Difficulty: Intermediate-Advanced | Prerequisites: Parts 1 and 2 (stereoisomer classification, R/S configuration, optical activity, specific rotation)
This section brings together the concepts from Parts 1 and 2 and applies them quantitatively. Enantiomeric excess (ee) and enantiomeric ratio let you describe how much of a mixture is one enantiomer vs. the other, which matters in pharmaceutical chemistry (where one enantiomer of a drug may be therapeutic and the other harmful) and in lab settings where you need to assess the selectivity of a reaction. The section also covers how to draw all possible stereoisomers of a compound and how different reactions (syn vs. anti addition, hydrogenation) produce specific stereochemical outcomes.
Enantiomeric excess tells you what fraction of a mixture is one enantiomer beyond the racemic background. The 2^n rule predicts the maximum number of stereoisomers for a compound with n stereocenters (fewer when meso forms exist). Syn addition delivers both groups to the same face of a double bond; anti addition delivers them to opposite faces, and each gives a predictable stereochemical outcome.
Enantiomeric excess (ee, % ee)
A measure of how far a mixture deviates from racemic. Calculated as: % ee = |R - S| / (R + S) x 100, or equivalently, % ee = (alpha observed / alpha pure) x 100. In simple terms, it tells you what percentage of the mixture is "extra" of one enantiomer beyond the 50:50 background.
Enantiomeric ratio
The ratio of one enantiomer to the other in a mixture, expressed as R:S or S:R (e.g. 85:15). Directly related to ee: once you know the ee, you can find the ratio, and vice versa.
Syn addition
Both new groups add to the same face of a double bond. This is the stereochemistry of catalytic hydrogenation (H2/Pd) and some oxidation reactions.
Anti addition
The two new groups add to opposite faces of a double bond. This is the stereochemistry of bromine addition (Br2) and halohydrin formation (e.g. with HBr in certain conditions).
Enantiomeric resolution
The process of separating a racemic mixture into its individual enantiomers. Typically involves forming diastereomeric salts with a chiral resolving agent, then separating by physical properties (e.g. selective crystallisation).
2^n rule
For a molecule with n stereocenters, the maximum number of stereoisomers is 2^n. The actual number may be lower if meso forms exist.
Two equivalent ways to calculate ee:
% ee = |R - S| / (R + S) x 100
% ee = (alpha observed / alpha pure) x 100
The second form is useful when you have measured optical rotations. The first is useful when you know the molar (or percentage) composition of the mixture.
Given: observed rotation = -47.7 degrees, pure rotation = -53 degrees.
% ee = 47.7 / 53 = 0.9 = 90%
This means 90% of the mixture is the excess enantiomer, and the remaining 10% is a racemic (50:50) portion.
Given: observed rotation = -22.4 degrees, pure rotation = -32 degrees.
% ee = 22.4 / 32 = 0.7 = 70%
The excess enantiomer accounts for 70% of the total beyond the racemic portion. To convert to a ratio:
100 - 70 = 30% is racemic. That racemic portion splits equally: 30 / 2 = 15% of each enantiomer.
The major enantiomer: 70 + 15 = 85%. The minor: 15%.
Enantiomeric ratio: 85:15.
Given: [alpha] pure = +78 degrees, ratio = 75(+) : 25(-).
Net excess: 75 - 25 = 50 parts of the (+) enantiomer.
As a fraction of total: 50 / 100 = 0.5.
[alpha] mixture = 0.5 x 78 = +39 degrees.
Alternatively: given [alpha] pure = 168.2 degrees, ratio = 60(-) : 40(+).
Net excess of (-): 60 - 40 = 20 parts.
Fraction: 20 / 100 = 0.2.
[alpha] mixture = 0.2 x (-168.2) = -33.6 degrees.
For a molecule with n stereocenters, draw up to 2^n stereoisomers by systematically varying R/S at each centre.
For two stereocenters (n = 2), 2^n = 4 possible stereoisomers. Label them by their configurations:
(2S,3S) and (2R,3R) are enantiomers (all centres flipped).
(2S,3R) and (2R,3S) are enantiomers (all centres flipped).
(2S,3S) and (2S,3R) are diastereomers (only one centre differs).
If the molecule has an internal plane of symmetry when some stereocenters have opposite configurations, two of the four "isomers" collapse into one meso compound. This reduces the count from four to three distinct stereoisomers.
Example from the chapter: 3,6-diethyl-4,5-octanediol. Two stereocenters, so 2^2 = 4 expected. But the (4R,5S) and (4S,5R) forms are the same meso compound, giving three stereoisomers total.
Syn addition (e.g. H2/Pd/C catalytic hydrogenation):
Both hydrogen atoms add to the same face of the double bond.
Produces cis product.
If the starting alkene is achiral and symmetric, hydrogenation gives a 50:50 mixture of enantiomers (racemic).
If the starting alkene is already chiral (has an existing stereocenter), the new stereocenter's configuration may be influenced by the existing one, and the product may not be racemic.
Anti addition (e.g. Br2 addition across a double bond):
The two bromine atoms add to opposite faces via a bromonium ion intermediate.
Produces trans product.
The stereochemical outcome depends on the symmetry of the starting alkene.
When H2 adds across a double bond, it creates a new stereocenter (or converts an sp2 carbon to sp3). Even if the spatial arrangement is preserved, the CIP priorities may shift because the substituents change. An (R) starting material can give what is labelled (S) product purely because the priority ranking changed, not because the geometry inverted.
Students often mix up enantiomeric excess and enantiomeric ratio. They are related but different: ee is the percentage excess of one enantiomer; the ratio is the actual proportion of each. A 90% ee corresponds to a 95:5 ratio, not 90:10.
Students frequently forget to account for the racemic portion when converting between ee and ratio. The racemic portion is split equally between both enantiomers.
Students sometimes assume that catalytic hydrogenation always gives a racemic product. It does when the alkene is symmetric and achiral. When the molecule has existing chirality, one face may be more accessible, leading to unequal products.
Students often confuse syn and anti addition. A useful mnemonic: hydrogenation (H2, same element, both from the same side) is syn; halogenation (Br2, bridged intermediate forces opposite sides) is anti.
Enantiomeric excess calculations are a staple of exams. You should be able to move fluently in both directions: given observed and pure rotations, find ee; given ee, find the enantiomeric ratio; given a ratio, find the specific rotation of the mixture.
Drawing all stereoisomers of a compound with multiple stereocenters is a common exam question. Use the 2^n rule, then check for meso compounds to reduce the count.
Syn vs. anti addition determines the stereochemistry of products. Expect exam questions that ask you to predict the stereochemical outcome of a reaction (e.g. hydrogenation of an alkene, bromination).
Questions about compound X having n stereocenters and a known [alpha] for the pure compound, then asking about the optical rotation or composition of a mixture, appear regularly.
Fill in the blank: For a compound with 3 stereocenters, the maximum number of stereoisomers is ______. Answer: 2^3 = 8.
True or false: A 90% ee means the mixture is 90:10 (major:minor enantiomer). Answer: False. 90% ee corresponds to a 95:5 ratio.
Fill in the blank: Catalytic hydrogenation (H2/Pd) is an example of ______ addition. Answer: Syn addition.
True or false: The specific rotation of a 50:50 racemic mixture is always zero. Answer: True.
Fill in the blank: Enantiomers are typically separated by a process called ______. Answer: Enantiomeric resolution.
Q: The specific rotation of pure (S)-ibuprofen is +54.5 degrees. You measure an observed rotation of +38.2 degrees on a sample (c = 1 g/mL, l = 1 dm). What is the % ee and the enantiomeric ratio?
A: % ee = 38.2 / 54.5 = 0.70 = 70%. Racemic portion = 100 - 70 = 30%, split equally: 15% each. Major (S) = 70 + 15 = 85%, minor (R) = 15%. Ratio: 85:15 (S:R).
Q: A mixture of enantiomers has a ratio of 75(+) : 25(-). The specific rotation of the pure (+) enantiomer is +78 degrees. What is the specific rotation of the mixture?
A: Net excess = 75 - 25 = 50 parts out of 100 = 0.5. [alpha] mixture = 0.5 x 78 = +39 degrees.
Q: Compound X has 3 asymmetric centres. Its pure RRS diastereomer has [alpha] = +20 degrees. A mixture gives [alpha] = -6 degrees. Which enantiomer is in excess and what is the ee?
A: The negative rotation means the SSR enantiomer (which has [alpha] = -20 degrees) is in excess. % ee = 6 / 20 = 0.3 = 30%. The ratio is 65:35 (SSR:RRS).
Q: An achiral, symmetrical alkene is treated with H2/Pd/C. How many products form, and what is the ratio of stereoisomers?
A: Syn addition creates a new stereocenter, but the catalyst can approach from either face equally. The result is a 50:50 racemic mixture of enantiomers.
Q: Draw all stereoisomers of a compound with two stereocenters where the molecule has a potential internal plane of symmetry. How many distinct stereoisomers exist?
A: The 2^n rule predicts four. However, one pair of "enantiomers" (where the two centres have opposite configurations) is actually a meso compound (one compound, not two). So three distinct stereoisomers: two enantiomers and one meso compound.
Enantiomeric excess calculations reappear in asymmetric synthesis (later chapters) and in pharmaceutical chemistry, where drug purity standards are defined in terms of ee. Syn and anti addition connect to reaction mechanisms (Chapters 7-11): understanding the stereochemical outcome of a reaction requires knowing the mechanism. The 2^n rule and meso compounds connect back to Part 1's symmetry analysis.
Enantiomeric excess, ee, % ee, enantiomeric ratio, specific rotation of mixtures, optical purity, racemic mixture, 2^n rule, maximum stereoisomers, syn addition, anti addition, catalytic hydrogenation, H2/Pd, bromination stereochemistry, bromonium ion, meso reduction, stereoisomer count, enantiomeric resolution, diastereomeric salt formation, selective crystallisation