Isomerism
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Isomerism — Key Facts for Makerere University (Uganda) Core concept: Isomers are compounds with the same molecular formula but different structural arrangements of atoms High-yield points: Structural (constitutional) isomerism vs stereoisomerism (geometric E/Z and optical R/S); identifying chiral centers; enantiomers vs diastereomers ⚡ Exam tip: If asked to draw all isomers of a formula, always check for: chain isomerism, positional isomerism, functional group isomerism, geometric (E/Z) isomerism, and optical isomerism
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Isomerism — Makerere University (Uganda) Study Guide
1. Classification of Isomerism
Isomerism
├── Structural (Constitutional) Isomers
│ ├── Chain isomerism
│ ├── Positional isomerism
│ ├── Functional group isomerism
│ └── Metamerism
└── Stereoisomers
├── Geometric (E/Z or cis-trans) isomerism
└── Optical isomerism (enantiomers and diastereomers)2. Structural (Constitutional) Isomers
Structural isomers have the same molecular formula but atoms are connected in different orders.
2.1 Chain Isomerism
Same molecular formula, different carbon skeleton (branching pattern).
Example — C₄H₁₀:
- Butane (CH₃–CH₂–CH₂–CH₃): straight chain
- 2-Methylpropane (CH₃–CH(CH₃)–CH₃): branched chain
Example — C₅H₁₂:
- Pentane: CH₃–CH₂–CH₂–CH₂–CH₃
- 2-Methylbutane (isopentane): (CH₃)₂CH–CH₂–CH₃
- 2,2-Dimethylpropane (neopentane): C(CH₃)₄
2.2 Positional Isomerism
Same molecular formula, same functional group, but in different positions on the carbon chain.
Example — C₃H₈O (alcohols):
- Propan-1-ol: CH₃–CH₂–CH₂OH
- Propan-2-ol: CH₃–CH(OH)–CH₃
Example — C₄H₈:
- But-1-ene: CH₂=CH–CH₂–CH₃
- But-2-ene: CH₃–CH=CH–CH₃
Example — C₃H₉N (amines):
- Propan-1-amine: CH₃–CH₂–CH₂NH₂
- Propan-2-amine: CH₃–CH(NH₂)–CH₃
- N-Methylethanamine: CH₃–CH₂–NH–CH₃
2.3 Functional Group Isomerism
Same molecular formula, different functional groups (different chemical families).
Important functional group isomer pairs:
| Formula | Isomer A | Isomer B |
|---|---|---|
| C₂H₆O | Ethanol (alcohol) | Methoxymethane (ether) |
| C₃H₈O | Propan-1-ol | Propan-2-ol |
| C₃H₆O | Propanal (aldehyde) | Propanone (ketone) |
| C₃H₆O₂ | Propanoic acid | Methyl ethanoate (ester) |
| C₄H₁₀O | Butan-1-ol | Butan-2-ol |
| C₄H₈O₂ | Butanoic acid | Methyl propanoate |
| C₇H₈O | Cresol (alcohol on ring) | Anisole (aryl ether) |
⚠️ Key difference:
- Alcohols (R–OH): –OH bonded directly to carbon, can H-bond
- Ethers (R–O–R’): –O– between two carbons, no H-bonding to own molecules
2.4 Metamerism
Special type of positional isomerism in ethers and esters where the alkyl groups on either side of the functional group differ.
Example — C₄H₁₀O ethers:
- CH₃–O–CH₂–CH₂–CH₃: 1-methoxypropane (methoxy + propyl)
- CH₃–CH₂–O–CH₂–CH₃: ethoxyethane (ethyl + ethyl)
- CH₃–CH₂–CH₂–O–CH₃: 1-methoxypropane (same as first, by symmetry)
Actually different: CH₃–O–CH(CH₃)₂ vs CH₃–CH₂–O–CH₂–CH₃ — same MW but different R groups on each side of O.
3. Stereoisomerism
Stereoisomers have the same molecular formula and the same connectivity, but different spatial arrangement of atoms.
3.1 Geometric (E/Z) Isomerism
Geometric isomerism occurs in alkenes when there is restricted rotation about the double bond. It requires:
- A C=C double bond
- Each double-bonded carbon must have two different substituents
Cis-Trans (E/Z) System:
E (Entgegen): Higher priority groups on OPPOSITE sides Z (Zusammen): Higher priority groups on SAME side
Assigning E vs Z using Cahn-Ingold-Prelog (CIP) Priority:
- At each double-bonded carbon, compare the two substituents
- Higher atomic number gets higher priority
- If the two substituents on the same carbon have identical first atoms, compare the next atoms along each chain
Example — 2-butene:
H H
\ /
C========C
/ \
CH₃ CH₃Both carbons have identical substituents (H and CH₃) → This is actually the SAME molecule (no E/Z) because flipping doesn’t change anything.
Example — 2-bromo-2-butene:
Br CH₃
\ /
C========C
/ \
CH₃ HOn C-2: Br > CH₃ (priority 1 and 2) On C-3: CH₃ > H (priority 1 and 2) Both priorities are on SAME side → Z isomer
⚠️ Common mistake: Thinking all alkenes show geometric isomerism. But-1-ene (CH₂=CH–CH₂–CH₃) has one carbon with two identical H atoms — no E/Z isomerism.
Geometric isomers have different physical properties:
| Property | Cis | Trans |
|---|---|---|
| Boiling point | Higher | Lower |
| Polarity | More polar | Less polar |
| Density | Higher | Lower |
| Melting point | Lower | Higher (more symmetrical) |
Example: cis-1,2-dichloroethene (b.p. 60°C) vs trans-1,2-dichloroethene (b.p. 48°C).
3.2 Optical Isomerism
Optical isomers (enantiomers) are non-superimposable mirror images that rotate plane-polarized light.
Key concepts:
- Chiral center (stereocenter): A carbon with four different substituents (e.g., CHBrClF)
- Achiral: Molecule superimposable on its mirror image (has a plane of symmetry or center of symmetry)
- Enantiomers: Non-superimposable mirror images; rotate light equally but in opposite directions
- Specific rotation: [α]ₜ = observed rotation / (concentration × path length)
Chiral molecules:
- Almost always contain at least one chiral center (asymmetric carbon)
- Cannot have a plane of symmetry
- Exist as pairs of enantiomers (d- and l- or R- and S-)
Acarbon that is NOT chiral (achiral):
- Has two or more identical substituents
- Example: CH₃CH₂OH (ethanol) — CH₂OH carbon has two H atoms
Example — Lactic acid:
COOH COOH
| |
H–C–OH ←→ HO–C–H
| |
CH₃ CH₃
(D)-lactic (L)-lactic
acid acidBoth carbons have four different groups (H, OH, COOH, CH₃) → chiral → enantiomers.
Racemic mixture: 50:50 mixture of two enantiomers; optically inactive (rotations cancel out). Labeled as (±) or dl-.
Example — Tartaric acid:
- Meso-tartaric acid: Has TWO chiral centers but is achiral (plane of symmetry) → optically inactive
- D- and L-tartaric acid: Enantiomers, optically active
- Racemic tartaric acid: 50:50 mixture of D and L, optically inactive
4. Chirality Beyond Carbon
Chiral molecules without carbon stereocenters:
- Allenes (cumulative double bonds): C=C=C with four different substituents on terminal carbons
- Spiranes: Two rings joined at one carbon
- Chiral轴: Biaryls with restricted rotation (atropisomerism)
5. Diastereomers
Diastereomers: Stereoisomers that are NOT mirror images of each other.
Diastereomer situations:
- Molecules with TWO (or more) chiral centers (but not identical substituents giving meso forms)
- Geometric isomers are diastereomers of each other
Example — 2,3-dichlorobutane:
CH₃ CH₃ CH₃ CH₃
| | | |
H–C–Cl vs H–C–Cl vs Cl–C–H vs Cl–C–Cl
| | | |
H–C–H H–C–H H–C–H H–C–H
| | | |
CH₃ CH₃ CH₃ CH₃Two chiral centers: 2,3 positions
- (2R,3R): enantiomer of (2S,3S)
- (2R,3S): identical to (2S,3R) by rotation (meso form!)
So there are only THREE stereoisomers: (R,R), (S,S) [enantiomeric pair] and the meso (R,S) [achiral].
6. Relationships Between Isomers
- Structural isomers: Same formula, different connectivity → different chemical properties
- Geometric isomers: Same formula, same connectivity, different spatial arrangement → different physical and some chemical properties
- Enantiomers: Mirror images, identical physical properties (except optical rotation), identical chemical properties (except with chiral reagents/environments)
- Diastereomers: Not mirror images → different physical properties, different chemical properties
7. Exam-Style Questions & Tips
Common exam question patterns at Makerere:
- “Draw all structural isomers of C₄H₁₀O and name them”
- “Explain why [compound] shows geometric isomerism but [compound] does not”
- “Define the term ‘chiral center’ and identify all chiral centers in [molecule]”
- “Draw the enantiomers of [compound] and label R/S”
- “State the relationship between [molecule A] and [molecule B] (identical, enantiomers, diastereomers, etc.)”
- “Calculate the number of stereoisomers possible for a compound with n chiral centers”
⚡ Exam tips:
- To find if a molecule is chiral: check for a plane of symmetry or center of symmetry
- For alkenes: if either carbon of the double bond has two identical substituents, there’s NO E/Z isomerism
- When counting stereoisomers: 2ⁿ formula (n = number of chiral centers), BUT subtract any meso forms
- Optical isomers have identical boiling points, densities, etc. — only optical rotation differs
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8. Advanced Stereochemistry Concepts
Cahn-Ingold-Prelog (CIP) Priority Rules — Full Detail
When assigning R/S or E/Z, CIP priority rules determine ranking of substituents:
-
Atomic number rule: Higher atomic number = higher priority
- I > Br > Cl > S > P > F > O > N > C > H
-
Multiple bond rule: Treat multiple bonds as duplicate attachments
- –CH=CH₂: C is attached to (C, C, H) [counts as C bonded to C twice]
- –CH₂–CH₃: C is attached to (C, H, H) [counts as C bonded to C once]
- So –CH=CH₂ > –CH₂–CH₃ for priority
-
If first atom is same, compare second atoms (and so on until difference found)
- –OH vs –NH₂: O (Z=8) > N (Z=7) → –OH wins
- –CH₂OH vs –CH₂NH₂: First atoms same (C); compare atoms attached: O > N → –CH₂OH wins
-
For stereochemistry (R/S): If lowest priority (4) is pointing toward you (wedge), then 1→2→3 clockwise = S (invert if tracing correctly)
R/S Assignment Method (with lowest priority in front/back)
Step-by-step:
- Identify the chiral center (atom with 4 different substituents)
- Assign priorities 1-4 to each substituent (1=highest, 4=lowest)
- Orient molecule so that lowest priority (4) is pointing AWAY from you (dashed bond)
- Trace from 1→2→3
- Clockwise = R (rectus); Counterclockwise = S (sinister)
- If lowest priority is toward you, reverse the result
Example — Glyceraldehyde (2,3-dihydroxypropanal):
CHO CHO
| |
HO–C–H HO–C–H
| |
H–C–OH H–C–OH
| |
CH₂OH CH₂OH
D-glyceraldehyde L-glyceraldehyde
R SPriority at C-2: OH (1) > CHO (2) > CH₂OH (3) > H (4) In D-glyceraldehyde: H is dashed (back), so 1→2→3 is clockwise → R
Thalidomide — A Cautionary Tale
Thalidomide exists as two enantiomers:
- (R)-thalidomide: Sedative/hypnotic drug
- (S)-thalidomide: Teratogen (causes severe birth defects) Problem: In the body, the two enantiomers interconvert (racemize), so even giving only the (R)-enantiomer led to birth defects. This illustrates why single-enantiomer drugs must be carefully studied.
Meso Compounds
A meso compound has chiral centers but is achiral overall (superimposable on its mirror image) due to an internal plane of symmetry.
Meso-tartaric acid:
HOOC–H–OH HOOC–H–OH
\ /
C–C
/ \
H–OH H–OH
| |
COOH COOH
meso-tartaric acid
(plane of symmetry through center C-C bond)Test for meso: Does the molecule have a plane of symmetry? If YES → meso (achiral).
Tautomerism (Brief Note — Connected to Isomerism)
Tautomers are constitutional isomers in rapid equilibrium:
- Keto-enol tautomerism: R–CH₂–COR ⇌ R–CH=C(OH)–R
- Imine-enamine tautomerism: R–CH₂–NH–R ⇌ R–CH=N–R (with H shift)
- Lactam-lactim tautomerism: In nucleic acid bases
Tautomerism is NOT considered stereoisomerism because atoms are reconnected.
Conformational Isomerism (Brief)
Conformational isomers (conformers/rotamers) arise from rotation about single bonds. They are NOT considered true isomers for the purpose of most organic chemistry exams because they interconvert rapidly at room temperature.
- Anti and gauche conformations of butane
- Chair and boat conformations of cyclohexane
For exam purposes, conformational isomers are usually treated as the same compound.
Number of Stereoisomers — Formula
For a molecule with n chiral centers:
- Maximum number of stereoisomers = 2ⁿ
- If the molecule has a plane of symmetry → some are meso → fewer than 2ⁿ
Examples:
| Molecule | Chiral Centers | 2ⁿ | Actual Stereoisomers | Reason |
|---|---|---|---|---|
| Butane-2,3-diol | 2 | 4 | 3 (pair + meso) | Has meso form |
| 2,3-dichlorobutane | 2 | 4 | 3 (pair + meso) | Same as above |
| 2,3,4-trihydroxybutanal | 3 | 8 | 4 | Some are meso |
| Glucose | 4 | 16 | 16 | Most are non-mesos |
Practice Problems
Q1: How many structural isomers are possible for C₃H₆Br₂? Draw and name them.
Q2: Which of the following alkenes shows geometric isomerism? For those that do, draw E and Z isomers. (a) CH₂=CH–CH₂–CH₃ (b) CH₃–CH=CH–CH₃ (c) CH₃–CH=CH–Cl (d) (CH₃)₂C=CH–CH₃
Q3: Identify all chiral centers in the following molecules: (a) 2-bromobutane (b) 2,3-dimethylbutane (c) 2-hydroxypropanoic acid (lactic acid) (d) alanine (2-aminopropanoic acid)
Q4: Assign R/S configuration to each chiral center in: (a) (S)-2-bromobutane (verify) (b) 3-chlorohexane
Q5: State the number of stereoisomers for: (a) 2,3-dichlorobutane (b) 2-chloro-3-bromobutane (c) 2,3-dimethylbutane
Q6: Draw the meso form of 2,3-dihydroxybutanedioic acid (tartaric acid) and explain why it is optically inactive.
Common Mistakes to Avoid
- Thinking all alkenes show E/Z isomerism: If either double-bonded carbon has two identical substituents, geometric isomers don’t exist.
- Confusing enantiomers with diastereomers: Enantiomers are mirror images; diastereomers are not. Geometric isomers are always diastereomers of each other.
- Forgetting to assign priorities correctly for E/Z: CIP rules must be applied strictly — don’t assume left=right or up=down.
- Thinking optical isomers have different physical properties: Enantiomers have identical physical properties (melting point, boiling point, density) — only optical rotation differs.
- Forgetting that meso compounds are ACHIRAL: Despite having chiral centers, meso compounds are not optically active.
- Miscounting chiral centers: A carbon is only chiral if it has FOUR different substituents. A carbon with two identical substituents (like CH₂ or C=O) is NOT chiral.
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