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CHAPTER 14: SPECTROSCOPY

CHEMICAL IDENTIFICATION

Historical Chemical Identification involved combustion analysis (to determine the molecular formula), simple chemical tests, degradation to smaller pieces and simple chemical tests on the resulting fragments. It was a very slow process and required painstaking detective work.

Maitotoxin is one of the largest natural products. It is very lethal (LD50 = 50 ng/kg). The structure was mostly determined by NMR experiments in 1993 and stereochemistry further determined in 19961.

1 Jones, M., Fleming, S.A., Organic Chemistry, 4th ed., Norton, 2010, pp. 745

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SPECTROSCOPY GENERALITIES

Wavelength Type of Radiation Effect on matter

10-10 m X-rays

10-8 m Ultraviolet

10-6 m Infrared

10-4 m Microwave

10-2 m Radio

NUCLEAR MAGNETIC RESONANCE (NMR) GENERALITIES

IMPORTANT ASPECTS OF AN NMR SPECTRUM

Chemical Shift: the frequency of each absorption (left or right position on spectrum, ppm number)

Integration: how many H atoms each grouping represents

Splitting: how many peaks are present in each grouping

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THE NMR PHENOMENON

NUCLEAR SPIN STATES

ABSORPTION

The a and b states are very close in energy, with only a slight excess in the a state. On a 60 MHz instrument, there are 1,000,000 nuclei in the b state and 1,000,009 nuclei in the a state.2

MAGNETIC RESONANCE IMAGING (MRI)

Clinical MRI instruments are 5,000-20,000 gauss (0.5-2.0 Tesla). The Earth’s magnetic field is 0.5 gauss.

I believe Enloe’s MRI is 14,092 gauss (H comes into resonance at 60 MHz).

In the MRI (NMR) instrument, hydrogen atoms from water undergo a “spin flip” and the instrument measures the amount of time it takes to return to normal (to relax). Water relaxes at different rates depending on its environment (tissue, tumors). The relaxation times are represented by colors or shades of grey.

2 Pavia, D.L., Lampman, G.M., Introduction to Spectroscopy, 4th ed., Brooks/Cole, 2009, pp. 111

α

β

α

β

α

β

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SPIN ACTIVE NUCLEI

1.   Not all atoms have “spin.” They must have odd mass, odd atomic # or both.

Commonly analyzed NMR active nuclei: 1H, 2H, 13C, 15N, 19F, 29Si, 31P, 119Sn

2.   Each spin active nuclei absorbs at a different frequency

NMR UNITS (FYI)

TETRAMETHYLSILANE (TMS) + UNITS

The internal standard TMS is sometimes used (meaning it is in the NMR solvent) to provide a reference absorption. The TMS absorption is defined as zero and other signals are referenced relative to this standard.

100 MHz instrument: signal at 1 ppm is 100 Hz greater than TMS 300 MHz instrument: signal at 1 ppm is 300 MHz greater than TMS.

Deuterium (2H)Carbon-13 (13C)Hydrogen (1H)

On a 300 MHz instrument:

300 MHz

75 MHz 46 MHz

TMS

H3C SiCH3

CH3

CH3

TMS

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NMR FEATURES NECESSARY FOR INTERPRETATION

SAMPLE PREPARATION AND SOLVENT

O

CH3CH2OH

CH3CH2OH

1 mL solvent

1 drop sample

O

solvent

sample

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CHEMICAL EQUIVALENCE

Chemically equivalent hydrogen atoms absorb in the same location.

How many signals should be present in the 1H NMR spectrum of each compound?

Structure

# signals

How many signals should ideally be present in the 1H NMR spectrum of each compound?

Structure

# signals

01234PPM

01234PPM

01234PPM

O O

O

0123PPM

0123PPM

O

0123PPM

H3C

CH3

Br

Cl

OH

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CHEMICAL SHIFT

ELECTRONIC SHIELDING

H C

H

H

C

O

C

H

H

Br

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CHEMICAL SHIFT TRENDS

1)   CORRELATION CHART

2)   PROXIMITY TO AN ELECTRONEGATIVE GROUP

Substance Shift

CH3Cl 3.06 ppm

CH2Cl2 CHCl3 CCl4

C C

H

H

O C

H

H

HN

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INTEGRATION

SIGNAL SPLITTING

012345PPM

012345PPM

H C

H

OH

H

H C

H

H

C

H

OH

H

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SIGNAL WITH ONE NEIGHBOR

SIGNAL WITH TWO NEIGHBORS

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SPLITTING SHORTCUT

What is the expected splitting (how many lines) for each signal?

Signal a b c d

Splitting

SIGNAL HEIGHTS

Pascal’s triangle describes the relative heights within a split signal.

singlet broad singlet doublet triplet quartet quintet sextet multiplet

s br. s d t q quint. sext. m

Br C

H

Br

C

H

H

Br

C C

H

H

C

H

C

H

O

O C

H

H

HH

H

Hba c d

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SIGNALS INVOLVED IN HYDROGEN BONDING

•   They can be seen over a broad range (ppm is concentration dependent).

•   The signals are often (but not always) broadened.

•   They are normally singlets (splitting is not seen even if they have neighbors).

DRAWING 1H NMR SPECTRUM FROM STRUCTURE

012345PPM

H C

H

H

C

O

O C

H

C

H

H

HH

4 3 2 1 0 ppm

CH3 CH2 OH CH3 CH2 OH

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OHCC

H H

H

C

CH

H

H

H

H

H

C

CC

C

CC C

C

O

H

HH

H

H

H

H

H

4 3 2 1 0 ppm

8 6 4 2 0 ppm 7 5 3 1

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DETERMINING STRUCTURE FROM 1H NMR SPECTRUM

COINCIDENTAL OVERLAP

EXAMPLE

RESOLUTION IMPROVES WITH MAGNET STRENGTH

Butte College 60 MHz Chico State 400 MHz and 300 MHz Biggest to date 1000 MHz (1 GHz)

012345678PPM

2H, t

3H, t

2H, q

3H, m

300 MHz

600 MHzResolution improves

The 3rd + 4th CH2�s overlap

ClCH2CH2CH2CH2CH3

O

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DEGREES OF UNSATURATION

MOLECULAR FORMULA

Alkane, C2H6 Alkene, C2H4 Alkyne, C2H2 C3H8

CALCULATING D.U.

1)   GENERAL FORMULA

 𝐷. 𝑈. =    & #( )&   –(#,)&

Ex. C6H8

Which of these structures are possible for C6H8?

2)   ATOMS OTHER THAN C

Convert other atoms into a C-H equivalent.

Halogens: Oxygen: Nitrogen:

Ex. C4H7NO

Which of these structures are possible for C4H7NO?

H C

H

H

C

H

H

H

C C

H

H H

H

C CH H H C

H

H

C

H

C

H

H

H

HC C

C

H H

H H

HH

H C

H

H

C

H

H

H

H C

H

H

C

Br

H

H

H C

H

H

C

H

H

H

H C

H

H

O C H

H

H

H C

H

H

C

H

H

H

H C

H

H

N H

H

NH2

HOH2N

O

HN

O

HN

O

NH

O

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APPROACH TO SOLVING 1H NMR PROBLEMS

1)   COMMON PATTERNS

Isopropyl group: Tert-butyl group:

Unknown C5H10O

H3CC

O

H3C CH3

CH2CH3H3C

CHCH3

Br

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Doubly substituted Aromatic:

Unknown C10H12O

012345678PPM

3H, s

2H, d 2H, d

OCH3 OCH3 OCH3

BrBr

Br

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2)   GENERAL APPROACH

1.   Calculate the degrees of unsaturation.

2.   Look at the integration of each signal. 2H is often a CH2 group, 3H is often a CH3 group.

3.   Look at the position of each signal to see what kind of H’s are present (aromatic, aldehyde), or what the H’s might be next to (O, Br, C=O).

4.   Look at how many peaks are in each signal (multiplicity) to determine neighbors. Start with simpler splittings like doublets or triplets.

5.   Try to connect pieces together as soon as possible. If there is only one signal that integrates to 1H and you find a signal that is neighboring 1H (because it is a doublet), put those pieces together.

6.   Keep building fragments of the structure, and check off the signals when you’ve already used them. Eventually the fragments should connect.

Unknown, C10H14

012345678PPM

3H, m

2H, t

1H, m

2H, quint.

3H, d3H, t

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PRACTICE PROBLEMS

Unknown A: C4H10O

Unknown B: C4H9Br

01234PPM

3H, s

1H, m

6H, d

01234PPM

2H, t 2H, quint.

2H, sext.

3H, t

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Unknown C: C9H10O2

Unknown D: C8H11N

012345678PPM

5H, m 2H, t2H, t

2H, s

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CARBON-13 NMR

DIFFERENCES BETWEEN 13C AND 1H

1.   13C nucleus absorbs at a different frequency than 1H nucleus (75 MHz on a 300 MHz magnet).

2.   Usable range is greater because the factors that affect chemical shift are felt through 1 fewer bond than with 1H.

3.   Only 1.08% of all carbon atoms are 13C while 99.98% of all hydrogen atoms are 1H. This makes 13C 6000× less sensitive than H: must use more concentrated samples.

4.   The instrument is set up so there is no splitting between 13C and 1H (every unique carbon shows up as a singlet).

5.   Integration is meaningless in 13C NMR.

6.   There is never any coincidental overlap.

Limonene (orange oil)

CDCl

O

CDCl3

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SIMILARITIES BETWEEN 13C AND 1H SPECTRA

1.   Chemical shifts parallel, are just more spread out in 13C NMR spectra.

2.   Symmetry considerations still apply.

How many total signals would we expect for each compound in their 13C NMR spectrum?

200 150 100 50 0 ppm

C

O

H

C

O

C

O

OH

C

O

OR

C

O

NH2

C

C=C

C C

C-O

C-Cl

C-NC-CH3

C-CH2-C

C CHC

C

Br

Cl

OBr Cl

Cl

Cl

Cl

Cl

Cl

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COMBINATION SPECTROSCOPY PROBLEMS

APPROACH

1.   Calculate D.U.

2.   Look briefly at IR for key functional groups (OH, NH, C=O, terminal alkyne)

3.   Look briefly at 13C NMR for general functional groups present (C=O, aromatics/alkenes), and check for symmetry (are there fewer signals than carbon atoms).

4.   Tackle the 1H NMR in the normal way.

PROBLEMS

Unknown A: C6H5NCl2

IR Spectrum of A: 13C NMR Spectrum of A:

1H NMR Spectrum of A:

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Unknown B: C5H9ClO2

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MORE COMBO PROBLEMS

Unknown C: C9H6O

13C NMR d: 175.0, 132.3, 128.4, 128.3, 122.7, 96.2, 88.2 ppm

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Unknown D: C4H8Cl2

01020304050PPM

01234PPM

1H

3H4H

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1H NMR Correlation Chart

The ranges below are approximate values for the bolded hydrogen atoms in the partial structures. Be aware that there will be cases where they will fall outside of their expected range.

These ranges represent the environment of the hydrogen atoms, not the quantity:

Br-CH3 will be found in the same location as Br-CH2-C, and Cl-CH2-C would also be found in the same range.

H on Alkyl Group H on Multiple Bonds C-CH3 0.7 – 1.3 ppm 1.7 – 2.7 ppm

C-CH2-C 1.2 – 1.4

4.6 – 5.7

1.4 – 1.7

6.5 – 8.0

Next to mild electron-withdrawing group

9.0 – 10.0

1.6 – 2.6 That can hydrogen bond

2.1 – 2.4

11.0 – 12.0

2.3 – 2.7

4.0 – 10.0

C-CH2-NH 2.2 – 2.9 C-CH2-OH 0.5 – 5.0

Next to strong electron-withdrawing group C-CH2-NH 0.5 – 5.0

C-CH2-O 3.2 – 3.8

C-CH2-Br 2.7 – 4.1

3.5 – 4.8

C-CH-Cl2 5.0 – 5.9

12 11 10 9 8 7 6 5 4 3 2 1 0 ppm

O

OH

HO

HH

C-CH

C-OH C-NH

CH-OCH-Cl

CH-Br

CHO

CH

OH

CH-Cl2

C CC H

H

H

C CHC

C

H

O

H

H

OH

O

OH

H OH

O

OCH

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