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Organic Chemistry - Stereoisomerism (HL)

Grade 11IBChemistry

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

🔑Concepts

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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.

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

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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.

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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.

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

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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.

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

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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) Grade 11 Notes & Examples