Difficulty: Advanced | Prerequisites: Resonance, acid-base chemistry, reaction mechanisms
Stereochemistry questions ask you to recognise the relationship between two structures (enantiomers, diastereomers, conformers, identical, or geometric isomers) and to predict the stereochemical outcome of reactions. Synthesis questions chain multiple reactions together, so you need a mental toolkit of reagents and what each one does to a functional group. The ability to work backwards from the product to the starting material is the single most useful skill for synthesis problems.
Conformers (conformational isomers)
Structures that differ only by rotation around a single bond. They are the same compound at different rotational angles. In simple terms, if you could twist one into the other without breaking any bonds, they are conformers.
Enantiomers
Non-superimposable mirror images. They have opposite configurations at every stereocentre (all R becomes S, all S becomes R). They have identical physical properties except for the direction they rotate plane-polarised light.
Diastereomers (diastereoisomers)
Stereoisomers that are not mirror images. They differ at some but not all stereocentres. Unlike enantiomers, diastereomers have different physical properties (melting point, boiling point, solubility).
Meso compound
A molecule with stereocentres but an internal plane of symmetry, making it achiral overall. Think of it as a molecule that looks like it should be chiral but cancels itself out.
Geometric isomers (cis/trans or E/Z)
Isomers that differ in the arrangement of groups around a double bond or ring. Cis = same side; trans = opposite sides.
Syn addition
Both new groups add to the same face of the double bond. OsO4 dihydroxylation is the classic syn addition.
Anti addition
The two new groups add to opposite faces of the double bond. Bromine addition (Br2) to an alkene is anti addition (via a bromonium ion intermediate).
Grignard reagent (RMgX)
An organomagnesium halide that acts as a strong nucleophilic carbon source. It attacks carbonyls (aldehydes, ketones, esters) to form new C-C bonds. After aqueous workup, you get an alcohol.
PCC (pyridinium chlorochromate)
A mild oxidising agent that converts primary alcohols to aldehydes (stops there, does not over-oxidise to carboxylic acids) and secondary alcohols to ketones.
OsO4 (osmium tetroxide)
A reagent for syn dihydroxylation of alkenes. Adds two OH groups to the same face of the double bond, giving a cis-diol.
mCPBA (meta-chloroperoxybenzoic acid)
A peracid used for epoxidation of alkenes. Converts a double bond into an epoxide (a three-membered ring containing an oxygen). The reaction is stereospecific: cis alkenes give cis epoxides and trans alkenes give trans epoxides.
NaBH4 (sodium borohydride)
A mild reducing agent that converts aldehydes and ketones to alcohols. It delivers hydride (H-) to the carbonyl carbon.
Tosylate (OTs)
The tosyl group (p-toluenesulfonyl) makes an excellent leaving group when attached to an oxygen. TsCl converts an alcohol's OH into OTs, turning a poor leaving group into a good one without changing the carbon skeleton or stereochemistry at that carbon.
EAS directing effects
Substituents already on an aromatic ring control where the next electrophile attaches. Electron-donating groups (OH, NH2, alkyl) are ortho/para directors. Electron-withdrawing groups (NO2, carbonyl) are meta directors. Halogens are deactivating but ortho/para directing.
The exam gives you pairs of structures and asks you to pick the best description of their relationship. Here is a systematic approach:
Check if the molecular formula is the same. If not, they are not isomers at all (different compounds).
Check connectivity. If the atoms are connected differently, they are constitutional (structural) isomers, not stereoisomers.
Check for rotation around single bonds. If you can convert one structure into the other purely by rotating around sigma bonds, they are conformers.
Check all stereocentres. Assign R/S to every stereocentre in both structures.
If every stereocentre is inverted (all R become S, all S become R), the structures are enantiomers.
If some but not all stereocentres are inverted, they are diastereomers.
If all stereocentres have the same configuration, they are identical. Check for an internal plane of symmetry; if present, the compound is meso.
Check for cis/trans isomerism around a double bond or ring. If they differ only in the arrangement around a C=C or a ring, they are geometric isomers (a subtype of diastereomers).
Pair 1: Two dibromobicyclo compounds (1,2-dibromocyclohexane derivatives)
When two substituents on a ring can be interconverted by a ring flip (chair-chair interconversion), the structures are conformers. If one has both Br atoms axial and the other has both equatorial, that is a conformational change, not a configurational one.
Answer: A (Conformers).
Pair 2: Two structures of 2,3-dibromobutane shown in Fischer-style projection
Assign R/S at each stereocentre. If the first structure is (2R,3R) and the second is (2S,3S), every centre is inverted, making them enantiomers. If one is (2R,3S) and the other is (2S,3R), and there is an internal mirror plane, both are the same meso compound.
Check the specific configurations shown. If both stereocentres flip, the answer is B (Enantiomers). If there is a plane of symmetry, the answer is D (Identical, meso).
Pair 3: Two structures of 2-butanol drawn as wedge-dash
If one shows (R)-2-butanol and the other shows (S)-2-butanol, they are enantiomers. If both show the same configuration, they are identical.
Answer: depends on the specific drawings. Assign R/S to determine the relationship.
OsO4 followed by NaHSO3/H2O adds two OH groups to the same face of the double bond (syn addition). For trans-2-pentene (CH3CH2-CH=CH-CH3, E configuration), both OH groups add from the same side.
Because the reaction creates two new stereocentres and the addition is syn, you get a pair of enantiomers (the reagent can add to either face of the double bond). Both products are the threo diol.
If the starting alkene were cis (Z), syn addition would give the meso diol (an internal plane of symmetry appears).
A Grignard reagent (RMgX) attacks a ketone from either face of the planar carbonyl. After aqueous workup, you get a tertiary alcohol.
For CH3CH2-C(=O)-CH(CH3)2 + CH3MgBr: the methyl nucleophile attacks the ketone carbonyl, and after H3O+ workup you get a tertiary alcohol with four different groups on the central carbon. If the carbon becomes a stereocentre, you get a racemic mixture (equal amounts of R and S) because the Grignard can attack from either face.
When HBr adds to an enol ether (a vinyl ether with OCH3 on the double bond), the proton adds to the less substituted carbon (Markovnikov, because the resulting carbocation is stabilised by the adjacent oxygen lone pairs). The Br then attacks the carbocation.
For a cyclohexene ring bearing OCH3 and CH3, protonation gives the most stable carbocation (consider which position puts the positive charge adjacent to oxygen for stabilisation). The bromide then attacks.
mCPBA converts an alkene to an epoxide. The reaction is stereospecific: cis alkenes give cis epoxides, trans alkenes give trans epoxides. For a cyclic alkene like cyclohexene, mCPBA gives the epoxide (a three-membered ring fused onto the cyclohexane ring). Both faces of the alkene are attacked, so you get a racemic mixture of enantiomeric epoxides.
Epoxides open under acidic conditions (H3O+) with anti stereochemistry: the nucleophile (water) attacks the more substituted carbon from the opposite side of the oxygen. This gives a trans-1,2-diol.
When followed by an oxidising agent like NaCr2O7, any alcohol can be further oxidised: secondary alcohols become ketones, primary alcohols become carboxylic acids.
PCC: primary alcohol to aldehyde, secondary alcohol to ketone. Stops at aldehyde for primary (no over-oxidation).
NaCr2O7 or Jones reagent: primary alcohol all the way to carboxylic acid, secondary alcohol to ketone.
NaBH4: aldehyde or ketone to alcohol (reduction).
Grignard (RMgX) + aldehyde, then H3O+: secondary alcohol.
Grignard (RMgX) + ketone, then H3O+: tertiary alcohol.
Grignard (RMgX) + ester: tertiary alcohol (two equivalents of R add).
TsCl: converts OH to OTs (a good leaving group) for subsequent SN2 with nucleophiles like NaCN, NaN3, etc.
Mg in Et2O: converts an alkyl halide (RBr) into a Grignard reagent (RMgBr).
HBr: electrophilic addition to alkenes (Markovnikov). Also cleaves ethers.
Br2: adds across double bonds (anti addition via bromonium ion). With light, radical substitution at benzylic or allylic positions.
NaNH2: a strong base that deprotonates terminal alkynes, also promotes E2 eliminations.
KOH/heat (or NaOH/heat): E2 elimination from alkyl halides to give alkenes. With excess strong base, double elimination gives alkynes.
Look at the product and ask: what bond was formed last? What functional group was the immediate precursor? Then work backwards step by step to the starting material.
Target: CH3CH2CH2-O-C(=O)-CH2CH3 (propyl propanoate, an ester).
Starting material: propene (CH3CH=CH2). Only organic reagents of two carbons or fewer allowed.
Retrosynthetic thinking:
An ester comes from an alcohol + a carboxylic acid (or acid chloride). So we need propanol and propanoic acid (or propanoyl chloride).
Propanol from propene: anti-Markovnikov hydration (hydroboration-oxidation: BH3/THF then H2O2/NaOH) gives 1-propanol. Or Markovnikov hydration (H2SO4/H2O) gives 2-propanol, which is not what we want here.
Propanoic acid from propanol: oxidise 1-propanol with a strong oxidant (NaCr2O7 or KMnO4) to get propanoic acid.
But we need two units of propanol (one for the alcohol part, one for the acid part). Since we start from propene, we can run two parallel sequences.
Ester formation: propanoic acid + SOCl2 gives propanoyl chloride, then propanoyl chloride + 1-propanol gives propyl propanoate.
Target: a benzene ring with a methyl ketone (COCH3) at one position and a methyl ester (CO2CH3) at another.
Starting material: PhCH2OH (benzyl alcohol).
Retrosynthetic thinking:
The methyl ester can come from oxidation of the CH2OH to a carboxylic acid (PhCO2H) then treatment with CH3OH/H+ (Fischer esterification) or via the acid chloride + CH3OH.
The methyl ketone requires Friedel-Crafts acylation: CH3COCl/AlCl3 on the ring.
Order matters: Friedel-Crafts does not work on strongly deactivated rings. The ester group is moderately deactivating, so it may be better to do the acylation before installing the ester, or to plan the steps carefully with directing effects in mind.
Several reaction chains appear in Question 6. Here are the key patterns:
Alcohol to aldehyde to secondary alcohol: CH3CH2CH2OH + PCC gives propanal (CH3CH2CHO). Then CH3MgX + propanal, followed by H2O workup, gives a secondary alcohol (4-carbon chain, 2-butanol).
Continuation with tosylation and cyanide: The secondary alcohol + TsCl gives the tosylate. Then NaCN (SN2 displacement) gives the nitrile, with inversion of configuration.
Alkene to Grignard to tertiary alcohol: An alkene + HBr gives an alkyl bromide (Markovnikov). The alkyl bromide + Mg/Et2O gives the Grignard. The Grignard + a ketone (CH3COCH3), then H2O, gives a tertiary alcohol.
Epoxidation and ring-opening chain: Alkene + mCPBA gives epoxide. Epoxide + H3O+/H2O opens the ring (anti, at the more substituted carbon) to give a trans-diol. The diol + NaCr2O7 oxidises the secondary alcohol to a ketone.
Tosylation then acetate displacement: Alcohol + TsCl gives tosylate. Tosylate + CH3CO2Na (sodium acetate) gives the ester via SN2.
Reduction: The resulting ketone + NaBH4 reduces back to a secondary alcohol.
Benzyl chloride to phenylacetic acid derivative: PhCH2Cl + NaCN gives PhCH2CN (SN2). Hydrolysis with H3O+/H2O gives PhCH2CO2H (phenylacetic acid). Then SOCl2 gives the acid chloride. The acid chloride + PhOH/pyridine gives the phenyl ester.
Styrene to alkyne: PhCH=CH2 + Br2 gives the vicinal dibromide. Double elimination with KOH/heat (excess) gives PhC triple-bond CH. Treatment with NaNH2 (deprotonation of the terminal alkyne to form the acetylide) then CH3Br (or CH3I) gives PhC triple-bond CCH3.
For Question 6d: O2N-C6H4 + CH3Cl/AlCl3 then Br2/light.
Step 1: Friedel-Crafts alkylation. The NO2 group is a meta director, so the CH3 goes to the meta position, giving 3-nitrotoluene.
Step 2: Br2 with light is radical (benzylic) bromination, not EAS. The bromine replaces a hydrogen on the methyl group (benzylic position), giving 3-nitrobenzyl bromide.
For Question 6, part with Mg/Et2O then CO2 then H2O:
The Grignard reagent (from the alkyl bromide + Mg) reacts with CO2 to give a carboxylate salt. Acidic workup gives the carboxylic acid.
Then SOCl2 converts the acid to the acid chloride.
The acid chloride + 2 CH3NH2 gives the N-methylamide (one equivalent of amine reacts with the acid chloride, the second equivalent acts as a base to neutralise the HCl produced).
Students often confuse enantiomers and diastereomers. The rule is simple: if every stereocentre is inverted, they are enantiomers. If only some are inverted, they are diastereomers.
A meso compound is not "optically inactive because it has no stereocentres." It does have stereocentres. It is optically inactive because its internal symmetry cancels out the rotation from each centre.
Students frequently forget that OsO4 gives syn addition while Br2 gives anti addition. This distinction is critical for predicting stereochemistry.
In synthesis problems, students often try to do Friedel-Crafts on a ring that bears a strongly deactivating group (NO2, for instance). Friedel-Crafts does not work on deactivated rings. Plan the order of steps accordingly.
⚠️ Stereochemical relationships (Question 4) are worth 6% and are pure pattern recognition. Practise assigning R/S quickly.
⚠️ Product prediction with stereochemistry (Question 5) is worth 14%. Syn vs. anti addition and the possibility of racemic mixtures are the usual traps.
⚠️ Multi-step reaction sequences (Questions 6 and 7) are worth a combined 24% of the exam. Knowing your reagent toolkit cold is essential.
⚠️ When the exam says "pay attention to possible stereoisomers," it is telling you that stereochemistry matters for full marks. Always consider whether the product has stereocentres and whether you get one enantiomer, a racemic mixture, or a meso compound.
True or false: OsO4 dihydroxylation gives anti addition of two OH groups. (False: it gives syn addition.)
Fill in the blank: A Grignard reagent attacking a ketone from either face gives a ______ mixture of the alcohol product. (Racemic.)
True or false: TsCl changes the configuration at the carbon bearing the OH group. (False: TsCl only replaces H in OH with OTs; the C-O bond is not broken, so configuration is retained.)
Fill in the blank: PCC oxidises a primary alcohol to an ______, while NaCr2O7 oxidises it all the way to a ______. (Aldehyde; carboxylic acid.)
True or false: Friedel-Crafts alkylation works well on nitrobenzene. (False: the NO2 group is strongly deactivating and Friedel-Crafts fails on deactivated rings.)
Q: Draw the product(s) of trans-2-pentene treated with OsO4/pyridine then NaHSO3/H2O. Assign stereochemistry.
A: Syn addition of two OH groups to either face of the trans double bond gives a pair of enantiomers: (2R,3R)- and (2S,3S)-pentane-2,3-diol (the threo diol). Both are formed in equal amounts (racemic mixture).
Q: What is the product of CH3CH2C(=O)CH(CH3)2 + CH3MgBr, followed by H3O+?
A: A racemic mixture of a tertiary alcohol: 2-methyl-3-ethyl-2-butanol (the CH3 from the Grignard adds to the carbonyl carbon). The product has a stereocentre and is formed as a 50:50 mixture of R and S.
Q: Outline a synthesis of propyl propanoate from propene, using only inorganic reagents and organic reagents of two carbons or fewer.
A: Step 1: Propene + BH3/THF then H2O2/NaOH gives 1-propanol (anti-Markovnikov hydration). Step 2: Oxidise a portion of 1-propanol with NaCr2O7 to give propanoic acid. Step 3: Propanoic acid + SOCl2 gives propanoyl chloride. Step 4: Propanoyl chloride + 1-propanol gives propyl propanoate.
Q: PhCH2Cl is treated with NaCN, then H3O+/H2O, then SOCl2, then PhOH/pyridine. Give the product at each stage.
A: Stage 1: PhCH2CN (benzyl cyanide, SN2 displacement). Stage 2: PhCH2CO2H (phenylacetic acid, nitrile hydrolysis). Stage 3: PhCH2COCl (phenylacetyl chloride). Stage 4: PhCH2CO2Ph (phenyl phenylacetate, nucleophilic acyl substitution with phenol as nucleophile and pyridine as base).
Stereochemistry connects to every mechanism: SN2 gives inversion, SN1 gives racemisation, syn addition preserves cis relationships, anti addition inverts them. The reagent toolkit here is also the toolkit for carbonyl chemistry and aromatic chemistry, so these notes overlap heavily with the reaction mechanisms document.
Acid-base chemistry (from the first set of notes) determines which species is the nucleophile, which is the leaving group, and whether a reaction is thermodynamically favoured.
Stereoisomers, enantiomers, diastereomers, meso compound, conformers, geometric isomers, cis-trans isomerism, E/Z nomenclature, R/S configuration, syn addition, anti addition, racemic mixture, OsO4 dihydroxylation, mCPBA epoxidation, Grignard reaction, PCC oxidation, Jones oxidation, NaBH4 reduction, tosylation, SN2 displacement, retrosynthetic analysis, Friedel-Crafts, EAS directing effects, multi-step synthesis, CHEM 2700