c nmr spectrachem213/213-7-18.pdf · 2018-10-26 · -ch3 >ch2-ch >c< q t d s} only 4 types...

13C NMR SPECTRA12C: I = 0 13C: I = ½99% 1.1%not active + low sensitivity relative to 1H

Therefore many scans (FT methods)

13CH313CH2Br

1% x 1% = .01% chance of being in same molecule

SO WE DO NOT SEE COUPLINGBETWEEN ADJACENT CARBONS

BUT DO SEE COUPLING TO HYDROGENS13CH3 is a 1J so is large, 125-250 Hz, see a quartethowever because 13C is only 1% does not affect 1H spec

p. 149

see coupling

suppress coupling

p. 150

scale now 200 ppm (C) ~ 15 x (H)

p. 150

-CH3 >CH2 -CH >C<

q t d s} only 4 types

We usually run proton decoupled spectra:

NOE, nuclear Overhauser enhancement effect transfers magnetization from H to directly bonded C greatersignal strength for C

chemically different carbons usually have different so # PEAKS = # of CHEMICALLY DIFFERENT CARBONS

However if we want to know this coupling information weneed to run a separate experiment

p. 150

-CH3 >CH2 -CH >C<

q t d s

} only 4 types

We usually run DEPT-135 experiments

Complicated pulse sequence but the NET EFFECT issimple and useful: peaks appearance governed bymultiplicity

+ve -ve +ve gone

This is much faster than simplyturning off the decoupler becausethe NOE effect still applies to DEPT

p. 150

A typical DEPT-135 spectrum: note phasing – CH2 are down and CH or CH3 are up

Proton decoupled 13C spectra : see only singlets

CH3-CH2-CHBr-CH3

12.1 (q) CH3

26.0 (q) CH3

34.2 (t) CH2

53.1(d) CH

p. 151

220 200 180 160 140 120 100 80 60 40 20 0 -20 ppm

RCCR

R-CO-R (ketones)

R-CO-H (aldehydes)

R-CO2H (acids)

R-CO2R (esters) F-C Cl-C, RS-C I-C

R2C=CH2

Aromatics

R-CN

RCH=CHR

H2C=CR2

Saturated alkanes

H2N-C (amines) NC-C

HO-C (alcohols) Br-C

O2N-C

R-O-C (ethers)

R-CO-C

Ar-C

R2C=CR-C

Carbon Chemical Shifts

Metal-C

Chemical Shifts

220 200 180 160 140 120 100 80 60 40 20 0 -20 ppm

>C=O -COO C=C C-O CH3

p. 151

ALKANES 10-50 ppm

CH3---CH2----CH2----CH2----14 23 ~30

CH3CH2CH2CH2CH2CH2CH3

Integrations are NOT reliable in C NMRrelaxation times are longer and more variable(if we wanted them to be reliable the experiment wouldtake a very long time to do!!)

Peak height is roughly correlated with number of attached H NOT so much with number of C

p. 152

ALKENES 100-160 ppmCH2=CH--

~115 ~140more substituted, more downfield

14 23 30

p. 152

ALKYNES 65-95 ppm

CC HC85 68s d

CC HC shielded

18, shielded from 304oC, so weak

p. 152

AROMATICS

128.5

X most X deshield this ipso C by ~10 ppm so at ~ 140 ppm

130

D: donor groups (OR, NR) deshield thiscarbon to ~ 155-165

and shield the ortho carbons to~ 115 pm

Aromatic shifts to remember: 160 140 130 115

p. 153

Full data is in table, yellow pages, Appendix p. A8Shifts are additive

Substituent C-1 (ipso) ortho meta para

OCH3 31.4 -14.4 1.0 -7.7

C

CC

C

CC

OCH3

OCH3

C = 128.5 + 31.4 + 1.0 = 160.9

p.153


OCH3 31.4 -14.4 1.0 -7.7

C = 128.5 + 31.4 + 1.0 = 160.9C

CC

C

CC

OCH3

OCH3

C = 128.5 -14.4 -14.4 = 99.7

p.153


OCH3 31.4 -14.4 1.0 -7.7

C = 128.5 + 31.4 + 1.0 = 160.9

C

CC

C

CC

OCH3

OCH3

C = 128.5 -14.4 -14.4 = 99.7C = 128.5 -14.4 -7.7 = 106.4C = 128.5 +1 +1 = 130.5

p.153

Patterns:X

d

dd

s

1s + 3d

X

X

sd

1s + 1d

X

Y

sd

ds

2s + 2d

X

X

X

X

X

Y

X

Y

sdd

d

sdd

dd

s

s

dd

s

dd

d

ds

1s + 3d

2s + 4d

1s + 2d

2s + 4d

p. 154

CARBONYL COMPOUNDS-CHO d 190-210>C=O s 200-220

ACID DERIVS -COOH, -COOR, -CON<, -COO-

all are singlets, and all at 160-180

27211

p. 155

Problem AA C5H11Cl p. 156

DBE = {(2x5 + 2)-(11 + 1) = 0only 4 lines in carbon spectrum, so two C’s identical

43t 42t 26d 22q-CH2- -CH2- >CH- -CH3

Now add these up = C4H8 – C5H11 = CH3 missing

Problem AA C5H11Cl

-CH2- -CH2- >CH-

-CH3-CH3 -Cl

The 26d is not down field enough to hold -Cl

so –Cl must be on a –CH2- which now becomes –CH2Cl

p. 156

Problem AA C5H11Cl

-CH2- >CH-

-CH3-CH3 -CH2Clso –Cl must be on a –CH2- which now becomes –CH2Cl

join bi or higher groups

-CH2-CH<

p. 156

Problem AA C5H11Cl p. 156

-CH3-CH3 -CH2Cl-CH2-CH< now add rest, BUT –CH3’s identical

ClCH2-CH2-CH(CH3)2

Are you happy with the chemical shifts?

Problem AB C4H8O2p. 157

1 DBE 171s 60t 21q 14q-CH3 -CH3>C<

acid(deriv)

-CH2-on-O-

Now add up to check C4H8 all H’s found, so deriv is an

ESTER -CH2OCO- CH3-CH2OCOCH314

Problem AC C6H8Op. 158

3DBE193d 152d 142d 131d 130d 19q-CHO -CH3ALKENE C’s

=CH =CH =CH =CH}C6H8O

put ½ ‘s together -CH=CH- -CH=CH-

these must be joined together -CH=CH-CH=CH-CH3-CH=CH-CH=CH-CHO

Problem AD C9H12p. 159

4DBE = aromatic

137s 127d 21qHighly symmetric=C< =CH- -CH3

benzene has 6C so C9 total so must be 3 (CH3) identical=9H

so must be 12H – 9H = 3H on ringso answer is atrimethyl-benzenewhich?

CH3

CH3 CH3mesitylene

5DBE, aromatic + 1200s 137s 133d 129d 128d 32t 8q>C=O -CH2- -CH3benzene ring

how do we tell what type of benzene?

1] pattern, 1s+3d=mono 2] add up rest and subtractC9H10O – C3H5O = C6H5 = mono

Problem AE C9H10Op. 160

>C=O -CH2- -CH3Ph-

join bifunctionals -CH2CO-

so is PhCH2COCH3 or CH3CH2COPh


PhCH2COCH3 or CH3CH2COPh

COCH2CH3

X X

Also, only the CH2 isfar enough downfieldto be next to C=O


Problem AF C5H8O2p. 161

2DBE 166s 130d 129t 60t 14q-COO- =CH =CH2 -OCH2- -CH3C5H8O2{

CH3- -CH=CH2 -COOCH2-

CH3COOCH2CH=CH2 or CH3CH2OCOCH=CH2

2DBE

195d 130s 115t 42d 15q-CHO =C< =CH2 >CH- -CH3 }C5H7O

C6H10Ox2 CH3>C=CH2

so we have (CH3)2 -CHO to go on the joined other units

>CH-C=CH2identical!

(CH3)2CH-C=CH2

CHO

IR:275016959509003050

E1, p. 162 C6H10O

E2, p. 162 C4H9N DBE = [(2x4 +2+1)-9]/2=1

Functional group = 30 amine = R3N

IR (cm-1): no bands at 3400, 2200 or 1700 cm-1

13C NMR: 50 t, 45q, 26t

-CH2N NCH3 -CH2-

E2, p. 162 C4H9N DBE=1 R3N

50t 45q 26t-CH2N< >N-CH3 -CH2- } C3H7

C4H9

? CH2

how many C’s on N ? 3, so must be –CH2N

NCH3

CH2

CH2

-CH2-

DBE=1

NCH3

3DBE

Find functional groups from IR

-OH 3300+2100 ? CHC

E3, p. 163 C6H8O

C6H8O 3DBE -OH CHC

137up 119x 86x 75up 59down 17up=CH- =C< CHC -OCH2 -CH3 }C6H7O

HOCH2- CH3- CHC >C=CH-

E3, p. 163 C6H8O

HOCH2- CH3- CHC >C=CH-

so what is next to the =CH ? if t, it is –CH2-

H

CH2OHCH3

CH2OH

HCH3

or

p. 163

E4, p. 164 C10H15NO2 DBE= 4so aromatic

IR = ? -NH2

O ? not >C=O

IR (cm-1): 3400 m, 3300 m

E4, p. 164 C10H15NO2 -NH2

161s 142s 104d 97d 55q 42t 38t benzene C6 -OCH3 -CH2- -CH2- }C9

13C NMR:

[-OCH3]2 identical

coupled

E4, p. 164 C10H15NO2 -NH2

161s 142s 104d 97d 55q 42t 38t benzene C6 -OCH3 -CH2- -CH2- }C9

13C NMR:

1H NMR:

6.5s (2H) 6.3s (1H) 3.8s (6H) 2.9t (2H) 2.7t (2H)

1.1 br s (2H) exchanges with D2O

[-OCH3]2 -CH2CH2- -NH2

to be identical, have to be on ring

CH2CH2NH2

CH3OOCH3

but that means

CH2CH2NH2

OCH3

OCH3

CH2CH2NH2

CH3O

CH3O

E4, p. 164 C10H15NO2 -NH2

CH2CH2NH2

OCH3

OCH3

CH2CH2NH2

CH3O

CH3O

= shielded = very shielded

161s 142s 104d 97d 55q 42t 38t 13C:

1H: all 3H are shielded, 1H at 6.3, 2H at 6.5

CH2CH2NH2

CH3O

CH3O

E4, p. 164 C10H15NO2 -NH2

E5, p. 165 C13H14O DBE = 74 of these aromatic

para-benzeneC6H4

IR: 825 3300, 2150 1680

CHC conjugated ketone

1H NMR: 7.8d (2H), 7.6d (2H), 2.9t (2H), 2.5s (1H), 1.7 pentet (2H), 1.4 sextet (2H), 0.9t (3H)

E5, p. 165 C13H14O

= 7DBECHCconj >C=O

199s 136s 132d 129d 127s 82s 79d

38t 26t 23t 14q-CH2- -CH2- -CH2- -CH3 } C13H14O

13C NMR

CHCconj >C=O

t 5 6 t-CH2- -CH2- -CH2- -CH3

CH3-CH2-t

-CH2-CH2-t

6 (or 5) 5(or6)

so CH3CH2CH2CH2-t 6 5 t



CHCconj >C=O CH3CH2CH2CH2-end groups!

O



CH2CH2CH2CH3

O

CH3CH2CH2CH2

O

In C NMR one aromatic peak is at 127(s)



Predicted = 126.6 Predicted = 148.3

E6, p. 166 C9H10O4 DBE=5, so aromatic +1

IR 3400-2400(br) 1680 = conj-COOH

167x 156x 130up 114x 104up 56up

-COOH benzene-OCH3

x2 (identical)

COOH

OCH3

OCH3

but which isomer?

COOH

OCH3

OCH3

COOH

CH3O OCH3

COOH

CH3O OCH3100d

115d115d

115d

130d

115d 13C

COOH

CH3O OCH3

COOH

CH3O OCH3d

t

d

s s

s

1H

shielded 103s

E6, p. 166 C9H10O4

ASSIGNMENT 6

c nmr spectrachem213/213-7-18.pdf · 2018-10-26 · -ch3 >ch2-ch >c< q t d s} only 4 types...

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