SME
Organic Chemistry I · Problem type
Specific rotation and enantiomeric excess
Use this for “A solution of butan-2-ol (0.100 g/mL) rotates light +0.945° in a 1.00 dm tube. Pure (S)-butan-2-ol has [α] = +13.5°. Find the ee.”
Diagram
Butan-2-ol, C₄H₁₀O: each corner and line end is a C, with its H left out.
Ranks 1–4 by atomic number at the first point of difference; a wedge points toward you, a dash away.
With rank 4 pointing away, 1 → 2 → 3 turns counterclockwise: S.
A mirror image has every stereocenter flipped: (S) 85% and (R) 15%.
Observed rotation α = 0.945°
Path length l = 1 dm
Concentration c = 0.1 g/mL
Specific rotation [α] = 9.45°
Pure (+) enantiomer’s specific rotation [α]₀ = 13.5°
Enantiomeric excess ee = 70%
This is an example. Type your own values to start a new problem.
α
Observed rotation
example · °
l
Path length
example · dm
c
Concentration
example · g/mL
[α]
Specific rotation
calculated · °
[α]₀
Pure (+) enantiomer’s specific rotation
example · °
ee
Enantiomeric excess
calculated · %
x₊
Share of the (+) enantiomer
calculated · %
Formulas
[
α
]
=
α
l
×
c
9.45
=
0.945
1 × 0.1
ee
=
100
×
[
α
]
÷
[
α
]₀
70
=
100
×
9.45
13.5
x₊
=
100 + ee
2
85
=
100 + 70
2
Assumptions
•
The specific rotation [α] is the rotation for 1 dm of path through 1 g/mL, so [α] = α ÷ (l × c). Pure liquids use their density for c.
•
Mirror-image enantiomers turn light equally far in opposite directions, so a racemic mixture, half of each, reads 0.
•
The (+) or (−) sign says nothing about R or S: those come from the CIP ranks. Here (S)-butan-2-ol is the (+) one, [α]₀ = +13.5°.
•
The excess is all one enantiomer and the rest is half each, so the (+) share is (100 + ee) ÷ 2.
Step-by-step
WE KNOW
α = 0.945°, l = 1 dm, c = 0.1 g/mL, [α]₀ = 13.5°
FIND
specific rotation ([α]), enantiomeric excess (ee), share of the (+) enantiomer (x₊)
Step 1 · Find specific rotation ([α])
Use [α] = α ÷ (l × c)
Divide the observed rotation by the path length times the concentration.
[
α
]
=
0.945
1 × 0.1
[
α
]
=
9.45°
Step 2 · Find enantiomeric excess (ee)
Use ee = 100 × [α] ÷ [α]₀
The sample turns light this fraction as far as the pure enantiomer does.
ee
=
100
×
9.45
13.5
ee
=
945
13.5
ee = 70%
Step 3 · Find share of the (+) enantiomer (x₊)
Use x₊ = (100 + ee) ÷ 2
The excess is all the (+) enantiomer; the rest is half (+) and half (−).
x₊
=
100 + 70
2
x₊
=
170
2
x₊ = 85%
Check
Put the numbers back in. Both sides should match.
9.45
=
0.945
1 × 0.1
✓
70
=
100
×
9.45
13.5
✓
85
=
100 + 70
2
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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