Difficulty: Advanced | Prerequisites: All previous Orgo II topics (substitutions, additions, carbonyl chemistry, aromatic reactions)
Stereochemistry runs through every reaction in Orgo II: you need to track configuration at chiral centres across each step of a synthesis. Spectroscopy (IR and NMR) is how you confirm what you have made. Multi-step synthesis ties everything together by asking you to plan a route from a given starting material to a target molecule, choosing the right reactions in the right order. This is where the course culminates.
R/S configuration uses Cahn-Ingold-Prelog priority rules to assign absolute stereochemistry, and each reaction mechanism determines whether configuration is retained, inverted or lost. IR and NMR spectroscopy let you identify functional groups, count distinct carbons and piece together molecular structure. Multi-step synthesis problems ask you to combine everything: choose the right reactions, in the right order, with the right stereochemical control, to build a target molecule from a given starting material.
R/S configuration
The absolute stereochemical designation at a chiral centre, assigned using Cahn-Ingold-Prelog (CIP) priority rules. R = clockwise path from priority 1 to 2 to 3 (with priority 4 pointing away). S = anticlockwise.
Cahn-Ingold-Prelog (CIP) priority rules
Rank substituents by atomic number at the point of attachment (highest atomic number = highest priority). If there is a tie, move to the next atom outward until the tie breaks. Place the lowest-priority group (4) away from the viewer, then trace 1 to 2 to 3.
Meso compound
A molecule with chiral centres but an internal plane of symmetry, making it optically inactive despite having stereocentres. In simple terms, the two halves of the molecule are mirror images of each other, so their rotations cancel.
Racemic mixture
A 50:50 mixture of R and S enantiomers. Optically inactive because the equal and opposite rotations cancel. Produced by SN1 reactions and any process that forms a planar intermediate.
Fischer projection
A two-dimensional representation of a chiral molecule. Horizontal lines represent bonds coming toward the viewer; vertical lines represent bonds going away. Used especially in carbohydrate chemistry.
IR spectroscopy
Measures the absorption of infrared light by molecular bonds. Each functional group absorbs at a characteristic frequency (measured in cm-1). The C=O stretch (1650 to 1750 cm-1) is the strongest and most diagnostic absorption.
1H NMR (proton NMR)
Measures the chemical environment of hydrogen atoms. Key information: chemical shift (where the signal appears, in ppm), integration (relative number of protons), and splitting pattern (n+1 rule: n equivalent neighbouring protons split the signal into n+1 peaks).
13C NMR
Measures distinct carbon environments. The number of signals equals the number of chemically non-equivalent carbons. No splitting or integration in routine spectra.
Retrosynthetic analysis
Working backwards from the target molecule to identify simpler precursors, using known reactions in reverse. The key skill in multi-step synthesis planning.
Assigning R/S
Rank the four substituents on the chiral centre by CIP priority (highest atomic number at the point of attachment = priority 1).
If two substituents start with the same atom, move outward until the tie breaks.
Orient the molecule so that priority 4 points away from the viewer.
Trace the path from 1 to 2 to 3: clockwise = R, anticlockwise = S.
Stereochemical outcomes by reaction type
SN2: inversion of configuration (backside attack).
SN1: racemisation (planar carbocation, attack from either face).
E2: anti-periplanar geometry determines which diastereomeric alkene forms.
Halogenation, halohydrin formation, epoxide opening: anti addition (groups add to opposite faces).
Hydroboration, catalytic hydrogenation, OsO4 dihydroxylation: syn addition (groups add to the same face).
Meso compounds and optical activity
A meso compound has stereocentres but an internal plane of symmetry, making it optically inactive.
A racemic mixture (50:50 R and S) is also optically inactive, but for a different reason: the rotations cancel rather than being absent.
Fischer projections
Horizontal bonds come toward the viewer; vertical bonds go away.
Rotating a Fischer projection 90° inverts the configuration. Rotating 180° preserves it.
Swapping any two groups inverts the configuration.
IR spectroscopy: key absorptions
Functional Group | cm-1 Range | Notes |
|---|---|---|
O-H (alcohol, carboxylic acid) | 3200 to 3600 | Broad; very broad for acids |
C-H (sp3) | 2850 to 3000 | Sharp |
C-H (sp2, aromatic) | 3000 to 3100 | Sharp |
C triple bond C / C triple bond N | 2100 to 2260 | Weak to medium |
C=O (general) | 1650 to 1750 | Strongest, most diagnostic |
Acid chloride C=O | ~1800 | Highest frequency carbonyl |
Ester C=O | ~1735 | - |
Ketone C=O | ~1715 | - |
Carboxylic acid C=O | ~1710 | Broad O-H overlap |
Amide C=O | 1650 to 1680 | Lowest frequency carbonyl |
C=C (alkene, aromatic) | 1450 to 1680 | Medium |
The C=O stretch is the single most useful absorption. Its exact position tells you which carbonyl derivative you have.
1H NMR basics
Chemical shift (delta, in ppm) tells you the electronic environment: alkyl ~0.9 to 1.5, adjacent to carbonyl ~2.0 to 2.5, adjacent to O or N ~3.5 to 4.0, aromatic ~6.5 to 8.5, aldehyde ~9 to 10.
Integration gives the relative number of protons producing each signal.
Splitting follows the n+1 rule: a proton with n equivalent neighbouring protons appears as n+1 peaks (e.g. a triplet has two equivalent neighbours).
13C NMR basics
The number of signals equals the number of chemically distinct carbons.
Symmetry reduces the signal count (e.g. para-disubstituted benzene shows fewer signals than an unsymmetric isomer).
Useful for confirming molecular symmetry and distinguishing constitutional isomers.
Terminal alkyne as a building block
Terminal alkyne + NaNH2 (strong base) removes the terminal H to form an acetylide anion.
Acetylide + RX (primary alkyl halide, SN2) extends the carbon chain.
The resulting internal alkyne can then be hydrated (Hg-catalysed, Markovnikov) to give a methyl ketone, or reduced to a cis-alkene (Lindlar) or trans-alkene (Na/NH3).
Alcohol synthesis strategies
Grignard + carbonyl: choose the right aldehyde/ketone and the right Grignard to build the target alcohol's carbon skeleton.
Hydroboration-oxidation: anti-Markovnikov alcohol from an alkene.
Reduction: NaBH4 or LiAlH4 to reduce a carbonyl to an alcohol.
Primary amine synthesis (avoiding over-alkylation)
Gabriel synthesis: phthalimide + base, then alkyl halide, then hydrazine (cleaves the phthalimide) to give the primary amine cleanly.
Azide method: alkyl halide + NaN3, then reduce (LiAlH4 or catalytic hydrogenation) to the primary amine.
Aryl halide from aniline: the Sandmeyer reaction
Aniline + NaNO2/HCl at 0°C gives a diazonium salt.
Diazonium salt + CuBr gives aryl bromide (Sandmeyer). Also works with CuCl or CuCN for aryl chloride or aryl nitrile.
Deuterium labelling
Grignard + D2O places deuterium at a specific carbon.
Alpha-hydrogens exchange with D2O/OD- for deuterium labelling adjacent to carbonyls.
Students often assign R/S without placing priority 4 away from the viewer first. If priority 4 is toward you, the apparent assignment is reversed: what looks clockwise is actually S, and vice versa.
A common error is assuming that all molecules with chiral centres are optically active. Meso compounds have chiral centres but are optically inactive due to their internal plane of symmetry.
In NMR, students confuse integration (number of protons) with splitting (number of peaks). Integration tells you how many protons produce the signal. Splitting tells you how many equivalent protons are on neighbouring carbons.
In synthesis problems, students often try to build the target forward from the starting material. Retrosynthetic analysis (working backwards from the target) is almost always more efficient.
⚠️ R/S assignment is tested on nearly every exam, often embedded in a reaction problem: "What is the configuration of the product?"
⚠️ Know which reactions give syn addition (hydroboration, H2/Pd, OsO4) and which give anti addition (Br2, epoxide opening, halohydrin). This determines the stereochemistry of the product.
⚠️ IR: if the exam gives you a spectrum, look at the C=O region first. Its position narrows the functional group immediately.
⚠️ Multi-step synthesis questions are typically worth the most marks. Practice retrosynthetic analysis: identify the key bond to form, then choose the reaction that forms it.
⚠️ The Sandmeyer reaction is a favourite synthesis question because it connects aromatic chemistry (EAS to put NH2 on the ring) to functional group interconversion (diazonium to halide).
Fill in the blank: In CIP priority rules, the atom with the highest ______ gets the highest priority. (Atomic number)
True or False: A meso compound is optically active. (False)
Fill in the blank: The C=O stretch in an IR spectrum of a ketone appears at approximately ______ cm-1. (~1715)
True or False: In 1H NMR, a proton with two equivalent neighbours appears as a triplet. (True, by the n+1 rule)
Fill in the blank: To avoid over-alkylation when making a primary amine, use the ______ synthesis. (Gabriel)
Q: A chiral secondary tosylate with S configuration is treated with NaCN. What is the configuration of the product, and what mechanism operates?
A: The product is a nitrile with R configuration. The mechanism is SN2 (backside attack on a secondary substrate with a good nucleophile), which inverts the stereochemistry.
Q: You have an IR spectrum showing a strong absorption at 1735 cm-1 and no broad O-H stretch. What functional group is present?
A: An ester. The C=O stretch at ~1735 cm-1 is characteristic of esters, and the absence of a broad O-H rules out a carboxylic acid.
Q: Propose a synthesis of 2-pentanone from 1-butyne.
A: Deprotonate 1-butyne with NaNH2, alkylate with CH3I (SN2) to give 2-pentyne, then hydrate with H2O/H2SO4/HgSO4 (Markovnikov) to give 2-pentanone (the methyl ketone).
Q: How would you convert aniline to bromobenzene?
A: Treat aniline with NaNO2/HCl at 0°C to form the diazonium salt, then add CuBr (Sandmeyer reaction) to replace the diazonium group with bromine.
Q: A compound shows three signals in its 13C NMR spectrum. It has the molecular formula C4H8O2. What is a likely structure?
A: The three signals indicate three distinct carbon environments, suggesting symmetry. A likely structure is ethyl formate (HCOOCH2CH3) or 1,4-dioxane, depending on the chemical shifts. Three distinct carbons in a four-carbon ester point to ethyl formate (the formate C, the OCH2 C, and the CH3 C).
Stereochemistry underpins every reaction in Orgo II: SN2 inversion, syn/anti addition, and enolate geometry all depend on spatial arrangement. Spectroscopy connects to every functional group you study, and IR/NMR problems often appear as a separate section on exams or as confirmation steps within synthesis problems. Multi-step synthesis is the integrative skill: it draws on substitutions, eliminations, additions, carbonyl chemistry, aromatic reactions and redox transformations all at once.
R configuration, S configuration, Cahn-Ingold-Prelog, CIP priority, chiral centre, stereogenic centre, enantiomer, diastereomer, meso compound, racemic mixture, optical activity, Fischer projection, specific rotation, inversion of configuration, retention of configuration, syn addition, anti addition, IR spectroscopy, infrared, C=O stretch, functional group identification, 1H NMR, proton NMR, chemical shift, integration, splitting, n+1 rule, coupling constant, 13C NMR, carbon NMR, DEPT, retrosynthetic analysis, retrosynthesis, disconnection approach, multi-step synthesis, Gabriel synthesis, phthalimide, azide reduction, Sandmeyer reaction, diazonium salt, terminal alkyne, acetylide, deuterium labelling, Orgo 2, organic chemistry II, CHEM 210