Optical isomerism

Chirality and the stereocentre. A carbon bonded to four different groups is a chiral centre (stereocentre, asymmetric carbon). A molecule built around such a carbon, with no internal symmetry to cancel the effect, is chiral: it is not superimposable on its own mirror image. The two non-superimposable mirror images are enantiomers — same connectivity, same bond lengths, same melting point and boiling point, differing only in the three-dimensional arrangement about the stereocentre (and hence in how they interact with other chiral things, including polarised light).

Reading the polarimeter. A polarimeter passes plane-polarised light through a sample tube and measures the angle by which the plane of polarisation has turned.

• Dextrorotatory (+ or d): rotates the plane clockwise (as seen facing the oncoming light).
• Laevorotatory (− or l): rotates the plane anticlockwise.

Two enantiomers of the same compound rotate light by exactly the same magnitude but opposite sign. The rotation observed on an instrument, $\alpha$, depends on how much material the light passes through and how concentrated it is, so chemists report a normalised quantity, the specific rotation:

where: $[\alpha]_\lambda^T$ is the specific rotation (degrees) at temperature $T$ and wavelength $\lambda$ (almost always the sodium D-line), $\alpha$ is the angle actually read off the polarimeter (degrees), $l$ is the path length of the sample tube in decimetres, and $c$ is the concentration in grams of solute per millilitre of solution. Because $[\alpha]$ is a fixed physical constant of a pure enantiomer (like a melting point), it lets you compare rotations measured on different tubes or concentrations on a common footing.

Racemic mixtures. A 50:50 mixture of the two enantiomers shows zero net rotation, because equal-and-opposite contributions cancel exactly. Such a mixture is a racemic mixture (racemate), written (±). Crucially, the individual molecules are still chiral — it is only the bulk mixture, not any single molecule, that is optically inactive.

R/S configuration. Each stereocentre also carries a structural label, R (rectus) or S (sinister), fixed by ranking the four attached groups using the CIP (Cahn–Ingold–Prelog) rules — highest atomic number at the first point of difference gets top priority — and reading the sense of decreasing priority with the lowest-priority group pointing away. R/S is a configurational label, not a rotation direction: R does not automatically mean dextrorotatory, and the actual sign of rotation must always be measured, never inferred from the letter.

$S_N2$ substitution on a chiral haloalkane goes through a single concerted transition state in which the incoming nucleophile attacks from the face directly opposite the leaving halide, pushing the other three groups through like an umbrella flipping in the wind. One enantiomer of starting material gives one enantiomer of product, but with the spatial arrangement inverted (Walden inversion) — the optical rotation of the product has the opposite sign relative to what simple retention would predict, and is fully correlated with the reactant's rotation.

$S_N1$ substitution first ionises the C–X bond completely, forming a planar, achiral carbocation at the stereocentre. Because the empty p-orbital is exposed equally on both faces, the nucleophile can attack from either side with roughly equal probability, so a single pure enantiomer of starting halide gives a nearly 1:1 mixture of both product enantiomers — i.e. a largely racemic product with observed rotation close to zero. Measuring how far the product's specific rotation falls below the fully-inverted (or fully-retained) value is direct, quantitative evidence for how much $S_N1$ character (versus clean $S_N2$) the reaction had.

Optically pure (R)-2-bromobutane has $[\alpha] = +23.1°$. A polarimeter tube of length $l = 1.00\ \text{dm}$ is filled with this pure compound at concentration $c = 0.850\ \text{g/mL}$. (a) Predict the observed rotation $\alpha$. (b) The same haloalkane is then hydrolysed under partial-$S_N1$ conditions; the unreacted 2-bromobutane recovered afterwards is loaded into an identical tube at the same concentration and gives an observed rotation of $+9.82°$. Find the optical purity and the enantiomer composition of the recovered sample.

Step 1. Rearrange equation (1) to find the observed rotation for the pure sample: $\alpha = [\alpha]\, l\, c = 23.1 \times 1.00 \times 0.850 = 19.635° \approx +19.6°$. This is what the polarimeter should read for the enantiomerically pure starting compound under these tube conditions.

Step 2. The recovered sample, measured under the identical $l$ and $c$, gives only $+9.82°$ instead of $+19.6°$ — some of the material has racemised because the substitution/ionisation partially proceeded through the planar carbocation pathway, converting some (R) into (S) even in the unreacted starting halide sampled after partial reaction.

Step 3. Optical purity (equivalently the enantiomeric excess, ee) is the fraction of the maximum possible rotation actually observed: $\text{optical purity} = \dfrac{\alpha_{\text{observed}}}{\alpha_{\text{pure}}} \times 100\% = \dfrac{9.82}{19.635}\times 100\% \approx 50.0\%$.

Step 4. Convert ee to enantiomer composition. If $R$ and $S$ are the mole percentages of each enantiomer, $R + S = 100$ and $R - S = \text{ee} = 50$. Solving simultaneously, $R = 75\%$ and $S = 25\%$. The recovered 2-bromobutane is 50.0% optically pure, corresponding to a 75:25 mixture of (R):(S) — i.e. the sample has partially, but not completely, racemised, consistent with a reaction that is a mix of $S_N2$ and $S_N1$ pathways rather than pure $S_N1$.

Competitive stretch

If instead the recovered sample had shown exactly zero rotation, optical purity would be $0\%$ ($R = S = 50\%$), the signature of complete racemisation and therefore a cleanly $S_N1$ pathway — the same reasoning used qualitatively in the Pass-1 discussion of 2-bromobutane hydrolysis, now made quantitative.

• Confusing R/S (a structural CIP label) with (+)/(−) (an experimentally measured rotation direction) — the two are independent; you cannot read the sign of rotation off the letter R or S.
• Assuming a racemic mixture means the molecules themselves are achiral — the molecules are still chiral; only the 50:50 bulk mixture is optically silent because the two rotations cancel.
• Forgetting that specific rotation requires both a path length correction and a concentration correction — plugging the raw polarimeter reading $\alpha$ into a comparison without dividing by $l\,c$ gives numbers that are not comparable between experiments.
• Treating every $S_N1$ reaction as giving exactly 0% optical purity — full racemisation is the idealised limiting case; real reactions often show partial retention of rotation because the leaving group can partially shield one face of the carbocation.

Ask the tutor: want to practice assigning R/S with CIP priority rules, or work through another specific-rotation/optical-purity calculation step by step?