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ORGANIC NMR INTERPRETATION

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Presentation on theme: "ORGANIC NMR INTERPRETATION"— Presentation transcript:

1 ORGANIC NMR INTERPRETATION

2 ALKANES AND ALKYL HALIDES
p. 101 ALKANES AND ALKYL HALIDES CH3—CH2—CH2—C d CH3—F CH3—O CH3—Cl CH3—Br CH3—I Dd ~ 2 ppm downfield CH3—CH2- CH3—X

3 Effect falls off with distance and is ~ 0 two C away
p. 102 Inductive effects CH3—CH2—CH2—C d CH3—CH2—CH2Cl d Effect falls off with distance and is ~ 0 two C away CH3—CH2—CH2Br d CH3—CH2—CH2I d Dd ½

4 CH3—CH2—CH2Cl 1.0 1.8 3.5 CH3—CH2—CH2Br 1.0 1.8 3.4 CH3—CH2—CH2I
p. 102 CH3—CH2—CH2Cl CH3—CH2—CH2Br CH3—CH2—CH2I Dd ½ CH3—CH2— d

5 Inductive effects are more or less additive
p. 103 CH3—CH2—CH2X CH3—CH2— d Dd ½ Each extra X adds ~ 2 ppm CH3—X CH2X CHX3 d Ballpark ONLY!! Inductive effects are more or less additive

6 Additional X next door has added about ½ ppm CH3—CH2—I d 1.8 3.2
I—CH2—CH2—I d Additional X next door has added about ½ ppm

7 In general, the more substituted, the more downfield
p. 103 In general, the more substituted, the more downfield CH3—I CH3—CH2—I (CH3)2CH—I d but additional alkyl groups are not as strong as –X CH3CH2Br = (CH3)2CHBr = CH3CHBr2 = 5.5 split by 6 = 6+1 split by 1 = 1+1 I—CH(CH3)2 6H 1H

8 More complex splittings
I—CH2—CH2—CH2—CH3 t ? ? t What about when neighbors are chemically different? If J’s are same, then can use splitting (# of lines) = total # of H neighbors + 1 Characteristic chain splitting in alkane chains J=7 J=7 J=0

9 More complex splittings
I—CH2—CH2—CH2—CH3 t t

10 ANISOTROPIC EFFECTS F Spherical atoms have same effect
in all directions p-electrons are above and below the plane of molecule so electron density is different above or below molecule than in plane

11 so alkenes and aromatics (and other p-bonds)
are not isotropic – they have effects that are different in different directions – we call them ANISOTROPIC p = Bp-electrons = Blocal so H feels B0 + Blocal so appear at low field

12 LOW FIELD aromatics & alkenes appear at - is shielded - (to lower ppm)
+ + is deshielded (to higher ppm) - p. 105

13 methyl on an aromatic ring, double bond or carbonyl ~2.3
aromatic hydrogens ~ d 7 p. 106 X=C—CH3 methyl on an aromatic ring, double bond or carbonyl ~2.3

14 Ph—CH2—CH2—(CH2)4—CH2—CH3 7.1-7.3 2.6 1.6 1.3 1.3 0.9
Ph—CH2—CH2—(CH2)6—CH2—CH3

15 Clearing up some terminology:
Downfield Deshielded Low field Greater d Upfield Shielded High field Smaller d

16 Electron Withdrawing Groups (EWG) deshield the ortho & para H’s,
o > p Benzene = 7.3 7.8 7.3 7.6 -CHO -COR -COOH -COOR -CN -NO2 -SO2 resonance effects +ve charge deshields: less electron density at the C and H

17 2:1:2 CHO d 10 EWG deshield 7.8 7.3 7.6 p. 108

18 Electron Donating groups (D:) SHIELD the ortho and para protons
o > p 6.8 7.2 7.0 Donating groups are X: (atoms with lone pairs but not halogens) e.g. -OH -OR -NH2 -NHR -NR2 -SR -R resonance effect dominates inductive effect negative charge shields: more electron density at C and H

19 p. 109 OCH3 CH3 on O + Ar ring ~ 3.8 7.2

20 HALOGENS Not easy to predict:
Lone pairs shield by resonance but deshield because of high electronegativity

21 Part of Table on manual page 110
Increments add to d 7.27 to predict shifts, e.g. Proton ortho to –CHO will be = 7.85

22 p. 110

23 Look carefully at peaks, they are doublets
3J ~ 8 Hz What about to other protons?

24 p. 111 H 3JORTHO H-H = ~8Hz H 4JMETA H-H = ~2Hz 5JPARA H-H = ~0 Hz H H

25 At high fields, can use trees to get patterns, IF chemical
shifts are far apart (called 1st order spectrum if Dd >> J) H= d (H) of d (H) H 8Hz 2Hz H= t (HH) of d (H) H 8,8 2 H= t (HH) of d (H) H H H= d (H) of d (H)

26 d ~ 10 O=C-H ALDEHYDES AND coupling constant to neighbors is small
- is shielded (to lower ppm) + is deshielded (to higher ppm) d ~ 10 AND coupling constant to neighbors is small

27 a Karplus showed relationship of J and a p. 112

28 Spectrum looks different on different instruments
60MHz 60Hz Spectrum looks different on different instruments J in Hertz is always independent of field 30Hz

29 ALKENE COUPLING CONSTANTS
16Hz 8Hz 2Hz

30 mono-substituted alkene R--CH=CH2
p. 114 mono-substituted alkene R--CH=CH2 Always 12 lines: d(JL)d(JM) d(JL)d(JS) d(JM)d(JS)

31 p. 115 Jcis Jtrans

32 p. 116

33 b to the substituent feels resonance effect
p. 117 Chemical shifts – much like aromatics ‘normal’ = 5.25 ppm Geminal (same C) always deshield by ~ 1ppm b to the substituent feels resonance effect

34 Manual, table page 117 Always deshield geminal Shield b Deshield b

35 4J = ~1Hz Cis Me 3J = ~7Hz Gem H Trans Ph 4J = ~1Hz
d cis to Me = – 0.07 = 4.96 d cis to Ph = – 0.28 = 5.33 See yellow pages A5 Cis Ph H Trans Me H

36 NMR time scale is ~ 10-2 – 10-3 sec fastest NMR can measure!
ALCOHOLS, AMINES, AMIDES AND ACIDS – exchangeable H’s Ar-CONHR and Ar-CONH2 More acidic means more d+ on H NMR time scale is ~ 10-2 – 10-3 sec fastest NMR can measure! p. 119

37 Since acidic H exchange between molecules occurs faster
p. 119 Since acidic H exchange between molecules occurs faster than the NMR time scale they DO NOT show coupling to any neighbours and are typically broadened

38 To prove which is –OH peak, add D2O and shake
ROH + D2O => ROD + HOD d ~ 5.2

39 Coupling visible Shape of peak depends upon temperature
(rate of exchange is affected by temperature) p. 120 exchange stopped Coupling visible slow exchange fast exchange HO—CH3

40 p. 121 Only a triplet Amines: RNH2 d 1-5; ArNH2 d 5-10

41 p. 121 Amides: d 5-10 Why 2?

42 p. 121 Acids: d 10-16 d = offset (2.0) = 11.85

43 You can now start Assignment 5


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