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

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

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

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

CH 3 CH

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

1 Doublet Pascal’s Triangle Singlet Triplet Quartet Quintet Sextet Septet n = 1 n = 2 n = 3 n = 4

:1 Doublet

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

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………

NH 2 Typical iso-propyl group pattern Septet & doublet CH 3 CH

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

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

OMe

HaHa HbHb Why is H a more shielded than H b ? Consider the resonance structures

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

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

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

OMe

HaHa HbHb HcHc HdHd