7.2: Calculating Degree of Unsaturation

Learning Objectives

After completing this section, you should be able to

• 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.
• draw all the possible isomers corresponding to a given molecular formula containing only carbon (up to a maximum of six atoms) and hydrogen.
• draw a specified number of isomers corresponding to a given molecular formula containing carbon, hydrogen, and possibly other elements, such as oxygen, nitrogen and the halogens.

Saturated and Unsaturated Molecules

A C-H molecule only containing single bonds with no rings is considered saturated, as no additional atoms can be added without removing any.

CH3CH2CH3

1-methyoxypentane

Unlike saturated molecules, unsaturated molecules contain double bond(s), triple bond(s) and/or ring(s). Because of its double bond, an alkene has fewer hydrogens than an alkane with the same number of carbons—C_nH_2n for an alkene versus C_nH_2n+2 for an alkane—and is therefore referred to as unsaturated. Ethylene, for example, has the formula C₂H₄, whereas ethane has the formula C₂H₆.

 

Calculating The Degree of Unsaturation (DoU) or Index of Hydrogen Deficiency (IHD)

For organic compounds containing only C and H atoms

• 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.
• if H = 2C + 2 the molecule is considered saturated, DoU is equal to 0.

Combining these mathematical concepts, the degree of unsaturation can be defined in the following equation:

DoU = [2C + 2 − H ] / 2 

C represents the total amount of carbons, and H represents the total amount of hydrogens.

Worked Example \(\PageIndex{1}\)

Example: Calculate the degree of unsaturation in CH₃CH₂CH_3, the molecular formula is C₃H₈

Solution

Applying the simplified equation: DoU = [2C + 2 − H ] / 2

DoU = (2*3 + 2 − 8) / 2 

DoU = 0 

DoU = 0 means that no unsaturation is observed in this molecule.

In general, each ring or double bond in a molecule corresponds to a loss of two hydrogens from the alkane formula C_nH_2n+2. Knowing this relationship, it’s possible to work backward from a molecular formula to calculate a molecule’s degree of unsaturation—the number of rings and/or multiple bonds present in the molecule.

As an example, for the molecular formula C₃H₄. The number of actual hydrogens needed for a compound with 3 carbon atoms to be saturated is 8. Calculated as 2C + 2 = (2*3 ) + 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 half the number of hydrogens the molecule needs to be classified as saturated (4 /2). Hence, the DoU is 2.

Worked Example \(\PageIndex{2}\)

Now let’s assume that we want to find the structure of an unknown hydrocarbon. A molecular weight determination yields a value of 82 amu, which corresponds to a molecular formula of C₆H₁₀.

Solution

Since the saturated C₆ alkane (hexane) has the formula C₆H₁₄, the unknown compound has two fewer pairs of hydrogens (H₁₄ − H₁₀ = H₄ = 2 H₂) so its degree of unsaturation is 2. The unknown therefore contains either two double bonds, one ring and one double bond, two rings, or one triple bond. There’s still a long way to go to establish its structure, but the simple calculation has told us a lot about the molecule.

DoU = [2C + 2 − H ] / 2

DoU = (2*6 + 2 − 10) / 2 

DoU= 4 / 2 = 2

Similar calculations can be carried out for compounds containing elements other than just carbon and hydrogen.

For Organic Compounds Containing Oxygen

• Organooxygen compounds (C, H, O)
  • Oxygen forms two bonds. When an oxygen atom is inserted into an alkane bond: C−C becomes C−O−C or C−H becomes C−O−H, and there is no change in the number of hydrogen atoms, so it doesn’t affect the formula of an equivalent hydrocarbon and can be ignored when calculating the degree of unsaturation. 

Worked Example \(\PageIndex{3}\)

Calculate the degree of unsaturation for the formula C₅H₈O.

Solution

C₅H₈O is equivalent to the hydrocarbon formula C₅H₈ and thus corresponds to two degrees of unsaturation. Using the equation: DoU = [2C + 2 − H ] / 2

DoU = (2*5 + 2 − 8) / 2 

DoU= 4 / 2 = 2

 

For Organic Compounds Containing Halogens

• Organohalogen compounds (C, H, X, where X = F, Cl, Br, or I)
  • A halogen substituent acts as a replacement for hydrogen in an organic molecule, so we can add the number of halogens and hydrogens to arrive at an equivalent hydrocarbon formula from which the degree of unsaturation can be found.
  • For example, the formula C₄H₆Br₂ is equivalent to the hydrocarbon formula C₄H₈ and thus corresponds to one degree of unsaturation.

For Organic Compounds Containing Nitrogens

• Organonitrogen compounds (C, H, N)
  • Nitrogen forms three bonds, so an organonitrogen compound has one more hydrogen than a related hydrocarbon. We therefore subtract the number of nitrogens from the number of hydrogens to arrive at the equivalent hydrocarbon formula. Again, you can convince yourself of this by seeing what happens when a nitrogen atom is inserted into an alkane bond: C−C becomes C−NH−C or C−H becomes C−NH₂, meaning that one additional hydrogen atom has been added. We must, therefore, subtract one from the number of hydrogen atoms per each nitrogen present in the molecule to arrive at the equivalent hydrocarbon formula.
  • For example, the formula C₅H₉N is equivalent to C₅H₈ and thus has two degrees of unsaturation.

As a Summary

• Ignore the number of oxygens.
• Add the number of halogens to the number of hydrogens.
• Subtract the number of nitrogens from the number of hydrogens.

Combined Equation:                             DoU = [2C + 2 − (H + X - N) ] / 2 

• 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

Exercise \(\PageIndex{1}\)

Calculate the degree of unsaturation in each of the following formulas, and then draw as many structures as you can for each:

1. C₄H₈
2. C₄H₆
3. C₃H₄

Answer

a. 1  b. 2  c. 2

Exercise \(\PageIndex{2}\)

Calculate the degree of unsaturation in each of the following formulas:

1. C₆H₅N
2. C₆H₅NO₂
3. C₈H₉Cl₃
4. C₉H₁₆Br₂ 
5. C₁₀H₁₂N₂O₃
6. C₂₀H₃₂ClN

Answer

a) 5   b) 5   c) 3   d) 1   e) 6   F) 5

Exercise \(\PageIndex{3}\)

Diazepam, marketed as an antianxiety medication under the name Valium, has three rings, eight double bonds, and the formula C₁₆H_?ClN₂O. How many hydrogens does diazepam have? (Calculate the answer; don’t count hydrogens in the structure.)

Answer

C₁₆H₁₃ClN₂O

Exercise \(\PageIndex{4}\)

How many degrees of unsaturation do the following compounds have?

Answer

a) 0   b) 1   c) 1   d) 2   e) 2   F) 2

Exercise \(\PageIndex{5}\)

Determine whether the following molecules are saturated or unsaturated. Then, determine the degrees of unsaturation for each of the following compounds.

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   b) 2   c) 0   d) 10   e) 1   F) 0

Exercise \(\PageIndex{6}\)

Calculate the degrees of unsaturation for the following molecular formulas:

a) C₉H₂₀ b) C₇H₈ c) C₅H₇Cl d) C₉H₉NO₄

Answer

a) 0   b) 4   c) 2   d) 6