Optical activity — specific rotation, racemic mixtures, resolution: Exercises

Two formulas power almost everything below. We restate them so no symbol is used before it is defined.

Look at the picture below before starting — it is the mental model behind every ee calculation.

Figure — Optical activity — specific rotation, racemic mixtures, resolution

Level 1 — Recognition

Exercise 1.1

Which of these can rotate plane-polarized light: (a) a single pure enantiomer, (b) a racemic mixture, (c) a meso compound, (d) an achiral molecule?

Recall Solution

Optical activity needs a chiral molecule that is not internally cancelled and not externally cancelled.

(a) Single pure enantiomer → YES. One "hand" only, nothing cancels it.
(b) Racemate → NO. Equal (+) and (−) cancel by external compensation.
(c) Meso → NO. It has an internal mirror plane (internal compensation); the molecule is achiral overall — see Meso compounds and internal compensation.
(d) Achiral → NO. No handedness to begin with. Answer: only (a).

Exercise 1.2

A polarimeter reads $α = - 8^{\circ}$ . State the sign name and the direction the plane was twisted (observer faces the oncoming light).

Recall Solution

A negative sign means laevorotatory (−) or (l): the plane rotates anticlockwise (to the left), as seen by an observer facing the incoming beam. So the light was twisted 8° to the left.

Level 2 — Application

Exercise 2.1

$0.8$ g of a sugar is dissolved to make $20$ mL of solution and read in a $10$ cm tube, giving $α = + 2.1 2^{\circ}$ . Find $[α]_{D}$ .

Recall Solution

Step 1 — concentration (WHY: formula demands g/mL): $c = \frac{0.8 g}{20 mL} = 0.04$ g/mL. Step 2 — path length (WHY: formula demands dm): $l = 10 cm = 1.0$ dm. Step 3 — apply formula: $[α]_{D} = \frac{+ 2.12}{0.04 \times 1.0} = + 5 3^{\circ}$ Answer: $+ 5 3^{\circ}$ .

Exercise 2.2

A pure liquid has density $ρ = 0.90$ g/mL and $[α]_{D} = + 2 0^{\circ}$ . It fills a $2.0$ dm tube (neat liquid, no solvent). What observed rotation $α$ appears?

Recall Solution

For a neat liquid, the "concentration" of the substance is just its density: $c = ρ = 0.90$ g/mL. Rearrange $α = [α] \cdot c \cdot l$ (WHY: we know the intrinsic property and the cell, so multiply back): $α = (+ 20) (0.90) (2.0) = + 3 6^{\circ}$ Answer: $+ 3 6^{\circ}$ .

Level 3 — Analysis

Exercise 3.1

A sample reads $[α]_{o b s} = + 4 5^{\circ}$ ; the pure (+) enantiomer has $[α]_{ma x} = + 6 0^{\circ}$ . Find the ee and the percentage of each enantiomer.

Recall Solution

Step 1 — ee (WHY: observed is the net twist; fraction of max = excess): $ee = \frac{+ 45}{+ 60} = 0.75 = 75%$ Step 2 — composition: let $x$ = fraction of (+). Then $2 x - 1 = 0.75 \Rightarrow x = 0.875$ . So 87.5% (+) and 12.5% (−). Check: the 12.5% (−) cancels 12.5% of the (+), leaving $87.5 - 12.5 = 75%$ pure (+) — exactly the ee. See the bar figure at the top of the page.

Exercise 3.2

Equal moles of the (+) and (−) forms are mixed, then $0.20$ mol extra of the (+) form is added, making $1.20$ mol total. If pure (+) gives $[α]_{ma x} = + 10 0^{\circ}$ , what is $[α]_{o b s}$ of the mixture?

Recall Solution

Step 1 — find fractions. Total $= 1.20$ mol. The (+) amount is $0.50 + 0.20 = 0.70$ mol... let's be careful: "equal moles" then "+0.20 extra (+)". Take equal parts $= 0.50$ mol each first $\Rightarrow 1.00$ mol, then add $0.20$ mol (+) $\Rightarrow$ (+) $= 0.70$ , (−) $= 0.50$ , total $= 1.20$ . $x = \frac{0.70}{1.20} = 0.5833 \dots$ Step 2 — ee: $ee = 2 x - 1 = 2 (0.5833) - 1 = 0.1667 = 16.67%$ . Step 3 — observed rotation: $[α]_{o b s} = ee \times [α]_{ma x} = 0.1667 \times 100 = + 16.6 7^{\circ}$ . Answer: $\approx + 16. 7^{\circ}$ . Shortcut check: only the excess $0.20$ mol out of $1.20$ is "unpaired," and $0.20/1.20 = 0.1667$ — same ee, as expected.

Level 4 — Synthesis

Exercise 4.1

A racemic acid is treated with one pure enantiomer of a chiral base to resolve it. (a) What kind of species forms, and why can they be separated? (b) After separating, how do you recover the free acids?

Recall Solution

(a) The (+)-acid and (−)-acid each bond to the same (−)-base, giving $(+) -acid + (-) -base \to (+, -) -salt, (-) -acid + (-) -base \to (-, -) -salt .$ These two salts are diastereomers — not mirror images of each other, so they have different melting points and solubilities. That difference lets you separate them by fractional crystallisation. (Enantiomers alone could never be separated this way — identical physical properties. See Enantiomers vs Diastereomers.) (b) Add a strong mineral acid (e.g. HCl) to each separated salt. This protonates and displaces the base, regenerating the free enantiopure acid, which you then isolate.

Exercise 4.2

A student measures $[α]_{o b s} = + 3 0^{\circ}$ for a sample and knows $[α]_{ma x} = + 15 0^{\circ}$ . They then wonder: what path length $l$ (dm) would make a $c = 0.10$ g/mL solution of this same sample read $α = + 3. 0^{\circ}$ on the polarimeter?

Recall Solution

Step 1 — the sample's own specific rotation is fixed at $[α]_{o b s} = + 3 0^{\circ}$ (its ee is baked in). We use this value in the workhorse formula, not $[α]_{ma x}$ . Step 2 — solve $α = [α]_{o b s} \cdot c \cdot l$ for $l$ : $l = \frac{α}{[ α ] _{o b s} \cdot c} = \frac{3.0}{30 \times 0.10} = 1.0 dm .$ Answer: $l = 1.0$ dm ( $= 10$ cm). (The value $[α]_{ma x} = 15 0^{\circ}$ is a distractor here — it only matters if you're asked for ee, which would be $30/150 = 20%$ .)

Level 5 — Mastery

Exercise 5.1

A compound has three stereocentres... (conceptual, no arithmetic). Given the tartaric-acid case: the (2R,3R) and (2S,3S) forms are enantiomers with $[α]_{ma x} = \pm 1 2^{\circ}$ , and (2R,3S) is meso. A bottle labelled "optically inactive tartaric acid" is opened. List every possibility for its contents and how you would distinguish them.

Recall Solution

Three ways to be inactive:

Racemate — a 1:1 mix of (2R,3R) and (2S,3S). External compensation. Resolvable into two active halves.
Meso (2R,3S) — a single achiral molecule. Internal compensation. Not resolvable; it's already one substance with no handedness.
(Not possible here, but in general) an achiral compound with no stereocentres at all. How to distinguish: attempt resolution (react with a single chiral base to form diastereomeric salts). Only the racemate separates into optically active fractions; the meso form yields a single kind of salt and stays inactive. See Meso compounds and internal compensation.

Exercise 5.2

A $50$ mL solution contains $2.0$ g of pure (+) enantiomer and $3.0$ g of pure (−) enantiomer of the same compound ( $[α]_{ma x} = + 8 0^{\circ}$ ). It is read in a $2.0$ dm tube. Find (a) the ee, (b) $[α]_{o b s}$ , and (c) the observed angle $α$ .

Recall Solution

Step 1 — net excess. The optically active part is only the excess of one enantiomer over the other: $3.0 - 2.0 = 1.0$ g of net (−). Total mass $= 5.0$ g. (a) ee $= \frac{∣3.0 - 2.0∣}{5.0} = \frac{1.0}{5.0} = 0.20 = 20%$ , and since (−) is in excess, the sample is laevorotatory. (b) $[α]_{o b s} = ee \times [α]_{ma x}$ , carrying the sign: $= 0.20 \times (- 80) = - 1 6^{\circ}$ . (c) Use $c$ = total concentration (the light meets all molecules): $c = \frac{5.0 g}{50 mL} = 0.10$ g/mL. $α = [α]_{o b s} \cdot c \cdot l = (- 16) (0.10) (2.0) = - 3. 2^{\circ} .$ Answers: (a) 20% ee, (−)-rich; (b) $[α]_{o b s} = - 1 6^{\circ}$ ; (c) $α = - 3. 2^{\circ}$ . Sanity check: the $2.0$ g (+) pairs off with $2.0$ g (−) to form an inactive racemate; only the leftover $1.0$ g (−) rotates — consistent with the 20% ee.

Recall Quick self-test

ee = 60% means what fraction is the major enantiomer? ::: $x = \frac{1 + 0.60}{2} = 0.80$ , i.e. 80% major / 20% minor. Convert $l = 25$ cm to dm. ::: $2.5$ dm. A racemate is resolvable; a meso form is not — true or false? ::: True.