H3 Chemistry 9813 · Study focus: H3 Chemistry: Apply R/S, optical activity and optical-purity reasoning
H3 Chemistry: Apply R/S, optical activity and optical-purity reasoning
Start from the governing chemical model, test it against evidence, then transfer the reasoning to an unfamiliar case.
Your success criteria
- Apply R/S, optical activity and optical-purity reasoning
- Use named chemical evidence.
- Transfer the governing reason to an unfamiliar case.
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.
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.
- 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.
- 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.
- 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.
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.
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.
- Calculate optical purity and composition for the +12° sample.
Open the feedback checkpoint after attempting
- Award 40% R excess, then R = (100 + 40)/2 = 70% and S = 30%.
Correct a towards-viewer trace
Suppose priorities 1→2→3 appear clockwise while group 4 is on a solid wedge towards the viewer.
The observed sense must be reversed, so the centre is S rather than R.
- Assign the configuration for the stated clockwise trace with priority 4 towards.
Open the feedback checkpoint after attempting
- Credit S and explicitly reverse because the lowest-priority group points towards the viewer.
Use the sign of a reference rotation
Pure R has [α] = +24° and a mixture gives [α]obs = −18° under the same conditions.
The magnitude gives 75% optical purity, while the negative sign shows S is in excess; composition is 87.5% S and 12.5% R.
- Determine optical purity, major enantiomer and composition for the −18° mixture.
Open the feedback checkpoint after attempting
- Award 75% S excess and solve S = 87.5%, 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.
- Repair: ‘Every R enantiomer is dextrorotatory.’
Open the feedback checkpoint after attempting
- 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.
- List the marking evidence for an R/S plus optical-purity response.
Open the feedback checkpoint after attempting
- Require CIP comparisons, view orientation, traced sense, signed ratio, named excess enantiomer and composition when requested.