SME
Organic Chemistry I · Problem type
Percent equatorial from the A-value
Use this for “A methyl group’s A-value is 7.3 kJ/mol. What percent of methylcyclohexane molecules hold the methyl equatorial at 298.15 K?”
Diagram
A ring flip turns every axial bond equatorial: the CH₃ axial in the first chair is equatorial in the second.
An axial group is crowded by the axial H two carbons away on each side (dotted).
K = 19.01, so 95.002% of the molecules are in the second chair.
This is an example. Type your own values to start a new problem.
A
A-value
example · kJ/mol
T
Temperature
example · K
K
Equatorial ÷ axial
calculated
%eq
Percent equatorial
calculated · %
Formulas
K
=
e
A
× 1,000 ÷ (8.314 ×
T
)
19.01
=
e
7.3 × 1,000 ÷ (8.314 × 298.15)
%eq
=
100
×
K
1 +
K
95.0024
=
100
×
19.01
1 + 19.01
Assumptions
•
A is the axial chair’s extra free energy: its group is crowded by the two axial H on the same side of the ring (1,3-diaxial strain).
•
The ring flips fast, so the two chairs are at equilibrium: K = equatorial ÷ axial = eA ÷ RT, with A in J/mol and R = 8.314 J/(mol·K).
•
A bulkier group has a larger A: about 7.3 kJ/mol for methyl and about 21 kJ/mol for tert-butyl, which is almost never axial.
Step-by-step
WE KNOW
A = 7.3 kJ/mol, T = 298.15 K
FIND
equatorial ÷ axial (K), percent equatorial (%eq)
Step 1 · Find equatorial ÷ axial (K)
Use K = eA × 1,000 ÷ (8.314 × T)
Change A to J/mol (× 1000), divide by RT, and raise e to that power.
K
=
e
7.3 × 1,000 ÷ (8.314 × 298.15)
K
=
e
7.3 × 1,000 ÷ 2,478.8191
K
=
e
7,300 ÷ 2,478.8191
K
=
e
2.945
K
=
19.01
Step 2 · Find percent equatorial (%eq)
Use %eq = 100 × K ÷ (1 + K)
For
every
1
axial
molecule
there
are
K
equatorial
ones:
K
out
of
every
1
+
K
.
%eq
=
100
×
19.01
1 + 19.01
%eq
=
100
×
19.01
20.01
%eq
=
1,901
20.01
%eq = 95.0024%
Check
Put the numbers back in. Both sides should match.
19.01
=
e
7.3 × 1,000 ÷ (8.314 × 298.15)
✓
95.0024
=
100
×
19.01
1 + 19.01
✓
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