Knowing the number of degrees of unsaturation in a molecule is useful because this number is related to how many multiple bonds or rings are present in an unknown compound. This morsel of information becomes very useful when you want to determine the structure of an unknown compound.
The degrees of unsaturation in a molecule are additive — a molecule with one double bond has one degree of unsaturation, a molecule with two double bonds has two degrees of unsaturation, and so forth. Just as the formation of a double bond causes two hydrogens to be lost, the formation of a ring also results in the loss of two hydrogens, so every ring in the molecule also adds one degree of unsaturation.
For every triple bond, two degrees of unsaturation are added to a molecule, because a molecule must lose four hydrogens to make a triple bond. Some examples of three-carbon molecules with different numbers of degrees of unsaturation are shown here.
To determine the number of degrees of unsaturation for any arbitrary structure, you sum all the individual elements of unsaturation in the molecule. The next figure shows a molecule that consists of one ring, one double bond, and one triple bond.
This molecule, therefore, has four degrees of unsaturation because the double bond and the ring each add one degree of unsaturation, and the triple bond adds two degrees, for a total of four. More important than determining the number of degrees of unsaturation from a molecular structure is being able to determine the number of degrees of unsaturation from a molecular formula.
The number of degrees of unsaturation can be determined from the molecular formula using the following equation. With this equation, the number of degrees of unsaturation can be determined for any hydrocarbon whose molecular formula is known. For compounds whose structure and formula are not known, chemists use an instrumental technique called mass spectrometry to determine the molecular formula of the compound. But what about molecules that contain atoms other than hydrogen and carbon?
In such cases, you need to convert these multi-atom molecular formulas into equivalent formulas that contain just carbon and hydrogen so they can be plugged into the preceding equation.
To do so, you use the following conversion factors:. For example, to determine the number of degrees of unsaturation in the formula C 8 H 6 F 3 NO 2you first make the proper substitutions for all atoms that are not hydrogen and carbon.
Fluorine is a halogen, so you add three hydrogen atoms to the molecular formula one for each F. The molecule contains one nitrogen, so you subtract one hydrogen from the molecular formula. The two oxygens in the molecule you ignore. Plugging this reduced formula into the preceding equation gives five degrees of unsaturation for the molecular formula C 8 H 6 F 3 NO 2. He received his PhD at the University of Maryland in He is currently a chemistry professor at Iowa State University.
How to Determine the Degrees of Unsaturation of a Molecule. The degrees of unsaturation for three-carbon molecules.Frequently, students of organic chemistry are asked -- in homework problems, on quizzes, and tests, etc.
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.
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 examples 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".
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:.
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've 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:. As such, we know that this formula must correspond to a compound having either one double bond or one ring as in propene and cyclopropane, shown previously.
Likewise, any compound with the formula C 5 H 8 must include one of the following four structural features:. Try to draw examples of hydrocarbons having each of these combinations. Thus, P should be treated as though it were N, and Si as though it were C.
You may have noticed that the number of oxygen atoms does not appear in the degree of unsaturation formula.
Thus, each of the following formulas has one degree of unsaturation:. Most organic compounds either consist of carbon and hydrogen only or consist of carbon, hydrogen, and oxygen only. In these cases, the degree of unsaturation formula simplifies to:. Although the number of oxygen atoms in a molecular formula has no effect on the degree of unsaturation, it certainly does have an effect on the number of possible isomers. Determine the number of possible constitutional isomers for the following formulas:.
If you use Beaker to solve this problem, notice that each of the structures drawn for each of the formulas has one degree of unsaturation: either one ring OR one double bond. The degree of unsaturation formula only works for neutral molecules, and only if all of the non-hydrogen atoms in molecule obey the octet rule! The examples shown on the following pages illustrate four sorts of problems you will encounter with charged molecules, or with molecules having octet violations.
And yet it has all single bonds and no rings! If a positive charge should be treated as though it removed one hydrogen ion from the formula, should a negative charge be treated as though it increased the number of hydrogen atoms in the molecular formula? Then apply the degree of unsaturation formula. Hypovalent species have an incomplete octet around at least one atom.Degree of unsaturation Index of Hydrogen Deficiency or IHD is a measure to how many hydrogen atoms a molecule is missing in order to be fully saturated.
A compound that does not have the maximum number of hydrogens per its structure is said to be unsaturated. A molecule that has the maximum number of hydrogen atoms is said to be saturated.
For example, hexane C6H14 is saturated but cyclohexane C6H12 is not. Hexane and cyclohexane have the exact same number of carbons, but cyclohexane actually has two fewer hydrogens than does hexane. Luckily, determining degrees of unsaturation AKA index of hydrogen deficiency, hydrogen deficiency index, etc. Degrees of Unsaturation general formula. IHD hexane formula. IHD cyclohexane formula. Are you ready for some good news?
It turns out that determining IHD from a structure is actually even easier than doing so for a molecular formula. IHD of hexane and cyclohexane. Okay, cool!
So, even though hexane and cyclohexane have the same number of carbons, cyclohexane has one degree of unsaturation while hexane is completely saturated. Cyclohexane and hexene would have the same IHD since a ring is a double bond equivalent. What would the IHD of cyclohexene be? Cyclohexene would have two degrees of unsaturation because it has a pi bond and a ring.
Molecular formula blank.
How many degrees of unsaturation are present in the molecules drawn below? Structure question blank. Molecular formula answered. Structure questions answered.
Benzene has four degrees of unsaturation, naphthalene seven degrees, 3-hexyne two degrees, and styrene has five degrees of unsaturation. Your professor is likely going to ask about molecules including heteroatoms like halides, oxygen, and nitrogen.
To learn how to calculate degrees of unsaturations for molecules with heteroatoms, check out my videos on Index of Hydrogen Deficiency. Hope this helps, and good luck studying!
Degree of Unsaturation
Johnny got his start tutoring Organic in when he was a Teaching Assistant. He now enjoys helping thousands of students crush mechanisms, while moonlighting as a clinical pharmacist on weekends.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.
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, these techniques require expensive instrumentation and are not always readily available. Fortunately, calculating the degrees of unsaturation provides useful information about the structure. The degree of unsaturation indicates the total number of pi bonds and rings within a molecule which makes it easier for one to figure out the molecular structure.
The molecular formula of a hydrocarbon provides information about the possible structural types it may represent. 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. As an example, for the molecular formula C3H4 the number of actual hydrogens needed for the compound to be saturated is 8. 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.Degree of Unsaturation aka Index of Hydrogen Deficiency
For example, consider compounds having the formula C5H8. The formula of the five-carbon alkane pentane is C5H12 so the difference in hydrogen content is 4. This difference suggests such compounds may have a triple bond, two double bonds, a ring plus a double bond, or two rings. Some examples are shown here, and there are at least fourteen others!
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. 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, C 2 H 5 OH, matches the number of hydrogens in ethane, C 2 H 6. Each row corresponds to a different combination.
When the DU is 4 or greater, the presence of benzene rings is very likely. The molecular formula for benzene is C 6 H 6.
This corresponds to benzene containing 1 ring and 3 double bonds. However, when given the molecular formula C 6 H 6benzene is only one of many possible structures isomers. The following structures all have DU of 4 and have the same molecular formula as benzene. However, these compounds are very rare, unlike benzene.
We will learn more about the reasons for benzen's high stability when we studey aromaticity in later chapters. Calculate degrees of unsaturation DoU for the following, and propose a structure for each. Calculate the degree of unsaturation DoU for the following molecules.
However, you can also determine the molecular formula and solve for the degrees of unsaturation by using the formula. Thus, the formula may give numerous possible structures for a given molecular formula.There are many ways one can go about determining the structure of an unknown organic molecule. In the lab, saturation may be thought of as the point when a solution cannot dissolve anymore of a substance added to it. As stated before, a saturated molecule contains only single bonds and no rings.
Hence, the DoB formula divides by 2. 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. Each row corresponds to a different combination. This corresponds to benzene containing 1 ring and 3 double bonds.
However, when given the molecular formula C 6 H 6benzene 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 Vollhardt, K. Organic Chemistry 5 th Ed. New York: W. C 10 H 6 N 4 Using the molecules from 1. Calculate the degrees of unsaturation for the following molecular formulas: a.
Thus, the formula may give numerous possible structures for a given molecular formula. Answers 1. Skip to main content. Chapter 5: Alkenes: Structure and Reactivity. Search for:.
How to Calculate the Degree of Unsaturation
Calculating Degree of Unsaturation 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.
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. Show Answer a. Exercises Questions 1. Calculate degrees of unsaturation DoU for the following, and propose a structure for each. Solutions 1. Show Answer. Show Answer 6 DoU.The Degree, or Elements, of Unsaturation tells you how may rings and multiple bonds are present in a compound if you know the molecular formula.
Most textbooks give you a formula to determine the Degree of Unsaturation. Here is an empirical method to do the same thing. This statement is true for both normal straight chain and branched alkanes. After all, both n-butane and isobutane are isomers same composition, different structures C 4 H When a hydrocarbon has the formula C n H 2nit is two hydrogen atoms shy of being completely saturated.
A hydrocarbon with a degree of unsaturation of 1 is either an acyclic non-cyclic alkene or a cycloalkane. You can imagine taking n-pentane above left and mentally removing a hydrogen atom from C 1 and C 2. What you have left, in addition to H 2is 1-pentene.
If the same mental exercise were conducted at C 1 and C 5 of n-pentane and the two radicals were joined, a cyclopentane ring would be formed. Convince yourself that each of the following hydrocarbons fits the formula C n H 2n. If a compound fits the formula C n H 2n-2 it has two degrees of unsaturation. Do the math. This compound may be a triple bond alkynetwo double bonds dienea double bond and a ring cycloalkeneor two ring bicyclic alkane.
The degree of unsaturation may also be employed in another way. Given a particular hydrocarbon structure for which you know the number of carbons and the degree of unsaturation, you can calculate the number of hydrogen atoms. Benzene has 6 carbon atoms and 4 degrees of unsaturation 1 ring and 3 double bonds. If you work backwards and double the degrees of unsaturation you have 8. The maximum number of hydrogens for a C 6 compound is The difference between 14 and 8 is 6.
Benzene is C 6 H 6.To utilize this method, one must understand the meaning of unsaturated and saturated compounds. In contrast, an example of an unsaturated compound includes an acyclic alkene that follows the formula C n H 2n two hydrogens less, result of double bond.
Similarly, cycloalkanes follow the same formula: C n H 2n. This leads to the concept of the degree of unsaturation, which can defined as the sum of numbers of rings and pi bonds present in a molecule. Each degree of unsaturation refers to a decrease in two hydrogens in the molecule, as a result of the presence of a pi bond or a ring.
A degree of unsaturation of 1 means that there is a decrease of two hydrogens in the molecular formula and that the resulting formula will CnH2n. Following the same logic, a degree of unsaturation value of 2 will give a molecular formula of C n H 2n There can be two pi bonds present, one pi bond and one ring present, or just two rings present.
Similarly, with a degree of unsaturation value of 3, the scenarios are as follows: three pi bonds present, two pi bonds and one ring present, or one pi bond and two rings present. Degrees of unsaturation can help us determine how an alkane molecule will act. The more saturated a chain is the less fluidity it has. This can have many implications for structural biochemistry because it changes the way a molecule behaves or its function by changing it's shape and fluidity.
Degree of unsaturation can illuminate structure by showing double or triple bonds that exist. This phenomena can also be described as hydrogen deficiency. The more rings and double bonds a molecule has the less saturated the molecule. It is important to remember what these double bonds and rings can do to function and behavior of a molecule. Usually when determining degrees of unsaturation molecules that are not carbons and hydrogens are ignored.
These can be molecules such as oxygen, nitrogen or any others. The examples given above apply to simple situations where there are no heteroatoms present. When these atoms are involved, the degree of unsaturation calculations require a little more work. A general equation can be used to determine the degree of unsaturation from a molecular formula: . This formula can be verified by drawing out various structures with an assortment of rings, pi bonds, halogens, nitrogens, and oxygens.
The presence of nitrogen increases the number of hydrogens required to reach saturation because nitrogen is trivalent. Adding an amine to the compound results in one extra hydrogen, since the nitrogen component is essentially taking away a C-H bond and adding two N-H bonds and a C-N bond.
For halogens, there is essentially one C-H bond being replaced with a C-X bond; therefore, the presence of halogens reducs the number of hydrogens required for saturation: compare ethane C 2 H 6 and chloroethane C 2 H 5 Cl.
Oxygen is divalent, and so it will have no effect on the calculation of degree of unsaturation. A simple example is finding the degree of unsaturation of the molecular formula C 4 H 8.
Therefore, the structure of the molecular formula of C 4 H 8 has either an alkene present 1 pi bondor it is a cyclobutane. A slightly more challenging example is a molecular formula of C 5 H 9 N. Thus, there can be two pi bonds present, one pi bond and one ring, or two rings present in the molecule. From Wikibooks, open books for an open world. Organic Chemistry Structure and Function 6th Edition.
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