Fall 2011
Chemistry 211
Double Bond Equivlents.
(CGWW pp.74-76)
Out of Class Applications:
ANALYSIS
A. Reading: CGWW pp. 74-76
B. Activities:
1. Use the method developed in step 14 of The Double Bond Equivalents Activity to predict,
without drawing structures, how many rings or multiple bonds might be present in
molecules with the following molecular formulas. Show your calculations.
C6H10
C8H18O
C12H22
Saturated C6 hydrocarbon
has 14 H's
Saturated C8 compound
Saturated C12 hydrocarbon
with 1 O has 18 H's O has no has 26 H's
effect.
This formula has 4 fewer H's
This formula has 4 fewer H's
This is a saturated
DE = #H diff/2 =2
compound
DE = #H diff/2 = 2
Possible 1 triple bond or 1
Possible 1 triple bond or 1 DE = #H diff/2 = 0
ring and 1 double bond or 2
ring and 1 double bond or 2
rings or 2 double bonds
No rings or double bonds rings or 2 double bonds
possible
C11H23NO2
C15H24
C20H35ClO
Saturated C11 compound
with 1 N (+1) and 2 O's (no
effect) has 25 H's.
This formula has 2 fewer H's
DE = #H diff/2 = 1
Saturated C15 hydrocarbon
has 32 H's
This formula has 8 fewer H's
DE = #H diff/2 = 4
Many possibilities: 2 triple
Possible 1 ring or 1 double bonds, or 1 triple bond and
bond
any combination of a total of
2 rings and/or double bonds,
etc.
Saturated C20 compound
with 1 Cl (-1) and 1 O (no
effect) has 41 H's.
This formula has 6 fewer H's
DE = #H diff/2 = 6
Many possibilities: 1 triple
bond and 1 ring or 1 double
bond, or any combination of a
total of 3 rings and/or double
bonds.
2
Double Bond Equivalents
2. Use the relationship developed in Step 12 of Class Group Activity #3 to write the molecular
formulas of the following structures without counting all of the atoms.
O
N
This structure has 16 C's.
A saturated HC with 16 C's has 34 H's
This structure has 3 double bonds
N
H
O
This structure has 16 C's, 2 N's and 2 O's.
A saturated compound with 16 C's, 2 N's
(+2) and 2 O's (no effect) has 36 H's
DE = 3
This structure has 5 rings and 4 double
bonds
So it has 6 fewer H's than the saturated
molecule.
DE = 9
molecular formula: C16H28
So it has 18 fewer H's than the saturated
molecule.
Molecular formula: C16H18N2O2
H
1
2 NH
3
4
This structure has 16 C's.
H
This structure has 17 C's and 1 N.
A saturated compound with 16 C's has 34
H's
A saturated compound with 17 C's and 1 N
(+1) has 37 H's
This structure has 2 rings and 1 double
bond
This structure has 4 rings (see numbers
above) and 1 double bond
DE = 3
DE = 5
So it has 6 fewer H's than the saturated
molecule.
So it should have 10 fewer H's than the
saturated molecule.
Molecular formula: C16H28
Molecular formula: C17H27N
C. Applicable Problems from CGWW: p. 76 Working with DBE’s for an unknown.