SME
Organic Chemistry II · Problem type
Anomers at equilibrium from rotations
Use this for “Pure α-D-galactose reads +150.7° and pure β reads +52.8°. At equilibrium it reads +80.2°. What percent is α?”
Chart
Rotation of the pure α anomer rα = +112.2°
Rotation of the pure β anomer rβ = +18.7°
Rotation at equilibrium req = +52.7°
This is an example. Type your own values to start a new problem.
req (°)
=
xα (%)
×
rα (°)
+
(1
−
xα (%)
)
×
rβ (°)
xβ
Share that is β
calculated · %
Formulas
req
=
(
xα
100
)
×
rα
+
(1
−
xα
100
)
×
rβ
+52.7
=
(
36.4
100
)
×
112.2
+
(1
−
36.4
100
)
×
18.7
xβ = 100 − xα
63.6 = 100 − 36.4
Assumptions
•
In water each anomer opens and recloses, so either pure anomer drifts to the same equilibrium mixture: mutarotation.
•
Only the two pyranose anomers count; the open chain and the furanoses are under 1% and left out.
•
Each anomer adds its rotation in proportion to its share, so req = xα rα + (1 − xα)rβ, with xα as a fraction and each r the specific rotation [α].
Step-by-step
WE KNOW
rα = +112.2°, rβ = +18.7°, req = +52.7°
FIND
share that is α (xα), share that is β (xβ)
Step 1 · Find share that is α (xα)
Use req = (xα ÷ 100) × rα + (1 − xα ÷ 100) × rβ
How far the equilibrium reading sits from the β reading toward the α reading is the share of α.
xα
=
100
×
req − rβ
rα − rβ
xα
=
100
×
52.7 − 18.7
112.2 − 18.7
xα
=
100
×
34
93.5
xα
=
3,400
93.5
xα = 36.4%
Step 2 · Find share that is β (xβ)
Use xβ = 100 − xα
Every molecule is one anomer or the other.
xβ = 100 − 36.4
xβ = 63.6%
Check
Put the numbers back in. Both sides should match.
+52.7
=
(
36.4
100
)
×
112.2
+
(1
−
36.4
100
)
×
18.7
✓
63.6 = 100 − 36.4 ✓
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