Calculate the degree of unsaturation in each of the...

Calculate the degree of unsaturation in each of the following formulas: (a) $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{~N}$ (b) $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NO}_{2}$ (c) $\mathrm{C}_{8} \mathrm{H}_{9} \mathrm{Cl}_{3}$ (d) $\mathrm{C}_{9} \mathrm{H}_{16} \mathrm{Br}_{2}$ (e) $\mathrm{C}_{10} \mathrm{H}_{12} \mathrm{~N}_{2} \mathrm{O}_{3}$ (f) $\mathrm{C}_{20} \mathrm{H}_{32} \mathrm{ClN}$

Calculate the degree of unsaturation in each of the following formulas:
(a) $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{~N}$
(b) $\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NO}_{2}$
(c) $\mathrm{C}_{8} \mathrm{H}_{9} \mathrm{Cl}_{3}$
(d) $\mathrm{C}_{9} \mathrm{H}_{16} \mathrm{Br}_{2}$
(e) $\mathrm{C}_{10} \mathrm{H}_{12} \mathrm{~N}_{2} \mathrm{O}_{3}$
(f) $\mathrm{C}_{20} \mathrm{H}_{32} \mathrm{ClN}$

Key Concepts

Degree of Unsaturation

The degree of unsaturation indicates the total number of rings and multiple bonds (pi bonds) in a molecule. It provides insight into the structure and possible isomerism of organic compounds by quantifying how many pairs of hydrogen atoms are missing from a fully hydrogenated (saturated) structure.

General Formula for Unsaturation Calculation

A common method of calculating the degree of unsaturation is to use the formula: DU = C - (H/2) + (N/2) + 1, with adjustments made for halogens (which are treated like hydrogens) and with oxygen and sulfur not affecting the calculation. This formula helps relate the actual number of atoms in a molecule to what would be expected in a saturated analog.

Impact of Heteroatoms on Unsaturation

Heteroatoms such as nitrogen and halogens require modifications to the standard degree of unsaturation formula. While nitrogen increases the effective hydrogen count by one half per atom, halogens are treated as hydrogen atoms, and oxygen typically has no direct effect. These adjustments are essential to accurately determine the unsaturation level of a molecule.

Transcript

00:02 So first of all, degree of unsaturation that is given by 2c, that is carbon atom minus hydrogen, this is h, minus x, this is for halogen, plus n, this is for nitrogen, plus n, this is for nitrogen, plus 2, divide by 2.

00:35 Okay so now we will put the values in this formula first of all we have c6 h5n okay so now degree of unsaturation that will be equal to two multiplied by there are six carbon atoms then minus we have five hydrogen atoms here then minus there is no halogen atom so this will be zero plus nitrogen atom so there is one nitrogen atom plus two divide by two so this will be equal to five okay so that is the degree of unsaturation for this compound five then b part c 6 h5 and no 2.

01:48 So degree of unsaturation, so that will be equal to, so we have 2 multiplied by 4 carbon, we have 6 carbon atoms here, minus for hydrogen atoms, we have 5 hydrogen atoms here, minus.

02:22 Now there is no halogen atom here, x, so there is no halogen atom.

02:27 And plus nitrogen atom so there is one nitrogen atom plus 2 divided by 2 so here we have to ignore the oxygen atoms okay so now solving this we will get again 5 then c part c8 h9 cl 3 so degree of unsaturation so that we have equals to so we have here eight carbon atoms two multiplied by eight minus nine hydrogen atoms are here minus halogen atoms so there are three halogen atoms okay three chlorine atoms so that represents for halogen plus there is no nitrogen an atom, okay? and this will be plus 2 divided by 2.

03:45 So solving this we will get this will be 3.

03:55 We have d part that is c9 h16 br2.

04:02 So degree of unsaturation that will be 2.

04:17 We have 2.

04:17 We have 2 multiple 2.

04:18 By atomic just carbon, okay, 2 multiplied by number of carbon atoms.

04:24 So there are nine carbon atoms, then minus, then we have hydrogen atoms...

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