Optical activity — specific rotation, racemic mixtures, resolution: Worked examples

Before we start, one reminder of the three quantities and their locked units (get these wrong and every answer is off by a factor of 10 or 1000):

The scenario matrix

Every problem this topic throws at you falls into one of these cells. The "Example" column tells you where we cover it.

#	Case class	What is unusual about it	Example
A	(+) rotation, find $[α]$	dextrorotatory, standard forward calc	Ex 1
B	(−) rotation, back-solve for $α$	sign must survive the algebra	Ex 2
C	Pure liquid (no solute)	there is no "concentration" — use density $ρ$	Ex 3
D	Zero / racemic reading	$α = 0$ , and what that does NOT prove	Ex 4
E	Partial mixture — optical purity & %	net of two enantiomers, enantiomeric excess	Ex 5
F	Unit-trap (cm, g/L given)	must convert BEFORE dividing	Ex 6
G	Solve for a cell dimension ( $l$ or $c$ )	rearrange the formula the other way	Ex 7
H	Limiting / degenerate ( $c \to 0$ , $l \to 0$ )	what happens at the edges	Ex 8
I	Word problem + exam twist	multi-step, resolution + composition	Ex 9

Case A — dextrorotatory, find the specific rotation

Worked example Example 1 (cell A)

A solution is made by dissolving $1.2$ g of a sugar in enough water to make $15$ mL. In a $25$ cm polarimeter tube it rotates light clockwise by $9. 9^{\circ}$ . Find $[α]_{D}$ .

Forecast: clockwise means the sign is (+). Guess the magnitude — will it be bigger or smaller than $9.9$ ? (The cell is $2.5$ dm long and $c$ is small, so $c l$ is smaller than 1, meaning $[α]$ comes out bigger than $α$ .)

Step 1 — concentration. $c = \frac{1.2 g}{15 mL} = 0.08$ g/mL. Why this step? $[α]$ is defined with $c$ in g/mL, so we form grams-over-millilitres directly.

Step 2 — path length. $l = 25 cm = 2.5$ dm. Why this step? The formula demands dm; divide cm by 10.

Step 3 — sign. Clockwise ⇒ $α = + 9. 9^{\circ}$ . Why this step? Dextrorotatory is defined as positive (facing the oncoming light).

Step 4 — apply. $[α]_{D} = \frac{+ 9.9}{0.08 \times 2.5} = \frac{+ 9.9}{0.20} = + 49. 5^{\circ}$ Why this step? Dividing by $c l$ strips out the sample size (how many molecules the light met), leaving the intrinsic twisting power that belongs to the compound itself.

Verify: units are $\frac{deg}{( g/mL ) ( dm )} = deg\cdotmL\cdotg^{- 1} dm^{- 1}$ ✓. Magnitude $49.5 > 9.9$ , matching our forecast that dividing by $0.20$ inflates the number ✓.

Case B — laevorotatory, back-solve for the observed angle

Case C — a pure liquid: there is no concentration, so use density

Case D — a zero reading, and the trap it sets

Worked example Example 4 (cell D)

Equal masses of $(+)$ -tartaric acid ( $[α]_{D} = + 1 2^{\circ}$ ) and $(-)$ -tartaric acid ( $[α]_{D} = - 1 2^{\circ}$ ) are dissolved together. What does the polarimeter read, and what does that reading prove?

Forecast: guess $α$ . Then guess: does "no rotation" mean the molecules are not chiral?

Step 1 — add the two contributions. With a $1 : 1$ mixture: $[α]_{n e t} = \frac{1}{2} (+ 12) + \frac{1}{2} (- 12) = 0^{\circ}$ Why this step? Each enantiomer twists by equal magnitude, opposite sign; halved amounts of each cancel exactly (external compensation).

Step 2 — so $α = 0$ for any $c$ and $l$ , since $[α]_{n e t} = 0$ . Why this step? Zero specific rotation makes $α = 0 \cdot c l = 0$ regardless of cell.

Step 3 — interpret carefully. A racemate is optically inactive but is built from chiral molecules that cancel. This is not the same as a meso compound (inactive by an internal mirror plane) or a truly achiral molecule. Why this step? "Inactive" has three causes — see Enantiomers vs Diastereomers. Only a racemate can be resolved back into active halves.

Verify: $\frac{1}{2} (12) + \frac{1}{2} (- 12) = 6 - 6 = 0$ ✓. The reading of $0^{\circ}$ therefore does not prove achirality — the mistake the parent note warns about.

Case E — a partial mixture: optical purity and composition

Worked example Example 5 (cell E)

A sample of an amino acid reads $[α]_{o b s} = - 2 1^{\circ}$ . The pure $(-)$ enantiomer has $[α]_{ma x} = - 3 0^{\circ}$ . Find the optical purity (enantiomeric excess) and the % of each enantiomer.

Forecast: the reading is a fraction of the max, so ee is between 0 and 100%. Guess it.

Step 1 — optical purity. $op = \frac{[ α ] _{o b s}}{[ α ] _{ma x}} = \frac{- 21}{- 30} = 0.70 = 70% ee$ Why this step? The observed rotation ( $[α]_{o b s}$ ) is the net; its fraction of the pure-enantiomer maximum ( $[α]_{ma x}$ ) is exactly the enantiomeric excess (the signs cancel, giving a positive fraction).

Step 2 — convert ee to composition. Let $x$ = fraction of the $(-)$ (majority) form. Excess of $(-)$ over $(+)$ equals ee: $x - (1 - x) = 0.70 \Rightarrow 2 x - 1 = 0.70 \Rightarrow x = 0.85$ Why this step? The minority $15%$ $(+)$ cancels an equal $15%$ of the $(-)$ ; the leftover $70%$ pure $(-)$ produces all the observed rotation.

Step 3 — state it. $85% (-)$ and $15% (+)$ . Why this step? We report both fractions explicitly so the composition, not just the ee, is the final answer — that is what the question asked for.

Verify: check the leftover-pure logic: $0.85 - 0.15 = 0.70 =$ ee ✓. And $0.70 \times (- 30) = - 21 = [α]_{o b s}$ ✓.

Case F — the unit trap: wrong units given on purpose

Worked example Example 6 (cell F)

A textbook lists $α = + 4. 0^{\circ}$ for a solution of $c = 50$ g/L in a 10 cm tube. A student writes $[α] = \frac{4.0}{50 \times 10} = 0.008$ . Find the correct $[α]_{D}$ and the student's error factor.

Forecast: the student used raw g/L and cm. Guess how far off they are.

Step 1 — convert concentration. $50$ g/L $= \frac{50 g}{1000 mL} = 0.05$ g/mL. Why this step? The definition is locked to g/mL; g/L is $1000 \times$ too many mL, so divide by 1000.

Step 2 — convert path length. $10$ cm $= 1.0$ dm. Why this step? Formula demands dm; divide cm by 10.

Step 3 — apply correctly. $[α]_{D} = \frac{+ 4.0}{0.05 \times 1.0} = + 8 0^{\circ}$ Why this step? Only after both conversions is the division meaningful.

Step 4 — the error factor. Student got $0.008$ ; correct is $80$ . Ratio $= 80/0.008 = 10, 000 = 1000 \times$ (from g/L) $\times 10 \times$ (from cm). Why this step? Confirms the two unit mistakes multiply — exactly the $10 \times$ and $1000 \times$ warnings from the parent note.

Verify: $\frac{80}{0.008} = 10000$ ✓; and $1000 \times 10 = 10000$ ✓.

Case G — solve for a cell dimension

Case H — limiting and degenerate cases

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

Worked example Example 8 (cells H)

Using $α = [α] c l$ with $[α]_{D} = + 66. 5^{\circ}$ (sucrose), describe the reading as (i) concentration $c \to 0$ , and (ii) tube length $l \to 0$ . Then read the straight-line graph above.

Reading the figure first: the horizontal axis is concentration $c$ (in g/mL), the vertical axis is the observed rotation $α$ (in degrees). The blue line is $α = [α] c l$ plotted at a fixed tube length $l = 1$ dm — a straight line. The yellow dot sits at the origin, marking where both degenerate limits land. The pink dot marks the worked point $c = 0.05$ g/mL. Notice the line is straight and passes through $(0, 0)$ : rotation is directly proportional to concentration, so halving $c$ halves $α$ .

Forecast: $α$ is a product of $c$ and $l$ . If either factor goes to zero, guess the reading.

Step 1 — dilute to nothing, $c \to 0$ . $α = 66.5 \times c \times l c \to 0 0^{\circ}$ Why this step? No molecules in the beam ⇒ nothing to twist the light. This is the yellow dot at the origin on the figure.

Step 2 — empty tube, $l \to 0$ . $α l \to 0 0^{\circ}$ Why this step? Zero path length means the light meets no solution — same zero reading. Both degenerate limits agree, landing at the same origin point.

Step 3 — read the slope. The graph's slope is $[α] l = 66.5$ deg per (g/mL). At the pink dot $c = 0.05$ g/mL the line gives $α = 66.5 \times 0.05 = 3.32 5^{\circ}$ . Why this step? Linearity ( $α \propto c$ ) is exactly the result $α = [α] c l$ from the parent note — a straight line through the origin, where the constant multiplying $c$ is the specific rotation $[α]_{D}$ we have been using all along (the "intrinsic twisting power").

Verify: slope $\times$ point: $66.5 \times 0.05 \times 1 = 3.32 5^{\circ}$ ✓, matching the marked pink dot in the figure. Both limits give $0$ because a product with a zero factor is zero ✓.

Case I — word problem with an exam twist

Worked example Example 9 (cell I)

A racemic acid ( $1.00$ g total) is resolved. From it you recover $0.60$ g of the pure $(+)$ -acid ( $[α]_{D} = + 2 5^{\circ}$ ) and $0.40$ g of the pure $(-)$ -acid. You redissolve only the recovered $(+)$ -acid to make $c = 0.30$ g/mL and read it in a $2.0$ dm tube. (a) What is $α$ for the $(+)$ fraction? (b) If instead you had recombined the original $0.60$ g $(+)$ and $0.40$ g $(-)$ into one solution, what net specific rotation would you measure, and what is its ee?

Forecast: part (a) is a clean forward calc. In part (b), the mixture is not $1 : 1$ , so guess: will it read zero or not?

Step 1 — part (a), forward calc. $α = [α] c l = (+ 25) (0.30) (2.0) = + 1 5^{\circ}$ Why this step? Pure single enantiomer ⇒ just multiply the known intrinsic value by the cell factors.

Step 2 — part (b), fractions of the mixture. Of the $1.00$ g total, the mass fraction of $(+)$ is $0.60$ and of $(-)$ is $0.40$ . Why this step? Net specific rotation is a mass-weighted average of the two enantiomers' specific rotations, so we first need each fraction.

Step 3 — part (b), weighted net. $[α]_{n e t} = 0.60 (+ 25) + 0.40 (- 25) = 15 - 10 = + 5^{\circ}$ Why this step? Each enantiomer contributes its specific rotation in proportion to its share; the unequal $0.60/0.40$ split does NOT cancel to zero (unlike the true $1 : 1$ racemate of Example 4).

Step 4 — part (b), enantiomeric excess. $ee = \frac{[ α ] _{n e t}}{[ α ] _{ma x}} = \frac{5}{25} = 0.20 = 20% (+)$ Why this step? ee is the net rotation as a fraction of the pure value; it should also equal the raw excess $0.60 - 0.40 = 0.20$ .

Step 5 — state the answers. (a) $α = + 1 5^{\circ}$ . (b) net specific rotation $[α]_{n e t} = + 5^{\circ}$ , ee $= 20%$ in favour of $(+)$ . Why this step? We collect both parts explicitly so nothing is left implied.

Verify: part (a) back-divide $\frac{15}{0.30 \times 2.0} = 25$ ✓. Part (b): $0.60 (25) + 0.40 (- 25) = 5$ ✓; ee via composition $0.60 - 0.40 = 0.20$ matches $5/25 = 0.20$ ✓ — the exam twist is that an unequal recovery leaves the sample optically active, unlike the true $1 : 1$ racemate of Example 4.

Recall Quick self-test (cover the right column)

Clockwise twist has which sign? ::: Positive, dextrorotatory (+) For a neat pure liquid, what do you use for $c$ ? ::: Its density $ρ$ in g/mL $[α]$ needs $l$ in what unit? ::: decimetres (dm), 1 dm = 10 cm What do the labels $D$ and $T$ on $[α]_{D}^{T}$ record? ::: $D$ = sodium D-line light (589 nm); $T$ = measurement temperature A zero polarimeter reading proves the molecule is achiral — true or false? ::: False; could be a racemate (external compensation) What does $[α]_{ma x}$ denote? ::: The specific rotation of the pure single enantiomer ee formula? ::: $ee = [α]_{o b s} / [α]_{ma x}$ If ee $= 60%$ , what is the majority %? ::: $80%$ (from $2 x - 1 = 0.6$ )

Connections

Parent topic
Polarimeter — the instrument that gives us $α$
Plane-polarized light and EM waves — why one plane of vibration exists
Chirality and stereocentres — the structural cause behind every nonzero reading
Enantiomers vs Diastereomers — behind Examples 4, 5, 9
Meso compounds and internal compensation — the other kind of "inactive"
R-S nomenclature (CIP rules) — naming the configurations we rotate