## Organic Chemistry 9th Edition

The number of Hydrogen atoms in the compounds are: (a) $C_{8}H_{?}O_{2}$ has 12 H atoms (b) $C_{7}H_{?}N$ has 13 H atoms (c) $C_{9}H_{?}NO$ has 13 H atoms
(a) In $C_{8}H_{?}O_{2}$, there are two rings and one double bond present, i.e. the degree of unsaturation in this compound is 3. Further, the equivalent hydrocarbon of this compound will have the formula $C_{8}H_{18}$. Subtracting the number of H atoms present in the original compound (let it be $x$) from the number of H atoms in the equivalent formula gives the degree of unsaturation, or, $H_{18}-H_{x} = H_{6}= 3H_{2}$ i.e. $H_{18}-H_{6}=H_{x}$ i.e. $H_{12}=H_{x}$ Thus $C_{8}H_{?}O_{2}$ has 12 H atoms. (b) $C_{7}H_{?}N$ has two double bonds or two degrees of unsaturation. The formula of the equivalent saturated hydrocarbon of this compound is $C_{7}H_{16}$. (let the number of H atoms be $x$ in the compound) Following the same method as above, $H_{16}-H_{x-1}=H_{4}=2H_{2}$ (we have subtracted 1 from the number of H atoms since the number of nitrogen atoms have to be subtracted from the number of hydrogen atoms to reach the equivalent formula for the unsaturated hydrocarbon) i.e. $H_{16}-H_{4}=H_{x-1}$ i.e $H_{12}=H_{x-1}$ or $H_{13}=H_{x}$ Thus $C_{7}H_{?}N$ has 13 H atoms. (c) $C_{9}H_{?}NO$ has one ring and three double bonds, i.e. 4 degrees of unsaturation. The formula of the equivalent hydrocarbon of this compound is $C_{9}H_{20}$. And, $H_{20}-H_{x-1}=H_{8}=4H_{2}$ i.e. $H_{20}-H_{x-1}=H_{8}$ i.e. $H_{20}-H_{8}=H_{x-1}$ $H_{12}=H_{x-1}$ or $H_{13}=H_{x}$ Thus $C_{9}H_{?}NO$ has 13 H atoms.