Calculate the degree of unsaturation in each of the following formulas, and then draw as many structures as you
can for each:
a) C4H8
b) C4H6
c) C3H4
Answer (1)
To calculate the degree of unsaturation (DoU) for hydrocarbons with formula [tex]C_{n} H_{m}[/tex] use: [tex]\text{DoU} = \frac{2C+2-H}{2}[/tex]a) [tex]C_{4} H_{8}[/tex] [tex]\text{DoU} = \frac{(2(4) + 2 + 0 - 8 - 0)}{2} = \frac{(8 + 2 - 8)}{2} = \frac{2}{2} = 1[/tex]Interpretation - 1 degree of unsaturation means either one double bond or one ring.Possible structuresButene isomers (1-butene, 2-butene) with one double bondCyclobutane (one ring)Methylcyclopropane (one ring)b) [tex]C_{4} H_{6}[/tex] [tex]\text{DoU} = \frac{(2(4) + 2 + 0 - 6 - 0)}{2} = \frac{(8 + 2 - 6)}{2} = \frac{4}{2} = 2[/tex]Interpretation - 2 degrees of unsaturation could be two double bonds, one triple bond, one ring plus one double bond, or two rings.Possible structures1,3-butadiene (two double bonds)Butyne isomers (one triple bond)Cyclobutene (one ring + one double bond)Bicyclobutane (two rings)c) [tex]C_{3} H_{4}[/tex] [tex]\text{DoU} = \frac{(2(3) + 2 + 0 - 4 - 0)}{2} = \frac{(6 + 2 - 4)}{2} = \frac{4}{2} = 2[/tex]Interpretation - 2 degrees of unsaturation.Possible structuresPropyne (one triple bond)Allene (two double bonds)Cyclopropene (one ring + one double bond)
