solving unknown structures using nmr organic structure analysis, crews, rodriguez and jaspars

Solving Unknown Structures Using NMR

Organic Structure Analysis, Crews, Rodriguez and Jaspars

Six Simple Steps for Successful Structure Solution

• Get molecular formula. Use combustion analysis, mass spectrum and/or 13C NMR spectrum. Calculate double bond equivalents.

• Determine functional groups from IR, 1H and 13C NMR • Compare 1H integrals to number of H’s in the MF. • Determine coupling constants (J’s) for all multiplets. • Use information from 3. and 4. to construct spin systems

(substructures) • Assemble substructures in all possible ways, taking account of

dbe and functional groups. Make sure the integrals and coupling patterns agree with the proposed structure.



USING 1H NMR DATAUNKNOWN B

• A compound shows an M+. in the EIMS at 154 m/z• Further fragments are at 121, 93, 71, 55 and 39 m/z• The IR shows bands at 3400 cm-1 (broad) & 1450 cm-1

Use the 1H and 13C data to determine the structure of the compound


d

s

d

t

s

tq

q

t

q

13C NMR DATAUNKNOWN B

A B C D E F GHI J


MOLECULAR FORMULA DETERMINATIONUNKNOWN B

(C)2 + (CH)2 + (CH2)3 + (CH3)3 = C10H17

s d t q

BE

AC

DFI

GHJ

= 137 Da

The M+. appears at 154 m/z, so there isa mass difference of 17 Da (= OH)

Therefore molecular formula = C10H18O (2 dbe)


1H NMR DATAUNKNOWN B

Integrals:

H

3H 2H3H 3H

5HONLY 17 H!


SUBSTRUCTURESUNKNOWN B

H3C

H

CH3

A

C 125 d, 132 sH 5.05 t&Me groups at H 1.5

H

H

H

B

C 112 t, 146 dH 5.00 dd, 5.15 dd, 5.85 dd

H3C OH

C

C 73 sH 1.05 s

H H H H

D'D

MF = C10H18O !


WORKING STRUCTURESUNKNOWN B

OH

Ha

Ha

Ha

OH

OH

A – Ha should be ddt

B – Ha should be ddt

C – Ha should be dd


MASS SPECTRAL FRAGMENTATIONUNKNOWN B

OH

Ha

O+mass spectral fragmentation

1m/z 99

H

HaOH

+O

O+

mass spectral fragmentation

2

m/z 113

m/z 85

H

H

Ha

OH +O

mass spectral fragmentation

m/z 71H

3

Fragments at: 121, 93, 71, 55 and 39 m/z


USING MASS SPECTRAL DATAUNKNOWN G

• A compound shows an M+. in the EIMS at 128 m/z• Further fragments are at 99, 83, 72 and 57 m/z• The IR shows bands at 1680 cm-1 (strong) &

bands at 1400 - 1500 cm-1

Use the 1H and 13C NMR and MS data to determine the structure of the compound


13C NMR DATAUNKNOWN G

A B C D E FG

d dd t

t t q


MOLECULAR FORMULA DETERMINATIONUNKNOWN G

(C)0 + (CH)3 + (CH2)3 + (CH3)1 = C7H12

s d t qABC

DEF

G

= 96 Da

The M+. appears at 128 m/z, so there isa mass difference of 32 Da (= O2)

Therefore molecular formula = C7H12O2 (2 dbe)


1H NMR DATAUNKNOWN G

Integrals:

HH

H H

3H

H H2H

H

12 H Total


SUBSTRUCTURES UNKNOWN G

C 140 d, 101 dH 6.15 d, 4.70 m

C 96 dH 4.9 tO

H

H

O

O

H

O

H H

C 64 tMe

4 Oxygens in substructures but only 2 in MF

MF = C7H12O2 !


WORKING STRUCTURES UNKNOWN G

O

O

O OO O

36*

62 20*

15

64

26

10*64

15

O O O

O O+.

+

m/z 83

m/z 99

+.O O

O O

m/z 72m/z 56

+.

13C Shift additivity data

MS Fragmentation

Retro Diels-Alder

Fragments are at 99, 83, 72 and 57 m/z


USING COSY DATAUNKNOWN H

• A compound shows an [M + H]+ in the FAB MS at 132 m/z• MW = 131 (Odd) therefore odd number of nitrogens• A further fragment is at 86 m/z• The IR shows bands at 3400cm-1 (broad) & 1640 cm-1 (broad)

Use the NMR data to determine the structure of the compound


d

s

d

t

t

13C NMR DATAUNKNOWN H

A B C D E


MOLECULAR FORMULA DETERMINATIONUNKNOWN H

(C)1 + (CH)2 + (CH2)2 + (CH3)0 = C5H6

s d t q

A BC

DE

= 66 Da

The MW is 131, so there isa mass difference of 65 Da (= NO3H3)

Therefore molecular formula = C6H9NO3 (2 dbe)


1H NMR DATAUNKNOWN H

Integrals:

HONLY 6 H!

H2H H H

D2O so noXH (OH, NH)

b c d d’ e e’


SUBSTRUCTURESUNKNOWN H

C 176 sIR band at 3400

C 70 dH 4.6 m

C 60 dH 3.9 m

MF = C5H9NO3 !

OH

O

A

OH

B

C

HN

ED


1H – 1H COSY NMR SPECTRUMUNKNOWN H

b c d d’ e e’

b-e/e’

c-e/e’

d-d’

d-e’

e-e’

b-d/d’


1H – 1H COSY NMR DATAUNKNOWN H

c

c-e/e’

e e'

b-e/e’

OHb

b-d/d’

d d'

Also 4-bond correlation d-e’Diastereotopic pairs d-d’ and e-e’


SUBTRUCTURESUNKNOWN H

c

e e'

OHb

d d'

HN

OH

O

MF = C5H9NO3

HN COOH

HO

Hc

Hb

HN COOH

HO Hb

Hc

1 2

Working structures:

12 % NOENO NOE


ASSIGNING NMR DATA TO A KNOWN STRUCTUREGUAIAZULENE

10 12

3

4

56

8

9

11

12

1314

15

7

MF = C15H18

Expect:(C)5

(CH)6

(CH2)0

(CH3)4


13C NMR DATAGUAIAZULENE

sA

sB

dJ

sD

dddEFG

sC

dH

dK

qqLM

qN

qO

sI

(C)5 + (CH)6 + (CH2)0 + (CH3)4 = C15H18


HSQC NMR DATAGUAIAZULENE

KK

k

LMLM

lm

NN

n

OO

o


HSQC NMR DATAGUAIAZULENE

EE

e

FF

f

GG

g

HH

h

JJ

j


1H NMR DATAGUAIAZULENE

Hg

He

Hf

Hj

Hh

Hk

3Hn

3Ho

3Hl

3Hm

Label spectrum according to HSQC:


HMBC NMR DATAGUAIAZULENE

We will need expansions:



g e f j h

HI

GFCDEB

A

C-gF-g

I-g

D-e

A-f

G-fD-j

I-j

B-h

C-h



g e f j h

O

NLM

KK-g K-f

N-h



n o lm

LM-lmK-lm

B-lm

I-oH-n

D-oC-n

A-n



Carbon Proton

A f, n

B h, l/m

C g, h, n

D e, j, o

E

F g

G f

H n

I g, j, o

J

K f, g, l/m

L m

M l

N h

O

1H-1H COSY data indicates that e and j are adjacent (J(e-j) = 4 Hz) as are f and h (J(f-h) = 11 Hz)

‘Obvious’ assignments:

KL

M

N/O

O/N

E/J

J/E F/H

H/F



Carbon Proton

A f, n

B h, l/m

C g, h, n

D e, j, o

E

F g

G f

H n

I g, j, o

J

K f, g, l/m

L m

M l

N h

O

KL

M

N/O

O/N

E/J

J/E H

F

G

Signal for f is a ddlong-range couplingto remaining proton g



Carbon Proton

A f, n

B h, l/m

C g, h, n

D e, j, o

E

F g

G f

H n

I g, j, o

J

K f, g, l/m

L m

M l

N h

O

KL

M

E/J

J/E H

F

G B

A

N

O

C

HMBC data can’t decide positions of E, JHMBC data can’t decide positions of D, I

How do we decide?


FINALISING THE ASSIGNMENTSGUAIAZULENE

KL

M

E

J H

F

G B

A

O

C

NNOE

Placing D (134 ppm) here

D

puts it in a similar environment to C (136 ppm)

This puts I (125 ppm) here

I

solving unknown structures using nmr organic structure analysis, crews, rodriguez and jaspars

Documents

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c nmr data unknown h

h hh d

h 3h2h

h nmr data unknown h

c data

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