I need to assign hydrogens to specific peaks and I am really struggling. Help would be greatly appreciated.
This is what the NMR looks like:
3 Answers
-
There are a total of 12 protons to be assigned.
The peaks at 4.57 ppm (1H) is assigned to the hydroxyl proton. The protons of the CH-OH group couple each other, forming doublets in the NMR spctrum.
The doublet at 5.98 ppm (1H) is assigned to the other proton in the couple.
Resonance of the phenyl ring towards a carbonyl group deshields the protons at the ortho- and para- positions. The doublet at 7.94 ppm (2H) is assigned to the protons attached ORTHO to the phenyl group attached to the carbonyl. The triplet at 7.56 ppm (1H) is assigned to the proton attached PARA to the phenyl group attached to the carbonyl.
The protons at the META positon are least deshielded and produce the triplet at 7.42 ppm (2H)
The multiplet at 7.29 to 7.40 ppm (5H) is assigned to the protons in the phenyl group attached to the CH-OH group.
-
Nmr Peak Assignment
-
It is all about which protons are an identical (ie the same) and which aren’t. That you could write the system of two-ethyl-1-butanol as (CH3CH2)2CHCH2OH – from this you will discover that there’s a set of two identical CH2 corporations (the ones in the brackets) and one other that is exclusive from these 2 (ie the one subsequent to the OH staff). The primary set will cut up the CH workforce into a pentent (as four an identical protons will do). The opposite set will then break up this pentet right into a triplet (coupling to 2 protons generates a triplet). The result is a 15 line sign, which in this case is referred to as a “triplet of pentets”.
