I have a question on the nmr of t-butylcyclohexanol.
The spec shows a peak at 4.05 which integrates to 1 and a peak at 3.52 which integrated to 2.52.
The 4.05 peak looks like a wide singlet with some irregularities toward the top.
The 3.52 peak looks a septet.
I need to assign protons to those peaks and than calculate the isomers using those integral values.
At first i thought that the 4.05 was the proton alcohol and the 3.52 was the protons on the carbon adjacent to the alcohol bearing carbon - this though would not help me figure out the ration of cis to trans.
I know the proton on the alcohol bearing carbon in the cis conformation has only 1 J value therefore apearing as a quintet and in the cis conformation has 2 J values thus apperaing as three triplets.
Im not sure if that information helps me becuase there is neither a quintet or three triplets that integrates to 1.
Please help or push my in the right direction.
Much thanks,
Aleks