Can a chemist please help with NMR?

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Can a chemist please help with NMR?

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NMR processing:

MDD

NMR assignment:

Backbone:

Side-chains:

NOEs:

Structure from NMR restraints:

Ab initio:

Fragment-based:

Rosetta-NMR (Robetta)

Template-based:

GeNMR

I-TASSER

Refinement:

Amber

Structure from chemical shifts:

Fragment-based:

Homology-based:

Torsion angles from chemical shifts:

Preditor

TALOS

Promega- Proline

Secondary structure from chemical shifts:

CSI (via RCI server)

TALOS

MICS caps, β-turns

d2D

PECAN

Flexibility from chemical shifts:

RCI

Interactions from chemical shifts:

HADDOCK

Chemical shifts re-referencing:

NMR model quality:

NOEs, other restraints:

Chemical shifts:

RDCs:

Pseudocontact shifts:

Protein geomtery:

SAVES2 or SAVES4

Quality Control Check

NMR spectrum prediction:

FANDAS

MestReS

V-NMR

Flexibility from structure:

Backbone S2

Methyl S2

B-factor

Molecular dynamics:

Gromacs

Amber

Antechamber

Chemical shifts prediction:

From structure:

Shiftx2

Sparta+

Camshift

CH3shift- Methyl

ArShift- Aromatic

ShiftS

Proshift

PPM

CheShift-2- Cα

From sequence:

Shifty

Camcoil

Poulsen_rc_CS

Disordered proteins:

MAXOCC

Format conversion & validation:

CCPN

From NMR-STAR 3.1

Validate NMR-STAR 3.1

NMR sample preparation:

Protein disorder:

DisMeta

Protein solubility:

Isotope labeling:

Solid-state NMR:

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09-06-2005, 07:25 AM

reba

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Answered: Can a chemist please help with NMR?

Please outline for me the chemical shift patterns of all the protons in NMR spectrum of SALICYLIC ACID. Thanks!

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Best Answer - Posted by romanwahoo

You should have 4 protons in the aromatic region between 6.5 and 8, which include two triples and two doublets. If you have the 1h nmr and you are just assigning the peaks, then simply figure out which carbons are more positively charged and which are more negatively charged. Since the two groups' effects are additive and not interfering, just go with the carboxylic acid, which is an electron withdrawing group. The carbon to which the -COOH is attached therefore is more negative, the next more positive, then more neg, and so forth. The protons attached to carbons that are more positive are more deshielded, so they have a higher shift. Using this method, you should be able to assign all four aromatic protons.The -COOH and the -OH are generally much harder to see, and they are concentration dependent. A broad peak around 6 though is usually indicative of the -OH peak. Because the oxygens in the -COOH can resonate charges, the carboxylic acid proton is usually shifted waaay downfield, around 14 or so, I believe, and which is usually not shown in a normal proton spectra.

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