Hydrocarbon B, \(\mathrm{C}_{6} \mathrm{H}_{6}\), gave an nmr spectrum with two signals: \(\delta 6.55\) and \(\delta 3.84\), peak area ratio \(2: 1\). When warmed in pyridine for three hours, B was quantitatively converted into benzene. Mild hydrogenation of \(B\) yielded \(C\), whose spectra showed the following: mass spectrum, mol. wt. 82 ; infrared spectrum, no double bonds; nmr spectrum, one broad peak at \(\delta 2.34\). (a)How many rings are there in \(\mathrm{C}\) ? (b) How many rings are there (probably) in B? How many double bonds in \(\mathrm{B}\) ? (c) Can you suggest a structure for \(\mathrm{B}\) ? for \(\mathrm{C}\) ? (d) In the nmr spectrum of B the downfield signal was a triplet. How must you account for these splittings?

Using the given information, we know that Hydrocarbon C has a molecular weight of 82 and has no double bonds according to its infrared spectrum. The molecular formula of Hydrocarbon C can be calculated using the degree of unsaturation formula: Degree of Unsaturation = (Number of carbons + 1 - Number of hydrogens / 2) Since we don't have any double bonds, the degree of unsaturation should be equal to the number of rings in the molecule. By calculating the difference in molecular weight between Hydrocarbon B (C6H6, molecular weight: 78) and Hydrocarbon C (molecular weight: 82), we find that: Difference in molecular weight = 82 - 78 = 4 This difference can be accounted for by adding two hydrogens to Hydrocarbon B, resulting in the molecular formula of Hydrocarbon C, which is C6H8. Degree of Unsaturation = (6 + 1 - 8 / 2) = 1 / 2 Since the degree of unsaturation is 1 / 2, this suggests that Hydrocarbon C has one ring.

As Hydrocarbon B is C6H6 and we have already determined that Hydrocarbon C has one ring, it is likely that Hydrocarbon B also has one ring. To determine the number of double bonds in Hydrocarbon B, we can use the degree of unsaturation formula once again: Degree of Unsaturation = (Number of carbons + 1 - Number of hydrogens / 2) = (6 + 1 - 6 / 2) = 3 Since we have already accounted for one ring, the remaining degree of unsaturation must be from double bonds. Therefore, Hydrocarbon B has one ring and three double bonds.

Based on the information given, Hydrocarbon B has one ring, three double bonds, and the NMR signals at δ 6.55 and δ 3.84 with a peak area ratio of 2:1. This information suggests that the structure of Hydrocarbon B is likely to be a 1,3,5-trisubstituted benzene ring. To account for the downfield triplet signal, it is possible that the three substituents on the benzene ring are protons, resulting in a 1,3,5-trimethylbenzene structure.

As Hydrocarbon C was obtained through mild hydrogenation of Hydrocarbon B, we know that C has one ring and no double bonds. Based on the molecular formula (C6H8) and the information about the NMR spectrum (one broad peak at δ 2.34), it is possible that Hydrocarbon C has a cyclohexane structure with the three methyl groups from Hydrocarbon B still attached. Therefore, the structure of Hydrocarbon C is likely to be 1,3,5-trimethylcyclohexane. Answers to the given questions: (a) Hydrocarbon C has one ring. (b) Hydrocarbon B probably has one ring and three double bonds. (c) The suggested structure for Hydrocarbon B is 1,3,5-trimethylbenzene, and for Hydrocarbon C is 1,3,5-trimethylcyclohexane. (d) The downfield triplet signal in the NMR spectrum of Hydrocarbon B can be accounted for by the three protons on the benzene ring, which create the 1,3,5-trimethylbenzene structure.