Fall 2010
Chemistry 211
Double Bond Equivalents
(Structural Information Obtainable from the Molecular Formulas)
(CGWW pp. 74-76)
SUMMARY of CLASS DISCUSSION
Learning Objectives:
1. To predict the number of rings and or multiple bonds required for a structure with a given molecular formula.
2. To quickly determine the number of hydrogen atoms present in a molecule from its bond-line structure.
3. To isolate variables and solve problems with symbolic data.
Exploration:
1. Note that all of the structures below have the same number of carbon atoms, however, they differ in the arrangement of the atoms, the number of rings
or multiple bonds.
C7H16
C7H16
C7H12
C7H12
C7H14
C7H14
C7H12
C7H14
C7H12
C7H16
C7H14
C7H14
2. Write the molecular formula for each molecule in 1 under the structure of the molecule.
See above
3. Based on your results in 2, what is the maximum number hydrogen atoms that can be present in a 7 carbon hydrocarbon? How does this number
compare to the number of carbons in the molecules? How many rings or double bonds are there in 7-carbon hydrocarbon with the maximum number of
hydrogen atoms? Circle or highlight these molecules above.
The highest number of H’s appears to be 16. That is two more than twice the number of carbon atoms in structures. (2*(#of C’s) + 2)
Structures with 16 H’s have all single bonds and no rings, saturated Hydrocarbons (HC’s). See circled structures above.
4. Organic molecules with the maximum number of hydrogen atoms are referred to as “saturated” because they are saturated with hydrogen. From the
relationship between carbon and hydrogen developed in 3. Predict the number of hydrogen atoms in saturated hydrocarbons with:
Using the relationship from # 3
2 carbons atoms
5 carbon atoms
25 carbon atoms
103 carbon atoms
2*2 + 2 = 6
2*5 + 2 = 12
2*25 + 2 = 50
2*103 + 2 = 208
2
Double Bond Equivalents
5. Re-draw the structures from 1. below in groups with common molecular formulas.
C7H14
C7H16
C7H16
C7H12
C7H14
C7H16
C7H12
C7H12
C7H14
C7H12
C7H16
C7H14
C7H14
C7H12
C7H14
6. Make some judgments as to the effects of rings or multiple bonds on molecular formulas of 7 carbon hydrocarbons.
Data Analysis:
All C7H16 molecules are saturated (have no rings or multiple bonds), C7H14 molecules have either 1 ring or 1 double bond, and C7H12 molecules
have either 2 double bonds, 1 ring and 1 double bond or 1 triple bond.
Claims:
Each double bond or ring present in a structure reduces the number of hydrogen atoms by 2 and each triple bond reduces the number of
hydrogen atoms by 4.
Warrant:
Structures with a ring or double bond have 2 fewer H’s in their molecular formulas than do corresponding saturated molecules. The
structure with a triple bond has 4 fewer H’s in its molecular formula than does the corresponding saturated molecule. Thus a triple bond
causes twice the effect of either a ring or a double bond. The effects of rings and double bonds appear to be additive since molecules with
either two double bonds or a ring and a double bond have 4 fewer H’s than do corresponding saturated molecules.
7. How did the grouping of structures required in 4 assist you in your analysis?
Grouping the structures helped in this analysis because it showed the relationship between the number of rings and double bonds and the
molecular formula by holding the molecular formula constant for each group of structures.
Double Bond Equivalents
3
8. Devise a method for using molecular formulas to predict, without drawing structures, the existence of rings or multiple bonds in 7 carbon
hydrocarbons.
a. Saturated 7 carbon HC’s have 16 H’s (See 3.)
b. As indicated in 6. HC’s with rings and/or multiple bonds have fewer hydrogen atoms than do saturated HC’s.
c. The difference between the number of H's in the molecular formula of any 7 carbon HC and the number of hydrogen atoms in a saturated
7 carbon HC can be used to predict the possible number of rings or multiple bonds. e.g. 2 fewer H's could result from 1 ring or 1 double
bond; 4 fewer H's could result from 1 triple bond, 1 ring and 1 double bond, 2 rings, or 2 double bonds; etc. So, we define the 2 H
difference as a “double bond equivalent”, DBE.
9. Can your method be easily modified to work for hydrocarbons with fewer than 7 carbon atoms? More than 7 carbon atoms? Demonstrate how or why
not.
The 16 H’s in a saturated C7 HC = 2*(the number of carbon atoms) + 2
This relationships can be visualized by considering a straight chain saturated HC in a dash structure:
H
H
H
H
H
H
H
H C
C
C
C
C
C
C H
H
H
H
H
H
H
H
+1
7*CH2
+1
= 2*n + 2
Where n= # of carbon atoms
So this 7 carbon molecule can be viewed as 7 carbons each with 2 H's and then two more on the ends -> 2*(the number of carbon atoms) +
2.
The 2n + 2 calculation for # of H’s in a saturated HC should work for a HC with any number of C’s and the effect of multiple bonds and
rings should be the same for any molecule.
The approach described in 8 c. should work for any HC molecule.
10. Below are listed molecular formulas for three hydrocarbons.
C10H20
C6H14
C9H16
Use your method developed in 9 to predict, without drawing structures, the number of rings and/or multiple bonds that must be present in molecules
with these molecular formulas. Explain your logic.
C10H20
C6H14
C9H16
The saturated HC 2n+2 H’s =22
The saturated HC 2n+2 H’s =14
The saturated HC 2n+2 H’s =20
Cpd has 20 H’s
Cpd has 14 H’s
Cpd has 16 H’s
Difference 2 H’s
Difference 0 H’s
Difference 4 H’s
Equivalent to 1 ring or 1 C=C
Equivalent to 0 ring or 0 C=C
Equivalent to a combination of a total of
2 rings + C=C’s
DBE = 1
DBE = 0
or 1 triple bond
DBE = 2
4
Double Bond Equivalents
11. Use the method developed in 9 to write the molecular formulas of the following structures without counting all of the atoms. Which atoms must be
counted? Which ones can be determined from your relationship? Explain the logic you used.
Note that the logic employed here is the reverse of the logic for answering # 10
Carbon atoms must be counted
13 carbons
saturated HC = 2n+2 = 28 H’s
compound has 2 rings & 1 C=C
DBE = 3
3 DBE’s = 6 H’s less than sat’d
Carbon atoms must be counted
14 carbons
saturated HC = 2n+2 = 30 H’s
compound has 2 rings & 3 C=C’s
DBE = 5
5 DBE’s = 10 H’s less than sat’d
C13H22
C14H20
12. Below are structures for several additional molecules with the same number of carbon atoms as those in 1 above but which also contain heteroatoms
(Atoms other than carbon or hydrogen). Write the molecular formula for each of these molecules under its structure and make a note of the number of
rings and double bonds in each.
O
O
O
Cl
H
N
N
Cl
O
I
Br
H
O
H
Br
H
H2N
Cl
O
Double Bond Equivalents
5
Re-draw the structures in groups with the same heteroatom present, write the molecular formula for each molecule under its structure and make a note
of the number of rings and double bonds in each.
Oxygen
Nitrogen
O
O
N
C
H
N
7
15
H C7H17N
no rings or double bonds one double bond
N
C7H16O
C7H14O
O
H
O
C7H14O
H
O
C7H16O
C7H14O
no rings or double bonds
one double bond
O
C7H14O
one ring
H2N C H N
7 15
one ring
Cl
H
Cl
C7H15Cl
C7H13Cl
C7H13Cl
Cl
Br
C7H15Br
I
C7H15Br Br
no rings or double bonds one double bond C H I
7 13
Halogens
one ring
Cl Br I
Note that within each group the number of hydrogen atoms decreases by 2 for each ring or double bond present. This is the same
relationship discovered for hydrocarbon molecular formulas in # 6.
13. After comparing the groups created in #12 with the hydrocarbon groups in #5 above:
a. Determine the effect that adding a halogen atom to a 7 carbon hydrocarbon has on the number of hydrogen atoms present in the molecule?
Explain.
Comparing halogen containing molecules with hydrocarbon molecules from 5 with the same number of carbon atoms and double
bond equivalents, we find that Halogen (univalent) atom containing molecules have 1 less hydrogen atom than the corresponding
hydrocarbon.
Effect of Cl, Br or I = decrease the number of H's by 1 for each halogen atom added. As noted in class, this relationship is most easily
seen by examining the saturated halogen containing compounds (circled above).
Note that the halogen atom simply replaces an H, decreasing to total number of H’s by 1.
H
C
H
C
H
H
Br
H
6
Double Bond Equivalents
b. Determine the effect that adding a nitrogen atom to a 7 carbon hydrocarbon has on the number of hydrogen atoms present in the molecule?
Explain.
Doing a comparison similar to that in part a. for nitrogen containing molecules, we find that nitrogen (trivalent atoms) containing
molecules have 1 more hydrogen atom than the corresponding hydrocarbon.
Effect of Nitrogen (trivalent) atoms = increase the number of H's by 1 for each N added.
Again, this relationship is most easily seen by examining the saturated halogen containing compounds (circled above). Visualizing the
nitrogen atom effect reveals that replacing an H with an N results in retaining the H (circled below) and adding both N & H (See the
oval below) to the molecular formula increasing the total number of H’s by 1
H
C
H
N
C
H
H
H
H
H
c. Determine the effect that adding an oxygen atom to a 7 carbon hydrocarbon has on the number of hydrogen atoms present in the molecule?
Explain.
Doing a comparison similar to that in part a. for oxygen containing molecules, we find that nitrogen (trivalent atoms) containing
molecules have 1 more hydrogen atom than the corresponding hydrocarbon.
Effect of Oxygen (divalent) atoms = no change in the number of H's in the molecular formula.
Again, this relationship is most easily seen by examining the saturated halogen containing compounds (circled above). Visualizing the
oxygen atom effect reveals that replacing an H with an O results in retaining the H (circled below) and adding only an O to the
molecular formula leaving the total number of H’s unchanged.
H
C
H
C
H
H
O
H
H
d. How did the grouping of structures required in 12 assist in your analyses in a -> c?
As before, grouping the structures helps to isolate variables. Each group held the new atom constant, illustrating that double bond
equivalents worked for each group. This leads to comparison of structures in 12 with appropriate hydrocarbon structures (same # of
DBE’s) resulting in a correct determination of the effect of each atom on the molecular formula.
14. Devise a general approach for determining the number of rings or multiple bonds in any molecule that contains one or more N, O or halogen atoms
from its molecular formula, without drawing any structures.
As proposed by the Alkene group in class, using the conclusions in 13., comparing the number of hydrogen atoms is a saturated hydrocarbon
with the same number of carbon atoms (CNH2N+2)) with those in the molecule gives a difference the must be corrected by adding 1 for each N
atom and subtracting 1 for each halogen atom. The corrected difference is then divided by 2 to get the # of DBE.
In our further discussion we explored other possible ways of solving the problem. Rather than correcting the difference as the alkenes
suggested, we could we could determine the number of H's in a saturated molecule containing halogens, oxygen and nitrogen atoms by
correcting the 2n + 2 for the presence of each halogen (-1 H) or each nitrogen (+1 H). This corrected number of H’s can then be compared to
the number of H's in the compound being evaluated (now both have the effects of the heteroatoms) as described in #8, to determine the
number of rings and/or multiple bonds.
Double Bond Equivalents
7
Alternatively we could modify the molecular formula of the compound being evaluated to remove the effect of each Halogen (+1H) or each
nitrogen (-1H) converting it to a corresponding hydrocarbon with the same number of DBE’s. The corrected value for the cpd can then be
compared to the number of H’s in the corresponding saturated hydrocarbon.
All three methods are equally effective in reaching a correct conclusion, so you should follow the one devised by your group as long as it
provided reproducibly correct determinations for you.
15. Below are listed molecular formulas for four new molecules.
C5H9N
C7H14O
C15H29Cl
C9H16NOBr
Without drawing any structures, what can you predict about the number of rings or multiple bonds that must be present in molecules with each of these
molecular formulas? Explain your logic. Using the modification of the corresponding saturated HC method.
C5H9N
C7H14O
C15H29Cl
C9H15OBr
Saturated C5
compound with 1 N
has (2N+2+1) 13 H's.
This formula has 4
fewer H's
DBE = 2
Possible 1 triple bond
or 1 ring and 1
double bond or 2
rings or 2 double
bonds
Saturated C7
compound with 1 O
has (2N+2 O has no
effect) 16 H's.
This formula has 2
fewer H's
DBE = 1
Possible 1 ring or 1
double bond
Saturated C15
compound with 1 Cl
has (2N+2-1) 31 H's.
This formula has 2
fewer H's
DBE = 1
Possible 1 ring or 1
double bond
Saturated C9
compound with 1 O
and 1 Br has
(2N+2-1) 19 H's.
This formula has 4
fewer H's
DBE = 2
Possible 1 triple bond
or 1 ring and 1
double bond or 2
rings or 2 double
bonds.
8
Double Bond Equivalents
16. Use the method developed in 14 to write the molecular formulas of the following structures without counting all of the atoms. Which atoms must be
counted? Which ones can be determined from your relationship from 12 above? Explain the logic you used.
NH
O
C
C
O
This structure has 15 C's and 2 O's.
A saturated compound with 15 C's and 2
O's has 32 H's (O has no effect)
This structure has 4 rings and 4 double
bonds.
DBE = 8
So it should have 16 fewer H's than the
saturated molecule.
Cl
This structure has 11 C's, 1 N and 1 Cl.
A saturated compound with 11 C’s, 1 N and
1 Cl has (2N+2+1-1) 24 H's
This structure has 1 triple bond, 1 ring and
2 double bonds.
DBE = 5
So it should have 10 fewer H's than the
saturated molecule.
Molecular formula: C11H14NCl
Molecular formula: C15H16O2
17. Reflector’s Report Discussion:
Identify the most important concepts you learned from this activity:
 The relationship between the number of hydrogen atoms in an molecular formula and the number of rings, double bonds and triple bonds.
 How to use the above relationship to determine the number of rings and multiple bond possible from a molecular formula and to
determine the number hydrogen atom in a molecule from its bond-line structure.
 How to organize symbolic data to determine patterns and relationships.
18. Strategy Analyst’s Report Discussion:
Look back at the process used to reach your method in 14. There were three organizing steps in each of the analyses. (Steps 2, 5 & 12) List them and
comment on the value of each operation as a general problem solving strategy.
These operations accomplished: Note that # 12 did both types of organization
 Steps 2 & 12: Converting all structures to a common unit (in this case molecular formulas).
 Steps 5 & 12: Grouping data into categories with a common characteristic (molecular formula in 5 and heteroatom present in 12).
Together these two types of reorganization steps allowed you to see trends in the data.
So in any problem solving situation, when trends in data are not readily apparent, it is often useful to group data according to various
variables to help reveal patterns in the data that might lead to a hypothesis.