)

1H 3H


Practice Compound J: C7H9N

(in CCl4)

2H 1H 2H 1H 3H


Practice Compound K: C7H9N

(liquid film)

5H 2H 2H


Practice Compound L: C4H8O

(liquid film)

1H 6H (doublet)

1H


Procedure: The assignment due next week is as follows: For each problem presented in the following pages (Compounds #1 thru 10),

(1) give the full structure of compound (showing all atoms and all bonds), (2) explain the NMR spectrum and show what peaks are due to which protons, also

explained the splitting of the peaks based on your structure, (3) point out the relevant peaks in the IR spectrum that helped you identify or confirm

the structure you have assigned for the compound. An example of how an analysis might be presented is shown below for Unknown #21, which has the molecular formula of C2H5O: The Index of Hydrogen Deficiency is zero. That eliminates the possibility of all double or triple bonds, or rings (eliminates all carbonyl compounds). The fact that it has one oxygen

tells us it must be either an alcohol or an ether. Its IR spectrum shows a strong, broad peak at 3500-3200 cm−1 indicative of the presence of an O−H. The only alcohol with two C's is ethyl alcohol. Its NMR confirms Unknown #21 is indeed ethyl alcohol.

HA at 1.2ppm is split by two HB into a triplet. HB at 3.7ppm is split by three HA into a quartet. (HB downfield because of adjacent O atom) HC is not split by HB and appears as a singlet. ANSWER: UNKNOWN #21 is ethyl alcohol.

¨

335

0 cm

−1

O−H

stre

tch

H C CH

HO

H

HH

A B C

In order to write comments on the spectra and turn them in you may either make photocopies of the pages (pp. 88 − 97) or tear them out of your lab manual.

1050

cm

−1

C−O

stre

tch


1. C3H8O

ppm

2H 1H 2H 3H

(liquid film)


2. C3H8O

ppm

1H 1H 6H

(liquid film)


3. C4H8O

(in CCl4 solution)

ppm

1H 2H 2H 3H


4. C4H8O

ppm

2H 3H 3H

(liquid film)


5. C4H8O2

ppm

2H 3H 3H

(liquid film)


6. C4H8O2

ppm

3H 2H 3H

(liquid film)


7. C9H12

ppm

5H 1H 6H

(liquid film)


8. C7H8O

ppm

5H 2H 1H

(liquid film)


9. C7H8O ppm

2H 3H 3H

(liquid film)


10. C8H11N

ppm

5H 1H 2H 3H

(liquid film)

experiment 11: nuclear magnetic resonance spectroscopyfaculty.ccbcmd.edu/~cyau/201 11 nmr...

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