Degree of Unsaturation in Organic Compounds: Rings, Bonds, and Groups, Slides of Molecular Chemistry

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Degree of Unsaturation

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

No hydrocarbon can contain a greater number of hydrogens than fits the formula C n

H

2n+ All such compounds are acyclic The number of hydrogens is always even.

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.

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.

Try the following formulas: C 6

H

6 C 7

H

10 C 10

H

8

norbornene naphthalene

DU

benzene Example

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

One example of an alkyl halide C 8

H

5 BrCl 2 1 DU - C 6 3 DU 1 DU - C 8 Br Br Cl Cl Cl 2

C

8

5 DU

H

H

H

H

H

H

5

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

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

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 )

C

8

H

8 (- Cl 2 + H 2 )

pyridine C 5

H

5

N

benzene C 6

H

6 (- N + CH) C 6

H

14

- C

6

H

6

= H

8

/ 2 = 4 DU

= 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

1 DU

1 DU

(6-membered ring)

1 DU

(5-membered ring)