Skip to main content
Chemistry LibreTexts

7.2: Calculating Degree of Unsaturation

  • Page ID
    67123
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\)

    Objectives

    After completing this section, you should be able to

    1. determine the degree of unsaturation of an organic compound, given its molecular formula, and hence determine the number of double bonds, triple bonds and rings present in the compound.
    2. draw all the possible isomers that correspond to a given molecular formula containing only carbon (up to a maximum of six atoms) and hydrogen.
    3. draw a specified number of isomers that correspond to a given molecular formula containing carbon, hydrogen, and possibly other elements, such as oxygen, nitrogen and the halogens.
    Key Terms

    Make certain that you can define, and use in context, the key terms below.

    • degree of unsaturation
    • saturated
    • unsaturated

    There are many ways one can go about determining the structure of an unknown organic molecule. Although, nuclear magnetic resonance (NMR) and infrared radiation (IR) are the primary ways of determining molecular structures, calculating the degrees of unsaturation is useful information since knowing the degrees of unsaturation make it easier for one to figure out the molecular structure; it helps one double-check the number of \(\pi\) bonds and/or cyclic rings.

    Saturated and Unsaturated Molecules

    In the lab, saturation may be thought of as the point when a solution cannot dissolve anymore of a substance added to it. In terms of degrees of unsaturation, a molecule only containing single bonds with no rings is considered saturated.

    CH3CH2CH3 chewiki_sat.bmp chewiki_sat2 (3).bmp 1-methyoxypentane

    Unlike saturated molecules, unsaturated molecules contain double bond(s), triple bond(s) and/or ring(s).

    CH3CH=CHCH3 chewiki_unsat1.bmp chewiki_unsat2.bmp chewiki_unsat3.bmp 3-chloro-5-octyne

    Calculating The Degree of Unsaturation (DoU)

    If the molecular formula is given, plug in the numbers into this formula:

    \[ DoU= \dfrac{2C+2+N-X-H}{2} \tag{7.2.1}\]

    • \(C\) is the number of carbons
    • \(N\) is the number of nitrogens
    • \(X\) is the number of halogens (F, Cl, Br, I)
    • \(H\) is the number of hydrogens

    As stated before, a saturated molecule contains only single bonds and no rings. Another way of interpreting this is that a saturated molecule has the maximum number of hydrogen atoms possible to be an acyclic alkane. Thus, the number of hydrogens can be represented by 2C+2, which is the general molecular representation of an alkane. As an example, for the molecular formula C3H4 the number of actual hydrogens needed for the compound to be saturated is 8 [2C+2=(2x3)+2=8]. The compound needs 4 more hydrogens in order to be fully saturated (expected number of hydrogens-observed number of hydrogens=8-4=4). Degrees of unsaturation is equal to 2, or half the number of hydrogens the molecule needs to be classified as saturated. Hence, the DoB formula divides by 2. The formula subtracts the number of X's because a halogen (X) replaces a hydrogen in a compound. For instance, in chloroethane, C2H5Cl, there is one less hydrogen compared to ethane, C2H6.

    For a compound to be saturated, there is one more hydrogen in a molecule when nitrogen is present. Therefore, we add the number of nitrogens (N). This can be seen with C3H9N compared to C3H8. Oxygen and sulfur are not included in the formula because saturation is unaffected by these elements. As seen in alcohols, the same number of hydrogens in ethanol, C2H5OH, matches the number of hydrogens in ethane, C2H6.

    The following chart illustrates the possible combinations of the number of double bond(s), triple bond(s), and/or ring(s) for a given degree of unsaturation. Each row corresponds to a different combination.

    • One degree of unsaturation is equivalent to 1 ring or 1 double bond (1 \( \pi \) bond).
    • Two degrees of unsaturation is equivalent to 2 double bonds, 1 ring and 1 double bond, 2 rings, or 1 triple bond (2 \( \pi \) bonds).
    DoU
    Possible combinations of rings/ bonds
     
    # of rings
    # of double bonds
    # of triple bonds
    1
    1
    0
    0
    0
    1
    0
    2
    2
    0
    0
    0
    2
    0
    0
    0
    1
    1
    1
    0

    Remember, the degrees of unsaturation only gives the sum of double bonds, triple bonds and/or rings. For instance, a degree of unsaturation of 3 can contain 3 rings, 2 rings+1 double bond, 1 ring+2 double bonds, 1 ring+1 triple bond, 1 double bond+1 triple bond, or 3 double bonds.

    Example 7.2.1: Benzene

    What is the Degree of Unsaturation for Benzene?

    Solution

    The molecular formula for benzene is C6H6. Thus,

    DoU= 4, where C=6, N=0,X=0, and H=6. 1 DoB can equal 1 ring or 1 double bond. This corresponds to benzene containing 1 ring and 3 double bonds.

    However, when given the molecular formula C6H6, benzene is only one of many possible structures (isomers). The following structures all have DoB of 4 and have the same molecular formula as benzene.

    References

    1. Vollhardt, K. P.C. & Shore, N. (2007). Organic Chemistry (5thEd.). New York: W. H. Freeman. (473-474)
    2. Shore, N. (2007). Study Guide and Solutions Manual for Organic Chemistry (5th Ed.). New York: W.H. Freeman. (201)

    Exercises

    Exercise \(\PageIndex{1}\)

    How many degrees of unsaturation do the following compounds have?

    7.3.1 how many deg unsat.svg

     

    Answer

    a) 0

    b) 1

    c) 1

    d) 2

    e) 2

    f) 2

    Exercise \(\PageIndex{2}\)

    Determine whether the following molecules are saturated or unsaturated.

    7.3.1 saturated or unsaturated.svg

    Answer

    a) unsaturated (Even though the rings only contain single bonds, rings are considered unsaturated.)

    b) unsaturated

    c) saturated

    d) unsaturated

    e) unsaturated

    f) saturated

    Exercise \(\PageIndex{3}\)

    Determine the degrees of unsaturation for each of the following compounds.

    7.3.2 saturated or unsaturated.svg

    Answer

    If the molecular structure is given, the easiest way to solve is to count the number of double bonds, triple bonds and/or rings. However, you can also determine the molecular formula and solve for the degrees of unsaturation by using the formula.

    a) 2 (2 rings)

    b) 2 (one double bond and the double bond from the carbonyl)

    c) 0 (no double bonds or rings)

    d) 10 (2(10) + 2 + 4 - 0 - 6)/2 = 10

    e) 1 (2(5) + 2 + 0 - 0 - 10)/2 = 1

    f) 0 (2(6) + 2 + 0 - 2 - 12)/2 = 0

    Exercise \(\PageIndex{4}\)

    Calculate the degrees of unsaturation for the following molecular formulas:

    a) C9H20 b) C7H8 c) C5H7Cl d) C9H9NO4

    Answer

    Use the formula to solve (O not involved in the formula)

    (a.) 0 (2(9) + 2 - 20)/2 = 0

    (b.) 4 (2(7) + 2 - 8)/2 = 4

    (c.) 2 (2(5) + 2 - 1 -7)/2 = 2

    (d.) 6 (2(9) + 1 - 7)/2 = 6

    Contributors and Attributions


    7.2: Calculating Degree of Unsaturation is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by LibreTexts.

    • Was this article helpful?