Question

Identify the compound shown in the Carbon-13 NMR from the list below, explain and provide reasoning why the compound is the one shown in the spectroscopy. MW is provided for A and B as 2 unknowns have the same amount of carbon signals. (Only A spec. is pictured)

-cycloheptanol

-1-methylcyclohexanol

-cyclohexanol

-cylopentanol

-4-methylcyclohexanol

Unknown A MW=68.1 -100 2017 EN JI M TTC NG IOCY61 311161 x parte

Answer 1

I don't know if I'm not understanding the question well, but the given molecular weight is too low to be any of the compounds on the list: cyclopentanol is the smallest molecule and it has a MW = 86.1 g/mol.

There are only 3 signals in the given NMR spectra. This means that there are 3 distinguishable carbons in the molecule (each distinguishable carbon generates a signal). Let's see how many distinguishable carbons are there in the molecules given in the list:

There are 4 distinguishable carbons in cycloheptanol, so it isn't the one from the spectrum.

5 distinguishable carbons in 1-methylcyclohexanol, so it is not the one in the spectrum.

4 in cyclohexanol, so it is not this one.

3 in cyclopentanol, it can be this one.

5 in 4-methycyclohexanol, it can't be this one.

Discarding, using only the number of signals, we can determine that of the given compounds, only cyclopentanol can give the shown spectrum.