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CHEM 344 Organic Chemistry Lab January 26 th & 27 th 2009 Structural Determination of Organic Compounds Lecture 3 – More NMR.

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Presentation on theme: "CHEM 344 Organic Chemistry Lab January 26 th & 27 th 2009 Structural Determination of Organic Compounds Lecture 3 – More NMR."— Presentation transcript:

1 CHEM 344 Organic Chemistry Lab January 26 th & 27 th 2009 Structural Determination of Organic Compounds Lecture 3 – More NMR

2 Review of Lecture 2 NMR – detailed info on molecular structure Different protons within the molecule give different signals in the 1 H-NMR spectrum # different protons = # signals Integration = # protons giving the signal Chemical shift value influenced by local structure of molecule (e.g. electronegative groups) Equivalent protons = same chemical shift

3 2 3 5.80 2.14 SiMe 4 CHCl 3 CH 3 CH 2 Typical 1 H-NMR spectrum from Lecture 2

4 4.42 2 1.58 3 CH 3 CH 2 Quartet Triplet Consider spin-spin splitting…….. n + 1 rule

5 1.12 12 3.64 2 CH 3 CH

6 3.64 1:6:15:20:15:6:1 Septet 1 6 15 20 15 6 1 Coupling constant J (Hz) – indicates strength of coupling J ~ 7 Hz for alkyl (sp 3 ) systems

7 1 Doublet Pascal’s Triangle 11 1 1 1 1 1 1 1 1 1 1 2 33 446 66 5510 2015 Singlet Triplet Quartet Quintet Sextet Septet n = 1 n = 2 n = 3 n = 4

8 1.12 12 3.64 2 1:1 Doublet

9 4 6 7.04 2.29 Equivalent H’s do not split each other HbHb HaHa Why isn’t H b a doublet? From Lecture 1………

10 2 2 9 2.42 0.87 1.04 NH 2 HaHa HbHb H’s on heteroatoms (O,N,S) do not couple Why is H b a singlet and not a triplet? From Lecture 1………

11 1.18 2.79 3.51 7.02 6.63

12 1.18 6 1 2 2.79 3.51 NH 2 Typical iso-propyl group pattern Septet & doublet CH 3 CH

13

14 6.63 7.02 22 HaHa HbHb Why is H a more shielded than H b ? Consider the substituents Consider resonance structures

15 H a more shielded than H b in structures A and C NH 2, OMe etc. are activating groups Activating groups direct e - density to the o and p positions

16 3.91 3 OMe

17 22 8.18 6.97 HaHa HbHb Why is H a more shielded than H b ? Consider the resonance structures

18 H b deshielded relative to H a in structures A and C -NO 2, -NR 3 +, -CF 3, -CO 2 R etc. are deactivating groups Deactivating groups reduce e - density at the o and p positions Remember that the OMe group will direct e - density to the H a protons

19 J ab = J ortho = 6 – 12 Hz J ab Coupling constants in aromatic systems

20 J ac = J para = 0 - 1 Hz J ortho > J meta > J para J ab ≈ J ad ≈ J bd = J meta = 1-3 Hz Coupling constants in aromatic systems

21 3.89 3 OMe

22 1 1 11 7.71 7.48 7.797.23 HaHa HbHb HcHc HdHd


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