Download presentation
Presentation is loading. Please wait.
Published bySawyer Mercer Modified over 10 years ago
1
Example from Wang group Work of Jiasheng Lu
2
Starting material
3
UV: Postulated Reaction +
4
H-NMR: Mesitylene
5
H-NMR: Aliphatic region * : Starting material * Me-A Me-B New compound
6
H-NMR: Aromatic region * : Starting material * * * *
7
COSY: Full SW Me
8
COSY expansion: Me vs aromatic Me H-C= Me Starting material Me H-C= New compound Me H-C= Methyl groups have long range coupling with ortho aromatic protons
9
COSY: aromatic 1 2 4 Py: 1 – 2 – 3 – 4 Ms 3 Other ring : A – B – C – D Not sure which proton is H-7 Start with a doublet (can be either H7 or H10) then go To the next proton to identify the suite. We will assign later with NOESY which one is H7 A B C D Most deshielded
10
NOESY: Choosing the Mixing time You cannot use too short mixing time: it takes time to develop NOE. It takes T 1 – too short mixing time would not show any cross peaks between protons! You cannot use too long mixing time. If you use too long mixing time, you can loose the labels of the chemical shifts that you record carefully during the evolution time (t 1 ) For small molecules, the best mixing time is ~ 80% of the T 1 value. (this is where the signal goes to null intensity in inversion recovery
11
Measuring relaxation time: Inversion recovery
12
Prepare for NOESY: evaluate relaxation time Py: 1 – 2 – 3 – 4 Other: A – B – C – D Ms 1 A 4 2 3 D C 12 ? Aromatic regionInversion recovery: Why different rate? Dipole Relaxation depends on the distance between protons and on the mobility of the molecule Ms is the smallest molecule: fast motion, not efficient T 1 longest relaxation time In Pyridine ring, H4 has faster relaxation: it has one ortho neighbor (H3) and its close to H7 on other ring. In fact, H7 must be very close to H4 because H3 and H2 (which has 2 ortho neighbors) relax slower than H4 HA relax fast at ~ same rate as H4: same type of neighbors: may be H7 ?
13
H-NMR Inversion recovery: Methyl region Ms Starting material New compound Ms CH2-12 Ms is the smallest molecule: fast motion, not efficient T 1 longest relaxation time
14
NOESY: Choosing the mixing time You cannot use too short mixing time: it takes time to develop NOE. It takes T 1 – too short mixing time would not show any cross peaks between protons! You cannot use too long mixing time. If you use too long mixing time, you can loose the labels of the chemical shifts that you record carefully during the evolution time (t 1 ) For small molecules, the best mixing time is ~ 80% of the T 1 value. (this is where the signal goes to null intensity in inversion recovery In our case, MIXING = 2 sec. will do good job according to T 1
15
NOESY : mix time 2s Methyl vs aromatic Me Ms Starting material Ms Py Py: 1 – 2 – 3 – 4 Other: A – B – C – D Ms 12 ? A D C B 1 4 2 3 Me New compound Me
16
NOESY :mix time 2s Aromatic Py: 1 – 2 – 3 – 4 Other: A – B – C – D Ms 1 A 4 D C 12 2 3 B A 7 – 8 – 9 – 10 C 12 D D
17
Whats left? Assign C13 1 J CH Use HSQC to assign carbons that are directly attached to protons 1 J CH n J CH Use HMBC to assign carbons that are further away from protons n J CH (HSQC : Heteronuclear Single Quantum Correlation) (HMBC: Heteronuclear Multiple Bond Correlation)
18
HSQC: aromatic Ms 1 7 4 10 9 12 2 3 8 C1: 136.2 C2: 115.5 C3: 127.9 C4: 121.5 C7: 124.5 C8: 119.5 Ms: 127.4 C9: 128.6 C10: 128.4 C12: 118.5
19
HSQC: Methyl Ms Starting material New compound Ms CH2 21.0 25.2 20.6 Ms 22.5 21.1
20
HMBC: Methyl Ms Starting material New compound Ms CH2 21.0 25.2 20.6 Ms 22.5 21.1 136.7 127.5 140.2 138.9
21
HMBC: aromatic Ms 1 7 4 10 9 12 2 3 8 C1: 136.2 C2: 115.5 C3: 127.9 C4: 121.5 C7: 124.5 C8: 119.5 C9: 128.6 C10: 128.4 C12: 118.5 22.5 21.1 136.7 127.5 140.2 138.9 2 C5: 142.3 6 C6: 118.8 11 9 C11: 145.6 10 11
Similar presentations
© 2024 SlidePlayer.com. Inc.
All rights reserved.