Move from molecular polarity and phase behavior through chromatography, extraction, recrystallization, and distillation, then constrain structures with IR, proton NMR, carbon NMR, and multi-spectrum inference.
Record geometry, phases, pH, density, distances, temperatures, bands, signals, integration, and absences.
02Operate
Use the one permitted model: vector sum, ratio, state, rank, recovery, or environment count.
03Intersect
Require every selected conclusion or candidate to satisfy the complete positive and negative evidence set.
04Bound
Do not turn a relative rank into an exact property, enrichment into purity, or a functional group into a unique structure.
The three official topic lines and named cues come from the ADA Organic Chemistry specification ↗. The nine-lesson order and original questions are DAT TRAIN study tools, not ADA weights.
Nine connected evidence lessons
Carry the conditions into the conclusion.
Every lesson includes a complete text or data equivalent. Spectrum visuals are generated from validated signal data and repeat every position, shape, multiplicity, and integration value in tables.
01
LESSON 1 · 19 MIN
Polarity
Add bond dipoles as vectors
Combine bond polarity with supplied three-dimensional geometry to decide whether molecular dipoles cancel.
ESSENTIAL QUESTIONWhich bond-dipole components cancel, and which remain after the geometry is respected?
Bond vectors become one molecular resultant
Bond vectors become one molecular resultant: complete evidence ledger
Evidence
Observation
Supported conclusion
Linear CO₂
Equal C═O vectors oppose at 180°
Zero net vector
Bent H₂O
One component cancels; another reinforces
Nonzero net vector
Asymmetric CH₃Cl
C–Cl contribution is not canceled
Qualitatively polar
The ledger preserves direction and symmetry so polar bonds are not mistaken for an automatically polar molecule.
ACCESSIBLE EVIDENCE MODEL · COMPLETE DATA OR SIGNAL TABLE INCLUDED
01
Separate bond from molecule
A polar bond contributes a vector; molecular polarity is the vector sum of every relevant contribution.
Mark bond direction
Do not stop at electronegativity
02
Use geometry
Linear, trigonal-planar, tetrahedral, and bent arrangements can reinforce or cancel otherwise similar bond dipoles.
Equal and opposite can cancel
Asymmetry often leaves a resultant
03
Keep the conclusion bounded
A qualitative structure can support zero versus nonzero dipole without supplying an exact experimental magnitude.
Classify before quantifying
Distortion needs a new vector sum
Worked example
Why can linear CO₂ be nonpolar even though each C═O bond is polar?
1
Assign one vector to each C═O bond.
2
Place the equal vectors 180° apart in the linear geometry.
3
Add the vectors and obtain a zero resultant.
ConclusionPolar bonds do not force a polar molecule when symmetry cancels their vector contributions.
Close the notes first
Retrieve the evidence rule.
01What must be combined with bond polarity to determine molecular polarity?
Three-dimensional molecular geometry.
Geometry fixes the direction of each bond-dipole vector.
02Can equal polar bonds cancel?
Yes, when symmetry makes their vectors sum to zero.
Dipoles add as vectors rather than simple scalars.
03What can a qualitative model safely conclude?
Whether the net dipole is zero or nonzero under the supplied geometry.
An exact value requires data or a validated calculation.
02
LESSON 2 · 21 MIN
Phase properties
Separate cohesion from packing
Rank bounded boiling or melting behavior using dispersion, permanent dipoles, hydrogen bonding, size, shape, and supplied packing evidence.
ESSENTIAL QUESTIONIs the comparison controlled by liquid cohesion, molecular contact, or solid packing?
Choose the property before choosing the dominant evidence
Choose the property before choosing the dominant evidence: complete evidence ledger
Boiling comparisons emphasize vaporization and liquid cohesion; melting comparisons must also carry solid-packing evidence.
ACCESSIBLE EVIDENCE MODEL · COMPLETE DATA OR SIGNAL TABLE INCLUDED
01
Inventory attractions
All molecules disperse; polar molecules add dipole interactions, and suitable donors plus acceptors can self-associate by hydrogen bonding.
Dispersion is never absent
Donor and acceptor roles are distinct
02
Control the comparison
Matched formulas or homologous series let size, shape, or hydrogen bonding be isolated more defensibly.
Name the controlled variables
Mass alone is not the mechanism
03
Treat melting separately
Melting depends strongly on crystal symmetry and packing, so the boiling order cannot simply be copied.
Liquid cohesion versus solid lattice
Do not promise exact temperatures
Worked example
Compare ethanol with dimethyl ether.
1
Hold molecular formula and size approximately constant.
2
Note that both disperse and are polar, but only ethanol has both an O–H donor and an acceptor.
3
Use ethanol self-hydrogen-bonding to support the higher boiling point.
ConclusionThe claim follows from a controlled interaction difference, not from formula or molar mass alone.
Close the notes first
Retrieve the evidence rule.
01Which intermolecular force acts in every molecular sample?
Dispersion forces.
Temporary dipoles occur in all electron clouds.
02Why can branching lower boiling point in a matched isomer pair?
It can reduce intermolecular contact surface.
Less contact can weaken cumulative dispersion.
03Why might melting order differ from boiling order?
Solid packing and symmetry can change independently of liquid cohesion.
The two phase changes probe different organization.
03
LESSON 3 · 21 MIN
Solubility
Carry solvent and protonation state
Predict a bounded solubility or phase preference from polarity, framework size, hydrogen bonding, ionic state, solvent, pH, and supplied pKa.
ESSENTIAL QUESTIONWhat form of the solute exists under these exact solvent and pH conditions?
Solubility begins with the actual solution state
Solubility begins with the actual solution state: complete evidence ledger
Evidence
Observation
Supported conclusion
Benzoic acid, pH 12
pH far above pKa 4.2
Benzoate; aqueous affinity enhanced
Aniline, pH 2
pH below anilinium pKa 4.6
Anilinium; aqueous affinity enhanced
Neutral naphthalene
Large nonpolar framework; no ionization
Organic phase favored, not exclusive
The same carbon skeleton can partition differently after protonation or deprotonation changes its charge.
ACCESSIBLE EVIDENCE MODEL · COMPLETE DATA OR SIGNAL TABLE INCLUDED
01
Compare displaced interactions
Dissolution must replace solute–solute and solvent–solvent attractions with sufficiently favorable solute–solvent interactions.
Name the solvent
Use the whole framework
02
Resolve acid–base state
A pH far from pKa can make an organic acid anionic or an amine cationic, often enhancing aqueous affinity.
Acid above pKa deprotonates
Base below conjugate-acid pKa protonates
03
Avoid absolute language
A phase preference is not zero solubility, complete extraction, or a quantitative distribution coefficient.
Dominant is not exclusive
Temperature and composition matter
Worked example
Place benzoic acid in aqueous base at pH 12.
1
Compare pH 12 with the supplied pKa 4.2.
2
Identify benzoate as the dominant anionic state.
3
Use charge hydration to support enhanced aqueous affinity.
ConclusionSolubility changes because protonation state changes, not because the carbon skeleton disappears.
Close the notes first
Retrieve the evidence rule.
01What happens to a carboxylic acid far above its pKa?
It is predominantly deprotonated.
The conjugate base is favored at high pH.
02Does one hydroxyl group guarantee high water solubility?
No.
A large nonpolar framework can dominate the balance.
03What does 'organic-phase favored' not mean?
It does not mean literally zero water solubility.
Partitioning is generally nonexclusive.
04
LESSON 4 · 22 MIN
Chromatography
Read retention only after naming the phases
Calculate condition-specific retention factors and interpret normal- or reverse-phase retention without overclaiming identity.
ESSENTIAL QUESTIONWhich phase does the analyte prefer under this exact chromatographic method?
Retention has a numerator, a denominator, and a method
Retention has a numerator, a denominator, and a method: complete evidence ledger
Evidence
Observation
Supported conclusion
TLC spot 3.0 cm
Solvent front 6.0 cm
Rf = 0.50
Normal-phase silica
Benzyl alcohol more polar than anisole
Benzyl alcohol retained more strongly
Reverse phase
Nonpolar stationary phase
The normal-phase polarity order cannot be copied
A measured ratio is exact for one run; polarity and identity conclusions remain conditional on the phase system and standards.
ACCESSIBLE EVIDENCE MODEL · COMPLETE DATA OR SIGNAL TABLE INCLUDED
01
Calculate the ratio
TLC Rf divides spot travel by solvent-front travel measured from the same origin on one plate.
Spot over front
Rf lies between zero and one
02
Name the mode
Polar silica in normal phase retains more-polar analytes; a nonpolar reverse-phase stationary phase changes that ordering logic.
Stationary phase first
Mode can reverse the trend
03
Limit identity claims
One retention value can support comparison, purity, or reaction-monitoring evidence but not unique identity without a valid standard or corroboration.
Co-spot when appropriate
One number is not a fingerprint
Worked example
A TLC spot travels 3.0 cm while the front travels 6.0 cm.
1
Confirm both distances use the same origin.
2
Divide 3.0 by 6.0.
3
Report Rf 0.50 only for the stated plate and solvent.
ConclusionThe calculation is exact for the measurements; the identity conclusion remains limited.
Close the notes first
Retrieve the evidence rule.
01How is TLC Rf calculated?
Spot distance divided by solvent-front distance.
Both distances are measured from the same origin.
02What retains strongly on normal-phase silica?
The more-polar analyte in the bounded comparison.
It interacts more strongly with the polar stationary phase.
03Can one matching Rf prove identity?
No.
Different compounds can share retention under one condition.
05
LESSON 5 · 24 MIN
Extraction + crystals
Separate phase choice from recovery
Locate extraction layers from density, move compounds by acid–base state, and audit the hot/cold solubility limit of recrystallization.
ESSENTIAL QUESTIONWhere is the target now, and how much can the stated equilibrium recover?
Locate the layer, then audit the recovery ceiling
Targetcircle marker · solid lineSoluble impuritytriangle marker · dashed line
The graph and ledger distinguish density-controlled layer position from temperature-controlled theoretical crystal recovery.Open the complete data table
Early fractions are enriched rather than automatically pure, and an azeotrope remains a boundary even with more fractionation.
ACCESSIBLE EVIDENCE MODEL · COMPLETE DATA OR SIGNAL TABLE INCLUDED
01
Rank volatility
The early vapor is enriched in the lower-boiling, more-volatile component—not necessarily the lighter molecule.
Use boiling evidence
Enriched is not pure
02
Select the column
A single volatile component or wide supplied gap can suit simple distillation; closer volatile components need more fractionation cycles.
Use the stated threshold
Fractional means more equilibration
03
Respect limits
A supplied azeotrope can limit ordinary distillation even with more theoretical plates.
Do not promise complete separation
State the supplied boundary
Worked example
Choose for cyclohexane and methylcyclohexane using a 25 °C threshold.
1
Identify two volatile components.
2
Calculate a 20.2 °C boiling-point gap.
3
Because the gap is below the supplied threshold, choose fractional distillation and expect early cyclohexane enrichment.
ConclusionThe course threshold is an explicit decision aid, not a universal phase-transition law.
Close the notes first
Retrieve the evidence rule.
01Which component enriches the early vapor?
The lower-boiling, more-volatile component.
It contributes more strongly to the vapor phase.
02Why use a fractionating column?
To create repeated vaporization–condensation cycles.
Those cycles improve separation of close-boiling components.
03Can ordinary fractionation always reach purity?
No, especially when an azeotropic limit is supplied.
The vapor and liquid composition relationship can impose a limit.
07
LESSON 7 · 23 MIN
Infrared
Use IR as positive and negative evidence
Match major IR absorption positions and shapes to bounded functional-group hypotheses while using missing expected bands to reject candidates.
ESSENTIAL QUESTIONWhich bond hypothesis explains both the bands that appear and the diagnostic bands that do not?
IR tests a functional-group hypothesis
A strong carbonyl-region band plus the absence of broad O–H evidence supports the bounded ketone hypothesis.Open the complete signal table
Bounded ketone infrared evidence: source signals
Signal
Position
Relative intensity
Shape
Multiplicity
Integration
C═O
1715 cm⁻¹
92
sharp
Not assigned
Not assigned
C–H
2960 cm⁻¹
42
sharp
Not assigned
Not assigned
IR tests a functional-group hypothesis: complete evidence ledger
Evidence
Observation
Supported conclusion
1715 cm⁻¹
Strong sharp absorption
Carbonyl-region evidence
3200–3600 cm⁻¹
No broad band
Alcohol O–H hypothesis disfavored
Complete IR only
No formula or NMR supplied
Functional group, not unique connectivity
A strong carbonyl-region band plus the absence of broad O–H evidence supports the bounded ketone hypothesis.
ACCESSIBLE EVIDENCE MODEL · COMPLETE DATA OR SIGNAL TABLE INCLUDED
01
Read region and shape
A broad O–H region, strong carbonyl region, or sharp nitrile region carries different bond-vibration evidence.
Position plus shape
Use adequate spectrum quality
02
Use absences
When a candidate requires a diagnostic band and the adequate spectrum lacks it, the absence can disconfirm that candidate.
Expected but missing matters
Check the whole candidate
03
Stop at functional groups
IR generally constrains bond and functional-group hypotheses; it rarely proves one complete connectivity by itself.
Group before identity
Combine techniques for structure
Worked example
Interpret a strong 1715 cm⁻¹ band with no broad O–H band.
1
Place 1715 cm⁻¹ in the carbonyl region.
2
Use the missing broad O–H feature to reject alcohol and acid candidates.
3
Within the bounded list, retain the ketone hypothesis.
ConclusionPositive and negative evidence together are stronger than one memorable peak.
Close the notes first
Retrieve the evidence rule.
01What does IR primarily identify?
Bond and functional-group evidence.
Absorbed frequencies correspond to molecular vibrations.
02What two properties of a band should be read together?
Its position and shape or breadth.
A broad O–H band and sharp nitrile band carry different evidence.
03Can IR alone usually prove full connectivity?
No.
Many molecules share the same functional groups.
08
LESSON 8 · 26 MIN
¹H + ¹³C NMR
Map equivalent nuclear environments
Use ¹H and broadband-decoupled ¹³C signal count, chemical shift, integration, and supplied first-order splitting to constrain structure.
ESSENTIAL QUESTIONHow many chemically distinct environments exist, and what does each signal reveal without overreading it?
One signal represents one modeled environment
Methyl acetate supplies two distinct three-proton environments but three carbon environments, including the carbonyl carbon.Open the complete signal table
Methyl acetate supplies two distinct three-proton environments but three carbon environments, including the carbonyl carbon.Open the complete signal table
One signal represents one modeled environment: complete evidence ledger
Evidence
Observation
Supported conclusion
¹H: 3.7 and 2.1 ppm
Two 3H singlets
Two nonequivalent methyl proton environments
¹³C: 172, 52, 20 ppm
Three broadband-decoupled signals
Three carbon environments
Equivalence
Three protons in each methyl share one environment
Signal count is not atom count
Methyl acetate supplies two distinct three-proton environments but three carbon environments, including the carbonyl carbon.
ACCESSIBLE EVIDENCE MODEL · COMPLETE DATA OR SIGNAL TABLE INCLUDED
01
Count environments
Chemically equivalent nuclei share signals, so signal count is not atom count.
Use symmetry
Separate proton and carbon spectra
02
Read proton evidence
¹H integration estimates relative proton count, while simple splitting can report supplied neighboring equivalent protons.
Integration is relative
Use n + 1 only inside its boundary
03
Add carbon evidence
Broadband-decoupled ¹³C NMR tracks distinct carbon environments and chemical-shift regions without ordinary proton splitting.
One line per modeled carbon environment
Do not treat ¹³C peak area as proton-like integration
Worked example
Explain methyl acetate's two ¹H signals and three ¹³C signals.
1
Group each methyl’s three equivalent protons into one signal.
2
Keep the acyl and methoxy methyl groups distinct, giving two proton environments.
3
Count acyl methyl, methoxy methyl, and carbonyl carbon as three carbon environments.
ConclusionEquivalence compresses multiple nuclei into shared signals without erasing distinct chemical environments.
Close the notes first
Retrieve the evidence rule.
01What does ¹H integration estimate?
Relative numbers of protons contributing to signals.
Area, not peak height alone, carries the proton ratio.
02What does broadband-decoupled ¹³C signal count track?
Distinct carbon environments.
Equivalent carbons share one resonance in the bounded model.
03When is the n + 1 rule safe here?
Only when the prompt excludes exchange, overlap, and higher-order complications.
The simplified rule has explicit limits.
09
LESSON 9 · 28 MIN
Multi-spectra
Intersect every supplied constraint
Combine supplied formula, IR, ¹H NMR, and ¹³C NMR evidence to compare candidates and use contradictions as decisive negative evidence.
ESSENTIAL QUESTIONWhich candidate satisfies every positive and negative constraint without importing unsupplied data?
A structure survives only the intersection
The ketone band, triplet–quartet–singlet proton pattern, four carbon environments, and supplied formula converge on 2-butanone while absent aldehyde and O–H evidence reject alternatives.Open the complete signal table
Bounded ketone infrared evidence: source signals
Signal
Position
Relative intensity
Shape
Multiplicity
Integration
C═O
1715 cm⁻¹
92
sharp
Not assigned
Not assigned
C–H
2960 cm⁻¹
42
sharp
Not assigned
Not assigned
The ketone band, triplet–quartet–singlet proton pattern, four carbon environments, and supplied formula converge on 2-butanone while absent aldehyde and O–H evidence reject alternatives.Open the complete signal table
2-Butanone proton NMR evidence: source signals
Signal
Position
Relative intensity
Shape
Multiplicity
Integration
CH₂
2.45 ppm
66
sharp
quartet
2
COCH₃
2.14 ppm
82
sharp
singlet
3
CH₃
1 ppm
78
sharp
triplet
3
The ketone band, triplet–quartet–singlet proton pattern, four carbon environments, and supplied formula converge on 2-butanone while absent aldehyde and O–H evidence reject alternatives.Open the complete signal table
2-Butanone carbon-13 NMR evidence: source signals
Signal
Position
Relative intensity
Shape
Multiplicity
Integration
C═O
209 ppm
64
sharp
Not assigned
Not assigned
CH₂
43 ppm
78
sharp
Not assigned
Not assigned
COCH₃
30 ppm
84
sharp
Not assigned
Not assigned
CH₃
8 ppm
72
sharp
Not assigned
Not assigned
A structure survives only the intersection: complete evidence ledger
Evidence
Observation
Supported conclusion
Formula C₄H₈O
Supplied, not inferred
One oxygen and one degree-of-unsaturation budget available
IR carbonyl; no broad O–H
Positive and negative functional-group evidence
Ketone candidate retained; alcohol rejected
¹H and ¹³C agree
Ethyl pattern, methyl singlet, four carbon signals
2-butanone survives every bounded constraint
The ketone band, triplet–quartet–singlet proton pattern, four carbon environments, and supplied formula converge on 2-butanone while absent aldehyde and O–H evidence reject alternatives.
ACCESSIBLE EVIDENCE MODEL · COMPLETE DATA OR SIGNAL TABLE INCLUDED
01
Inventory the inputs
Treat formula, IR groups, proton environments, integration, splitting, and carbon environments as separate constraints.
Do not infer an unsupplied formula
Keep techniques distinct
02
Intersect candidates
A candidate survives only if it satisfies all supplied constraints; one incompatible absence can disqualify it.
Positive and negative evidence
Check the least memorable feature
03
State a bounded result
A uniquely surviving candidate is supported within the supplied set and data quality, not universally proven by one technique.
Candidate set matters
Name the evidence limit
Worked example
Distinguish 2-butanone from butanal using formula C₄H₈O.
1
Both formulas and carbonyl evidence remain compatible initially.
2
Check for an aldehyde proton near 9–10 ppm.
3
Its absence plus a triplet–quartet–singlet pattern supports 2-butanone in the bounded set.
ConclusionThe correct candidate wins by complete constraint intersection, not carbonyl recognition alone.
Close the notes first
Retrieve the evidence rule.
01Should a molecular formula be inferred when none is supplied?
No.
IR and basic NMR evidence do not necessarily determine formula.
02Can a missing expected signal be decisive?
Yes, when the spectrum is adequate and the candidate requires it.
Negative evidence can contradict a structure.
03What must the chosen candidate satisfy?
Every supplied positive and negative constraint.
Partial matches are insufficient.
Randomized evidence retrieval
Choose the complete conclusion.
All thirty-six evidence records are represented. Questions and five-choice answer sets shuffle without detaching conditions, measurements, constraints, or scientific boundaries.
36 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
Evidence supports; it does not overrule the boundary.
This route covers all three official Chemical & Physical Properties topic lines and every frozen evidence objective. The ADA does not publish topic-line or cue-level quotas, so DAT TRAIN does not invent them.
The validators enforce the bounded evidence operations and complete candidate intersections. They do not establish experimental safety, quantitative purity, yield, a molecular formula that was not supplied, or multistep route feasibility, which belongs to synthesis practice.