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Degree of Unsaturation

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Frequently, students of organic chemistry are asked -- in homework problems, on quizzes, and tests, etc. -- to draw one or more legitimate structural formulas for a compound having some particular molecular formula. This process can be simplified considerably if one understands that a molecular formula dictates not only the number and type of atoms that must appear in the structural formula, but also the number and types of bonds that must be present. This page explores the use of the "degree of unsaturation" formula as a way to take the guesswork out of a molecular formula -- to -- structural formula problem.

Strongly Related Topics

Somewhat Related Topics

Glossary Terms
constitutional isomers isomers saturated

Degree of Unsaturation

A molecule with only single bonds is said to be "saturated." The presence of multiple bonds introduces what is known as "unsaturation":

Notice in the example above that for each new pi bond introduced, the number of hydrogen atoms decreased by two. Thus, each pi bond introduces one "degree of unsaturation."

Note also, however, the following cases:

There are no pi bonds in cyclopropane, yet its chemical formula is identical to that of propene. Thus, a ring is said to introduce one degree of unsaturation, just like a pi bond. We can compute the total degree of unsaturation for a molecule by adding:

(# of double bonds) + (2 x # of triple bonds) + (# of rings)

Note that a triple bond counts for two degrees of unsaturation, because a triple bond contains two pi bonds.

Quite often, a large number of isomeric compounds can be drawn from the same molecular formula. It is a remarkable fact, however, that regardless of which isomer you have drawn, it must have the same degree of unsaturation as any other isomer -- molecular formula dictates the degree of unsaturation, according to the formula below:

Degree of Unsaturation = [2 + (2 x #Carbons) + #Nitrogens - #Hydrogens - #Halogens]/2

Any compound C3H6 must have one degree of unsaturation, because [2+(2x3)-6]/2=1. As such, we know that this formula must correspond to a compound either one double bond or one ring (as in propene and cyclopropene, shown previously). Likewise, any compound with the formula C5H8 must include one of the following four stuctural features:

  • two double bonds, as in
  • one double bond and one ring, as in
  • two rings, as in
  • one triple bond, as in

Self-test question #1

What are the possible combinations of rings, double bonds, and/or triple bonds that could exist for a compound having three degrees of unsaturation. Try to draw examples of hydrocarbons (species containing only carbon and hydrogen) having each of these combinations.


Related reading in textbook (McMurry, Organic Chemistry, 4th ed.)

  • Chapter 6, Section 2,(pages 180-183): Calculating a Molecule's Degree of Unsaturation

Related Computer-Based Learning Materials

  • Links to Related Chem TV Files
    • Volume 1, Topic 2: Bonding in Organic Compounds

  • Links to Related xxx Files

  • Links to Related yyy Files

  • Links to Related zzz Files

  • Related Beaker Menu Functions
    • Redraw, "Line Segment"
    • Analysis, "Structural Isomers"
    • Analysis, "Molecular Formula"
    • Analysis, "Unsaturation Number

Links to Related Internet Resources

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This page was prepared by Matthew Stabinsky of the Penn State University, Schuylkill Campus, Fall 1996-Spring 1997

Send questions, comments, or suggestions to:
Dr. Thomas H. Eberlein
Copyright © 1996 Thomas H. Eberlein

Version 1.1.7, 3/17/97