A compound, $\mathrm{C}_{20} \mathrm{H}_{30}$, can be hydrogenated with platinum metal and hydrogen to give a compound $\mathrm{C}_{20} \mathrm{H}_{38}$. How many double bonds (DB) and rings (R) does the original compound have? (The original compound has no triple bonds.)

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Quizplus · Accepted Answer

The original compound gains 8 hydrogens to become $\mathrm{C}_{20} \mathrm{H}_{38}$, indicating it has 4 double bonds (since hydrogenation adds 2 hydrogens per double bond). Using the formula for degrees of unsaturation, $DU = C + 1 + N/2 - H/2 - X/2$, where $C$ is the number of carbons, $N$ is the number of nitrogens (0 in this case), $H$ is the number of hydrogens, and $X$ is the number of halogens (0 in this case), the degrees of unsaturation for the original compound is 5 ($DU = 20 + 1 - 30/2 = 5$). Since 4 degrees of unsaturation come from double bonds, the remaining 1 degree of unsaturation indicates the presence of 1 ring.

A Compound C20H30\mathrm{C}_{20} \mathrm{H}_{30}C20​H30​ , Can Be Hydrogenated with Platinum Metal and Hydrogen to Metal

A Compound $\mathrm{C}_{20} \mathrm{H}_{30}$ , Can Be Hydrogenated with Platinum Metal and Hydrogen to Metal