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2.4 Degree of Unsaturation
The task of drawing isomers for a particular formula can be made easier by com- paring the number of hydrogens with the number of carbons in the formula. Such a comparison allows the calculation of the degree of unsaturation, which furnishes useful information about possible structures that will fit the formula (such as whether double bonds can be present) and provides a starting point for drawing isomers. The isomers shown in Figure 2.4 have the maximum number of hydrogens pos- sible for a compound with six carbons. The maximum number of H’s can easily be calculated from the number of C’s present. To see how the formula arises, consider a straight, or linear, chain of some number of C’s. Each C has two H’s bonded to it, with the exception that the two end C’s have one additional H, because they are bonded to only one C. Therefore, for n carbons, the maximum number of hydrogens is 2 n 2. The general formula for a hydrocarbon with the maximum number of hy- drogens is C n H (^2) n 2. As an example, C 5 H 12 has the maximum number of H’s for 5 C’s [2(5) 2 12]. Now let’s see what happens to the number of hydrogens in a compound if a double bond is present. To form a double bond, a hydrogen must be removed from each of two adjacent carbons. Therefore, a compound with one double bond has two fewer H’s than the maximum.
Similarly, to form a ring from a chain of carbons, one H must be removed from both end C’s so that they can be bonded together. A compound with a ring also has two fewer H’s than the maximum number.
Any compound whose formula has two hydrogens less than the maximum number possible (2 n 2) must contain one double bond or one ring. The total number of mul- tiple bonds plus rings is called the degree of unsaturation (DU) and is equal to 1 for this case. The DU of a compound can be calculated by using the following formula:
DU
(Maximum possible H’s) (Actual H’s) 2
H±C±C±C±C±C±C±H
H
W
W
H
H
W
W
H
H
W
W
H
H
W
W
H
H
W
W
H
H
W
W
H
±
±
±
±
±
±
± ± C
C
C H
H H
H H
H
H
H
H
H
H
H
C
C
C
C 6 H 14 C 6 H 12
Remove two H’s and form ring
H±C±C±C±C C
H
W
W
H
H
W
W
H
H
W
W
H
W
H
H
H
H±C±C±C±C±C±H
H
W
W
H
H
W
W
H
H
W
W
H
H
W
W
H
H
W
W
H
Remove 2 H’s and form CœC
C 5 H 12 C 5 H 10
2.4 DEGREE OF UNSATURATION 39
For a hydrocarbon with the formula C n H x the degree of unsaturation is
DU
where n actual C’s present x actual H’s present The DU is very useful when drawing isomers. Let’s look at a simple example. Suppose we are asked to draw the constitutional isomers with the formula C 5 H 10. The DU for C 5 H 10 is [(2(5) 2) 10]/2 1. Therefore, this compound must have one double bond or one ring. Any compound that has five C’s and one double bond or one ring will fit the formula. Although C 5 H 12 has only three isomers, having two fewer H’s actually increases the number of isomers, because there is now a double bond or a ring to vary. As structures become more complex, drawing them as a line or Kekulé struc- ture, in which each bond is shown as a line, becomes more time consuming. The method of grouping together atoms that are bonded to the same atom to give a condensed structure, which was presented in Chapter 1, can be used, but even this becomes tedious. An even faster method, called a skeletal structure, shows the C – C bonds as lines. Each line is assumed to have a C at each end unless another atom is shown. Hydrogens on the C’s are not shown. (However, if a carbon is shown with a C, the hydrogens on it must also be shown as H’s.) All other atoms (N, O, Cl, and so on) must be shown, along with any hydrogens bonded to them. Unshared pairs of electrons may or may not be shown. Sometimes the various methods are mixed to emphasize a particular feature. Figure 2.5 shows four of the possible iso- mers of C 5 H 10 O using these various methods.
PROBLEM 2.
Calculate the DU for these formulas and draw two constitutional isomers for each: a) C 10 H 22 b) C 9 H 16 c) C 6 H (^6)
PROBLEM 2.
Convert these structures to skeletal structures:
d) CH 3 CHCH 2 NHCH 2 CH 2 CH 3
CH 3
W
c) CH 3 CH 2 CCH 2 CH(CH 3 )
O
X
b)
± ±
±
C H
O
H
H
C
C
W
H±C±H
W
H
C
C
H
W
C
H
a) H±C±C±±±C±±±CœC±C±H
H
W
W
H
H
W
W
H
H
W
W
H
H
W
H±C±H
W
W
H
H
W
H
W
(2 n 2) x 2
40 CHAPTER 2 ORGANIC COMPOUNDS : A FIRST LOOK
Important Convention
42 CHAPTER 2 ORGANIC COMPOUNDS : A FIRST LOOK
Click Coached Tutorial Problems to practice Drawing Skeletal Structures and Recognizing Isomerism.
HCPCCH 2 CH 2 CHCH 3
OH
W
OH
CH 2
CH 2
CH 2
H 2 C
H 2 C
O
X
C
O
CH 2
CH 3
HCœœCH
H 2 C
CH
O
W
±^ O ±CH^3
or
or
or
Figure 2.
T HREE ( OF MANY ) CONSTITUTIONAL ISOMERS OF C 6 H 10 O.
PROBLEM 2.
Determine whether these structures represent the same compound or isomers:
The DU can also be calculated for formulas that have atoms other than C and H. Be- cause halogens are monovalent, they are counted as hydrogens in the DU calculation. For example, C 5 H 11 Cl is counted as C 5 H 12 and has DU 0 —it is saturated. Oxygen is divalent. If an O is added to a structure, it can, for example, be inserted between a C and H or between two C’s without changing the number of hydrogens. Therefore, we can ignore oxygens when performing a DU calculation. The DU for C 6 H 10 O is 2, the same as that for C 6 H 10. This compound must have two double bonds, one triple bond (a triple bond contributes 2 to the DU), one double bond and one ring, or two rings. Although the presence of oxygen is ignored in calculating the DU, oxygen can, of course, be involved in the features, double bonds or rings, that contribute to the DU. Figure 2.6 shows several isomers for C 6 H 10 O.
f ) CH 3 CH 2 CHCH 2 CH 2 CH 3
CH 3
W
CH 3 CH 2 CH 2 CH 2 CHCH 3
CH 3
W
e) CH 3 CHCH 2 CH 2 CH 3
CH 3
W
CH 3 CH 2 CH 2 CHCH 3
CH 3
W
c)^ d)
a)^ b)
Finally, let’s consider the effect of nitrogen on a DU calculation. Nitrogen is triva- lent. If an N is added to a structure by inserting it between two atoms, one H must also be added to satisfy the third valence of the N. Therefore, each nitrogen that is present in a compound increases the maximum number of H’s by one. For example, the maxi- mum number of H’s for C 10 H 15 N is 2(10) 2 1 23. The DU is (23 15)/2 4. You must be very careful when using a shorthand method to show structures. Be- ginning organic chemistry students commonly forget about hydrogens or do not recog- nize other features of such structures. It is often a good idea to redraw the structures more completely, showing each carbon and all the hydrogens, until you are very com- fortable with them and can automatically picture all the features of the molecule.
PRACTICE PROBLEM 2.
Calculate the DU for C 8 H 13 BrO.
Solution
Br counts as an H and O can be ignored, so calculate as though the formula were C 8 H 14. DU ^12 [(2)(8) 2 14] 2.
PROBLEM 2. Calculate the DU for these formulas and draw two constitutional isomers for each: a) C 10 H 20 O b) C 6 H 9 N c) C 7 H 14 F (^2) d) C 6 H 8 ClN e) C 9 H 15 NO
PROBLEM 2. Determine the DU for each of these structures:
e)
CO 2 H
OCCH 3
X
O
Acetylsalicylic acid (aspirin)
CH 2 ±CH±C±OH
d)
N W H Tryptophan (an amino acid)
NH 2
W
O
X
c) (^) Br
O
OCH 2 CH 3 b) (^) N
O
X
a)^ ±
C
2.4 DEGREE OF UNSATURATION 43
Click Coached Tutorial Problems to practice Determining Degrees of Unsaturation.
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