Proton NMR: Proton NMR spectrum tells us… 1) Number of different proton environments from
number of peaks. 2) Types of proton environment from chemical shift values.
3) Relative number of proton in each environment given from integration traces/ ratio numbers
of relative peak areas. 4) Number of non- equivalent protons adjacent to given proton from spinspin splitting pattern, using n + 1 rule.
Use radio waves region of electromagnetic spectrum
Equivalent: If protons equivalent/ same chemical environment/ symmetrical, absorb at same
chemical shift, increasing size of peak. Must be same distance from other atoms to be same
environment.
2 proton environments here.
4 environments here.
Spin- Spin Coupling: A proton NMR peak can split. Caused by proton’s spin interacting with spin
states of nearby protons in different environments. The number of peaks is (n+1) one greater
than (n) number of adjacent protons. Does not work if adjacent is in same environment.
Can get quintet, hextet, heptet and so on.
Heptet as shown on the LHS, often caused by 2 CH3 groups.
Multiplet: When there are two adjacent protons on either side with different environments, the
middle proton environment responsible for peak will be a multiplet.
Aromatic Protons: In aromatic region of chemical shift, there will be weird splitting, but assume
a multiplet and an aromatic.
Benzene- 1- 4 disubstituted benzene produces two doublet peaks (a quartet) in aromatic
chemical shift range. The number of protons of the benzene should be 4.
Structures either side of benzene- do not include benzene as part of spin- spin pattern.
Do not discuss benzene splitting unless stated otherwise.
Must say how many ‘aromatic H’s’ in structure or write C6Hn.
When Not Bonded to Carbon: Protons may not be bonded to carbon atoms, but be –OH or –NH
protons instead where bonded to O and N. Happens in–OH, phenols, -COOH and NH2. All these
NMR peaks can occur at any chemical shift in range, except –COOH carboxylic acid.
Proton Exchange: In technique called proton exchange, normal NMR run. Then small amount of
deuterium oxide D2O added to sample and second NMR run. Compare two spectra. Deuterium
replaces OH and NH protons, so these peaks disappear.
Distance from Atoms: If a CH group is adjacent to electronegative atoms like O or N, then
increases chemical shift of CH peak.
When peak area is 6, indicates 2 methyl groups. Peak area 9 = 3 methyl groups in same
environment (symmetrical).
For every peak, 3 things needed-
– Peak at – ppm indicating –. Write the full structure as shown on data sheet. State an exact
number not a range for ppm.
– No. of C or H in environment = number of peaks.
– Splitting. E.g. singlet indicates adjacent C bonded to no H atoms OR Triplet indicates adjacent
C is bonded to 2 H.
Throughout question remember to write and discuss ALL pieces of information given, not just
the H NMR spectre. E.g. the Mr, the number of environments, carbon NMR, empirical formula,
mass spec and IR. As some marks are dedicated to them specifically.
Check structure against number of environments/ Mr etc.
If an O, treat them as if nothing next to the carbon. O does not show up in adjacent C’s spectra.
Ester made up of a C-O and C=O in H NMR.
Properly draw out structures on data sheet- makes more sense. The peak may be a CH2
individually group but data sheet says it’s a benzene with an adjacent CH group attached. So it’s
a benzene with CH2 group next to it since has relative peak area of 2. Data sheets shows the
hydrogens in in bold.
Peak area gives how many protons attached to structure shown on data sheet.
Similarly a C=O (or C-O) and the CH2 next to it is part of same environment so that CH2 not
involved in splitting as same environment.
If D2O, added, ignore NH2 and OH when writing the splitting or relative peak area. But they will
show up in H NMR range for HC- N and HC-O, even when D2O added, since H not directly bonded
to N or O.
If D2O not added, remember the H directly bonded in OH and NH can be any range. So when see
a peak somewhere that gives something unusual like an alkene, most likely an OH or NH- if
questions says there is an OH or NH.
June 2015- given complicated molecule. First decide what in spectrum is in known part of
molecule, so leaves the rest for unknown. There is a benzene peak with 7 hydrogens- 3 of those
are in benzene in known part. So there is ANOTHER benzene with 5 hydrogens in unknown. R1
and R2 are two different groups need to work out.
If the benzene- CH one (the one lower down in spectrum), if that’s split into quartet, then
attached to CH is CH3.
If peak not between 1 and 2. Not a R- CH group.
Follow the ranges on data sheet quite strictly.
Pay attention if question says that the peak at 3.7 is actually meant to be 2.7, only done it so it
doesn’t clash.
Watch out two peaks may give the same part in compound.
Here only two singlet peaks. The C – H is bonded to a C which is bonded to no
protons. CH3 x 3 are bonded to a C which is bonded to no protons, only carbons.
The C(CH3)2, always looking from the H3 perspective, the H3 are bonded to C
which is bonded to a carbon so produces singlet peak.
And always looking at what adjacent carbon is bonded to for spin- spin coupling.