Question

In solution, dimedone (5,5-dimethylcyclohexan-1,3-dione) is a mixture of keto (A) and enol (B & C) isomers The 1H NMR is Shown below, In the sample the two enol isomers are equilibrating very fast compared to the NMR time scale .Furthermore, the enol isomers are indistinguishable by 1H NMR. Assign the signals in the spectrum that correspond to the keto form. Assign the signals in the spectrum that correspond to the enol form. Use the integrations to determine the percent of keto form present in the mixture.

Answer 1

The below NMR (estimated) spectrum should be the typical spectrum for keto form with intensity 9: 4: 2 (from lower delta to higher) Peak at around 3.4splits into two peaks, when it converted from keto to enol. Peak at around 2.6 shifted to lower delta, when it converted from keto to enol. Choose the chemical shift value which is changing when keto to enol conversion happens. Percentage of Keto form = (Keto integration times 100) /(Keto integration + Enol integration) = (2.28 times 100)/3.99 = 57.14 % You can also choose the Enol OH and alkene integration with CH_2 integration. = (1.13 times 100)/(0.44 + 0.42 + 1.13) = 56.78 % Both are coming approximately in the same range of 57 %.