D.M. Collard 2007

9-2

TOPIC 9. STRUCTURE DETERMINATION(chapter 9 and part of chapter 2)

9-6

OBJECTIVES

1. Use combustion analysis to determine empirical formula.2. Determine molecular weight (and molecular formula) from mass

spectrometry.2. Calculate number of rings and double bonds from molecular formula.3. Determine functional groups present from infrared spectroscopy.4. Use 1H and 13C NMR spectroscopy to identify other structural features. 5. Combine conclusions from individual techniques to determine the structure

of organic compounds.

Problems in this section will be restricted to the following classes of aliphatic and aromatic compounds:

Alkanes, alkenes, alkynes, alkyl halides, alcohols, ethers, aldehydes, ketones, carboxylic acids, esters, acyl chlorides, anhydrides, amides, amines, nitriles

D.M. Collard 2007

9-7

???

COMBUSTION ANALYSIS

CxHyOz + O2 xCO2 + (y/2)H2O

Measure mass of CO2 and H2O formed from a known mass of compound;data cited as mass% of each element present. Generally do not measure mass%O.e.g., C H OMass% 54.51 9.09

Mole Ratio

=

=

Empirical formula

Be careful when rounding, elemental analyses provided ±0.4 mass% - do not round by more than 0.05 to 0.1. Multiply ratios before rounding.

9-8EMPIRICAL AND MOLECULAR FORMULAS;

DETERMINATION OF “SODAR”Molecular formula is an integral number of times the empirical formula.

C2H4OSum of double bonds and rings (“SODAR”)

For C,H,O: (2#C + 2 – #H) / 2

For C,H,O,N,Hal: (2#C + 2– #H – #Hal + #N) / 2Each Hal replaces a HEach N adds an extra HS,O: no effect on calculation

If SODAR calculated from empirical formula is not a positive integral (or 0), this cannot be the molecular formula.

C2H4O MOLECULAR FORMULA COULD BE: ???

D.M. Collard 2007

9-12If SODAR ≥4: consider the possible presence of a benzene ring

OO

9-13

MASS SPECTROMETRY

Theory and Experiment

M + e– (70 eV) M+ + 2e–

M+ lower mass ions

S:9.12-9.18

D.M. Collard 2007

9-14Data Availablee.g., mass spectrum of ethane, CH3CH3

Ions also fragment, which can give further clues about the structure (beyond scope of this course).

Knowledge of molecular weight (from mass spec) and empirical formula (from combustion analysis) allows determination of molecular formula.

0 10 20 30m/e

abun

danc

e

9-15

???

Empirical formula: C2H4O

Mass spectrum: M+ m/e = 88

MOLECULAR FORMULA IS:

D.M. Collard 2007

9-16SPECTROSCOPY:

THE INTERACTION OF LIGHT AND MATTER

HighestMost dangerous

Very dangerous

Microwaveoven, rotationof molecules

IR - bondsvibrating

Tanning raysskin cancer

What wesee

Lowest energyradio/TVNMR

High energy - kills life

S:9.1-9.2

9-17

INFRARED SPECTROSCOPY

Molecular Vibrations: Stretching and Bending

Analogy to masses and springs:

Spring constant (strength) , Frequency, υ

Masses , υ

Prob:2.29,43,45,

46

S:2.16

D.M. Collard 2007

9-18Electromagnetic RadiationE = hν c = λν

For molecular vibrations, Frequency, ν = 6 x 1012 to 1.25 x 1014 HzWavelength, λ = 50 to 2.5 μmDefine wavenumber,

= 200 to 4000 cm-1

Instrumentation SpectrumA simple scheme

IR source

sample

referencedetector

ν =1

λ (in cm)

wavenumber / cm-14000 600%

tran

smitt

ance

9-20Selected Infrared AbsorptionsFunctional Group Range cm-1 Intensity and shapesp3C─H 2850-2960 medium to strong; sharpsp2C─H 3010-3190 medium to strong; sharpspC─H about 3300 medium to strong; sharpC═C 1620-1660 weak to medium; sharpC≡C 2100-2260 weak to medium; sharpN─H 3300-3500 medium; broadO─H (very dilute) about 3600 medium; sharpO─H (H-bonded) 3200-3550 strong; broadO─H (carboxylic acid) 2500-3000 medium; very broadC─O 1050-1150 medium to strong; sharpC═O (aldehyde,ketone) 1690-1740 very strong; sharpC═O (ester) 1735-1750 very strong; sharpC═O (carboxylic acid) 1710-1780 strong; sharpC═O (amide) 1630-1690 strong; sharpC≡N 2200-2260 medium; sharp

D.M. Collard 2007

9-21Validating resonance theory

C═O (ester) 1735-1750 cm-1

C═O (amide) 1630-1690 cm-1

Which is the stronger bond?

Can we understand this using resonance theory?

N

O

O

O

N

O

O

O

9-22Infrared spectroscopy indicates the presence of particular bonds in a sample. The combination of bonds indicates which functional groups are present.You do not need to memorize the data in Table 2.7 (or the previous slide) for this class. However, you must develop experience at interpreting IR spectra to be able to determine the presence of particular functional groups.

Wavenumber / cm-14000 600

Ether

Carboxylic acid

Ester

Aldehyde or ketone

Alcohol

C-OC=OO-H

D.M. Collard 2007

9-23Hexane

4000 3000 2000 1500 1000 500Wavenumber (cm-1)

Tran

smitt

ance

(%) 100

50

0

9-241-Hexene

4000 3000 2000 1500 1000 500Wavenumber (cm-1)

Tran

smitt

ance

(%) 100

50

0

CH

CH2

D.M. Collard 2007

9-25

4000 3000 2000 1500 1000 500Wavenumber (cm-1)

Tran

smitt

ance

(%) 100

50

0

1-HexyneC

CH

9-26

4000 3000 2000 1500 1000 500Wavenumber (cm-1)

Tran

smitt

ance

(%) 100

50

0

Toluene

CH3

D.M. Collard 2007

9-27Butyl methyl ether

4000 3000 2000 1500 1000 500Wavenumber (cm-1)

Tran

smitt

ance

(%) 100

50

0

O

9-28

4000 3000 2000 1500 1000 500Wavenumber (cm-1)

Tran

smitt

ance

(%) 100

50

0

1-Hexanol

OH

D.M. Collard 2007

9-29

4000 3000 2000 1500 1000 500Wavenumber (cm-1)

Tran

smitt

ance

(%) 100

50

0

2-HexanoneO

9-30

4000 3000 2000 1500 1000 500Wavenumber (cm-1)

Tran

smitt

ance

(%) 100

50

0

Hexanoic Acid

OH

O

D.M. Collard 2007

9-31

4000 3000 2000 1500 1000 500Wavenumber (cm-1)

Tran

smitt

ance

(%) 100

50

0

Methyl Propionate

9-34

???

4000 3000 2000 1500 1000 500Wavenumber (cm-1)

Tran

smitt

ance

(%) 100

50

0

C4H8O2

Structure contains:

Draw eight possible structuresconsistent with this data.

D.M. Collard 2007

9-36

4000 3000 2000 1500 1000 500Wavenumber (cm-1)

Tran

smitt

ance

(%) 100

50

0

Problem: C8H8O2. What functional group(s) is(are) present? What other structural features?

9-37Problem: Determine the structure of the compound with the following empirical formula and IR spectrum.Empirical formula: C4H8OIR; peak at 1720 cm-1; no broad peak at 3300, no strong peaks at 1000 to 1200 cm-1, no strong peaks at 1600-1650 cm-1

D.M. Collard 2007

9-39

Important infrared adsorptions

Wavenumber / cm-14000 600

1700

1000-1200 C–O

2900 sp3C–H

3100 sp2C–H

3500, O-H

COOH

1600 C=C

9-40HYDROGEN (PROTON) NUCLEAR MAGNETIC

RESONANCE (1H NMR) SPECTROSCOPYNuclear Spin (1H, 13C, 19F, 31P)

S:9.3-9.4

N

N

S

S

S

N

D.M. Collard 2007

9-41The Energy of the Spin States of Hydrogen Nuclei in a Magnetic Field

Bo

EBo=0

The frequency ν required to flip the spin state from α to β is called resonance frequency

α

β

9-42An early 1H NMR spectrum

...not much chemical information

A modern 1H NMR spectrum

…lots of chemical informationE=hν

E=hν

abso

rptio

n

D.M. Collard 2007

9-44

Ha

Hb

If the applied magnetic field is held constant, the protons Ha and Hbabsorb irradiation with different frequencies.

B

E

hνab

s orp

tion

The spectrum

9-45Preview: Types of Information available from a 1H NMR spectrum

A 1H nuclear magnetic resonance spectrum contains information about the:

(a) number of different types of proton

(b) relative number of each type of proton

(c) proximity to functional groups

(d) the number of adjacent protons

HCH

Cl CH

HH

D.M. Collard 2007

9-46(a) The number of signals in the spectrum is the number of types of proton.

1H NMR spectrum of methane CH4

one peak ⇒ one type of proton

1H NMR spectrum of ethane CH3–CH3

one peak ⇒ one type of proton

abso

rptio

nab

s orp

tion

hν

hν

9-47Protons that are related to each other by a rotation or plane of symmetry are identical (chemical shift equivalent)

HCH

H CH

HH

H

HH

H H

H

CH3CH2CH2CHCH3

CH3

CH3

HH

H H

H

Cl

ClH

H H

H

CH3

HCl

H H

H

Hax

Hax

HeqHeq

Hax

Hax

Hax

HaxHeq

Heq

Heq

Heqlow temperature: 2 signals

high temperature (room T): 1 averaged signal

H3C C CH3

OH

H

D.M. Collard 2007

9-481H nuclear magnetic resonance spectrum of methanol…

CH3–OH

two peaks ⇒ two types of proton

abso

rptio

n

hν

9-51

CH3H3C

HH

H H

H3C OH

(b) The relative area of each peak (the integration) corresponds to the relative number of each type of proton

1H NMR spectrum of methanol

integral ratio : ⇒ ratio of types of proton

1H NMR spectrum of p-xylene

integral ratio : ⇒ ratio of types of proton

abso

rptio

nab

s orp

tion

hν

hν

D.M. Collard 2007

9-53

(c) The chemical shift indicates the environment of the proton

1H NMR spectrum of methanol

S:9.5-9.6

chemical shift, δ /

abso

rptio

n

9-54Shielding and Deshielding

Bo Bind Ha

B

E

Bo-Bind Boshielded

abso

rptio

n

hν

hν

D.M. Collard 2007

9-55The δ-scale

The frequency at which a proton resonates is measured relative to the frequency for the protons of tetramethylsilane, TMS (which resonates at a relatively low frequency), and cited on the δ (delta) scale in parts per million of the frequency at which TMS resonates.

δ = x 106 [ppm] νproton - νTMS

νTMS

H3C Si

CH3

CH3

CH3

electronegativity C=2.5 Si=1.7

9-56

Field Strength υTMS / Hz υCH4 / Hz δCH4 / ppmB / T2.33 100,000,000 100,000,1004.66 200,000,000 200,000,2007.00 300,000,000 300,000,300

- The resonance frequency (in Hz) depends on magnet strength.

- The chemical shift (δ) is independent of magnet strength.

TMS

Bo

E Bo=0

α

β

D.M. Collard 2007

9-57Effect of Structure on Chemical Shift (δ scale, ppm)

CH3-CH3 CH3- N(CH3)2 CH3-OCH3 CH3-F0.9 2.2 3.2 4.3

CH3-Cl3.1

CH3-Br2.7

CH3-I2.2

CHCl3 CH2Cl2 CH3Cl7.3 5.3 3.1

Ha

B

E

Bo-Bind Boshielded

hν

Bo-Binddeshielded

9-58

You will be provided with a copy of Table 9.1 on exams. This provides approximate ranges for values of chemical shifts for particular types of protons. Remember that protons adjacent to two (or more) electron withdrawing groups will appear further downfield than a proton adjacent to only one.

Approximate [1H]-NMR chemical shifts for various substitutent groups Expected range

Type of proton Chemical shift (δ) (CH3)4Si 0.00 CH3-C-R (sp3) 0.9 - 1.8 -CH2-C-R (sp3) 1.1 - 2.0 -CH-C-R (sp3) 1.3 - 2.1 H-C-N 2.2 - 2.9 H-C-O 3.3 - 3.7 H-C-Cl 3.1 - 4.1 H-C-Br 2.7 - 4.1 H-C-C=O 2.1 - 2.5 H-C-C=C 1.6 - 2.6 H-C-Ar 2.3 -2.8 H-C=O (sp2) 9 - 10 H-C=C (sp2) 4.5 - 6.5 H-Ar (sp2) 6.5 - 8.5 H-C C (sp) 2.5 H-N (amine) 1 - 3 H-OR (alcohol) 0.5 - 5 H-OAr (phenol) 6 - 8 H-O2CR (acid) 10 - 13

D.M. Collard 2007

9-59

H H

H

H H

H

H

H

H

H

H

δ=7.3 δ=5.3

9-60

Other peaks in spectrum:

δ / ppm10 5 0

TMSH2OCDCl3

D.M. Collard 2007

9-64

(d) The multiplicity of a signal indicates the number of adjacent protons

1H NMR spectrum of ethanol

S:9.8

9-65Spin-Spin Coupling

The magnetic field experienced by a proton is effected by the magnetic field generated by each adjacent proton.

Bo

If Hb spin ↑, need to apply larger E (=hν)to achieve resonance of HaIf Hb spin ↓, need to apply smaller E (=hν)to achieve resonance of Ha

Ha

B

E

Bo+↓ Bo+↑C C

Hb Ha

D.M. Collard 2007

9-66

Bo

To achieve resonance of HaIf Hb spins ↑ + ↑, need to apply higher νIf Hb spins ↑ + ↓, or, ↓ + ↑ signal appears at δaIf Hb spins ↓ + ↓, need to apply lower ν

C C

Hb

Hb

Ha

9-67Coupling is only observed between non-equivalent protons.

The signal for a proton with N adjacent equivalent protons will be split into a multiplet consisting of N+1 lines.

The relative area of each peak within a multiplet can be determined from Pascal’s triangle

D.M. Collard 2007

9-70Some Common Sets of Multiplets

Et–X a b

i-Pr–X ab

t-Bu–X a

CH2 CH3

CCH3

CH3

H

C CH3

CH3

CH3

a b

aba

a aa

X

X

X

9-71

H3C CHCl2

D.M. Collard 2007

9-72Problems: How can you get a pentet?

9-73Multiplets of Multiplets

Jab = Jbc Jab > Jbc

C C C

HbHa Hc

D.M. Collard 2007

9-74Geminal protons

Rings

Vinyl groups

R

RC

H

H

Ha

Hb

R

Hc

Ha

Hb

R

Hc

9-75Summary: Types of information available from 1H NMR spectrum

Number of signals Number of types of proton

Integral of signals Relative number of each type of proton

Chemical Shift Electronic environment

Multiplicity Number of adjacent protons

D.M. Collard 2007

9-76Problems: Predict the appearance (chemical shift, integral and multiplicity) of the 1H NMR signals of the following compounds.

(a) 2-butanone, CH3COCH2CH3

(b) methyl 3-chloropropanoate, ClCH2CH2CO2CH3

9-77Problems: Provide structures consistent with the following data.

(a) Compound A: C2H4Cl2, one singlet in 1H NMR spectrum

(b) Compound B: C5H12O2, two singlets in 1H NMR spectrum

D.M. Collard 2007

9-78C4H6O2

SODAR = 1IR: 1720 cm-1 » C=O

no peak at 3300 cm-1 » no O-H ???

O H

O

O

O

O H

O

O

O

HO

O

H

O

O

O

O

HO

O

1H NMR

9-81

13C NMR SPECTROMETRY

Introduction13C (not 12C) has nuclear spin

However, 13C is only present at 1.1% abundance

- Signals are weak- Spectra usually acquired without multiplicity information

- Larger range of chemical shifts (0 to >200 ppm)

S:9.10

D.M. Collard 2007

9-82

δ / ppm

200 100 50 0150

9-83Types of Information available from a 13C NMR spectrum

A 13C nuclear magnetic resonance spectrum contains information about the:(a) number of different types of carbon

- Each peak corresponds to a different type of carbon(b) type of carbons and proximity to functional groups

- Chemical shift provides information about the typeof carbon present

Unlike 1H NMR spectra, simple 13C NMR spectra do not provide information about the:

relative number of each type of protonor

number of adjacent protons or carbons

D.M. Collard 2007

9-84

USING DATA TO DETERMINE STRUCTURE

1. Write down conclusions from each individual technique- Use combustion analysis to determine empirical formula, and mass

spectrometry to give molecular weight (and molecular formula)- Calculate number of rings and double bonds from molecular formula

(SODAR)- Determine presence of functional groups present from infrared

spectroscopy.- Use 1H and 13C NMR spectroscopy to identify other structural features.

2. Use these conclusions to determine the structure3. Check your answer (predict the spectra; do your predicted spectra match the

data provided?)

9-85WORK PROBLEMS!!In the remaining lectures we will cover problem solving approaches for

determining organic structures from spectral (and other) information. Additional problem appear in the following sources:

Solomons 9.29-38www.nd.edu/~smithgrp/structure/workbook.html

(just do the “easy” problems from this site. The answers to these problems are NOTavailable from this site; do not bother Prof. Smith. The answers to the easy problems, along with answers to the problems in these notes are posted on the “Notes and HWebs” page of the course web site)

D.M. Collard 2007

9-86USING SPECTRAL INFORMATION AND OTHER DATA TO DETERMINE THE

STRUCTURE OF AN UNKNOWN COMPOUND

Elemental Analysis: C, 54.51; H, 9.09Mass spectrometry: m/e M+, 88

A. If you are provided with the elemental analysis (mass% C,H,O...), determine the empirical formula i.e., ratio of atoms, C : H : O : ... = mass% C/12 : mass %H/1 : mass% O/16 : ...

empirical formula:

REMEMBER: The molecular formula might be an integral times the empirical formula

If you are provided with the molecular weight (from mass spectrometry), determine the molecular formula empirical formula weight = molecular weight = molecular formula: Determine the number of sites of unsaturation (i.e., SODAR = π bonds + rings): SODAR = (2#C + 2 - #H - #Hal + #N) / 2 ; SODAR=

9-87Infrared Spectrum

4000 3000 2000 1500 1000 500frequency – cm-1

100

50

0

B. Bonds vibrate at certain frequencies. Determine the bonds present (and hence the functional groups) by considering the infrared spectrum

i. Does your compound contain oxygen Y ? N If there is no oxygen, go to part Bii.

(a) C=O bonds vibrate at approximately 1700 cm-1, usually giving rise to a strong absorption. Is there a strong absorption around 1700 cm-1? Some tables list frequencies for specific carbonyl-containing functional groups.

Y ? N Conclusion:

(b) C-O bonds vibrate at approximately 1200-1000 cm-1; these peaks are usually strong. Is there a strong absorption at 1000-1200 cm-1?

Y ? N Conclusion:

(c) O-H bonds give rise to a broad vibration band; in alcohols this appears at approximately 3300 cm-1, but in carboxylic acids this band is very broad and might also appear between 2800 and 3200 cm-1 (which overlaps with sharp C-H bond vibrational bands). If nitrogen is present, remember that N-H bonds also vibrate at 3000-3500 cm-1. Is there a broad absorption in the 3500-3000 cm-1 range?

Y ? N Conclusion:

abso

rban

ce -

%

D.M. Collard 2007

9-88

(d) Based on the presence or absence of vibrational bands arising from covalent bonds to oxygen, you should be able to predict the functional groups present in the molecule. Circle the combination of bonds present and the corresponding functional group:

Bonds Present Absent Functional Group BondsPresent Absent Functional Group C=O, O-H, C-O - acid C-O C=O, O-H ether C=O, C-O O-H ester C=O O-H, C-O aldehyde or ketone O-H, C-O C=O alcohol C=O, N-H O-H, C-O 1° or 2° amide

ii. Other common vibration bands to look for are those for C≡C and C=C. If nitrogen is present, look for C�N and N-H vibration bands. Are any of these bands present?

Y ? N Conclusion:

iii. sp2 C-H bonds (in arenes and alkenes) give rise to a sharp vibrational band at 3100-3000 cm-1. Is there a sharp absorption at 3100-3000 cm-1?

Y ? N Conclusion: iv. Draw the functional groups indicated by the infrared spectrum (i.e., structures from i(d) , ii

and iii)

4000 3000 2000 1500 1000 500

frequency – cm-1

abso

rban

ce -

%

100

50

0

9-89

C. Consider the 13C NMR spectrum...

i. Each type of carbon gives rise to a separate singlet How many types are carbon are there?

ii. The 13C chemical shift gives you a clue about the type of carbon (i.e., carbonyl, aromatic, alkenes) and the presence of electronegative substituents.

Conclusion:

200 150 100 50 0δ / ppm

13C NMR Spectrum

D.M. Collard 2007

9-90

D. Certain common features give rise to recognizable patterns in the 1H NMR spectrum...

i. Ethyl groups always give rise to a quartet (2H) downfield from a triplet (3H). Is there a quartet downfield from a triplet?

Y N Conclusion:

ii. iso-Propyl groups always give a septet (1H) downfield from a doublet (6H). Remember, you might not see the weak outside peaks in the septet. Is there a septet downfield from a doublet?

Y N Conclusion:

iii. tert-Butyl groups always give a singlet (9H) at ca. δ1.2. Is there a singlet at ca. δ1.2?

Y N Conclusion:

iv. If there are ethyl or isopropyl groups: consider the chemical shift of the downfield signal, it gives you a clue about the substituent (i.e., “X” in CH3CH2-X or (CH3)2CH-X).

Conclusion:

1H NMR Spectrum

9-91

v. Aromatic protons appear at δ6.5-8.5. Is there a signal at δ6.5-8.5?

Y N Conclusion:

vi. If there are aromatic protons: consider the multiplicity and integration, they give you clues about the substitution pattern?

Conclusion:

vii. Certain functional groups that you have already identified by infrared give rise to specific signals in the 1H NMR, such as acid (-CO2H), aldehyde (-CHO), alcohol (-OH), amine (-NH). Do signals corresponding to these functional groups appear in the 1H NMR spectrum?

Y N Conclusion:

viii. Are there any other singlets or multiplets which you have not assigned? If so, consider the multiplicity and integration. Here is a list of common isolated structural features (i.e., without neighboring protons) and their corresponding signals (multiplicity and integration). Do any of these appear in the spectrum? (Circle the combination of peaks and the structural subunit). You will also have to consider vinylic protons in D(x).

CH-CH d(1) d(1) CH2-CH2 t(2) t(2) CH-CH3 q(1) d(3) CH-CH2 t(1) d(2) CH3 s(3)

ix. If you have identified additional fragments in part viii: consider the chemical shift of the downfield signal, it gives you a clue about substituents.

Conclusion:

D.M. Collard 2007

9-92

ix. If any of these multiplets appear: consider the chemical shift of the downfield signal, it gives

you a clue about the substituents on the fragment (Remember that substitutent effects are cumulative).

Conclusion: x. Vinylic protons appear at δ4.5-6.5. Are there signals at δ4.5-6.5 which you have not yet

accounted for?

Y N Conclusion:

xi. If there are vinylic protons: consider the multiplicity and integration, they give you clues about the substitution pattern.

Conclusion:

xii. Draw the substructures derived from the appearance of the 1H NMR spectrum (i-xi).

9-93

E. If you are provided with the molecular formula: do you need to account for any other

atoms in the molecule other than those determined in parts B,C and D?

Y N Remaining atoms: F. ...Bring it all together

Draw the functional groups derived from IR (B), substructures derived from NMR (C and D), and any other atoms left over in the molecular formula (E).

Bringing it all together

D.M. Collard 2007

9-94

G. FINAL STRUCTURE Combine these fragments in such a way that the valency of each atom (H=1, C= 4, O=2, N=3) and the number of sites of unsaturation, and formula are satisfied. Remember to consider symmetrical structures.

Check your answer: predict the spectra of your final structure, it should be the same as those provided

9-95

WORK AS MANY PROBLEMS AS

POSSIBLE…

D.M. Collard 2007

9-96

4000 3000 2000 1500 1000 500

13C NMR 1H NMR

IR 100

50

0

trans

mitt

ance

Wavenumber / cm-1

Mass Spec

Elemental Analysis: C, 54.51; H, 9.09

9-97

4000 3000 2000 1500 1000 500

13C NMR 1H NMR

Empirical Formula: C4H5

IR 100

50

0

trans

mitt

ance

Wavenumber / cm-1

Mass Spec

2-lines

Elemental Analysis: C, 90.38; H, 9.47

D.M. Collard 2007

9-99

4000 3000 2000 1500 1000 500

13C NMR 1H NMR

Empirical Formula: C5H7

Mass Spec: M+ m/e= 134

IR 100

50

0

trans

mitt

ance

Wavenumber / cm-1

9-100

4000 3000 2000 1500 1000 500

13C NMR 1H NMR

Empirical Formula: C5H5O

Mass Spec: M+ m/e= 162

IR 100

50

0

trans

mitt

ance

Wavenumber / cm-1

H2O

D.M. Collard 2007

9-102

4000 3000 2000 1500 1000 500

13C NMR 1H NMR

Empirical Formula: C5H10O2

Mass Spec: M+ m/e= 102

IR 100

50

0

trans

mitt

ance

Wavenumber / cm-1

H2O

9-103

4000 3000 2000 1500 1000 500

13C NMR 1H NMR

Empirical Formula: C4H10O

Mass Spec: M+ m/e= 74

IR 100

50

0

trans

mitt

ance

Wavenumber / cm-1

D.M. Collard 2007

9-105

4000 3000 2000 1500 1000 500

13C NMR 1H NMR

Empirical Formula: C6H14O

Mass Spec: M+ m/e= 102

IR 100

50

0

trans

mitt

ance

Wavenumber / cm-1

9-107

4000 3000 2000 1500 1000 500

13C NMR 1H NMR

Empirical Formula: C3H8O2

Mass Spec: M+ m/e= 76

IR 100

50

0

trans

mitt

ance

Wavenumber / cm-1

D.M. Collard 2007

9-108

4000 3000 2000 1500 1000 500

13C NMR 1H NMR

Empirical Formula: C8H9Cl

Mass Spec: M+ m/e= 140 and 142 (3:1 ratio)

IR 100

50

0

trans

mitt

ance

Wavenumber / cm-1

2-lines

9-110

4000 3000 2000 1500 1000 500

13C NMR 1H NMR

Empirical Formula: C7H9N

Mass Spec: M+ m/e= 107

IR 100

50

0

trans

mitt

ance

Wavenumber / cm-1

D.M. Collard 2007

9-111

4000 3000 2000 1500 1000 500

13C NMR 1H NMR

Empirical Formula: C7H9N

Mass Spec: M+ m/e= 107

IR 100

50

0

trans

mitt

ance

Wavenumber / cm-1

9-112

4000 3000 2000 1500 1000 500

13C NMR 1H NMR

Empirical Formula: C4H4O

Mass Spec: M+ m/e= 136

IR 100

50

0

trans

mitt

ance

Wavenumber / cm-1

2-lines

D.M. Collard 2007

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4000 3000 2000 1500 1000 500

13C NMR 1H NMR

Empirical Formula: C8H9Br

Mass Spec: M+ m/e= 184 and 186 (1:1 ratio)

IR 100

50

0

trans

mitt

ance

Wavenumber / cm-1

2-lines

9-114

4000 3000 2000 1500 1000 500

13C NMR 1H NMR

Empirical Formula: C4H5O

Mass Spec: M+ m/e= 138

IR 100

50

0

trans

mitt

ance

Wavenumber / cm-1

D.M. Collard 2007

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CHAPTER 9 ON EXAM 5

Types of Questions- Use data from elemental analysis, infrared spectroscopy and nuclear

magnetic resonance spectrometry (1H and 13C) to determine the structure of organic compounds.

- In class assignment- Take home assignment (more complex problems)

Preparing for Exam 4:- Work as many problems as possible. - Work in groups.- Do the “Learning Group Problem” at the end of the chapter.- Work through the practice exam

CHEM2311/2380 DETERMINATION OF ORGANIC STRUCTURES

This handout introduces the process of using spectroscopic data to determine the structure of organic compounds containing common functional groups and structural units (e.g., alkanes, arenes, alkenes, alkynes, alkyl halides, alcohols, aldehydes, ketones, carboxylic acids, esters, amines, amides, nitriles). You need to ask these questions when faced with any structure determination problem. In addition to chemical knowledge, you must apply some reasoning skills to arrive the correct answer. Fill in all of the blanks. By the time you get to Section F you will have a few (3-5) fragments which can only be put together in a limited number of ways to arrive at the right answer in Section G. When you have proposed a structure try to predict its spectra: They should match the data provided if this is the right answer!

A. If you are provided with the elemental analysis (mass% C,H,O...), determine the empirical formula i.e., ratio of atoms, C : H : O : ... = mass% C/12 : mass %H/1 : mass% O/16 : ...

empirical formula:

REMEMBER: The molecular formula might be an integral times the empirical formula

If you are provided with the molecular weight (from mass spectrometry), determine the molecular formula

empirical formula weight = molecular weight = molecular formula: Determine the number of sites of unsaturation (i.e., SODAR = π bonds + rings): SODAR = (2#C + 2 - #H - #Hal + #N) / 2 ; SODAR=

B. Bonds vibrate at certain frequencies. Determine the bonds present (and hence the functional groups) by considering the infrared spectrum

i. Does your compound contain oxygen Y ? N If there is no oxygen, go to part Bii.

(a) C=O bonds vibrate at approximately 1700 cm-1, usually giving rise to a strong absorption. Is there a strong absorption around 1700 cm-1? Some tables list frequencies for specific carbonyl-containing functional groups.

Y ? N Conclusion:

(b) C-O bonds vibrate at approximately 1200-1000 cm-1; these peaks are usually strong. Is there a strong absorption at 1000-1200 cm-1?

Y ? N Conclusion:

(c) O-H bonds give rise to a broad vibration band; in alcohols this appears at approximately 3300 cm-1, but in carboxylic acids this band is very broad and might also appear between 2800 and 3200 cm-1 (which overlaps with sharp C-H bond vibrational bands). If nitrogen is present, remember that N-H bonds also vibrate at 3000-3500 cm-1. Is there a broad absorption in the 3500-3000 cm-1 range?

Y ? N Conclusion:

(d) Based on the presence or absence of vibrational bands arising from covalent bonds to oxygen, you should be able to predict the functional groups present in the molecule. Circle the combination of bonds present and the corresponding functional group:

Bonds Present Absent Functional Group BondsPresent Absent Functional Group C=O, O-H, C-O - acid C-O C=O, O-H ether C=O, C-O O-H ester C=O O-H, C-O aldehyde or ketone O-H, C-O C=O alcohol C=O, N-H O-H, C-O 1° or 2° amide

ii. Other common vibration bands to look for are those for C≡C and C=C. If nitrogen is present, look for C≡N and N-H vibration bands. Are any of these bands present?

Y ? N Conclusion:

iii. sp2 C-H bonds (in arenes and alkenes) give rise to a sharp vibrational band at 3100-3000 cm-1. Is there a sharp absorption at 3100-3000 cm-1?

Y ? N Conclusion:

© D.M. Collard & H.M. Deutsch, 2000

iv. Draw the functional groups indicated by the infrared spectrum (i.e., structures from i(d) , ii and iii)

C. Consider the 13C NMR spectrum...

i. Each type of carbon gives rise to a separate singlet How many types are carbon are there?

ii. The 13C chemical shift gives you a clue about the type of carbon (i.e., carbonyl, aromatic, alkenes) and the presence of electronegative substituents.

Conclusion:

D. Certain common features give rise to recognizable patterns in the 1H NMR spectrum...

i. Ethyl groups always give rise to a quartet (2H) downfield from a triplet (3H). Is there a quartet downfield from a triplet?

Y N Conclusion:

ii. iso-Propyl groups always give a septet (1H) downfield from a doublet (6H). Remember, you might not see the weak outside peaks in the septet. Is there a septet downfield from a doublet?

Y N Conclusion:

iii. tert-Butyl groups always give a singlet (9H) at ca. δ1.2. Is there a singlet at ca. δ1.2?

Y N Conclusion:

iv. If there are ethyl or isopropyl groups: consider the chemical shift of the downfield signal, it gives you a clue about the substituent (i.e., “X” in CH3CH2-X or (CH3)2CH-X).

Conclusion:

v. Aromatic protons appear at δ 6.5-8.5. Is there a signal at δ6.5-8.5?

Y N Conclusion:

vi. If there are aromatic protons: consider the multiplicity and integration, they give you clues about the substitution pattern?

Conclusion:

vii. Certain functional groups that you have already identified by infrared give rise to specific signals in the 1H NMR, such as acid (-CO2H), aldehyde (-CHO), alcohol (-OH), amine (-NH). Do signals corresponding to these functional groups appear in the 1H NMR spectrum?

Y N Conclusion:

viii. Are there any other singlets or multiplets which you have not assigned? If so, consider the multiplicity and integration. Here is a list of common isolated structural features (i.e., without neighboring protons) and their corresponding signals (multiplicity and integration). Do any of these appear in the spectrum? (Circle the combination of peaks and the structural subunit). You will also have to consider vinylic protons in D(x).

CH-CH d(1) d(1) CH2-CH2 t(2) t(2) CH-CH3 q(1) d(3) CH-CH2 t(1) d(2) CH3 s(3)

ix. If you have identified additional fragments in part viii: consider the chemical shift of the downfield signal, it gives you a clue about substituents.

Conclusion:

© D.M. Collard & H.M. Deutsch, 2000

ix. If any of these multiplets appear: consider the chemical shift of the downfield signal, it gives you a clue about the substituents on the fragment (Remember that substitutent effects are cumulative).

Conclusion:

NOTE OF CAUTION: you will observe other multiplets if protons couple to more than three neighbors, or to two different types of neighbors

(consider X-CHx-CHy-CHz-X and X-CHx-CHy-CHz-Y)

x. Vinylic protons appear at δ 4.5-6.5. Are there signals at δ 4.5-6.5 which you have not yet accounted for?

Y N Conclusion:

xi. If there are vinylic protons: consider the multiplicity and integration, they give you clues about the substitution pattern.

Conclusion:

xii. Draw the substructures derived from the appearance of the 1H NMR spectrum (i-xi).

E. If you are provided with the molecular formula: do you need to account for any other atoms in the molecule

other than those determined in parts B,C and D?

Y N Remaining atoms: F. ...Bring it all together

Draw the functional groups derived from IR (B), substructures derived from NMR (C and D), and any other atoms left over in the molecular formula (E).

G. FINAL STRUCTURE

Combine these fragments in such a way that the valency of each atom (H=1, C= 4, O=2, N=3) and the number of sites of unsaturation, and formula are satisfied. Remember to consider symmetrical structures.

Check your answer: Predict the spectra of your final structure - they should be the same as those provided

© D.M. Collard & H.M. Deutsch, 2000