4.1.7 · D5General Organic Chemistry (GOC)
Question bank — Optical activity — specific rotation, racemic mixtures, resolution
True or false — justify
True or false: A molecule that rotates plane-polarized light must be chiral.
True — rotation requires the molecule to be non-superimposable on its mirror image; an achiral molecule interacts identically with both circular components of the light and produces zero net twist.
True or false: If a solution shows zero rotation, the molecules in it must be achiral.
False — zero rotation can come from a chiral molecule too, if it is a racemate (external compensation) or if it is meso (internal compensation); "inactive" is not the same as "achiral". See Meso compounds and internal compensation.
True or false: Enantiomers have identical specific rotation.
False — they have equal magnitude but opposite sign of ; e.g. vs . Everything else physical (mp, bp) is identical in an achiral environment.
True or false: A racemate can be separated by very careful fractional crystallisation of the racemate itself.
False — the two enantiomers have identical solubility in an achiral solvent, so they co-crystallise; you must first convert them to diastereomers, which differ in solubility. See Enantiomers vs Diastereomers.
True or false: The observed angle is a fixed property of a compound.
False — scales with concentration and path length (); only , obtained by dividing those out, is an intrinsic property.
True or false: Doubling the tube length doubles the observed rotation.
True — since and are held fixed, is directly proportional to ; twice the path means twice the molecules the light meets.
True or false: A dextrorotatory compound must have the R configuration.
False — the sign of rotation () is an experimental property and has no fixed link to the R/S label, which comes from R-S nomenclature (CIP rules). Either sign can pair with either descriptor.
True or false: A meso compound can be resolved into two optically active halves.
False — a meso compound is a single achiral molecule (internal mirror plane), not a mixture; there is nothing to separate. Only a racemate is resolvable.
True or false: A 1:1 mixture of two diastereomers is optically inactive.
False — diastereomers are not mirror images and have unrelated rotations, so their contributions do not cancel; the mixture is generally active. Only equal-and-opposite enantiomers cancel.
Spot the error
" with cm and g/L." — what's wrong?
The units are wrong for the definition: must be in dm ( dm) and in g/mL ( g/mL). Using cm and g/L throws the answer off by a factor of .
"Since the racemate has , its molecules are not chiral." — spot the flaw.
The molecules are chiral; the zero is external compensation — every twist from one enantiomer is cancelled by a from its mirror partner. Chirality is a property of the molecule, inactivity a property of the mixture.
"Optical purity = , so 50% purity means 50% of one enantiomer." — find the mistake.
50% optical purity is the enantiomeric excess, not the fraction of one enantiomer. Solving gives , i.e. 75% of the major enantiomer and 25% of the minor.
"We resolved the racemic acid by adding a racemic base." — what breaks?
A racemic (chiral-but-mixed) helper produces mixtures of salts that are still mirror-image pairs, so they remain inseparable. The helper must be a single pure enantiomer to create genuine diastereomeric salts.
"The two salts formed in resolution are enantiomers, so we can crystallise them apart." — error?
They are diastereomers, not enantiomers — they share the same fixed helper configuration but differ at the resolved centre, so their physical properties differ and that is why crystallisation works.
" has no temperature or wavelength dependence, so we can drop the and labels." — why is this careless?
Rotation genuinely depends on both wavelength and temperature, which is exactly why carries the sodium D-line ( nm) and temperature as fixed reference conditions; dropping them makes two measurements incomparable.
Why questions
Why do we divide by both and rather than just one of them?
Both independently set "how many molecules the light meets" — more path or more concentration each add twist (). Dividing out both isolates the per-molecule twisting power .
Why can't ordinary distillation separate two enantiomers?
In an achiral environment enantiomers have identical boiling points, solubilities and densities; distillation exploits differences that simply don't exist here. A chiral influence must be introduced first.
Why does adding a single chiral base make separation possible?
It converts the two enantiomeric acids into two diastereomeric salts with genuinely different solubilities and melting points, so fractional crystallisation can now discriminate between them.
Why is the sign of rotation reported experimentally rather than predicted from structure?
The magnitude and sign of rotation arise from subtle electronic interactions with light and cannot be read off the R/S descriptor; they must be measured in a Polarimeter.
Why does an enzyme succeed at resolution where a solvent fails?
The enzyme is itself chiral, so it presents a handed environment in which the two enantiomers react at different rates — one is consumed while the other survives, effectively separating them.
Why does plane-polarized light, and not ordinary light, reveal chirality?
Ordinary light wiggles in all planes at once, so any net twist averages away; polarizing it to one plane gives a definite direction whose rotation can be seen and measured. See Plane-polarized light and EM waves.
Edge cases
What is for a perfectly racemic sample?
Exactly — the equal and opposite contributions cancel, giving , regardless of concentration or path length.
What rotation does a meso compound show, and why is it different from a racemate's zero?
Zero as well — but by internal compensation (one half of the molecule cancels the other via an internal mirror plane), so it is a genuinely achiral, non-resolvable single substance.
If a sample of pure single enantiomer somehow reads , what could explain it?
Either the concentration/path length was effectively zero, or the true rotation exceeded a full turn and read back to a multiple of — measuring at a second concentration or path length distinguishes real zero from a wrapped-around angle.
What happens to optical purity as a sample approaches 100% of one enantiomer?
Optical purity , meaning ; the enantiomeric excess equals 1 and essentially all molecules are the single active form.
What is the enantiomeric excess of a mixture that is 60% (+) and 40% (−)?
— the 40% (−) cancels 40% of the (+), leaving 20% effectively pure (+) that produces all the observed rotation.
If you resolve a racemate but leave a trace of the chiral helper behind, why might rotation readings drift?
The leftover helper is itself optically active and contributes its own rotation, biasing ; the helper must be fully removed (e.g. by regenerating the free acid) before the sample's true can be read.
Recall Quick self-test before you leave
Inactive substance — three possible reasons? ::: Achiral molecule, meso (internal compensation), or racemate (external compensation) — distinguished by resolvability. Which of those three can be resolved into active halves? ::: Only the racemate. What single change makes enantiomers separable? ::: Converting them to diastereomers via a single pure chiral reagent.