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Degree of Unsaturation How to determine the number of rings and multiple bonds in a compound from its molecular formula.

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Presentation on theme: "Degree of Unsaturation How to determine the number of rings and multiple bonds in a compound from its molecular formula."— Presentation transcript:

1 Degree of Unsaturation How to determine the number of rings and multiple bonds in a compound from its molecular formula

2 No hydrocarbon can contain a greater number of hydrogens than fits the formula C n H 2n+2 All such compounds are acyclic The number of hydrogens is always even.

3 They may be straight (normal) chains as in n-butane (C 4 H 10 ) or they may be branched as in isobutane (C 4 H 10 ) CH 3 CH 2 CH 2 CH 3 CH 3 CH(CH 3 )CH 3

4 then a ring, cyclobutane (C 4 H 8 ), is formed with the loss of H 2 If two hydrogen atoms on non-adjacent carbons of n-butane (C 4 H 10 ) are removed This process is not necessarily a chemical reaction but rather a conceptual device.

5 then If two hydrogen atoms on adjacent carbons of n-butane (C 4 H 10 ) are removed a double bond is formed with the loss of H 2. In this case, the alkene, 1-butene (C 4 H 8 ), is formed.

6 A compound with the molecular formula C 4 H 8 is either an alkene (olefin) or cycloalkane. How to determine the Degree of Unsaturation (DU) of this compound? C 4 H 10 - C 4 H 8 = H 2 most saturated C 4 compound divided by 2= 1 DU

7 Try the following formulas: C6H6C6H6 C 7 H 10 C 10 H 8 3 7 norbornene naphthalene4 DU benzene Example

8 How is the Degree of Unsaturation of a hydrocarbon containing halogen, or other monovalent atom, determined? Every halogen in a hydrocarbon replaces a hydrogen.

9 Replace each halogen with hydrogen and then compare this hydrocarbon with the most saturated hydrocarbon. C 8 H 18 - C 8 H 8 = H 10 /2 = 5 DU The alkyl halide C 8 H 5 BrCl 2 becomes C8H8C8H8

10 One example of an alkyl halide C 8 H 5 BrCl 2 1 DU - C 6 3 DU 1 DU - C 8 Br Cl Cl 2 C8C8 5 DU H H H H H H5H5

11 How is the Degree of Unsaturation of a hydrocarbon containing oxygen, or other divalent atom, determined?

12 C4H8OC4H8O 1 DU C4H8C4H8 Ignore divalent atoms!

13 2,4-Dichlorophenoxyacetic acid benzene ring - 4 DU carbonyl - 1 DU

14 C 8 H 6 Cl 2 O 3 2,4-Dichlorophenoxyacetic acid C 8 H 18 - C 8 H 8 =H 10 / 2 = 5 DU C 8 H 6 Cl 2 (drop O 3 ) C8H8C8H8 (- Cl 2 + H 2 )

15 How is the Degree of Unsaturation of a hydrocarbon containing nitrogen, or other trivalent atom, determined? Substitute CH for every N.

16 pyridine C5H5NC5H5N benzene C6H6C6H6 (- N + CH) C 6 H 14 - C 6 H 6 =H 8 / 2 = 4 DU

17 = H 16 / 2 = 8 DU Cocaine - C 17 H 21 NO 4 C 17 H 21 NO 4 C 17 H 21 N C 18 H 22 C 18 H 38 - C 18 H 22 4 DU 1 DU (6-membered ring) 1 DU (5-membered ring)

18 The End Degree of Unsaturation F. E. Ziegler 2003


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