Build a conditional reaction library across substitution, elimination, pi-bond chemistry, alcohols, ethers and epoxides, radicals, aromatics, carbonyls, acid derivatives, and alpha-carbonyl reactions—then retrieve it from original questions.
Reconcile carbon sources, bond changes, charge state, regiochemistry, stereochemistry, and the evidence boundary.
The One-Step topic and named reaction families come from the ADA Organic Chemistry specification ↗. The ten-cluster teaching order and original questions are DAT TRAIN study tools, not ADA weights.
Ten connected reaction lessons
Keep the reaction as one conditional mapping.
For every family, state the required change before scanning reagents. The product is accepted only after atoms, bonds, product state, and selectivity reconcile.
01
LESSON 1 · 24 MIN
Substitution
Choose substitution from substrate and access
Distinguish bounded SN2 and SN1 mappings, select compatible nucleophile and solvent conditions, and audit carbon count and configuration.
ESSENTIAL QUESTIONCan the nucleophile reach the leaving-group carbon directly, or do the conditions support ionization first?
Substitution keeps the leaving-group carbon in view
C3H7Br+NaCNNaCN · DMSO→C4H7N+NaBr
The cyanide carbon becomes the fourth product carbon; it is not a spectator.
Substitution keeps the leaving-group carbon in view: product audit
Check
Before
After
Bond
C–Br
C–C≡N
Carbon count
3 in substrate + 1 in cyanide
4 in nitrile
Boundary
Primary electrophile
SN2-compatible product
ACCESSIBLE REACTION MODEL · COMPLETE EQUATION AND PRODUCT-AUDIT TABLE INCLUDED
01
Classify the reaction center
Methyl and primary centers favor accessible backside substitution; tertiary centers block SN2 and may ionize in a polar-protic medium.
Inspect the carbon bearing the leaving group
Do not classify from nucleophile name alone
02
Match partner and medium
Strong unhindered nucleophiles pair with polar-aprotic SN2 conditions, while weak neutral nucleophiles can capture a supported carbocation in solvolysis.
Nucleophile and solvent are one condition set
A strong base can introduce elimination competition
03
Audit the product
Substitution exchanges one sigma-bond partner. Track reagent-derived atoms, leaving-group loss, and inversion or loss of stereospecificity.
Recount carbon after cyanide addition
Reassign CIP labels after changing a substituent
Worked example
Convert 1-bromopropane to butanenitrile in one operation.
1
The target replaces C–Br with a C–C≡N bond and gains the cyanide carbon.
2
A primary bromide, CN⁻, and a polar-aprotic solvent form a compatible SN2 set.
3
The product audit finds four carbons, no bromine, and the nitrile attached at the original leaving-group carbon.
ConclusionNaCN in DMSO satisfies the required bond change and carbon ledger; a tertiary-solvolysis reagent set does not.
Close the notes first
Retrieve the conditional map.
01What substrate feature blocks SN2 most strongly?
Crowding at the leaving-group carbon, especially a tertiary center.
Backside approach must reach that carbon.
02What stereochemical event belongs to a clean SN2 step?
Inversion at the reacting stereogenic center.
Backside displacement reverses the local tetrahedral geometry.
03Why can cyanide change the product carbon count?
Its carbon becomes part of the product skeleton.
Reagent-derived atoms belong in the atom map.
02
LESSON 2 · 24 MIN
Elimination
Resolve substitution–elimination competition
Use substrate, base size, heat, and anti geometry to select E1 or E2 conditions and predict a bounded major alkene.
ESSENTIAL QUESTIONWhich accessible beta hydrogen can leave under the complete condition set?
Elimination removes two sigma partners and creates pi bond order
C4H9Br+C2H5ONaNaOEt · EtOH · heat→C4H8+C2H6O+NaBr
The displayed equation is an atom ledger; major-alkene choice still requires beta-site and anti-geometry analysis.
Elimination removes two sigma partners and creates pi bond order: product audit
Check
Before
After
Break
β-C–H and C–Br
H transferred to base; Br leaves
Form
C–C single bond
C=C double bond
Boundary
Strong small base
E2, anti geometry required
ACCESSIBLE REACTION MODEL · COMPLETE EQUATION AND PRODUCT-AUDIT TABLE INCLUDED
01
Separate E2 from E1
E2 requires a strong base and anti-periplanar C–H/C–LG geometry; E1 begins with supported ionization and is not stereospecific.
Strong base points toward E2
Polar-protic heat can support E1 for tertiary substrates
02
Compare beta sites
Small bases often favor the more substituted alkene, while bulky bases can favor the more accessible, less substituted alkene.
Name the base before invoking Zaitsev
Accessibility can overturn substitution preference
03
Respect ring geometry
Cyclohexane E2 requires a trans-diaxial hydrogen and leaving group in the reacting chair.
Ring-flip before deciding
Equatorial leaving group is not the reacting E2 geometry
Worked example
Predict the major alkene from 2-bromo-2-methylbutane with potassium tert-butoxide and heat.
1
The tertiary bromide cannot undergo SN2 and the reagent is a strong bulky base.
2
E2 removes the most accessible beta hydrogen while C–Br breaks.
3
That bond ledger forms the less substituted terminal alkene, 2-methyl-1-butene.
ConclusionThe bulky-base condition makes the Hofmann alkene the bounded major product; ‘most substituted always wins’ is not valid here.
Close the notes first
Retrieve the conditional map.
01What geometry must an E2 C–H and C–LG pair have?
Anti-periplanar.
The concerted orbital alignment is part of the E2 contract.
02What cyclohexane arrangement supplies that geometry?
A trans-diaxial beta hydrogen and leaving group.
Adjacent axial bonds can align anti.
03When can a bulky base change regioselectivity?
When it reaches a less hindered beta hydrogen more readily.
Access can favor the less substituted alkene.
03
LESSON 3 · 26 MIN
Alkenes and Alkynes
Read addition as a three-part code
Predict alkene and alkyne products by linking reagent identity to bond-order change, regiochemistry, stereochemistry, and reduction endpoint.
ESSENTIAL QUESTIONWhat adds, where does it add, from which face, and where must the multiple bond stop?
Partial alkyne reduction must stop at the requested alkene
C4H6+H2Lindlar catalyst→C4H8
The poisoned catalyst lowers one unit of alkyne bond order and delivers the cis alkene rather than the alkane.
Partial alkyne reduction must stop at the requested alkene: product audit
Check
Before
After
Bond order
C≡C
C=C
Hydrogen
One H₂ molecule
One H added to each alkyne carbon
Stereo
Planar surface delivery
Syn addition; cis alkene
ACCESSIBLE REACTION MODEL · COMPLETE EQUATION AND PRODUCT-AUDIT TABLE INCLUDED
01
Lower bond order deliberately
Addition consumes one unit of pi bond order and forms new sigma bonds; partial alkyne reduction must stop at C=C.
Record C=C or C≡C before and after
Do not let a poisoned catalyst reduce to alkane
02
Pair regio and stereo rules
Hydroboration–oxidation gives anti-Markovnikov alcohol placement with syn addition; bromination gives anti vicinal dibromide formation.
Regio asks where
Stereo asks how faces relate
03
Distinguish alkyne reducers
Lindlar catalyst delivers the cis alkene, while sodium in liquid ammonia gives the trans alkene.
Lindlar: syn and cis
Dissolving metal: anti and trans
Worked example
Stop 2-butyne at trans-2-butene.
1
The target lowers C≡C to C=C and adds exactly two hydrogens.
2
A trans endpoint requires anti delivery rather than a poisoned catalytic surface.
3
Sodium in liquid ammonia supplies the dissolving-metal reduction map.
ConclusionNa/NH₃(l) gives trans-2-butene; H₂/Lindlar would give the cis stereoisomer.
The oxygen-equivalent notation is an atom-balanced summary; the named mild anhydrous reagent controls the aldehyde endpoint.
Oxidation endpoint follows alcohol class and conditions: product audit
Check
Before
After
Carbon
Terminal CH₂OH
Terminal CHO
Hydrogen
C–H and O–H present
Two hydrogens removed
Boundary
Mild and anhydrous
Aldehyde stop
ACCESSIBLE REACTION MODEL · COMPLETE EQUATION AND PRODUCT-AUDIT TABLE INCLUDED
01
Classify the alcohol carbon
Primary alcohols can stop at aldehydes under mild anhydrous oxidation or continue to acids under strong aqueous oxidation; secondary alcohols give ketones.
Primary versus secondary first
Water content changes the primary endpoint
02
Activate before eliminating
Acid protonates OH so water can leave during dehydration; heat favors alkene formation.
OH alone is a poor leaving group
Track catalyst regeneration
03
Audit carbon and oxygen
Oxidation changes C–O bond order without adding carbon, while dehydration removes water and creates C=C.
Keep the carbon skeleton
Distinguish oxidation from elimination
Worked example
Convert 1-butanol to butanal without continuing to butanoic acid.
1
The target keeps four carbons and changes terminal CH₂OH to CHO.
2
A mild anhydrous oxidant is required to stop at the aldehyde.
3
Dess–Martin periodinane in dichloromethane matches that endpoint.
ConclusionThe reagent context—not the word oxidation alone—separates butanal from the stronger aqueous acid endpoint.
Close the notes first
Retrieve the conditional map.
01What does a secondary alcohol form on ordinary oxidation?
A ketone.
The alcohol carbon loses H while C–O becomes C=O.
02Why can a primary aldehyde be overoxidized in water?
The aqueous strong-oxidant pathway supports the carboxylic-acid endpoint.
Conditions control how far oxidation proceeds.
03What bond is created in dehydration?
A carbon–carbon double bond.
Loss of beta H and activated OH increases C–C bond order.
05
LESSON 5 · 25 MIN
Ethers and Epoxides
Choose the ether bond that can actually break
Plan Williamson synthesis and ether or epoxide cleavage by identifying the attacked carbon, ring strain, protonation state, and atom source.
ESSENTIAL QUESTIONWhich carbon is accessible to the nucleophile under the stated acid–base regime?
Epoxide opening accounts for the ring oxygen and the nucleophile
C3H6O+H2OH⁺ catalyst→C3H8O2
The simplified hydration equation shows the atom ledger; an unsymmetrical epoxide still needs condition-specific regiochemistry.
Epoxide opening accounts for the ring oxygen and the nucleophile: product audit
Check
Before
After
Ring
Two strained C–O bonds
One C–O ring bond cleaved
Oxygen
Epoxide O + water O
Two alcohol oxygens
Stereo
Nucleophile approaches opposite broken bond
Anti opening
ACCESSIBLE REACTION MODEL · COMPLETE EQUATION AND PRODUCT-AUDIT TABLE INCLUDED
01
Build ethers through SN2
Williamson synthesis pairs an alkoxide with a methyl or primary electrophile; a tertiary electrophile instead invites elimination.
Put crowding on the electrophile ledger
Join fragments through oxygen
02
Cleave the feasible side
Strong acid activates an ether, but backside attack occurs at methyl or suitable sp³ carbon—not an ordinary aromatic sp² carbon.
Preserve aryl–oxygen when methyl can be attacked
Track both cleavage fragments
03
Reverse epoxide regiochemistry with medium
Basic opening attacks the less substituted carbon; a tertiary-substituted protonated epoxide can direct a weak nucleophile toward the more substituted carbon. Both open anti.
Base: less substituted
Acidic tertiary case: more substituted
Worked example
Open 1,2-epoxypropane with methoxide in methanol.
1
The strong nucleophile operates under basic conditions.
2
It attacks the less substituted epoxide carbon from the backside and breaks that C–O ring bond.
3
Protonation leaves OH on C2 and OCH₃ on terminal C1.
ConclusionThe product is 1-methoxypropan-2-ol; its added methoxy carbon must appear in the carbon ledger.
Close the notes first
Retrieve the conditional map.
01What electrophile class is safest for Williamson synthesis?
Methyl or primary.
Those centers permit SN2 rather than strong-base elimination.
02Where does a strong nucleophile attack an unsymmetrical epoxide under basic conditions?
Backside attack occurs opposite the breaking C–O bond.
06
LESSON 6 · 23 MIN
Radical Reactions
Keep radical initiation and propagation separate
Recognize light- or initiator-dependent radical mappings, rank abstraction sites, and distinguish allylic substitution from addition across a double bond.
ESSENTIAL QUESTIONWhich bond is homolyzed, which radical is propagated, and what feature controls site choice?
Allylic bromination preserves the alkene
C6H10+Br2NBS · hν→C6H9Br+HBr
NBS/light replaces an allylic hydrogen; this is not the Br₂ addition product across C=C.
Allylic bromination preserves the alkene: product audit
Check
Before
After
Initiation
No radical chain
Light supplies radical initiation
Site
Allylic C–H
Allylic C–Br
Alkene
One C=C
One C=C retained
ACCESSIBLE REACTION MODEL · COMPLETE EQUATION AND PRODUCT-AUDIT TABLE INCLUDED
01
Require an initiation cue
Light, heat, or a peroxide can create radicals; fishhook bookkeeping and a propagation cycle distinguish the pathway from polar chemistry.
No radical without a source
Propagation must regenerate a chain carrier
02
Compare radical sites
Bromination is selective and favors formation of the more stable radical, while chlorination is less selective.
Major is not exclusive
Count available hydrogen environments
03
Protect the alkene ledger
NBS replaces an allylic H while retaining C=C; HBr with peroxide adds across C=C with anti-Markovnikov regiochemistry.
NBS: allylic substitution
HBr/ROOR: radical addition
Worked example
Predict cyclohexene with NBS and light.
1
Light supplies radical initiation and NBS maintains low bromine concentration.
2
An allylic hydrogen is abstracted to form a resonance-stabilized allylic radical.
3
Bromine replaces that allylic H while the ring alkene remains.
ConclusionThe bounded product is 3-bromocyclohexene, not a vicinal dibromide from addition across C=C.
Close the notes first
Retrieve the conditional map.
01What does NBS bromination preserve?
The alkene while replacing an allylic hydrogen.
Its chain reaction targets the allylic radical site.
02Which propane monobromination product is major?
2-bromopropane.
The secondary radical pathway is favored.
03What regiochemistry does peroxide-promoted HBr addition give?
Anti-Markovnikov.
The radical chain is stabilized when Br adds to the less substituted endpoint.
07
LESSON 7 · 25 MIN
Aromatic Reactions
Substitute the ring and restore aromaticity
Select electrophilic aromatic substitution conditions, preserve the aromatic ring, and use existing substituents to bound regiochemistry.
ESSENTIAL QUESTIONWhich electrophile is generated, and where can substitution occur before aromaticity is restored?
Aromatic substitution restores the ring after electrophile capture
C6H6+HNO3H₂SO₄→C6H5NO2+H2O
The net equation replaces ring hydrogen with nitro and restores aromaticity.
Aromatic substitution restores the ring after electrophile capture: product audit
Check
Before
After
Electrophile
HNO₃/H₂SO₄
Nitrating species generated
Ring
Aromatic benzene
Aromatic nitrobenzene
Net change
Aryl C–H
Aryl C–NO₂
ACCESSIBLE REACTION MODEL · COMPLETE EQUATION AND PRODUCT-AUDIT TABLE INCLUDED
01
Generate the electrophile
Lewis or Brønsted acid partners activate bromination, nitration, or acylation reagents; the aromatic ring then substitutes rather than simply adding.
Br₂ needs FeBr₃ in the standard map
HNO₃/H₂SO₄ generates the nitrating electrophile
02
Restore the ring
The sigma complex loses H+ to restore aromaticity, so the net change replaces aromatic C–H with C–E.
Substitution preserves the ring pi system
Do not leave a nonaromatic addition product
03
Apply directing effects conditionally
Methoxy directs ortho/para, and steric comparison can favor para, but exact ratios require data.
Name the existing substituent
Direction and activation are related but distinct
Worked example
Convert benzene to acetophenone without rearranging an alkyl carbocation.
1
The target replaces aryl H with an acetyl group and retains a carbonyl.
2
Acetyl chloride plus AlCl₃ generates a resonance-stabilized acyl electrophile.
3
Aqueous workup releases acetophenone after aromatic substitution.
ConclusionFriedel–Crafts acylation adds the two-carbon acetyl fragment while preserving the ring and carbonyl.
Close the notes first
Retrieve the conditional map.
01Why is benzene bromination substitution rather than addition?
Loss of H restores aromaticity after electrophile capture.
The net ring pi system is preserved.
02What activates Br₂ for the standard benzene reaction?
FeBr₃.
The Lewis acid helps generate the electrophilic brominating species.
03Where does methoxy direct nitration?
Ortho and para.
Lone-pair donation stabilizes those sigma-complex pathways.
08
LESSON 8 · 26 MIN
Aldehydes and Ketones
Map nucleophiles and redox at the carbonyl carbon
Choose aldehyde/ketone reduction, oxidation, carbon addition, or olefination and account for the carbonyl oxygen and every added carbon.
ESSENTIAL QUESTIONDoes the target keep C=O, reduce it, oxidize it, or replace it with a new carbon bond?
Carbonyl reduction retains oxygen and lowers bond order
C6H10O+H2NaBH₄ · then H₂O→C6H12O
Hydride adds at carbonyl carbon and the oxygen becomes OH after protonation; no carbon is added.
Carbonyl reduction retains oxygen and lowers bond order: product audit
Check
Before
After
Bond order
C=O
C–O
Hydrogen
Carbonyl carbon has no H
C–H and O–H added
Skeleton
Six carbons
Six carbons
ACCESSIBLE REACTION MODEL · COMPLETE EQUATION AND PRODUCT-AUDIT TABLE INCLUDED
01
Name the carbonyl endpoint
NaBH₄ reduces aldehydes or ketones to alcohols; aldehyde oxidation gives an acid; Wittig replaces C=O with C=C.
Write the target functional group first
Do not call every carbonyl reagent a reduction
02
Track carbon nucleophiles
A Grignard reagent forms a new C–C bond at the carbonyl carbon and is protonated only during a later aqueous workup.
Add the reagent carbon to the count
Keep water out until the addition is complete
03
Audit oxygen fate
Hydride reduction retains oxygen as OH, oxidation adds the acid OH relationship, and Wittig transfers oxygen to a phosphorus coproduct.
Follow oxygen explicitly
Product appearance alone is not a complete ledger
Worked example
Convert ethanal to 2-propanol using a one-carbon nucleophile.
1
The target adds one carbon and changes C=O to a secondary alcohol.
2
Methylmagnesium bromide forms the new C–C bond under dry ether conditions.
3
Hydronium workup protonates oxygen after, not before, carbon addition.
ConclusionThe ordered Grignard sequence reconciles all three product carbons and the secondary-alcohol endpoint.
Close the notes first
Retrieve the conditional map.
01What does NaBH₄ make from a ketone?
A secondary alcohol.
Hydride adds to carbonyl carbon and oxygen is protonated.
02When is water added to a Grignard reaction?
After carbon–carbon bond formation.
An earlier proton source would quench the reagent.
03What does a methylene Wittig reagent replace C=O with?
C=CH₂.
The ylide carbon becomes the new alkene carbon.
09
LESSON 9 · 25 MIN
Carboxylic Acids and Derivatives
Follow nucleophilic acyl substitution and product state
Predict acid-derivative substitution, hydrolysis, and reduction while retaining the acyl carbon, leaving-group fragment, charge state, and ordered workup.
ESSENTIAL QUESTIONWhich group leaves the acyl carbon, and what protonation state exists under the stated conditions?
Acyl substitution retains the carbonyl and exchanges the leaving group
C2H3OCl+C2H6Opyridine→C4H8O2+HCl
Alcoholysis joins the ethoxy oxygen to the acyl carbon; pyridine manages the acid ledger in the stated laboratory map.
Acyl substitution retains the carbonyl and exchanges the leaving group: product audit
Check
Before
After
Acyl bond
C–Cl
C–OEt
Carbonyl
C=O
C=O retained
Carbon
2 acyl + 2 alcohol
4 in ester
ACCESSIBLE REACTION MODEL · COMPLETE EQUATION AND PRODUCT-AUDIT TABLE INCLUDED
01
Attack the acyl carbon
Alcoholysis and aminolysis replace an acid-chloride leaving group while retaining the carbonyl.
Addition then elimination at the acyl center
Account for acid-neutralizing equivalents
02
Keep hydrolysis state honest
Base-promoted ester hydrolysis ends at carboxylate plus alcohol unless an acid workup is explicitly supplied.
Do not draw CO₂H in strong base
Track both ester fragments
03
Distinguish substitution from reduction
LiAlH₄ reduces the ester carbonyl and cleaves the acyl–alkoxy relationship, producing alcohols only after aqueous workup.
NaBH₄ is not the default ester reducer
Workup follows dry hydride delivery
Worked example
Hydrolyze ethyl ethanoate with aqueous NaOH and heat, with no acid workup.
1
Hydroxide attacks the acyl carbon and the ethoxy group leaves through the substitution ledger.
2
The two-carbon acyl fragment remains deprotonated under basic conditions.
3
The ethoxy fragment is protonated to ethanol.
ConclusionProducts are sodium ethanoate and ethanol; neutral ethanoic acid would require an added acid workup.
Close the notes first
Retrieve the conditional map.
01What bond changes in acid-chloride alcoholysis?
Acyl C–Cl is replaced by acyl C–OR.
The carbonyl is retained during nucleophilic acyl substitution.
02What is the carboxylic product state after saponification without acid workup?
Carboxylate salt.
The medium is basic.
03What reduces an ester to primary alcohol products in this library?
LiAlH₄ followed by aqueous workup.
The strong hydride and ordered workup reach the alcohol endpoint.
10
LESSON 10 · 27 MIN
Alpha-Carbonyl Chemistry
Treat the alpha carbon as a condition-controlled nucleophile
Choose alpha halogenation, enolate alkylation, aldol addition, or condensation and audit carbon sources, carbonyl retention, and dehydration state.
ESSENTIAL QUESTIONWhich alpha hydrogen is removed, what electrophile is captured, and does the product stop before or after dehydration?
Aldol addition joins two carbonyl fragments before dehydration
2C2H4Odilute OH⁻ · cold⇌C4H8O2
Cold conditions stop at the beta-hydroxy aldehyde; heat would change the endpoint through dehydration.
Aldol addition joins two carbonyl fragments before dehydration: product audit
Check
Before
After
Carbon
2 + 2
4
New bond
Separate ethanal molecules
Alpha C–carbonyl C bond
Endpoint
Cold dilute base
Beta OH retained
ACCESSIBLE REACTION MODEL · COMPLETE EQUATION AND PRODUCT-AUDIT TABLE INCLUDED
01
Generate the alpha nucleophile
Enol or enolate formation makes the alpha carbon nucleophilic; acid/base conditions and reagent order determine which model applies.
Alpha means adjacent to C=O
Base first, electrophile second for directed alkylation
02
Separate substitution from condensation
Alpha halogenation or alkylation replaces one alpha H; aldol chemistry creates a C–C bond between carbonyl partners.
Retain the carbonyl in alpha substitution
Count both carbonyl reactants in aldol
03
Stop at the stated endpoint
Cold dilute base supports beta-hydroxy aldol addition, while heat can eliminate water to an alpha,beta-unsaturated carbonyl.
Cold: beta OH retained
Heat: conjugated C=C after dehydration
Worked example
Differentiate ethanal aldol addition from condensation.
1
Two ethanal molecules supply four total product carbons through a new alpha-to-carbonyl C–C bond.
2
Cold dilute base stops at 3-hydroxybutanal with a beta OH.
3
Heating promotes water loss and forms conjugated but-2-enal.
ConclusionTemperature and dehydration evidence distinguish two valid endpoints; they are not interchangeable answer drawings.
Close the notes first
Retrieve the conditional map.
01Where is the alpha carbon?
Directly adjacent to the carbonyl carbon.
Its hydrogen can be removed to form an enol or enolate.
02What must happen before methyl iodide in directed enolate alkylation?
The base must form the enolate.
Reagent order protects the intended carbon nucleophile.
03What feature distinguishes aldol condensation from addition?
Loss of water and formation of a conjugated C=C bond.
The beta-hydroxy group is removed during dehydration.
Randomized reaction retrieval
Choose the complete mapping.
All forty rules are represented. Questions and five-choice answer sets shuffle without detaching the validated substrate, condition, product, or selectivity contract.
40 PRACTICE QUESTIONS
Retrieve before you review.
Question order and all five answer options are shuffled when you begin. The correct answer stays attached to the same underlying choice.
Scope and score notice
One operation, bounded evidence.
This route covers the official One-Step synthesis topic through ten explicit reaction clusters. The ADA does not publish cue-level quotas, so DAT TRAIN does not invent them.
The validators enforce conditional reaction identity, required workups, carbon-source accounting, bond-change signatures, and stated selectivity. They do not establish laboratory yield, safety, a unique mechanism, or multistep route feasibility.