# Use of Bond Equations in Solving Problems

```Use of Bond Equations in
Solving Problems
Ray A. Gross Jr
Prince George’s Community College
1
How many bonds are
present in benzene, C6H6?
2
Consider Water, H2O
1
2
1
H O H
normal
covalence
B = 1/2(1 + 2 + 1) = 2
B = 1/2NVV
3
For C6H6
• B = &frac12;[(6 x 4) + (6 x 1)]
• B = &frac12;(24 + 6)
• B = 15
H
H
H
H
H
H
4
Alternative Solution
Hydrocarbons
• Total Bonds = Bonds (corresponding
saturated, acyclic compound) minus
number of p bonds and rings in the
structure.
5
C6H14
B = 19
C6H6
B = 15
Lose one bond for every p bond or ring.
Let B = bond deficiency, BD.
BD = p + r
B = Bsat - BD
6
BD for Hydrocarbons
Alkane: CnH2n + 2
Hydrocarbon: CnH2n + 2 - 2p - 2r
h = 2n + 2 - 2p - 2r
p + r = n +1 - h/2
BD = n + 1 - h/2
7
B = Bsat - BD
B = 3n + 1 - [n + 1 - h/2]
B = 2n + h/2
C6H6
B = 2(6) + 6/2 = 15
8
BD for All Compounds
BD = 1 – &frac12;(X – N – 2C – 3P – 4S)
X = number of monovalent atoms
N = number of trivalent atoms
C = number of tetravalent atoms
P = number of pentavalent atoms
S = number of hexavalent atoms
9
Taxol, C47H51NO14 , has 7 rings.
How many s bonds does it have?
p + r = 1 – &frac12;(X – N – 2C – 3P – 4S)
p + r = 1 – &frac12;[(51 – 1 – 2(47)] = 23
p = 23 – 7 = 16
B = &frac12;[47(4) + 51 + 3 + 14(2)] = 135
s = 135 – 16 = 119
10
How many bonds are in HPO4-2?
B = &frac12;[1 + 5 + 4(2) – 2] = 6
O
H O P OO11
What’s all of this good for?
• Increases students knowledge of
structure and bonding.
• Possibly use B to classify
reactions.
12
Derive an Eqn for Btot in an
Alkyne.
B = Bsat – BD
B = 3n + 1 – 2
B = 3n - 1
13
Dehydration of Cyclohexanol
O H
H
B = -2
+
B = 2
C6H12O
C6H10
B = 19
B = 17
H O
H
14
```