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.

Organic Structure Analysis, Crews, Rodriguez and Jaspars

Organic Structure Analysis, Crews, Rodriguez and Jaspars

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

Organic Structure Analysis, Crews, Rodriguez and Jaspars

d

s

d

t

s

tq

q

t

q

13C NMR DATAUNKNOWN B

A B C D E F GHI J

Organic Structure Analysis, Crews, Rodriguez and Jaspars

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)

Organic Structure Analysis, Crews, Rodriguez and Jaspars

1H NMR DATAUNKNOWN B

Integrals:

H

3H 2H3H 3H

5HONLY 17 H!

Organic Structure Analysis, Crews, Rodriguez and Jaspars

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 !

Organic Structure Analysis, Crews, Rodriguez and Jaspars

WORKING STRUCTURESUNKNOWN B

OH

Ha

Ha

Ha

OH

OH

A – Ha should be ddt

B – Ha should be ddt

C – Ha should be dd

Organic Structure Analysis, Crews, Rodriguez and Jaspars

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

Organic Structure Analysis, Crews, Rodriguez and Jaspars

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

Organic Structure Analysis, Crews, Rodriguez and Jaspars

13C NMR DATAUNKNOWN G

A B C D E FG

d dd t

t t q

Organic Structure Analysis, Crews, Rodriguez and Jaspars

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)

Organic Structure Analysis, Crews, Rodriguez and Jaspars

1H NMR DATAUNKNOWN G

Integrals:

HH

H H

3H

H H2H

H

12 H Total

Organic Structure Analysis, Crews, Rodriguez and Jaspars

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 !

Organic Structure Analysis, Crews, Rodriguez and Jaspars

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

Organic Structure Analysis, Crews, Rodriguez and Jaspars

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

Organic Structure Analysis, Crews, Rodriguez and Jaspars

d

s

d

t

t

13C NMR DATAUNKNOWN H

A B C D E

Organic Structure Analysis, Crews, Rodriguez and Jaspars

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)

Organic Structure Analysis, Crews, Rodriguez and Jaspars

1H NMR DATAUNKNOWN H

Integrals:

HONLY 6 H!

H2H H H

D2O so noXH (OH, NH)

b c d d’ e e’

Organic Structure Analysis, Crews, Rodriguez and Jaspars

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

Organic Structure Analysis, Crews, Rodriguez and Jaspars

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’

Organic Structure Analysis, Crews, Rodriguez and Jaspars

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’

Organic Structure Analysis, Crews, Rodriguez and Jaspars

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

Organic Structure Analysis, Crews, Rodriguez and Jaspars

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

Organic Structure Analysis, Crews, Rodriguez and Jaspars

13C NMR DATAGUAIAZULENE

sA

sB

dJ

sD

dddEFG

sC

dH

dK

qqLM

qN

qO

sI

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

Organic Structure Analysis, Crews, Rodriguez and Jaspars

HSQC NMR DATAGUAIAZULENE

KK

k

LMLM

lm

NN

n

OO

o

Organic Structure Analysis, Crews, Rodriguez and Jaspars

HSQC NMR DATAGUAIAZULENE

EE

e

FF

f

GG

g

HH

h

JJ

j

Organic Structure Analysis, Crews, Rodriguez and Jaspars

1H NMR DATAGUAIAZULENE

Hg

He

Hf

Hj

Hh

Hk

3Hn

3Ho

3Hl

3Hm

Label spectrum according to HSQC:

Organic Structure Analysis, Crews, Rodriguez and Jaspars

HMBC NMR DATAGUAIAZULENE

We will need expansions:

Organic Structure Analysis, Crews, Rodriguez and Jaspars

HMBC NMR DATAGUAIAZULENE

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

Organic Structure Analysis, Crews, Rodriguez and Jaspars

HMBC NMR DATAGUAIAZULENE

g e f j h

O

NLM

KK-g K-f

N-h

Organic Structure Analysis, Crews, Rodriguez and Jaspars

HMBC NMR DATAGUAIAZULENE

n o lm

LM-lmK-lm

B-lm

I-oH-n

D-oC-n

A-n

Organic Structure Analysis, Crews, Rodriguez and Jaspars

HMBC NMR DATAGUAIAZULENE

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

Organic Structure Analysis, Crews, Rodriguez and Jaspars

HMBC NMR DATAGUAIAZULENE

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

Organic Structure Analysis, Crews, Rodriguez and Jaspars

HMBC NMR DATAGUAIAZULENE

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?

Organic Structure Analysis, Crews, Rodriguez and Jaspars

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