Apply R/S, optical activity and optical-purity reasoning

A tetrahedral centre with four different substituents can have two non-superimposable mirror-image configurations.

  • GCE A-Level H3 Chemistry 9813-2027
On this page

Learning objectives

  • Apply R/S, optical activity and optical-purity reasoning

Separate structure from rotation sign

A tetrahedral centre with four different substituents can have two non-superimposable mirror-image configurations.

R/S describes configuration from CIP priorities; the sign and magnitude of optical rotation must be measured and cannot be predicted from the letter R or S.

Configuration, activity and purity

With priority 4 directed away, a clockwise 1→2→3 sequence is R and an anticlockwise sequence is S; reversing one pair of groups inverts the configuration.

Optical purity, or enantiomeric excess, is [α]obs/[α]pure × 100%; the sign identifies which reference enantiomer is in excess and the magnitude gives the excess percentage.

Build a complete CIP assignment

Rank directly attached atoms by atomic number, give a heavier isotope higher priority, and resolve ties by comparing ordered attached-atom lists; treat multiple bonds as bonds to duplicate atoms.

If priority 4 points towards the viewer, reverse the apparent clockwise/anticlockwise result; enantiomers have opposite configurations at every stereogenic centre, whereas diastereomers are stereoisomers that are not mirror images.

For R/S, the most reliable routine is rank 1–4, turn group 4 away, trace 1→2→3 and reverse the result only if group 4 points towards you. Optical rotation is then a separate experimental calculation, not another way of assigning R or S.

Calculate an enantiomeric excess

For [α]obs = +12° and pure R = +30° under identical conditions, optical purity is (+12/+30) × 100% = 40% in favour of R.

An R excess of 40% corresponds to 70% R and 30% S because the two percentages sum to 100% and differ by 40 percentage points.

Try this

Calculate optical purity and composition for the +12° sample.

Check your answer

A strong answer gives 40% R excess, then R = (100 + 40)/2 = 70% and S = 30%.

Correct a towards-viewer trace

Supplied view: priorities 1→2→3 appear clockwise, and group 4 is on a solid wedge towards the viewer.

Recall which way group 4 must point for the traced direction to be read directly, and adjust if it points the other way.

Try this

Assign the configuration for the stated clockwise trace with priority 4 towards.

Check your answer

The direct reading applies only with group 4 pointing away. Here it points towards the viewer, so the clockwise trace must be reversed: the centre is S, not R.

Use the sign of a reference rotation

Supplied data: pure R has [α] = +24°; a mixture gives [α]obs = −18° under the same conditions.

Use the magnitude ratio for optical purity and the sign for the excess enantiomer, then solve for the percentage of each enantiomer.

Try this

Determine optical purity, major enantiomer and composition for the −18° mixture.

Check your answer

Optical purity = 18 ÷ 24 = 75%. The negative sign means S (which has [α] = −24°) is in excess. S − R = 75% and S + R = 100%, so S = 87.5% and R = 12.5%.

Do not translate R into positive

R and S follow a structural priority convention; (+) and (−) are experimental rotation signs.

A racemic 50:50 mixture contains both enantiomers yet has zero net rotation because equal opposite rotations cancel.

Try this

Better reasoning: ‘Every R enantiomer is dextrorotatory.’

Check your answer

Reject the claim: rotation sign requires measurement against a stated enantiomer under fixed conditions.

Audit configuration and composition separately

For R/S, show four priorities, the direction of group 4, the 1→2→3 sense and any reversal; for optical purity, retain the sign and state reference conditions.

Next transfer mirror-image and superimposability tests to square-planar and octahedral coordination geometries.

Write structural and optical conclusions on separate lines: first justify R/S from CIP geometry, then calculate enantiomeric excess from the stated rotations and convert it to the two percentages if asked.

Try this

List the marking evidence for an R/S plus optical-purity response.

Check your answer

Your answer should include CIP comparisons, view orientation, traced sense, signed ratio, named excess enantiomer and composition when requested.

Apply R/S, optical activity and optical-purity reasoning scientific representation

The fixed projection names every substituent, page position and towards/away bond; each numerical step includes symbols, signs, percentages and the named enantiomer.

About 5 minutes

Key visual: Apply R/S, optical activity and optical-purity reasoning. A fixed wedge–dash model keeps the configuration decision separate from the signed optical-purity calculation.
  1. Fixed 2-butanol projection at C2: place CH₃ up and CH₂CH₃ down in the plane, OH on a solid wedge to the right and H on a hashed wedge to the left; priorities OH > CH₂CH₃ > CH₃ > H, with H away, trace clockwise to give R.
  2. Configuration record: list priorities 1–4, state whether group 4 is away or towards, trace 1→2→3 and record any required reversal before writing R or S.
  3. Purity record: [α]obs/[α]pure × 100%, retain the sign to name the excess enantiomer, then solve major = (100 + ee)/2 and minor = (100 − ee)/2.

Text alternative: The fixed projection names every substituent, page position and towards/away bond; each numerical step includes symbols, signs, percentages and the named enantiomer.