Isomerism and Stereochemistry
Structural isomerism, geometric stereoisomerism (cis-trans, E/Z), chirality, enantiomers, diastereomers, optical activity, Fischer projections, R/S configuration.
Complete Theory
4Isomers are different compounds that have the same molecular formula but a different arrangement of atoms. Structural (constitutional) isomerism is divided into three categories:
- Chain isomerism: carbon atoms are connected differently (linear vs branched chain). Example: C₅H₁₂ has three isomers: n-pentane (linear), isopentane (branched), neopentane (highly branched). The number of isomers grows rapidly: C₆H₁₄ has 5 isomers, C₁₀H₂₂ has 75.
- Position isomerism: same atoms and same chain, but a functional group or substituent occupies a different position. Example: C₃H₈O can be 1-propanol (CH₃CH₂CH₂OH) or 2-propanol (CH₃CH(OH)CH₃).
- Functional group isomerism: same atoms but different functional group. Example: C₂H₆O can be ethanol (CH₃CH₂OH, alcohol) or dimethyl ether (CH₃OCH₃, ether). Compounds with the same formula but different functions have radically different chemical and physical properties (ethanol boils at 78°C, dimethyl ether at −24°C).
Stereoisomers have the same connectivity between atoms but a different spatial arrangement. Geometric (cis-trans) isomerism occurs when rotation around a bond is prevented:
The CIP rules for assigning priority:
- C=C double bonds: the bond prevents rotation. If the two substituents on each carbon of the double bond are different, two isomers exist: cis (same side) and trans (opposite sides). Example: maleic acid (cis) and fumaric acid (trans), with very different properties (melting point, solubility, reactivity).
- Cycloalkanes: in rings, rotation around C—C bonds is prevented. Substituents can be on the same side of the ring plane (cis) or on opposite sides (trans).
The CIP rules for assigning priority:
- Compare the atomic numbers of the atoms directly bonded to the double-bond carbon. Higher Z → higher priority.
- If the first atoms are identical, proceed along the chains to the first point of difference.
- Atoms with multiple bonds are counted as duplicated or triplicated atoms (e.g., C=O counts as C bonded to O, O, C).
An object is chiral if it is not superimposable on its mirror image (like the right and left hands). In organic chemistry, the most common chirality is due to a stereogenic center (asymmetric carbon): a carbon atom bonded to four different substituents.
Two enantiomers are non-superimposable mirror images:
Real-world applications: thalidomide — one enantiomer was an effective sedative, the other caused severe fetal malformations. Today, the FDA requires stereochemical characterization of every new chiral drug. The pharmaceutical industry increasingly produces enantiomerically pure drugs (e.g., omeprazole, escitalopram).
Two enantiomers are non-superimposable mirror images:
- Same molecular formula, same connectivity, but different absolute configuration
- Same physical properties in an achiral environment (melting point, boiling point, solubility)
- Different interaction with plane-polarized light: one rotates the plane of polarization clockwise (dextrorotatory, +), the other counterclockwise (levorotatory, −). A 1:1 mixture is racemic and shows no optical activity.
- Different interaction with other chiral compounds (e.g., enzymes, biological receptors): often one enantiomer is therapeutically active, the other inactive or toxic.
Real-world applications: thalidomide — one enantiomer was an effective sedative, the other caused severe fetal malformations. Today, the FDA requires stereochemical characterization of every new chiral drug. The pharmaceutical industry increasingly produces enantiomerically pure drugs (e.g., omeprazole, escitalopram).
Fischer projections represent chiral molecules in 2D, standard for carbohydrates and amino acids:
Real-world applications: specific rotation is a fast, non-destructive method for evaluating the enantiomeric purity of a sample. Chiral stationary phase chromatography (chiral HPLC) separates enantiomers by exploiting their different interaction with a chiral selector.
- The carbon chain is vertical, with the most oxidized carbon at the top (for sugars: CHO at top for aldoses)
- Horizontal bonds are coming toward the viewer (out of the plane)
- Vertical bonds are behind the plane (away from the viewer)
- A 180° rotation of the projection in the plane leaves the configuration unchanged; a 90° rotation inverts it
- Assign priorities to the four substituents based on atomic number (1 = highest priority, 4 = lowest)
- Orient the molecule so that the substituent with priority 4 is pointing away from the viewer (at the top in a Fischer projection)
- Observe the order 1 → 2 → 3: if clockwise → R (rectus), if counterclockwise → S (sinister)
Real-world applications: specific rotation is a fast, non-destructive method for evaluating the enantiomeric purity of a sample. Chiral stationary phase chromatography (chiral HPLC) separates enantiomers by exploiting their different interaction with a chiral selector.
Worked Examples
2Example 1Enantiomers of Lactic Acid
Given
Lactic acid: CH₃CH(OH)COOH
Central carbon C* bonded to: H, OH, CH₃, COOH
Find
Identify the chiral carbon
Draw the two enantiomers in Fischer projection
Assign the R/S configuration
Step-by-step solution
1The central carbon (C₂) is bonded to four different groups: —H, —OH, —CH₃, —COOH. It is therefore a stereogenic center (asymmetric carbon). The molecule exists as a pair of enantiomers.
2In Fischer projection: arrange the chain vertically with COOH at the top (most oxidized carbon) and CH₃ at the bottom. On the horizontal line, place H on the left and OH on the right — this is the projection of the R enantiomer. For the S enantiomer, swap H and OH (or draw the mirror image).
3R/S assignment for the enantiomer with H left and OH right: priority (1) OH (Z=8), (2) COOH (C bonded to O,O,O), (3) CH₃ (C bonded to H,H,H), (4) H (Z=1). Orient with H (priority 4) away. Going 1→2→3 is clockwise → R. The other enantiomer is S.
✓ Final result: C₂ chiral (4 different substituents); R-(−)-lactic acid (dextrorotatory), S-(+)-lactic acid (levorotatory)
Example 2cis-trans Isomerism of Butenedioic Acid
Given
Butenedioic acid: HOOC—CH=CH—COOH
Molecular formula C₄H₄O₄
Find
Draw the two geometric isomers
Assign the E/Z nomenclature
Compare physical properties
Step-by-step solution
1The two carbons of the double bond each have the same substituents (H and COOH). cis isomer: the two COOH groups are on the same side of the double bond → maleic acid. trans isomer: the COOH groups are on opposite sides → fumaric acid.
2For E/Z nomenclature: on each carbon, compare CIP priorities. COOH (C bonded to O,O,O) > H (Z=1). In maleic acid, the two higher priority groups (COOH) are on the same side → Z (Z)-butenedioic. In fumaric acid they are on opposite sides → E (E)-butenedioic.
3Properties: maleic acid (Z) has melting point 130°C, is soluble in water, and upon heating dehydrates to form maleic anhydride. Fumaric acid (E) melts at 287°C, is less soluble, and does not form an anhydride. The difference is due to the different spatial arrangement affecting intermolecular hydrogen bonds.
✓ Final result: Maleic acid (Z): cis, mp 130°C; fumaric acid (E): trans, mp 287°C
Exercises with Solutions
4Exercise 1Structural isomerismMedium
Problem to solve
Write and name all structural isomers of pentane C₅H₁₂.
Given data
C₅H₁₂
Step-by-step solution
1Linear chain isomer: n-pentane CH₃CH₂CH₂CH₂CH₃ (5-carbon chain). Boiling point: 36°C.
2Isomer with a methyl branch on C₂: 2-methylbutane (isopentane) CH₃CH(CH₃)CH₂CH₃. Main chain of 4 C with CH₃ at position 2. Bp: 28°C.
3Isomer with two methyl branches on C₂: 2,2-dimethylpropane (neopentane) C(CH₃)₄. Main chain of 3 C with two CH₃ at position 2. Bp: 9.5°C. Increased branching reduces London forces, lowering the boiling point.
✓ Final answer: n-pentane, 2-methylbutane, 2,2-dimethylpropane (3 isomers)
Exercise 2Position isomerismMedium
Problem to solve
Write the formulas and names of all position isomers of dichlorobenzene C₆H₄Cl₂.
Given data
C₆H₄Cl₂ (benzene with 2 Cl atoms)
Step-by-step solution
1The benzene ring has 6 equivalent positions. Two chlorine atoms can occupy different relative positions: positions 1,2 (adjacent) → 1,2-dichlorobenzene (ortho).
2Positions 1,3 (one carbon apart) → 1,3-dichlorobenzene (meta).
3Positions 1,4 (opposite) → 1,4-dichlorobenzene (para). These are three positional isomers, with different melting points: ortho −17°C, meta −24°C, para 53°C.
4No other positional isomers exist: other combinations (1,5 = 1,3; 1,6 = 1,2) are equivalent due to benzene ring symmetry.
✓ Final answer: 1,2- (ortho), 1,3- (meta), 1,4- (para) dichlorobenzene
Exercise 3R/S ConfigurationHard
Problem to solve
Assign the absolute R/S configuration to 2-butanol CH₃CH(OH)CH₂CH₃.
Given data
2-butanol: CH₃—*CH(OH)—CH₂CH₃
Step-by-step solution
1Carbon C₂ is bonded to four substituents: (a) —OH, (b) —CH₂CH₃, (c) —CH₃, (d) —H. They are all different → stereogenic center.
2Assign CIP priorities: (1) OH (Z=8), (2) CH₂CH₃ (C bonded to C,H,H), (3) CH₃ (C bonded to H,H,H), (4) H (Z=1).
3In the Fischer projection (vertical chain, OH left or right), place the H (priority 4) at the top (away from the viewer). If the order 1→2→3 is clockwise: R; counterclockwise: S.
4The configuration depends on which enantiomer we consider: (R)-(−)-2-butanol and (S)-(+)-2-butanol are an enantiomeric pair.
✓ Final answer: C₂ chiral: (R)-(−)-2-butanol and (S)-(+)-2-butanol (enantiomeric pair)
Exercise 4Diastereoisomerism of tartrateVery Hard
Problem to solve
Tartaric acid HOOC—CH(OH)—CH(OH)—COOH has 2 stereogenic centers. How many stereoisomers exist? Explain and identify the meso compound.
Given data
Tartaric acid: HOOC—CH(OH)—CH(OH)—COOH (2 C* centers)
Step-by-step solution
1With 2 stereogenic centers one would expect stereoisomers. However, if the two centers have the same identical substituents (H, OH, COOH, CH(OH)COOH), the molecule can have an internal plane of symmetry.
2The four possible stereoisomers: (R,R), (S,S), (R,S), and (S,R). (R,R) and (S,S) are enantiomers (non-superimposable mirror images) — these are (+)-tartaric acid and (−)-tartaric acid, respectively.
3(R,S) and (S,R) are actually the same molecule (superimposable by rotation): meso-tartaric acid. It has an internal plane of symmetry (between C₂ and C₃) that makes it achiral despite having two stereogenic centers. It shows no optical activity.
4Total: 3 stereoisomers — one enantiomeric pair (R,R)/(S,S) and one achiral meso compound (R,S).
✓ Final answer: 3 stereoisomers: (R,R) and (S,S) enantiomers, (R,S) meso achiral
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