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Thursday, August 20, 2026

Organic Compounds and Functional Groups UPCAT Reviewer: Lesson and Practice

TEACHER ABI UPCAT SCIENCE

Organic Compounds and Functional Groups

Recognize common organic structures and use functional groups to predict classification, properties, reactions, and polymer behavior.

5-10 minute lesson27 original questionsAdaptive practiceSaves progress

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Find the functional group before naming the compound

Organic compounds are built mainly from carbon frameworks. A functional group is a recognizable arrangement of atoms that strongly influences a compound’s properties and reactions.

alkene: C = Calcohol: −OHaldehyde: terminal −CHOketone: internal > C = Ocarboxylic acid: −COOHester: −COO−amine: −NH₂

Hydrocarbons contain C and H only

Alkanes have single bonds; alkenes contain at least one C = C double bond.

Position matters

A terminal carbonyl with H is an aldehyde; a carbonyl between carbon groups is a ketone.

Functional groups affect polarity

Alcohols and carboxylic acids interact with water more strongly than similar-sized hydrocarbons.

Same formula can hide different structures

Structural isomers have different connectivity and may have different properties.

Polymers come from repeating monomers

Alkene monomers can join through addition polymerization.

DO IT FAST

Circle O, N, and multiple bonds first

1. Circle every oxygen and nitrogen.

2. Mark C = C or C≡C bonds.

3. Check whether a carbonyl is terminal or internal.

4. Match the visible group before considering the carbon-chain name.

Why it works

The carbon chain may change while the same functional group preserves a recognizable property and reaction pattern.

WORKED EXAMPLES

Five forms you should recognize

1. Alcohol or acid

Problem: Compare CH₃CH₂OH and CH₃COOH.

CH₃CH₂OH contains −OH and is an alcohol. CH₃COOH contains −COOH and is a carboxylic acid.

2. Aldehyde or ketone

Problem: Compare CH₃CHO and CH₃COCH₃.

CH₃CHO has a terminal −CHO group, so it is an aldehyde. CH₃COCH₃ has an internal carbonyl, so it is a ketone.

3. Property from structure

Problem: Why does ethanol boil at a higher temperature than ethane?

Ethanol’s O−H group permits hydrogen bonding; ethane has only weaker dispersion forces.

4. Addition polymer

Problem: Identify the monomer for the repeat −CH₂−CH₂−.

CH₂ = CH₂ → [−CH₂−CH₂−]ₙ

The ethene double bond opens as monomers join.

5. Evidence and limits

Problem: An unknown hydrocarbon decolorizes bromine solution.

Conclusion: A reactive multiple bond is suggested. The observation alone does not identify the exact carbon chain.

COMMON TRAPS

Check before you commit

Calling every compound with oxygen an alcohol
Confusing an aldehyde with a ketone
Treating molecular formula as a complete structural formula
Calling an alkene saturated
Assuming all plastics behave or degrade identically
Predicting water solubility without considering polarity and molecular size

FIVE-FORM SKILL CHECK

Do you need the lesson-or just practice?

One original question in each form recommends your next step. It does not yet verify mastery.

CHOOSE YOUR PRACTICE

Work at the level you need.

Foundations

Build the core procedure with immediate explanations.

Core Practice

Use mixed forms with less scaffolding.

UPCAT-Style Transfer

Apply the competency in unfamiliar representations.

FRESH MASTERY CHECK

Ready to verify this competency?

A score of 5/5 verifies mastery. An unsuccessful attempt loads a different five-form bank.

QUICK ANSWERS

Organic Compounds and Functional Groups FAQ

How much naming should I memorize?

Prioritize recognizing the carbon count, saturation, and functional group. Long-name memorization is less useful than structure-property reasoning.

Can compounds have more than one functional group?

Yes. Amino acids, for example, contain both amino and carboxyl groups.

Why do structural isomers behave differently?

Different atom connections change molecular shape, polarity, and intermolecular interactions.

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Electrochemistry UPCAT Reviewer: Lesson and Practice

TEACHER ABI UPCAT SCIENCE

Electrochemistry

Track electron transfer, identify electrodes and agents, calculate cell potential, and interpret galvanic cells, electrolysis, plating, and corrosion.

5-10 minute lesson27 original questionsAdaptive practiceSaves progress

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Electrons leave the anode and arrive at the cathode

Oxidation is loss of electrons; reduction is gain of electrons.

AN OX: oxidation at the anodeRED CAT: reduction at the cathodeelectron flow: anode → cathode

These electrode reactions are always true. What changes between galvanic and electrolytic cells is the electrode sign and the direction of energy conversion.

Galvanic cell

A spontaneous redox reaction produces electrical energy; anode is negative and cathode is positive.

Electrolytic cell

Electrical energy drives a nonspontaneous change; anode is positive and cathode is negative.

Salt bridge carries ions

Electrons travel through the wire; ions migrate through the electrolyte and bridge.

Oxidizing agent is reduced

It accepts electrons from the species it oxidizes.

Cell potential

Using reduction potentials: E°cell = E°cathode−E°anode.

DO IT FAST

Use AN OX → wire → RED CAT

Write the oxidation half-reaction at the left and the reduction half-reaction at the right:

ANODE: atoms lose e⁻ → e⁻ flow → CATHODE: ions gain e⁻

Then check mass: a reactive metal anode often dissolves, while metal ions can plate onto the cathode.

Why it works

This single map prevents the most common mix-ups involving electrode names, electron direction, and which species gains or loses mass.

WORKED EXAMPLES

Five forms you should recognize

1. Zinc-copper cell

Problem: In Zn|Zn²⁺ || Cu²⁺|Cu, identify oxidation, reduction, and electron direction.

anode: Zn→Zn²⁺+2e⁻cathode: Cu²⁺+2e⁻→Cu

Electrons travel from the zinc electrode to the copper electrode.

2. Cell potential

Problem: E°red(Cu²⁺/Cu) = +0.34 V and E°red(Zn²⁺/Zn) = −0.76 V.

E°cell = 0.34−(−0.76) = +1.10 V

The positive standard potential supports a spontaneous galvanic reaction as written.

3. Oxidation number

Problem: Find Mn in KMnO₄.

(+1) + Mn + 4(−2) = 0Mn = +7

4. Electroplating

Problem: A metal coating is deposited on an object.

The object is the cathode because metal ions gain electrons there and become solid metal.

5. Sacrificial protection

Problem: Magnesium is attached to an iron pipeline.

Magnesium oxidizes more readily and supplies electrons, so it corrodes in place of the iron.

COMMON TRAPS

Check before you commit

Sending electrons through the salt bridge
Saying reduction happens at the anode
Changing the anodecathode reaction rule between cell types
Calling the oxidized species the oxidizing agent
Adding reduction potentials without reversing the anode term
Assuming a positive ion plates at the anode

FIVE-FORM SKILL CHECK

Do you need the lesson-or just practice?

One original question in each form recommends your next step. It does not yet verify mastery.

CHOOSE YOUR PRACTICE

Work at the level you need.

Foundations

Build the core procedure with immediate explanations.

Core Practice

Use mixed forms with less scaffolding.

UPCAT-Style Transfer

Apply the competency in unfamiliar representations.

FRESH MASTERY CHECK

Ready to verify this competency?

A score of 5/5 verifies mastery. An unsuccessful attempt loads a different five-form bank.

QUICK ANSWERS

Electrochemistry FAQ

Are anode and cathode signs always the same?

No. Their reaction roles stay fixed, but their signs differ between galvanic and electrolytic cells.

What moves through the salt bridge?

Ions, not electrons, move to maintain electrical neutrality.

Why does a positive E°cell matter?

Under standard conditions it indicates that the redox reaction is thermodynamically favored in the written direction.

RELATED COMPETENCIES

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Thermochemistry UPCAT Reviewer: Lesson and Practice

TEACHER ABI UPCAT SCIENCE

Thermochemistry

Track heat between system and surroundings, read reaction-energy diagrams, and calculate calorimetry, bond-energy, and Hess’s-law changes.

5-10 minute lesson27 original questionsAdaptive practiceSaves progress

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Track where energy goes and keep the sign consistent

The system is the reaction being studied; everything else is the surroundings.

In an exothermic reaction, the system releases heat and ΔH is negative. In an endothermic reaction, the system absorbs heat and ΔH is positive.

ΔH = Hproducts−Hreactantsq = mcΔT

In an ideal calorimeter, heat lost by one part equals heat gained by the other.

Exothermic: ΔH<0

Products lie below reactants on an energy diagram.

Endothermic: ΔH>0

Products lie above reactants.

Breaking bonds absorbs energy

Forming bonds releases energy.

Catalysts change activation energy

They do not change ΔH or the initial and final states.

Hess’s law adds reaction steps

Reverse a reaction to reverse ΔH; multiply the equation to multiply ΔH.

DO IT FAST

Use SIGN → SCALE → SURROUNDINGS

SIGN: Did the system release heat (negative) or absorb heat (positive)?

SCALE: If the chemical equation is multiplied or reversed, change ΔH the same way.

SURROUNDINGS: In calorimetry, qreaction and qsolution have opposite signs.

For bond energies:

ΔH≈bonds broken−bonds formed

Why it works

Most thermochemistry errors are sign errors, scaling errors, or confusion between heat gained by the solution and heat lost by the reaction.

WORKED EXAMPLES

Five forms you should recognize

1. Exothermic evidence

Problem: A reaction warms its container.

Reasoning: Heat moved from the reaction system to the surroundings, so the process is exothermic and ΔH is negative.

2. Calorimetry

Problem: How much heat does 100 g water absorb when it warms by 5°C? Use c = 4.18 J per gram per °C.

q = 100(4.18)(5) = 2090 J

3. Read an energy diagram

Problem: Reactants are at 100 kJ, the peak at 160 kJ, and products at 40 kJ.

activation energy = 160−100 = 60 kJΔH = 40−100 = −60 kJ

4. Bond energies

Problem: For H₂ + Cl₂→2HCl, breaking bonds requires 679 kJ and forming bonds releases 862 kJ.

ΔH≈679−862 = −183 kJ

5. Hess’s law

Problem: A→B is + 40 kJ and B→C is −70 kJ.

A→C: +40 + (−70) = −30 kJ

The intermediate B cancels when the steps are added.

COMMON TRAPS

Check before you commit

Giving exothermic reactions a positive ΔH
Forgetting qreaction and qsurroundings have opposite signs
Using final temperature instead of ΔT
Saying a catalyst changes ΔH
Reversing an equation without reversing ΔH
Treating bond breaking as an energy-releasing process

FIVE-FORM SKILL CHECK

Do you need the lesson-or just practice?

One original question in each form recommends your next step. It does not yet verify mastery.

CHOOSE YOUR PRACTICE

Work at the level you need.

Foundations

Build the core procedure with immediate explanations.

Core Practice

Use mixed forms with less scaffolding.

UPCAT-Style Transfer

Apply the competency in unfamiliar representations.

FRESH MASTERY CHECK

Ready to verify this competency?

A score of 5/5 verifies mastery. An unsuccessful attempt loads a different five-form bank.

QUICK ANSWERS

Thermochemistry FAQ

How is this different from Physics Heat and Thermodynamics?

This page centers on energy changes in chemical reactions. The Physics reviewer treats heat transfer, temperature, thermal expansion, and thermodynamic systems more broadly.

Can an exothermic reaction still need activation energy?

Yes. It may release energy overall but still require an initial barrier to be overcome.

Why can temperature remain constant during melting?

Absorbed energy changes intermolecular arrangement rather than average kinetic energy during the phase change.

RELATED COMPETENCIES

Continue your mathematics review.

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Mastery results save to your Teacher Abi study profile.

Return to Student Hub View UPCAT Coverage

Limiting Reactants and Percent Yield UPCAT Reviewer: Lesson and Practice

TEACHER ABI UPCAT SCIENCE

Limiting Reactants and Percent Yield

Determine which reactant limits product formation, calculate theoretical and actual yields, and interpret excess reactants and experimental results.

5-10 minute lesson27 original questionsAdaptive practiceSaves progress

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The limiting reactant sets the theoretical yield

A balanced equation gives the mole ratio in which reactants are consumed. The limiting reactant runs out first and therefore determines the greatest possible amount of product. Any other reactant supplied beyond the required ratio is in excess.

percent yield = actual yield ÷ theoretical yield × 100%

Theoretical yield comes from stoichiometry. Actual yield is what the experiment produces.

Convert before comparing

When amounts are in grams, convert them to moles before using equation coefficients.

Predict product twice

Calculate the product possible from each reactant; the smaller result identifies the limit.

Excess means leftover

Subtract the amount consumed from the amount initially available.

Theoretical yield is a maximum

It is predicted from the limiting reactant under the model.

Actual yield is measured

Loss, side reactions, incomplete reaction, or purification can lower it.

DO IT FAST

Use PRODUCT FROM A versus PRODUCT FROM B

1. Balance the equation.

2. Convert each reactant to moles.

3. Independently calculate the product each reactant could form.

4. The smaller product is the theoretical yield; its source is the limiting reactant.

5. Only then calculate excess remaining or percent yield.

Why it works

Choosing the smaller mass or coefficient is unreliable. Comparing predicted product places both reactants on the same basis.

WORKED EXAMPLES

Five forms you should recognize

1. Identify the limiting reactant

Problem: For 2H₂ + O₂→2H₂O, 4 mol H₂ reacts with 1 mol O₂.

Four moles H₂ could form 4 mol H₂O. One mole O₂ can form only 2 mol H₂O.

Conclusion: O₂ is limiting and the theoretical yield is 2 mol H₂O.

2. Find excess remaining

Problem: In the same reaction, how much H₂ remains?

One mole O₂ consumes 2 mol H₂.

H₂ remaining = 4−2 = 2 mol

3. Start from masses

Problem: For 2H₂ + O₂→2H₂O, react 8 g H₂ with 32 g O₂.

8 g H₂÷2 g per mole = 4 mol H₂32 g O₂÷32 g per mole = 1 mol O₂

O₂ limits and forms 2 mol H₂O, which has a mass of 2(18) = 36 g.

4. Calculate percent yield

Problem: Theoretical yield is 22 g CO₂, but 17.6 g is collected.

percent yield = 17.6÷22×100% = 80%

5. Interpret more than 100%

Problem: A student reports 108% yield.

Interpretation: The collected material may be wet or impure, or a measurement error occurred. The result does not mean the reaction created matter beyond the theoretical prediction.

COMMON TRAPS

Check before you commit

Selecting the reactant with the smaller mass
Comparing moles without applying the coefficient ratio
Calculating yield from the excess reactant
Using actual yield in place of theoretical yield
Forgetting to subtract consumed excess reactant
Accepting a yield above 100% without checking impurities or error

FIVE-FORM SKILL CHECK

Do you need the lesson-or just practice?

One original question in each form recommends your next step. It does not yet verify mastery.

CHOOSE YOUR PRACTICE

Work at the level you need.

Foundations

Build the core procedure with immediate explanations.

Core Practice

Use mixed forms with less scaffolding.

UPCAT-Style Transfer

Apply the competency in unfamiliar representations.

FRESH MASTERY CHECK

Ready to verify this competency?

A score of 5/5 verifies mastery. An unsuccessful attempt loads a different five-form bank.

QUICK ANSWERS

Limiting Reactants and Percent Yield FAQ

Can the reactant with more moles be limiting?

Yes. A reaction may require several moles of it for every mole of the other reactant.

Why is actual yield often below theoretical yield?

Reactions may be incomplete, side reactions may occur, and product may be lost during transfer or purification.

How is this different from the Stoichiometric Mole Ratios reviewer?

That reviewer practices balanced-equation conversions. This one adds competing reactants, leftovers, and experimental yield.

RELATED COMPETENCIES

Continue your mathematics review.

SAVE AND CONTINUE

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Mastery results save to your Teacher Abi study profile.

Return to Student Hub View UPCAT Coverage

Mole Concept UPCAT Reviewer: Lesson and Practice

TEACHER ABI UPCAT SCIENCE

Mole Concept

Move accurately among mass, moles, particles, atoms, formula units, and gas volume while explaining why each conversion factor is used.

5-10 minute lesson27 original questionsAdaptive practiceSaves progress

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The mole is a counting bridge

One mole contains 6.02×10²³ representative particles. The representative particle depends on the substance: atoms for a monatomic element, molecules for a molecular substance, and formula units for an ionic compound.

moles = mass ÷ molar massparticles = moles × 6.02×10²³

For school-level problems at STP, one mole of an ideal gas is commonly approximated as 22.4 L.

Molar mass links grams and moles

Add the atomic masses shown in the chemical formula.

Avogadro’s number links moles and particles

Multiply going from moles to particles; divide going back.

Subscripts count particles within particles

One CO₂ molecule contains two oxygen atoms.

Ionic compounds use formula units

NaCl is counted in formula units, not discrete molecules.

Gas volume needs stated conditions

Use 22.4 L per mole only when the problem specifies the school-standard STP model.

DO IT FAST

Use the mole road map

grams ⇄ moles ⇄ particles

Grams to moles: divide by molar mass.

Moles to grams: multiply by molar mass.

Moles to particles: multiply by 6.02×10²³.

Particles to moles: divide by 6.02×10²³.

Only after reaching molecules or formula units should you use a formula subscript to count particular atoms or ions.

Why it works

Moles keep extremely large particle counts manageable and allow chemical formulas and equations to describe measurable laboratory amounts.

WORKED EXAMPLES

Five forms you should recognize

1. Mass to moles

Problem: How many moles are in 18 g H₂O? Use H = 1 and O = 16.

molar mass H₂O = 2(1) + 16 = 18 g per molemoles = 18 g÷18 g per mole = 1 mol

2. Moles to particles

Problem: How many molecules are in 0.50 mol O₂?

0.50 mol×6.02×10²³ = 3.01×10²³ molecules

3. Particles to mass

Problem: Find the mass of 1.204×10²⁴ CO₂ molecules.

moles = (1.204×10²⁴)÷(6.02×10²³) = 2 molmass = 2 mol×44 g per mole = 88 g

4. Count atoms inside molecules

Problem: How many hydrogen atoms are in 3.01×10²³ CH₄ molecules?

Each molecule has four H atoms.

4(3.01×10²³) = 1.204×10²⁴ H atoms

5. Gas volume at STP

Problem: How many moles occupy 11.2 L at STP using 22.4 L per mole?

moles = 11.2÷22.4 = 0.50 mol

Why: The stated STP model supplies the volume-to-mole conversion.

COMMON TRAPS

Check before you commit

Multiplying by molar mass when converting grams to moles
Calling ionic formula units molecules
Forgetting formula subscripts when counting atoms
Using Avogadro’s number as a molar mass
Using 22.4 L per mole without stated gas conditions
Rounding too early in a multistep conversion

FIVE-FORM SKILL CHECK

Do you need the lesson-or just practice?

One original question in each form recommends your next step. It does not yet verify mastery.

CHOOSE YOUR PRACTICE

Work at the level you need.

Foundations

Build the core procedure with immediate explanations.

Core Practice

Use mixed forms with less scaffolding.

UPCAT-Style Transfer

Apply the competency in unfamiliar representations.

FRESH MASTERY CHECK

Ready to verify this competency?

A score of 5/5 verifies mastery. An unsuccessful attempt loads a different five-form bank.

QUICK ANSWERS

Mole Concept FAQ

What is a representative particle?

It is the entity being counted: an atom, molecule, formula unit, ion, or another specified particle.

Why is NaCl not counted in molecules?

Solid NaCl forms an ionic lattice rather than separate NaCl molecules.

How is this different from stoichiometry?

This page builds conversions for one substance. Stoichiometry uses balanced-equation mole ratios to connect different substances.

RELATED COMPETENCIES

Continue your mathematics review.

SAVE AND CONTINUE

Your progress stays on this browser.

Mastery results save to your Teacher Abi study profile.

Return to Student Hub View UPCAT Coverage