nuclear magnetic resonance spectroscopy · 2! 8.0!!h-ch2-ch-h chemnmr 1h estimation 1.69 3.34 br...

1

Nuclear Magnetic Resonance Spectroscopy: Purpose: Connectivity, Map of C-H framework Four Factors of Proton NMR (PMR OR 1H NMR): 1. Symmetry: Number of chemically different protons (symmetry) as shown by

number of signals 2. Chemical Shift Value (X-axis): provides information regarding what kind of

chemical environment the proton is in (electron-rich or electron-poor). 3. Integration: Electronic measurement of peak areas: provides you with the number

of chemically equivalent protons producing each signal on the spectrum. 4. Splitting Patterns: Complexity of peak patterns - information is gained regarding

the connectivity of neighboring protons. General Notes:

1. We usually study one type of nucleus at a time. In Organic Chemistry, we focus on the hydrogen-1 isotope (1H) and the Carbon-13 (13C) isotope as they are the most widely used nuclei in the spectroscopy of organic compounds.

2. The position on the chart at which a nucleus absorbs is called its “chemical shift”. The X-axis of a proton NMR spectrum ranges from 0-10 (or as high as 14) ppm. The unit “ppm” stands for “parts per million” and is a standardized unit for the signal frequency.

3. The area under the peaks is calculated by the instrument and the ratio of protons per peak is determined based on the peak area. Integration.

4. On most of the unknown spectra, a calibration peak will be visible. There are two possibilities for calibration: a. Tetramethylsilane (TMS) – External standard that was added to all samples for Qual Lab. The 12 hydrogens of TMS are assigned a value of 0.0000 ppm.

b. Deuterated chloroform, CDCl3, contains a trace amount of CHCl3 – and that proton has a chemical shift value of 7.2600 ppm, thus acting like an internal standard. Deuterated solvents must be used to prevent interference of a molecule’s spectrum with that of the solvent. Deuterium is the H2 isotope and has no magnetic field generation.

Symmetry: How many different types of protons? isobutyl chloride: (CH3)2-CH-CH2-Cl

CH3

SiH3C CH3

CH3

2

3,3-dimethyl-1-butanol: HO-CH2-CH2-C(CH3)3

How many signals are visible here:

0123PPM

0123PPM

012345PPM

0.06 0.06 0.06 0.12 0.18 0.36

3

Answer:

Symmetry in Aromatic rings: How many types of aromatic protons? Monosubstituted:

Disubstituted – Para

Disubstituted – Meta

Disubstituted – Ortho

New in 332: Trisubstituted aromatic rings Trisubstituted – 1,2,3-

012345PPM

0.06 0.06 0.06 0.12 0.18 0.36

X

H H

HH

H

X

H H

HH

X

X

H H

HH

Y

H

X X

HH

H

H

X Y

HH

H

H

H X

XH

H

H

H X

YH

H

4

Trisubstituted – 1,2,4-

Trisubstituted – 1,3,5-

Chemical Shift – Electronic Environments Protons on sp3 C (no EWG near by) 0-2 ppm Protons next to unsaturations 1.8-3 ppm Benzylic Alpha to carbonyls

Protons on sp3 C with O, N or X attached 3-5 ppm

Protons on sp2C (C=C):

Aromatic 6.5-8.5 ppm

Protons of CHO (aldehydes) 8.5-10 ppm

Y

X X

HH

H

Y

X Y

HH

H

X

H X

HH

X

X

H H

ZY

H

Y

H H

XX

H

X

H H

XX

H

CH C

CH

O

CH

O

H

CH

O

5

Protons of CO2H (carboxylic acids) 10-14 ppm

Protons of Hydroxyl groups:

Aliphatic (alcohols) 2-5 ppm Aromatic (phenols) 6-10 ppm

Protons of Amine groups:

Aliphatic (amines) 0.5-3 ppm Aromatic (anilines) 1-5 ppm

Protons on Amides 5.5-8 ppm

What types of protons are in the molecule, as determined by the values on the X-axis?

COH

O

OH OH

NH2 NH2

NH2

O

012345PPM

0.06 0.06 0.06 0.12 0.18 0.36

6

How do you know that the one on the far left is an OH?

Reminder about alcohols (amine and carboxylic acid protons): “exchangeable protons” Under normal, slightly acidic conditions of CDCl3, these protons “hop on and off” on a regular basis. This distorts the peak of the proton, broadening it so it is not as sharp in appearance as the other protons.

In the lab, an easy way to distinguish an alcohol proton from one that is not exchangeable is to add deuterated water (D2O) to the sample. Deuterium does not have a nucleus that can generate a magnetic field so it will not show up on the NMR spectrum. If deuterium substitutes for a proton, the signal will disappear!

Example: Consider the proton NMR of isobutyl alcohol:

012345PPM

0.06 0.06 0.06 0.12 0.18 0.36

0-2 ppm alkyl group H's

3-5 ppm H-sp3C-(O,N,X)

2-5 ppm H-O

H+R O H R O HH

R OH

-H+

R O HD+

R O HD - H+

R OD

signal

no signal

0123PPM

7

Now view deuterated isobutyl alcohol: What’s missing?

Integration Ratio – Ratio of Protons for the signals on a 1H NMR spectrum

• Peak areas are electronically measured by instrument • To determine the ratio of protons producing signals, divide each peak area by

smallest area to determine whole number ratio of peaks Total of Protons should equal number of protons in molecular formula Determine the integration values for each signal in the following:

Splitting Patterns – Spin-Spin Coupling Splitting Patterns are caused by the interaction of the magnetic fields of neighboring nuclei. In Proton NMR, only the magnetic fields of adjacent protons can interact. • Tells you how many protons are on adjacent carbon atoms • Always remember that splitting can only occur between NON-Equivalent Protons

(CH3CH2CH2CH3)

0123PPM

012345PPM

0.06 0.06 0.06 0.12 0.18 0.36

8

• N+1 Rule – The signal of a proton (or set of equivalent protons), with N number of protons on an adjacent carbon atom, will be split into N+1 peaks

Predict the NMR spectrum for bromoethane:

CH3-CH2-Br

Spectrum:

Normally, you will be viewing proton NMR spectra. You need to analyze the splitting patterns of peaks to determine the molecular fragments of a molecule you are attempting to solve. Splitting Patterns: Singlet (symbolized as “s”): N+1 = 1 therefore N = 0, or no neighboring protons. Doublet (symbolized as “d”): N+1 = 2 therefore N = 1, where a CH or OH may be in the adjacent position

0123PPM

ChemNMR 1H Estimation

1.74

3.78

1.74

Br

Estimation quality is indicated by color: good, medium, rough

01234PPM

Protocol of the H-1 NMR Prediction:

Node Shift Base + Inc. Comment (ppm rel. to TMS)

CH 3.78 1.50 methine 0.34 2 alpha -C 1.94 1 alpha -BrCH3 1.74 0.86 methyl 0.83 1 beta -Br 0.05 1 beta -CCH3 1.74 0.86 methyl 0.83 1 beta -Br 0.05 1 beta -C

1H NMR Coupling Constant Prediction

shift atom index coupling partner, constant and vector

3.78! 2! 1! 6.8!!H-C-CH2-H! 3! 6.8!!H-C-CH2-H1.74! 1! 2! 6.8!!H-CH2-C-H1.74! 3! 2! 6.8!!H-CH2-C-H


0.89 0.89

0.890.89


0123PPM



CH3 0.89 0.86 methyl 0.15 3 beta -C -0.12 general correctionsCH3 0.89 0.86 methyl 0.15 3 beta -C -0.12 general correctionsCH3 0.89 0.86 methyl 0.15 3 beta -C -0.12 general correctionsCH3 0.89 0.86 methyl 0.15 3 beta -C -0.12 general corrections



0.89! 10.89! 30.89! 40.89! 5

9

Triplet (symbolized as “t”): N+1 = 3 therefore N = 2. A CH2 may be adjacent or two symmetrical CH groups Doublet-Doublet (symbolized as “dd”): There are two CH’s on either side that are chemically different or perhaps one CH and one OH. WARNING: resembles a triplet. Quartet (symbolized as “q”): N+1 = 4 therefore N = 3. A CH3 is adjacent. Doublet-Triplet (symbolized as “dt”): On one side there is a CH (or OH) and on the other side there is a CH2. WARNING: resembles a quartet. Quintet (no symbol): N+1 = 5 therefore N = 4. Two CH2 groups are adjacent Doublet-Quartet (symbolized as “dq”): On one side there is a CH (or OH) and on the other side there is a CH3. WARNING: Kind of resembles a quintet (not really – the center three peaks are all much higher than the outer two but there’s still five…) Sextet (no symbol): N+1 = 6 therefore N = 5. There is a CH3 on one side and a CH2 on the other side.


1.69

3.34

Br


01234PPM



CH2 3.34 1.37 methylene 0.00 1 alpha -C 1.97 1 alpha -BrCH3 1.69 0.86 methyl 0.83 1 beta -Br



3.34! 2! 1! 8.0!!H-CH-CH2-H1.69! 1! 2! 8.0!!H-CH2-CH-H


1.69

3.34

Br


01234PPM



CH2 3.34 1.37 methylene 0.00 1 alpha -C 1.97 1 alpha -BrCH3 1.69 0.86 methyl 0.83 1 beta -Br



3.34! 2! 1! 8.0!!H-CH-CH2-H1.69! 1! 2! 8.0!!H-CH2-CH-H


0.90

1.31

1.77

3.68

Cl


01234PPM



CH2 3.68 1.37 methylene 2.05 1 alpha -Cl -0.04 1 beta -C 0.30 general correctionsCH2 1.77 1.37 methylene 0.24 1 beta -Cl -0.04 1 beta -C 0.20 general correctionsCH2 1.31 1.37 methylene 0.00 1 alpha -C -0.04 1 beta -C -0.02 general correctionsCH3 0.90 0.86 methyl 0.10 1 beta -C-R -0.06 general corrections



3.68! 4! 3! 7.1!!H-CH-CH-H1.77! 3! 4! 7.1!!H-CH-CH-H! 2! 7.1!!H-CH-CH-H1.31! 2! 3! 7.1!!H-CH-CH-H! 1! 8.0!!H-CH-CH2-H0.90! 1! 2! 8.0!!H-CH2-CH-H


0.90

1.31

1.77

3.68

Cl


01234PPM



CH2 3.68 1.37 methylene 2.05 1 alpha -Cl -0.04 1 beta -C 0.30 general correctionsCH2 1.77 1.37 methylene 0.24 1 beta -Cl -0.04 1 beta -C 0.20 general correctionsCH2 1.31 1.37 methylene 0.00 1 alpha -C -0.04 1 beta -C -0.02 general correctionsCH3 0.90 0.86 methyl 0.10 1 beta -C-R -0.06 general corrections



3.68! 4! 3! 7.1!!H-CH-CH-H1.77! 3! 4! 7.1!!H-CH-CH-H! 2! 7.1!!H-CH-CH-H1.31! 2! 3! 7.1!!H-CH-CH-H! 1! 8.0!!H-CH-CH2-H0.90! 1! 2! 8.0!!H-CH2-CH-H

10

Septet (no symbol): N+1 = 7 therefore N = 6. Two CH3 are adjacent, one on either side. May be hard to see side peaks. Octet (no symbol): N+1 = 8 therefore N = 7. Two CH3s and a CH are attached on three sides of a CH. Harder to see! Nonet (no symbol): N+1 = 9 therefore N = 8. Two CH3s and a CH2 are attached on three sides of a CH. Hardest to see! Multiplet (symbolized as “m”): Unable to count peaks and not recognizable as any shown above. Unable to determine adjacent number of protons. Still have chemical shift and integration ratio! Aromatic Systems: Splitting Patterns Monosubstituted Aromatic Rings: Total of five aromatic protons No actual need to evaluate splitting patterns – if the aromatic region integrates to 5, the ONLY possible structure is the monosubstituted aromatic ring.


1.74

3.78

1.74

Br


01234PPM



CH 3.78 1.50 methine 0.34 2 alpha -C 1.94 1 alpha -BrCH3 1.74 0.86 methyl 0.83 1 beta -Br 0.05 1 beta -CCH3 1.74 0.86 methyl 0.83 1 beta -Br 0.05 1 beta -C



3.78! 2! 1! 6.8!!H-C-CH2-H! 3! 6.8!!H-C-CH2-H1.74! 1! 2! 6.8!!H-CH2-C-H1.74! 3! 2! 6.8!!H-CH2-C-H


7.14

7.04

7.14

7.06

7.06 2.34


012345678PPM



CH 7.06 7.26 1-benzene -0.20 1 -C 0.00 general correctionsCH 7.06 7.26 1-benzene -0.20 1 -C 0.00 general correctionsCH 7.14 7.26 1-benzene -0.12 1 -C 0.00 general correctionsCH 7.14 7.26 1-benzene -0.12 1 -C 0.00 general correctionsCH 7.04 7.26 1-benzene -0.19 1 -C -0.03 general correctionsCH3 2.34 0.86 methyl 1.49 1 alpha -1:C*C*C*C*C*C*1 -0.01 general corrections



7.06! 6! 1! 7.5!!H-C*C-H! 4! 1.5!!H-C*C*C-H! 2! 1.5!!H-C*CH*C-H7.06! 4! 3! 7.5!!H-C*C-H! 6! 1.5!!H-C*C*C-H! 2! 1.5!!H-C*CH*C-H7.14! 3! 4! 7.5!!H-C*C-H! 2! 7.5!!H-C*C-H! 1! 1.5!!H-C*CH*C-H7.14! 1! 6! 7.5!!H-C*C-H! 2! 7.5!!H-C*C-H! 3! 1.5!!H-C*CH*C-H7.04! 2! 3! 7.5!!H-C*C-H! 1! 7.5!!H-C*C-H! 4! 1.5!!H-C*CH*C-H! 6! 1.5!!H-C*CH*C-H2.34! 7


0.91

2.143.45

1.57

0.90

0.91

Cl


01234PPM



CH 3.45 1.50 methine 1.98 1 alpha -Cl -0.02 2 beta -C -0.01 1 beta -CCH 2.14 1.50 methine 0.34 2 alpha -C 0.31 1 beta -Cl -0.01 1 beta -CCH2 1.57 1.37 methylene 0.00 1 alpha -C 0.24 1 beta -Cl -0.04 1 beta -CCH3 0.91 0.86 methyl 0.10 1 beta -C-R 0.05 1 beta -C -0.10 general correctionsCH3 0.91 0.86 methyl 0.10 1 beta -C-R 0.05 1 beta -C -0.10 general correctionsCH3 0.90 0.86 methyl 0.10 1 beta -C-R -0.06 general corrections



3.45! 3! 2! 7.0!!H-C-C-H! 4! 7.0!!H-C-CH-H2.14! 2! 3! 7.0!!H-C-C-H! 1! 6.8!!H-C-CH2-H! 6! 6.8!!H-C-CH2-H1.57! 4! 3! 7.0!!H-CH-C-H! 5! 8.0!!H-CH-CH2-H0.91! 1! 2! 6.8!!H-CH2-C-H0.91! 6! 2! 6.8!!H-CH2-C-H0.90! 5! 4! 8.0!!H-CH2-CH-H


0.91

2.25

3.26

0.91

Br


01234PPM



CH2 3.26 1.37 methylene 1.97 1 alpha -Br -0.08 2 beta -CCH 2.25 1.50 methine 0.34 2 alpha -C 0.41 1 beta -BrCH3 0.91 0.86 methyl 0.10 1 beta -C-R 0.05 1 beta -C -0.10 general correctionsCH3 0.91 0.86 methyl 0.10 1 beta -C-R 0.05 1 beta -C -0.10 general corrections



3.26! 3! 2! 7.0!!H-CH-C-H2.25! 2! 3! 7.0!!H-C-CH-H! 1! 6.8!!H-C-CH2-H! 5! 6.8!!H-C-CH2-H0.91! 1! 2! 6.8!!H-CH2-C-H0.91! 5! 2! 6.8!!H-CH2-C-H

11

Disubstituted Aromating Rings: total of four aromatic protons Para: 1,4 disubstituted:

If disubstituted “para” with the same group, all four protons are the same (equivalent) and no splitting would occur. You would see a singlet, integrating to four. 1,4-dibromobenzene:

If disubstituted “para” with two different groups, the plane of symmetry would result in two TYPES of aromatic protons (integration of two for each signal) and splitting patterns of doublets for each.

012345678PPM

0.55 0.33

XH H

HHY

doublet

doublet

XH H

HHX

singlet

012345678PPM

CH3

Br

Br

12

1-bromo-4-chlorobenzene:

Meta: 1.3-disubstituted

If disubstituted “meta” with the same group, the system is symmetrical, with three types of protons, with one being isolated and appearing as a singlet. 1,3-dibromobenzene

If disubstituted “meta” with two different groups, no plane of symmetry exists. All four protons are different. Four signals result. One will still be isolated from the rest and appear as a singlet. The others will be split.

01234567PPM

H

X X

HH

H

singlet

triplet2 H+ doublet

H

X Y

HH

H

singlet

doubletdoublet

tripletor dd

0123456789PPM

Cl

Br

Br Br

13

1-chloro-3-hydroxybenzene

Enlarged Aromatic region:

Note the “W” coupling occurring… Ortho: 1,2-disubstituted

If disubstituted “ortho” with the same group, a plane of symmetry would produce two types of protons and each will appear as a doublet.

012345678PPM


7.05

7.22

6.83

7.15 5.35Cl OH

012345678PPM

H

H X

XH

H

doubletdoublet H

H X

YH

H

doublet

doublet

triplet or dd

triplet or dd

Cl OH

7.05

7.22

6.83

7.15 5.35Cl OH

14

1,2-dibromobenzene:

If disubstituted “ortho” with two different groups, no plane of symmetry exists so all four aromatic protons are different and would causing all signals to be split. 1-bromo-2-hydroxybenzene:

Again, “W” coupling is occurring…

012345678PPM

012345678PPM


7.22

6.84

6.90

7.43

5.35

Br

OH


012345678PPM



OH 5.35 5.00 aromatic C-OH 0.35 general correctionsCH 7.43 7.26 1-benzene 0.17 1 -Br -0.17 1 -O 0.17 general correctionsCH 6.84 7.26 1-benzene -0.11 1 -Br -0.53 1 -O 0.22 general correctionsCH 6.90 7.26 1-benzene -0.11 1 -Br -0.44 1 -O 0.19 general correctionsCH 7.22 7.26 1-benzene -0.06 1 -Br -0.17 1 -O 0.19 general corrections



5.35! 87.43! 6! 4! 7.5!!H-C*C-H! 1! 1.5!!H-C*CH*C-H6.84! 2! 1! 7.5!!H-C*C-H! 4! 1.5!!H-C*CH*C-H6.90! 4! 6! 7.5!!H-C*C-H! 1! 7.5!!H-C*C-H! 2! 1.5!!H-C*CH*C-H7.22! 1! 2! 7.5!!H-C*C-H! 4! 7.5!!H-C*C-H! 6! 1.5!!H-C*CH*C-H

Br

Br

Br

OH

7.22

6.84

6.907.43

5.35

Br

OH

15

Now: Trisubstituted Aromatic Rings: Total of three aromatic protons 1, 2, 3-Systems:

With symmetry, there are only two types of protons. Ex: 1,3-dibromo-2-chlorobenzene

Without symmetry: all protons are different – three signals. Ex: 1-bromo-2,3-dichlorobenzene

Y

X X

HH

Hdoublet

triplet

Y

X Y

HH

Hdoubletdoublet

triplet or dd

012345678PPM

012345678PPM

Br BrCl

Br ClCl

16

1, 2, 4-Systems: No Symmetry. Ever.

Ex: 4-bromo-1,2-dichlorobenzene

1, 3, 5-Systems: Always singlets (Amount of symmetry changes number of signals)

XH X

HHX

singlet

doublet

doublet

012345678PPM

YH H

XXH

YH H

ZXH

XH H

XXH

1 singlet 2 singlets 3 singlets

ClCl

Br

17

0123456789PPM

0123456789PPM

Br

Cl

Br

BrBr

Br

18

The goal for proton NMR is to be able to interpret and reconstruct the molecule’s structure. Now, combine splitting patterns with the integration (shown in brackets below). What molecular fragment does each of the following represent? a. [1H] q b. [2H] dt c. [1H] septet Determine the fragments, and the structure, for the following NMR spectrum of an alcohol, if the molecular formula is C6H14O:

0123456PPM

012345PPM

0.06 0.06 0.06 0.12 0.18 0.36

HO

CH3

Br

19

Assembling your fragments requires up to three pieces of information: Integration (tells you the number of protons in the actual signal), Splitting Pattern (tells you how many protons are next door) and Chemical Shift. Chemical Shift Integration Splitting Pattern Fragments Things to remember: 1. Begin by first identifying the number of signals and where on the X-axis they exist (Chemical Shift). Be flexible! Sometimes multiple functional groups are present and protons aren’t always where you expect them to be. 2. Calculate the integration ratio. Identify how many of each type you have. Recognize the number of CH3 versus CH2, etc. that you have in your molecule (integration ratio). Keep in mind that signals in the aromatic region are NEVER CH3 or CH2’s. 3. Splitting patterns are the toughest to interpret. Singlets, doublets and triplets are usually straightforward although triplets are often doublet-doublets in disguise (1 CH2 neighbor versus 2 CH neighbors). Quartets (CH3 neighbor) are often Doublet-Triplets in disguise (telling you there is a CH and a CH2 set of neighbors). Check back with your integration values to see what options are possible. Assign something as a multiplet if you don’t really know what it is. Seek help if necessary! Don’t flounder lost until the last moment… Once the pieces are built, cross off (mentally or literally?) the extra’s that appear multiple times (CH3-CH2 and CH2-CH3, for instance). Narrow down to the correct number of carbons in your molecular formula. But for heaven’s sake, don’t violate the octet rule!

20

Proton NMR Problems – CHEM 332 a. C2H4Br2

b. C3H7I

c. C6H14O - IR: 3200-3500, 1050 cm-1

0.11

0123456PPM

0.33

01234PPM

0.30 0.30 0.45

01234PPM

0.540.180.060.06

21

d. C9H13N - IR: “V”-shaped peak @3200-3500 cm-1

e. C5H10O2 - IR: 1745, 1240 cm-1

f. C9H10O2 - IR: 1695, 2730, 1210, 820 cm-1

012345678PPM

0.30 0.120.120.18 0.06

0123456PPM

0.11 0.33 0.66

0246810PPM

0.08 0.16 0.16 0.17 0.25

22

g. C6H12O - IR: 1712 cm-1

h. C9H10O - IR: 1685 cm-1

i. C7H13N - IR: 2210 cm-1

0123PPM

0.12 0.18 0.12 0.12 0.18

012345678PPM

0.55 0.22 0.33

0123PPM

0.36 0.180.120.12

23

j. C7H9N - IR: “W”-shaped peak @3200-3500 cm-1

k. C5H12O - IR: “U shaped peak @3200-3500, 2984, 2895, 1104 cm-1

l. C10H12O2 - IR:3065, 2983, 2901, 1745, 1604, 1565, 1462, 1280, 745, 695 cm-1

012345678PPM

0.210.150.07 0.21

01234PPM

0.360.120.12 0.060.06

012345678PPM

0.180.120.120.30

24

m. C7H9N IR: “W”-shaped peak @3200-3500, 3042, 2941, 2896, 1602, 1556, 1473, 792, 695 cm-1

n. C9H13N IR: “V”-shaped peak @3200-3500, 3054, 2986, 2897, 1601, 1562, 1451, 745, 697 cm-1

012345678PPM

0.210.150.07 0.21

012345678PPM

0.30 0.120.120.18 0.06

nuclear magnetic resonance spectroscopy · 2! 8.0!!h-ch2-ch-h chemnmr 1h estimation 1.69 3.34 br...

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