Organic Chemistry & Material Design — Chapter Practice, Grade 11

Use carbon structure, polymer properties, and life-cycle evidence to evaluate materials.

Problem 1

How many H atoms are in acyclic alkane C₆H…?
Hints

Write the relationship first; substitute with consistent units.

Answer and explanation

14 H atoms

2×6+2=14.

Problem 2

Why is an alcohol group not hydroxide ion?
Answer and explanation

The O–H group is covalently part of a neutral molecule, not a separate OH⁻ ion.

Response criteria

  • State a scientifically accurate conclusion.
  • Show the calculation or mechanism that supports it.
  • Include units, conditions, or limits where relevant.

Problem 3

Find hydrogen count for a simple acyclic alkane with seven carbon atoms.
Hints

List the given quantities, select a relationship, and check units before substituting.

Answer and explanation

16 H atoms

2×7+2=16.

Problem 4

What two products form in complete methane combustion?
Answer and explanation

CO₂ and H₂O; balanced CH₄+2O₂→CO₂+2H₂O.

Response criteria

  • State a scientifically accurate conclusion.
  • Show the calculation or mechanism that supports it.
  • Include units, conditions, or limits where relevant.

Problem 5

Identify the functional group in CH₃COOH. Show the reasoning, not just the conclusion.
Hints

A molecular formula alone does not specify structure.

Answer and explanation

A carboxyl group; the compound is a carboxylic acid.

Problem 6

Why can ethanol hydrogen-bond as a donor while dimethyl ether cannot donate through an O–H bond?
Hints

State what the supplied observations or conditions establish, and separate that from any assumption.

Answer and explanation

Ethanol contains O–H; the ether has oxygen but no O–H bond, though it can accept suitable hydrogen bonds.

Problem 7

Isomers share a formula while differing in structure or spatial arrangementIsomers share a formula while differing in structure or spatial arrangementStructural isomers: same totals, different connectionsCH₃–CH₂–OHCH₃–O–CH₃Both: C₂H₆OA molecular formula alone does not determine structure or behavior.
Original selected condensed-isomer comparison; no claim that all isomers have equally large property differences.
Interpret the supplied visual. Are the two C₂H₆O examples merely two rotations of one molecule?
Answer and explanation

No. Oxygen has different bonded neighbors, so their connectivity differs and they are structural isomers.

Structural isomers share a molecular formula but differ in connectivity, as ethanol and dimethyl ether do. Stereoisomers retain connectivity but differ in spatial arrangement. Restricted rotation or a chiral environment can make spatial differences chemically relevant; not every rotated drawing is a new stereoisomer. For alkene geometric isomerism, check whether each double-bond carbon has two different substituents before assigning a distinct arrangement. A two-dimensional formula can omit stereochemical information needed to predict behavior.

No. Oxygen has different bonded neighbors, so their connectivity differs and they are structural isomers.

Response criteria

  • Interpret labels, structures, axes, units and directions correctly.
  • Explain the chemistry relationship or mechanism.
  • Respect stated assumptions and distinguish observation from inference.

Problem 8

Two molecules are CH₃CH₂OH and CH₃OCH₃. Show their shared molecular formula, identify their structural-isomer relationship and distinguish their hydrogen-bond donor roles.
Answer and explanation

Both are C₂H₆O but have different connectivity, so they are structural isomers. Ethanol has an O–H donor; dimethyl ether lacks O–H and does not supply that donor pattern, although its oxygen can accept suitable interactions.

Both are C₂H₆O but have different connectivity, so they are structural isomers. Ethanol has an O–H donor; dimethyl ether lacks O–H and does not supply that donor pattern, although its oxygen can accept suitable interactions.

Response criteria

  • Use the stated measurements or conditions, with correct units.
  • Connect the result to the chemical mechanism or evidence relationship.
  • Identify the assumption, model limit or further evidence requested.

Problem 9

Repeat-unit mass 44 g/mol and 200 units: estimated chain mass?
Hints

Write the relationship first; substitute with consistent units.

Answer and explanation

8800 g/mol

44×200=8800.

Problem 10

Does fragmenting into microplastic establish biodegradation?
Answer and explanation

No. The polymer can remain chemically present in smaller particles.

Response criteria

  • State a scientifically accurate conclusion.
  • Show the calculation or mechanism that supports it.
  • Include units, conditions, or limits where relevant.

Problem 11

Repeat-unit mass is 50 g/mol and an ideal chain has 300 units. Estimate chain molar mass ignoring ends.
Hints

List the given quantities, select a relationship, and check units before substituting.

Answer and explanation

15000 g/mol

50×300=15000.

Problem 12

What should a compostability comparison specify?
Answer and explanation

Temperature, moisture, biological conditions, timescale, test criteria, and access to appropriate processing.

Response criteria

  • State a scientifically accurate conclusion.
  • Show the calculation or mechanism that supports it.
  • Include units, conditions, or limits where relevant.

Problem 13

How can cross-linking change a polymer's behavior? Show the reasoning, not just the conclusion.
Hints

Polymer properties depend on chain structure, additives and processing.

Answer and explanation

Links between chains can restrict movement, often increasing rigidity or elastic network behavior depending on structure.

Problem 14

Why should a biodegradable label be checked against actual disposal conditions?
Hints

State what the supplied observations or conditions establish, and separate that from any assumption.

Answer and explanation

Degradation depends on temperature, moisture, organisms and timescale; a label does not guarantee rapid breakdown in every environment.

Problem 15

A material’s life-cycle claim needs a defined service and system boundaryA material’s life-cycle claim needs a defined service and system boundaryEvidence boundaryQuestion before comparing materialsServiceWhat function, durability and failure conditions?ProductionWhich energy/material inputs are included?End of lifeIs collection and the actual processing routeavailable?Environmental outcomeWhat emissions, leakage and breakdown conditions?A label such as bio-based does not settle the full impact account.
Original comparison-boundary organizer, not a ranking of current commercial products or a claim about a specific recycling system.
Interpret the supplied visual. Why does biodegradable not prove that an item will break down rapidly in every ocean, landfill or backyard condition?
Answer and explanation

Breakdown depends on material, organisms, temperature, moisture and time; the label requires specified test conditions and does not establish every disposal outcome.

A life-cycle comparison evaluates a defined service across selected production, use and end-of-life stages. Bio-based, biodegradable and recyclable describe different claims: feedstock origin, breakdown under specified conditions, and recovery/process feasibility. None automatically establishes the others or the lowest total impact. Mass alone can be a misleading comparison if products have different lifetimes or functions. Identify local collection, contamination, energy and leakage evidence before turning a material label into an environmental conclusion.

Breakdown depends on material, organisms, temperature, moisture and time; the label requires specified test conditions and does not establish every disposal outcome.

Response criteria

  • Interpret labels, structures, axes, units and directions correctly.
  • Explain the chemistry relationship or mechanism.
  • Respect stated assumptions and distinguish observation from inference.

Problem 16

A part needs load index ≥70. P scores load 80, input 40 and 20 cycles; Q scores load 65, input 24 and 30 cycles; R scores load 90, input 30 and 10 cycles. Select a feasible option by input per cycle and name a missing criterion.
Answer and explanation

Q fails the load constraint. Among feasible P and R, P uses 40/20=2 input units/cycle versus R’s 30/10=3, so P is preferred under this indicator. Cost, failure variability, repair, emissions or end-of-life evidence could change the broader decision.

Q fails the load constraint. Among feasible P and R, P uses 40/20=2 input units/cycle versus R’s 30/10=3, so P is preferred under this indicator. Cost, failure variability, repair, emissions or end-of-life evidence could change the broader decision.

Response criteria

  • Use the stated measurements or conditions, with correct units.
  • Connect the result to the chemical mechanism or evidence relationship.
  • Identify the assumption, model limit or further evidence requested.

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