krit.club logo

Organic Chemistry - Stereoisomerism (HL only)

Grade 12IBChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

Stereoisomers are compounds with the same structural formula but different arrangements of atoms in space. They are divided into conformational isomers (rotation about σ\sigma bonds) and configurational isomers.

Configurational isomers are further subdivided into cis-trans (or E/ZE/Z) isomers and optical isomers.

Cis-trans isomerism occurs due to restricted rotation, typically around a C=CC=C double bond or within a cyclic structure. ZZ (zusammen) refers to high-priority groups on the same side, while EE (entgegen) refers to high-priority groups on opposite sides.

The Cahn-Ingold-Prelog (CIP) priority rules determine E/ZE/Z notation based on the atomic number (ZZ) of the atoms directly attached to the double-bonded carbons.

Optical isomerism occurs in chiral molecules that contain an asymmetric carbon atom (chiral center) bonded to four different groups.

Enantiomers are non-superimposable mirror images of each other. They have identical physical and chemical properties in an achiral environment but rotate plane-polarized light in opposite directions.

A racemic mixture (racemate) contains equimolar amounts of two enantiomers and is optically inactive because the rotations cancel out.

Diastereomers are stereoisomers that are not mirror images of each other. They occur when a molecule has two or more chiral centers and differ in physical and chemical properties.

📐Formulae

Maximum number of stereoisomers=2n\text{Maximum number of stereoisomers} = 2^n (where nn is the number of chiral centers)

Specific Rotation [α]=αc×l\text{Specific Rotation } [\alpha] = \frac{\alpha}{c \times l}

Priority Rule: I>Br>Cl>F>O>N>C>H\text{Priority Rule: } I > Br > Cl > F > O > N > C > H

💡Examples

Problem 1:

Determine the E/ZE/Z configuration for 11-bromo-11-chloro-prop-11-ene.

Solution:

EE-configuration

Explanation:

On carbon 1, the groups are BrBr (Z=35Z=35) and ClCl (Z=17Z=17), so BrBr is high priority. On carbon 2, the groups are CH3CH_3 (carbon Z=6Z=6) and HH (Z=1Z=1), so CH3CH_3 is high priority. Since the high priority groups (BrBr and CH3CH_3) are on opposite sides of the double bond, it is the EE isomer.

Problem 2:

Identify the number of chiral centers and calculate the maximum number of stereoisomers for 2,32,3-dihydroxybutanedioic acid (tartaric acid).

Solution:

2 chiral centers; 22=42^2 = 4 potential stereoisomers (though one is a meso compound).

Explanation:

The molecule has chiral centers at C2C_2 and C3C_3. Using the formula 2n2^n, where n=2n=2, we expect 4 stereoisomers. However, because of internal symmetry in one form (the meso form), there are actually only 3 distinct stereoisomers: (+), (-), and meso.

Problem 3:

A sample of pure (+)(+)-limonene has a specific rotation of +125.6+125.6^\circ. A mixture of enantiomers shows a rotation of +62.8+62.8^\circ. Calculate the percentage of the (+)(+)-enantiomer in the mixture.

Solution:

75%75\%

Explanation:

Optical purity (enantiomeric excess) = +62.8+125.6×100=50%\frac{+62.8}{+125.6} \times 100 = 50\%. An eeee of 50%50\% means 50%50\% of the mixture is the pure (+)(+) isomer and the remaining 50%50\% is a racemic mixture (25%(+)25\% (+) and 25%()25\% (-)). Total (+)(+) enantiomer = 50%+25%=75%50\% + 25\% = 75\%.

Stereoisomerism (HL only) - Revision Notes & Key Formulas | IB Grade 12 Chemistry