344 Organic Chemistry Laboratory Fall 2013 Lecture 2 More 1 H-NMR - - PowerPoint PPT Presentation

Lecture 2 More1H-NMR Spectroscopy June 18 2013

344

Organic Chemistry Laboratory Fall 2013

1.18 2.79 3.51 7.02 6.63

Finished Lecture 1 here

1.18 6 1 2 2.79 3.51

NH2

Typical iPr group pattern

Septet & doublet CH3 CH

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 (sp3) systems

Derivation of splitting diagrams

1) Two different coupling constants in a simple alkyl chain (Jab much larger than Jbc) triplet of triplets

Hb

Jab

Jab > Jbc

C Y C Ha Ha Hb Hb C Hc Hc X

Jbc

apply strongest coupling first!

Derivation of splitting diagrams

2) Two identical coupling constants in a simple alkyl chain (Jab equal or almost equal to Jbc)

Hb

Jab Jbc

Jab = Jbc

C Y C Ha Ha Hb Hb C Hc Hc X Overlap of peaks in each signal due to similarity of coupling constants Jab and Jbc Pentet observed! This is the splitting pattern predicted by the n+1 rule for Hb when n = 4

Derivation of splitting diagrams

2) Two identical coupling constants in a simple alkyl chain (Jab equal or almost equal to Jbc) Hb

Jab Jbc

Jab = Jbc

C Y C Ha Ha Hb Hb C Hc Hc X

When coupling constants are equal, splitting results in n + 1 peaks (i.e. a “normal” splitting pattern) THIS IS THE MOST COMMON CASE FOR A SIMPLE ALKYL GROUP

For Hb: n = 4 n + 1 = 5 1:4:6:4:1 pentent

Signals from benzene ring protons

Jab = Jortho = 6 – 12 Hz Jab

Coupling constants in aromatic systems

NH2 Ha Hb Hb Ha Me Me

Jab

6.63 7.02 2 2

Ha Hb

NH2 Ha Hb Hb Ha Me Me

Why is Ha more shielded than Hb?

Consider the substituents Consider resonance structures (when applicable)

Jab Jab

• NH2, -NR2, -OMe, -OH, etc. are common electron-donating groups.

Electron-donating groups increase e- density at the o and p C-atoms. This shields the H-atoms at those positions relative to benzene H-atoms. Ha shielded relative to Hb which can be rationalized by resonance effects. (Shielding/deshielding by alkyl groups is usually minor.)

• 0.28
• 0.28
• 0.21
• 0.21

NBO charges B3LYP/6-31G(d)

3.91 3

OMe

OMe NO2 Hb Ha Ha Hb

2 2 8.18 6.95

Ha Hb

OMe NO2 Hb Ha Ha Hb

Why is Ha more shielded than Hb? Consider the resonance structures

• r…..why is Hb so deshielded?

What is this?

Hb deshielded relative to Ha which can be rationalized by resonance effects.

• NO2, -NR3

+, -CF3, -CO2R etc. are common electron-withdrawing groups.

Electron-withdrawing groups reduce e- density at the o and p C-atoms. This deshields the H-atoms at those positions relative to benzene H-atoms. Recall, -OCH3 is electron-donating which helps shield the Ha atoms.

• 0.32
• 0.27
• 0.19
• 0.20

NBO charges B3LYP/6-31G(d)

1.93 3

Me

1 1 1 2 2

Ha R Hb Hc

Jac = Jtrans = 12 – 18 Hz Jab = Jcis = 6 – 12 Hz Jbc = Jgem = 1 – 3 Hz

Jtrans > Jcis > Jgem

Coupling constants in alkene systems

Ha is trans to Hc Ha is cis to Hb

7.23 7.11 6.61 5.70 5.17 2 2 1 1 1 But which is which?

Alkene protons (by chem. shift and integration)

Consider J values

Aromatic ring protons (by chem. shift and integration)

Derivation of splitting diagrams

3) Splitting in alkene systems (Jac > Jab)

Ha

Jac Jab X Ha Hb Hc Jac = Jtrans Jab = Jcis

Doublet of doublets

Hb-Hc geminal coupling not shown

Same concept as example 1)

Jtrans > Jcis

apply strongest coupling first!

7.23 7.11 6.61 5.70 5.17 2 2 1 1 1

Ha He Hd

Doublet of doublets with Jtrans & Jcis

Me He Hd Hd He Ha Hb Hc

5.70 ppm

5.17 ppm 17.2 Hz 10.5 Hz

Scale now in Hz

Jtrans > Jcis

Me He Hd Hd He Ha Hb Hc

Hc Hb Ha

Jtrans

Jcis

Jtrans

Me He Hd Hd He Ha Hb Hc

Jac Jab Jbc Jbc

trans + gem coupling cis + gem coupling

Hb Hc

3.89 3

OMe

Jac = Jpara = 0 - 1 Hz

Jortho > Jmeta > Jpara

Jab ≈ Jad ≈ Jbd = Jmeta = 1-3 Hz = 6 – 12 Hz

OMe NO2 Ha Hb Hc Hd

Coupling constants in aromatic systems

1 1 1 1 7.71 7.48 7.79 7.23

OMe Ha NO2 Hb Hc Hd

Write down the relationships between the protons!

Hc is ortho to Hb

• rtho to Hd

(para to Ha)

7.5 ppm 6.8 ppm 7.2 ppm 7.7 ppm

Derivation of splitting diagrams

4) Splitting in aromatic systems (Jcb = Jcd)

Ha-Hc para coupling not shown

Hc

Jcb Jcb = Jortho Jcd = Jortho X Ha Y Hb Hc Hd Jcd Jcd Overlap of peaks in each signal due to similarity of coupling constants Jcb and Jcd Same concept as example 2) Doublet of doublets but…….. central peaks overlap to give appearance of 1:2:1 triplet (ITS NOT A REAL TRIPLET!!!) Can also apply to Ha, just draw diagram using meta coupling constants Hc is ortho to Hb

• rtho to Hd

(para to Ha)

As practice, draw the splitting diagrams for Hb and Hd

• list relationships of each proton
• apply n+1 rule to these relationships
• factor in the coupling constants

OMe Ha NO2 Hb Hc Hd

1 1 1 1 7.71 7.48 7.79 7.23

Ha Hb Hc Hd

OMe Ha NO2 Hb Hc Hd

Hb is most deshielded (proximity to NO2 group) Hd is most shielded (proximity to OMe group)

7.5 ppm 6.8 ppm 7.2 ppm 7.7 ppm NMR Chemical Shifts B3LYP/6-31G(d)

NH O Me O OH N-Acetylanthranillic acid Experiment Information Inset Spectrum 7.0 - 8.9 ppm TMS impurity impurity HCCl3 “missing” OH peak Integration values

How many different types of H-atoms? Indicated by how many groups of signals What types of H-atoms? Indicated by the chemical shift of each signal How many H-atoms of each type are there? Indicated by the integration of the signal for each group What is the connectivity of the molecule? Indicated by the splitting pattern and coupling constant of each signal

Practice and Ask Questions!

Strategy and Tactics for solving NMR spectra

What other evidence do you have? Use GC-MS, 13C-NMR, IR, melting point etc. as complimentary information