Module 7: Organic Chemistry

Master the chemistry of carbon compounds—from nomenclature and reaction mechanisms to polymers and materials—with systematic problem-solving strategies for organic synthesis.

Pillar 1 of 3

Organic Foundations

Master systematic nomenclature for organic compounds, identify structural and functional group isomers, and understand substitution reactions of alkanes and addition reactions of alkenes including Markovnikov's rule.

Nomenclature

IUPAC nomenclature provides a systematic way to name organic compounds. For HSC Chemistry, you must be able to name compounds from C1 to C8 and apply functional group priority rules.

Alkane Stem Names

The parent chain (longest continuous carbon chain) determines the stem name:

Carbons	Prefix	Alkane
1	meth-	methane
2	eth-	ethane
3	prop-	propane
4	but-	butane
5	pent-	pentane
6	hex-	hexane
7	hept-	heptane
8	oct-	octane

Memory tip: "My Elephant Plays Piano, But Prefers Heavy Orchestra" (Meth, Eth, Prop, Pent, But, Hex, Hept, Oct - note But and Pent reversed in the mnemonic to help remember the spelling)

Functional Group Priority

When multiple functional groups are present, the highest priority group determines the suffix. Lower priority groups become prefixes.

Priority order (highest to lowest):

Carboxylic Acid (-COOH) → suffix: -oic acid
Amide (-CONH₂) → suffix: -amide
Aldehyde (-CHO) → suffix: -al
Ketone (C=O) → suffix: -one
Alcohol (-OH) → suffix: -ol
Amine (-NH₂) → suffix: -amine (or prefix: amino-)
Alkene (C=C) → suffix: -ene
Alkyne (C≡C) → suffix: -yne
Haloalkane (F, Cl, Br, I) → prefix: fluoro-, chloro-, bromo-, iodo-
Alkyl groups (CH₃-, C₂H₅-) → prefix: methyl-, ethyl-

Example: If a molecule contains both -OH and C=C, the alcohol takes priority for the suffix (-ol), and the alkene is indicated by inserting "en" before the -ol: hex-2-en-1-ol

Naming Rules

Step 1: Find the longest carbon chain containing the highest priority functional group. This is the parent chain.

Step 2: Number the chain from the end that gives the lowest number to the highest priority functional group (or to the first point of difference if groups have equal priority).

Step 3: Name substituents with their position numbers. List alphabetically (ignoring di-, tri-, etc.).

Step 4: Assemble the name:

[position]-[substituent(s)]-[parent stem]-[position]-[suffix]

Important rules:

Use hyphens between numbers and words: 2-methylpropane
Use commas between numbers: 2,3-dimethylbutane
Alphabetize substituents: ethyl before methyl
Use di-, tri-, tetra- for multiple identical groups
No space in the final name: it's one word

Examples:

CH₃CH₂CH₂CH₃ = butane

CH₃CH(CH₃)CH₃ = 2-methylpropane

CH₃CH₂OH = ethanol

CH₃COCH₃ = propanone (acetone)

CH₃CH₂COOH = propanoic acid

Worked Example

Name the compound: CH₃CH(OH)CH₂CH₂CH₃

Step 1: Identify the longest chain:

The longest continuous carbon chain has 5 carbons: pent-

Step 2: Identify functional groups:

There is an -OH group (alcohol), so the suffix will be -ol

The compound is a pentanol

Step 3: Number the chain:

Number from the end closest to the -OH group

CH₃-CH(OH)-CH₂-CH₂-CH₃

Numbering left to right: 1-2-3-4-5

The -OH is on carbon 2

Step 4: Assemble the name:

Parent: pentan

Position of -OH: 2

Suffix: -ol

Final name: pentan-2-ol

(Also acceptable: 2-pentanol)

Practice Question

Name the following compound and explain your reasoning: CH₃CH₂CH(Cl)CH₂CH₃

Show Answer

Step-by-step solution:

1. Longest chain: The longest continuous carbon chain has 5 carbons, so the parent is pentane.

2. Functional groups: There is a chlorine atom attached, making this a chloroalkane. Halogens are named as prefixes: chloro-

3. Numbering: Since there's no high-priority functional group, number from the end that gives the substituent the lowest number.

From left: CH₃(1)-CH₂(2)-CH(Cl)(3)-CH₂(4)-CH₃(5) → Cl is on carbon 3

From right: CH₃(5)-CH₂(4)-CH(Cl)(3)-CH₂(2)-CH₃(1) → Cl is still on carbon 3

Either way, the chlorine is on carbon 3.

4. Final name: 3-chloropentane

Explanation: We use the prefix "chloro" to indicate the chlorine substituent, the number "3" to show its position, and "pentane" as the parent alkane. The name is written as one word with a hyphen between the number and the name: 3-chloropentane.

Isomerism

Isomers are compounds with the same molecular formula but different structural arrangements. Understanding the types of isomers is essential for predicting properties and reactions.

Chain Isomers (Structural Isomers)

Definition: Isomers that differ in the arrangement of the carbon skeleton. The carbon chain has a different branching pattern.

Example: C₅H₁₂ (three chain isomers)

1. Pentane (straight chain):

CH₃-CH₂-CH₂-CH₂-CH₃

2. 2-methylbutane (one branch):

CH₃-CH(CH₃)-CH₂-CH₃

3. 2,2-dimethylpropane (two branches):

CH₃-C(CH₃)₂-CH₃

Properties: Chain isomers have different physical properties (boiling points, melting points) because branching affects molecular shape and surface area. More branching generally means lower boiling point due to reduced surface contact between molecules.

Key point: Same molecular formula (C₅H₁₂), different carbon skeleton arrangement.

Position Isomers

Definition: Isomers that have the same carbon skeleton but differ in the position of a functional group or substituent.

Example: C₃H₈O (two position isomers - alcohols)

1. Propan-1-ol (-OH on carbon 1):

CH₃-CH₂-CH₂-OH

2. Propan-2-ol (-OH on carbon 2):

CH₃-CH(OH)-CH₃

Another example: C₄H₈ (alkenes)

1. But-1-ene (double bond between C1 and C2):

CH₂=CH-CH₂-CH₃

2. But-2-ene (double bond between C2 and C3):

CH₃-CH=CH-CH₃

Properties: Position isomers often have different chemical properties because the functional group is in a different environment. For example, propan-1-ol can be oxidized to an aldehyde, while propan-2-ol oxidizes to a ketone.

Key point: Same molecular formula, same carbon skeleton, different functional group position.

Functional Group Isomers

Definition: Isomers that have different functional groups, resulting in completely different chemical properties.

Common examples:

1. Alcohols vs Ethers (C₂H₆O):

Ethanol (alcohol): CH₃-CH₂-OH

Methoxymethane (ether): CH₃-O-CH₃

Very different properties: ethanol is polar and forms hydrogen bonds, while ether cannot hydrogen bond as effectively

2. Aldehydes vs Ketones (C₃H₆O):

Propanal (aldehyde): CH₃-CH₂-CHO

Propanone (ketone): CH₃-CO-CH₃

Different reactivity: aldehydes are easily oxidized, ketones are not

3. Carboxylic Acids vs Esters (C₃H₆O₂):

Propanoic acid: CH₃-CH₂-COOH

Methyl ethanoate: CH₃-COO-CH₃

Acid is acidic and corrosive, ester is neutral and often fragrant

Properties: Functional group isomers have dramatically different chemical and physical properties because they contain different functional groups. They belong to different homologous series.

Key point: Same molecular formula, completely different functional groups.

Worked Example

Draw all structural isomers of C₄H₁₀ and identify which type of isomerism is present. Name each isomer.

Type of isomerism:

These will be chain isomers (structural isomers) because C₄H₁₀ is an alkane with no functional groups—the only way to create isomers is by changing the carbon skeleton.

Isomer 1: Straight chain (4 carbons in a row)

Structure: CH₃-CH₂-CH₂-CH₃

Name: Butane

Isomer 2: Branched chain (3 carbons in main chain, 1 branch)

Structure: CH₃-CH(CH₃)-CH₃

(A methyl group branches off the middle carbon)

Name: 2-methylpropane (also called isobutane)

Total isomers:

C₄H₁₀ has 2 structural isomers

You cannot make more isomers by numbering differently—these are the only two distinct structures for C₄H₁₀.

Practice Question

Draw and name three different isomers of C₃H₈O. Identify what type of isomerism each pair represents.

Show Answer

Three isomers of C₃H₈O:

Isomer 1: Propan-1-ol (primary alcohol)

Structure: CH₃-CH₂-CH₂-OH

The -OH group is on carbon 1 (end of chain)

Isomer 2: Propan-2-ol (secondary alcohol)

Structure: CH₃-CH(OH)-CH₃

The -OH group is on carbon 2 (middle of chain)

Isomer 3: Methoxymethane (ether)

Structure: CH₃-O-CH₃

An oxygen atom is between two methyl groups (ether functional group)

Types of isomerism:

Propan-1-ol vs Propan-2-ol: Position isomers

They have the same carbon skeleton (3 carbons in a row) and the same functional group (-OH), but the -OH is in different positions.

Propan-1-ol vs Methoxymethane: Functional group isomers

They have the same molecular formula but completely different functional groups: one is an alcohol (-OH) and the other is an ether (C-O-C). They have very different properties.

Propan-2-ol vs Methoxymethane: Functional group isomers

Same reasoning as above—different functional groups.

Alkanes

Alkanes are saturated hydrocarbons containing only single C-C and C-H bonds. They undergo substitution reactions where hydrogen atoms are replaced by other atoms or groups.

Properties of Alkanes

General formula: C_nH_2n+2 (for straight-chain and branched alkanes)

Bonding: All carbon atoms are sp³ hybridized with tetrahedral geometry (bond angles ≈109.5°). Only single bonds present (sigma bonds only).

Physical properties:

Non-polar molecules (C-H bonds have very low polarity)
Weak intermolecular forces (dispersion forces only)
Low boiling points compared to polar molecules of similar mass
Insoluble in water, soluble in non-polar solvents
Boiling point increases with chain length (more dispersion forces)
Boiling point decreases with branching (less surface contact)

Chemical properties:

Relatively unreactive (strong C-C and C-H bonds)
Do not react with acids, bases, or oxidizing agents under normal conditions
Burn in oxygen (combustion) to produce CO₂ and H₂O
Undergo substitution reactions with halogens in presence of UV light

Why alkanes are unreactive: The C-H bond is strong (413 kJ/mol) and non-polar. The electron density is evenly distributed, providing no reactive sites for electrophiles or nucleophiles to attack.

Substitution Reactions

Definition: A reaction where one atom or group is replaced by another atom or group. For alkanes, hydrogen atoms are substituted by halogen atoms.

General equation:

\text{Alkane} + \text{Halogen} \xrightarrow{\text{UV light}} \text{Haloalkane} + \text{Hydrogen halide}

Example - Chlorination of methane:

\mathrm{CH_4} + \mathrm{Cl_2} \xrightarrow{\text{UV}} \mathrm{CH_3Cl} + \mathrm{HCl}

Methane + Chlorine → Chloromethane + Hydrogen chloride

Critical requirement: UV light or sunlight

Without UV light, alkanes do not react with halogens at room temperature. UV light provides the activation energy to break the Cl-Cl bond, initiating the reaction through a free radical mechanism.

Further substitution is possible:

CH₃Cl + Cl₂ → CH₂Cl₂ + HCl (dichloromethane)

CH₂Cl₂ + Cl₂ → CHCl₃ + HCl (trichloromethane, chloroform)

CHCl₃ + Cl₂ → CCl₄ + HCl (tetrachloromethane)

Controlling the product: Use excess alkane to favor monosubstitution, or excess halogen to drive multiple substitution.

Free Radical Mechanism (Extension)

The substitution occurs via a free radical chain reaction with three stages:

1. Initiation: UV light breaks the Cl-Cl bond homolytically (each atom takes one electron)

Cl₂ → 2Cl• (chlorine radicals)

2. Propagation: Chain reaction producing products and regenerating radicals

CH₄ + Cl• → •CH₃ + HCl

•CH₃ + Cl₂ → CH₃Cl + Cl•

(The Cl• radical is regenerated, allowing the chain to continue)

3. Termination: Radicals combine, ending the chain

Cl• + Cl• → Cl₂

•CH₃ + Cl• → CH₃Cl

•CH₃ + •CH₃ → C₂H₆

Why UV light is essential: The Cl-Cl bond (243 kJ/mol) is strong enough that thermal energy at room temperature cannot break it. UV light provides photons with sufficient energy to cause homolytic fission, generating the chlorine radicals needed to initiate the chain reaction.

Worked Example

Write a balanced equation for the reaction between propane and bromine in the presence of UV light. Name the organic product formed in the first substitution.

Equation:

$\mathrm{C_3H_8} + \mathrm{Br_2} \xrightarrow{\text{UV}} \mathrm{C_3H_7Br} + \mathrm{HBr}$

Or more explicitly:

$\mathrm{CH_3CH_2CH_3} + \mathrm{Br_2} \xrightarrow{\text{UV}} \mathrm{CH_3CH_2CH_2Br} + \mathrm{HBr}$

Product name:

The main organic product is 1-bromopropane

(Also called bromopropane since the position of Br can also be on carbon 2, forming 2-bromopropane as a minor product)

Important notes:

1. UV light must be shown above the arrow

2. Multiple products are possible: both 1-bromopropane and 2-bromopropane form

3. HBr is produced as a by-product

4. Further substitution can occur with excess Br₂

Practice Question

Explain why alkanes are unreactive under normal conditions but react with halogens in the presence of UV light. What role does UV light play in the reaction?

Show Answer

Why alkanes are unreactive:

Alkanes contain only C-C and C-H single bonds, which are both strong and non-polar. The C-H bond has a bond energy of approximately 413 kJ/mol, making it difficult to break under normal conditions. Additionally, the even distribution of electron density in alkane molecules means there are no partial positive or negative charges for other reagents to attack. This makes alkanes resistant to reaction with acids, bases, oxidizing agents, and even halogens at room temperature.

Why UV light enables reaction with halogens:

Halogens like Cl₂ and Br₂ have covalent bonds that are strong enough (Cl-Cl: 243 kJ/mol, Br-Br: 193 kJ/mol) that they don't break spontaneously at room temperature. UV light provides high-energy photons that can break the halogen-halogen bond through homolytic fission—each atom takes one electron from the shared pair, creating two highly reactive free radicals:

Cl₂ → 2Cl•

Role of UV light:

UV light acts as an initiator for the free radical chain reaction. The chlorine radicals (Cl•) are extremely reactive because they have an unpaired electron. These radicals can abstract hydrogen atoms from alkanes, breaking the C-H bond and forming new radicals that continue the chain reaction:

Cl• + CH₄ → •CH₃ + HCl

•CH₃ + Cl₂ → CH₃Cl + Cl•

Once initiated, the reaction becomes self-sustaining through the propagation steps, where new radicals are constantly generated. Without UV light, the initial radicals never form, so the reaction cannot proceed. This is why you must always indicate "UV light" or "sunlight" above the arrow when writing halogenation equations for alkanes—it's an essential condition, not just a detail.

Alkenes

Alkenes are unsaturated hydrocarbons containing at least one C=C double bond. They are much more reactive than alkanes and undergo addition reactions across the double bond.

Properties of Alkenes

General formula: C_nH_2n (for alkenes with one double bond)

Bonding: The C=C double bond consists of one sigma (σ) bond and one pi (π) bond. Carbon atoms in the double bond are sp² hybridized with trigonal planar geometry (bond angles ≈120°). The pi bond is weaker than the sigma bond.

Physical properties:

Non-polar (similar to alkanes)
Slightly higher boiling points than corresponding alkanes (double bond creates a more rigid structure)
Insoluble in water, soluble in non-polar solvents
Restricted rotation around C=C bond (unlike single bonds)

Chemical properties:

Much more reactive than alkanes
The pi bond is the reactive site (weaker than sigma bond, ~260 kJ/mol vs ~350 kJ/mol)
Undergo addition reactions (pi bond breaks, sigma bonds form)
React with H₂, X₂, HX, and H₂O
Can be oxidized and can polymerize

Why alkenes are reactive: The pi bond electrons are more exposed (above and below the molecular plane) than sigma bond electrons, making them accessible to electrophiles. Breaking the pi bond releases energy, which drives addition reactions.

Addition Reactions

Definition: A reaction where two atoms or groups add across the C=C double bond. The pi bond breaks and two new sigma bonds form.

General pattern:

\text{Alkene} + \text{X-Y} \rightarrow \text{Product (single bond)}

1. Hydrogenation (Addition of H₂):

Converts alkenes to alkanes

Requires catalyst: Ni, Pt, or Pd

Example:

\mathrm{C_2H_4} + \mathrm{H_2} \xrightarrow{\text{Ni}} \mathrm{C_2H_6}

Ethene + Hydrogen → Ethane

Used industrially to convert vegetable oils (unsaturated) to margarine (saturated)

2. Halogenation (Addition of X₂):

Adds two halogen atoms across the double bond

No catalyst needed, occurs at room temperature

Example:

\mathrm{C_2H_4} + \mathrm{Br_2} \rightarrow \mathrm{C_2H_4Br_2}

Ethene + Bromine → 1,2-dibromoethane

Test for unsaturation: bromine water decolorizes (brown → colorless) when alkene is present

3. Hydration (Addition of H₂O):

Adds H and OH across the double bond to form an alcohol

Requires acid catalyst (H₂SO₄ or H₃PO₄) and heat

Example:

\mathrm{C_2H_4} + \mathrm{H_2O} \xrightarrow{\mathrm{H_2SO_4}} \mathrm{C_2H_5OH}

Ethene + Water → Ethanol

Industrial method for producing ethanol

4. Hydrohalogenation (Addition of HX):

Adds H and halogen across the double bond

Example:

\mathrm{C_2H_4} + \mathrm{HBr} \rightarrow \mathrm{C_2H_5Br}

Ethene + Hydrogen bromide → Bromoethane

Markovnikov's Rule

The rule: In the addition of HX or H₂O to an unsymmetrical alkene, the hydrogen atom adds to the carbon of the double bond that already has more hydrogen atoms.

Simple version: "The rich get richer" - hydrogen goes to the carbon with more hydrogens.

Why it matters: For symmetrical alkenes like ethene (CH₂=CH₂), it doesn't matter which carbon gets H and which gets X - you get the same product either way. But for unsymmetrical alkenes like propene (CH₃-CH=CH₂), there are two possible products, and Markovnikov's rule predicts which one forms predominantly.

Example: Hydration of propene

Propene structure: CH₃-CH=CH₂

Left carbon of double bond: has 1 H (the CH group)

Right carbon of double bond: has 2 H (the CH₂ group)

By Markovnikov's rule: H adds to the CH₂ carbon (which already has more H)

Therefore: OH adds to the CH carbon

Product: CH₃-CH(OH)-CH₃ = propan-2-ol (major product)

NOT CH₃-CH₂-CH₂OH (propan-1-ol, which is the minor product)

Another example: Hydrochlorination of but-1-ene

But-1-ene: CH₃-CH₂-CH=CH₂

Double bond between C3 and C4

C3 has 1 H, C4 has 2 H

H adds to C4, Cl adds to C3

Product: CH₃-CH₂-CHCl-CH₃ = 2-chlorobutane (major)

Mechanism explanation (extension): The rule works because addition proceeds through the most stable carbocation intermediate. Secondary and tertiary carbocations are more stable than primary carbocations. When H⁺ adds to the carbon with more H, it forms a carbocation on the carbon with more alkyl groups (more substituted), which is more stable.

Worked Example

Predict the major product when but-2-ene (CH₃-CH=CH-CH₃) reacts with: (a) hydrogen in the presence of a nickel catalyst, (b) bromine water. Name each product.

(a) Hydrogenation:

Reaction: $\mathrm{CH_3CH{=}CHCH_3} + \mathrm{H_2} \xrightarrow{\text{Ni}} \mathrm{CH_3CH_2CH_2CH_3}$

The double bond is converted to a single bond by adding H₂

Product: Butane

Note: Markovnikov's rule doesn't apply here because H₂ adds symmetrically—one H to each carbon of the double bond.

(b) Bromination (bromine water):

Reaction: $\mathrm{CH_3CH{=}CHCH_3} + \mathrm{Br_2} \rightarrow \mathrm{CH_3CHBrCHBrCH_3}$

One Br adds to each carbon of the double bond

Product: 2,3-dibromobutane

Observation: The brown color of bromine water disappears (decolorizes), confirming the presence of an alkene.

Note: Markovnikov's rule doesn't apply because Br₂ adds symmetrically.

Practice Question

Using Markovnikov's rule, predict the major product when propene (CH₃-CH=CH₂) reacts with HBr. Write the equation and name the product. Explain your reasoning.

Show Answer

Step 1: Identify the double bond carbons

Propene structure: CH₃-CH=CH₂

Left carbon of double bond (C2): bonded to 1 hydrogen (part of CH group)

Right carbon of double bond (C3): bonded to 2 hydrogens (CH₂ group)

Step 2: Apply Markovnikov's rule

"Hydrogen adds to the carbon with more hydrogens"

The CH₂ carbon already has 2 H atoms, so the H from HBr adds there

The Br adds to the other carbon (the CH carbon)

Step 3: Draw the product

H adds to C3: CH₃-CH-CH₃ (with H on the right carbon)

Br adds to C2: CH₃-CHBr-CH₃

Equation:

$\mathrm{CH_3CH{=}CH_2} + \mathrm{HBr} \rightarrow \mathrm{CH_3CHBrCH_3}$

Product name:

2-bromopropane

Explanation:

According to Markovnikov's rule, in the addition of HBr to an unsymmetrical alkene, the hydrogen atom attaches to the carbon of the double bond that already has more hydrogen atoms. In propene, C3 (the CH₂ group) has more hydrogen atoms than C2 (the CH group), so H adds to C3 and Br adds to C2, forming 2-bromopropane as the major product.

The alternative product, 1-bromopropane (CH₃-CH₂-CH₂Br), would form if Br added to C3 instead, but this is the minor product because it violates Markovnikov's rule. The major product is always the one predicted by Markovnikov's rule.

Pillar 2 of 3

Functional Groups

Explore alcohol classification and oxidation, compare fossil fuels with biofuels, synthesize esters through esterification, and understand how soaps and detergents work through micelle formation and emulsification.

Alcohols

Alcohols contain the hydroxyl (-OH) functional group bonded to a saturated carbon atom. They are classified by the number of carbon atoms bonded to the carbon bearing the -OH group, which determines their oxidation behavior.

Classification of Alcohols

Alcohols are classified as primary (1°), secondary (2°), or tertiary (3°) based on how many carbon atoms are directly bonded to the carbon bearing the -OH group.

Primary (1°) alcohol: The carbon with -OH is bonded to ONE other carbon (or zero carbons for methanol)

General structure: R-CH₂-OH

Example: Ethanol CH₃-CH₂-OH

The carbon with -OH is bonded to one other carbon (the CH₃)

Secondary (2°) alcohol: The carbon with -OH is bonded to TWO other carbons

General structure: R₂-CH-OH

Example: Propan-2-ol CH₃-CH(OH)-CH₃

The carbon with -OH is bonded to two other carbons (two CH₃ groups)

Tertiary (3°) alcohol: The carbon with -OH is bonded to THREE other carbons

General structure: R₃-C-OH

Example: 2-methylpropan-2-ol (CH₃)₃C-OH

The carbon with -OH is bonded to three other carbons (three CH₃ groups)

Key point: Count the carbons bonded to the C-OH carbon, NOT the total number of carbons in the molecule. Ignore hydrogens when classifying.

Oxidation of Alcohols

Alcohols can be oxidized using oxidizing agents such as acidified potassium dichromate (K₂Cr₂O₇/H⁺) or acidified potassium permanganate (KMnO₄/H⁺). The color change from orange (Cr₂O₇²⁻) to green (Cr³⁺) indicates oxidation has occurred.

Primary alcohol oxidation (2 steps possible):

Step 1: Primary alcohol → Aldehyde

\text{R-CH}_2\text{-OH} \xrightarrow{[\text{O}]} \text{R-CHO} + \text{H}_2\text{O}

Example: Ethanol → Ethanal

CH₃-CH₂-OH → CH₃-CHO

Step 2: Aldehyde → Carboxylic acid (with excess oxidizing agent)

\text{R-CHO} \xrightarrow{[\text{O}]} \text{R-COOH}

Example: Ethanal → Ethanoic acid

CH₃-CHO → CH₃-COOH

Overall: Primary alcohol → Aldehyde → Carboxylic acid

To stop at the aldehyde stage, use gentle heating and distill the aldehyde as it forms (it has a lower boiling point). For full oxidation to the acid, use excess oxidizing agent and reflux.

Secondary alcohol oxidation (1 step only):

Secondary alcohol → Ketone

\text{R}_2\text{CH-OH} \xrightarrow{[\text{O}]} \text{R}_2\text{C=O} + \text{H}_2\text{O}

Example: Propan-2-ol → Propanone

CH₃-CH(OH)-CH₃ → CH₃-CO-CH₃

Ketones cannot be oxidized further under normal conditions (no H on the carbonyl carbon)

Tertiary alcohol oxidation:

No reaction under normal conditions

Tertiary alcohols resist oxidation because the C-OH carbon has no hydrogen atoms to remove

The dichromate solution remains orange (no color change)

Summary:

1° alcohol → Aldehyde → Carboxylic acid

2° alcohol → Ketone (stops here)

3° alcohol → No reaction

Dehydration of Alcohols

Dehydration: Removal of water (H₂O) from an alcohol to form an alkene.

Conditions: Concentrated sulfuric acid (H₂SO₄) or phosphoric acid (H₃PO₄) catalyst, heat (~170°C)

General equation:

\text{Alcohol} \xrightarrow{\text{conc. H}_2\text{SO}_4, \text{ heat}} \text{Alkene} + \text{H}_2\text{O}

Example: Dehydration of ethanol

\mathrm{CH_3CH_2OH} \xrightarrow{\text{conc. H}_2\text{SO}_4, 170°\text{C}} \mathrm{CH_2{=}CH_2} + \mathrm{H_2O}

Ethanol → Ethene + Water

Mechanism: The acid protonates the -OH group, turning it into a good leaving group (H₂O). Water leaves, forming a carbocation, then a hydrogen is removed from an adjacent carbon to form the double bond.

Note: This is the reverse of hydration of alkenes. Dehydration forms alkenes from alcohols, while hydration forms alcohols from alkenes.

Fermentation

Fermentation: The anaerobic breakdown of glucose by yeast to produce ethanol and carbon dioxide.

Equation:

\mathrm{C_6H_{12}O_6} \xrightarrow{\text{yeast}} 2\mathrm{C_2H_5OH} + 2\mathrm{CO_2}

Glucose → Ethanol + Carbon dioxide

Conditions:

Anaerobic (absence of oxygen)
Warm temperature (~30-40°C, optimal for yeast enzyme activity)
Yeast enzyme (zymase) acts as biological catalyst
Aqueous solution

Process: Yeast cells contain enzymes that catalyze the breakdown of glucose. The process stops when ethanol concentration reaches about 15% because the alcohol denatures the yeast enzymes.

Applications:

Production of alcoholic beverages (beer, wine, spirits)
Biofuel production (bioethanol from crops)
Bread making (CO₂ makes dough rise, alcohol evaporates during baking)

Comparison with hydration of ethene: Fermentation is a biological process using renewable glucose from plants, while hydration uses ethene from petroleum (non-renewable). Fermentation is slower but more sustainable.

Worked Example

Classify the following alcohols as primary, secondary, or tertiary. Then predict the oxidation product(s) for each: (a) Butan-1-ol: CH₃CH₂CH₂CH₂OH, (b) Butan-2-ol: CH₃CH(OH)CH₂CH₃, (c) 2-methylpropan-2-ol: (CH₃)₃COH

(a) Butan-1-ol: CH₃CH₂CH₂CH₂OH

Classification: Primary (1°) alcohol

Reason: The carbon bearing the -OH group is bonded to only ONE other carbon

Oxidation products:

First oxidation: Butanal (CH₃CH₂CH₂CHO) - an aldehyde

Further oxidation: Butanoic acid (CH₃CH₂CH₂COOH) - a carboxylic acid

(b) Butan-2-ol: CH₃CH(OH)CH₂CH₃

Classification: Secondary (2°) alcohol

Reason: The carbon bearing the -OH group is bonded to TWO other carbons

Oxidation product:

Butanone (CH₃COCH₂CH₃) - a ketone

No further oxidation occurs; ketones are resistant to oxidation

Classification: Tertiary (3°) alcohol

Reason: The carbon bearing the -OH group is bonded to THREE other carbons

Oxidation product:

No reaction

Tertiary alcohols cannot be oxidized under normal conditions. The dichromate solution would remain orange.

Practice Question

Explain why primary alcohols can be oxidized to two different products (aldehyde and carboxylic acid), while secondary alcohols can only be oxidized to one product (ketone). Include the structural reason for this difference.

Show Answer

Primary alcohols - two oxidation steps:

When a primary alcohol is oxidized, it first loses two hydrogen atoms (one from the -OH and one from the C-OH carbon) to form an aldehyde. The aldehyde has the structure R-CHO, where the carbonyl carbon still has ONE hydrogen atom attached.

Because this hydrogen is still present on the carbonyl carbon, the aldehyde can be oxidized further. It gains an oxygen atom and loses the hydrogen, forming a carboxylic acid (R-COOH).

Key point: The aldehyde intermediate has a H on the carbonyl carbon, making further oxidation possible.

Secondary alcohols - one oxidation step:

When a secondary alcohol is oxidized, it loses two hydrogen atoms to form a ketone with the structure R₂C=O. The carbonyl carbon in a ketone is bonded to TWO carbon atoms and has NO hydrogen atoms.

Because there is no hydrogen on the carbonyl carbon, further oxidation cannot occur under normal conditions. To oxidize a ketone would require breaking strong C-C bonds, which needs very harsh conditions and would destroy the molecule.

Key point: Ketones have NO hydrogen on the carbonyl carbon, so they resist further oxidation.

Structural explanation:

Oxidation of organic compounds typically involves removing hydrogen atoms from a carbon and adding oxygen or oxygen-containing groups. For oxidation to occur easily, there must be a hydrogen atom available on the carbon being oxidized.

- Primary alcohol (R-CH₂-OH): Has 2 H on C-OH carbon → oxidizes to aldehyde (R-CHO) which has 1 H → oxidizes to carboxylic acid (R-COOH) which has 0 H

- Secondary alcohol (R₂CH-OH): Has 1 H on C-OH carbon → oxidizes to ketone (R₂C=O) which has 0 H → cannot oxidize further

- Tertiary alcohol (R₃C-OH): Has 0 H on C-OH carbon → cannot oxidize at all

Fuels & Biofuels

Understanding the differences between fossil fuels and biofuels is essential for evaluating sustainable energy options. The comparison focuses on energy density, carbon neutrality, and environmental impact.

Fossil Fuels

Definition: Fuels formed from the remains of ancient organisms over millions of years. Main examples: coal, petroleum (crude oil), natural gas.

Composition: Primarily hydrocarbons (compounds of carbon and hydrogen). Petrol, diesel, and kerosene are refined from crude oil.

Combustion: Fossil fuels combust completely in excess oxygen to produce CO₂ and H₂O, releasing energy:

\text{Hydrocarbon} + \mathrm{O_2} \rightarrow \mathrm{CO_2} + \mathrm{H_2O} + \text{Energy}

Example for octane (component of petrol):

2\mathrm{C_8H_{18}} + 25\mathrm{O_2} \rightarrow 16\mathrm{CO_2} + 18\mathrm{H_2O}

Advantages:

High energy density (release large amounts of energy per kg)
Easy to transport and store (liquid or gas)
Well-established infrastructure for extraction and distribution
Relatively inexpensive (currently)

Disadvantages:

Non-renewable resource (finite supply)
Release stored carbon, increasing atmospheric CO₂ (greenhouse gas)
NOT carbon neutral (add net CO₂ to atmosphere)
Produce pollutants (SO₂, NOₓ, particulates) causing acid rain and health problems
Incomplete combustion produces toxic CO

Bioethanol as a Biofuel

Definition: Ethanol (C₂H₅OH) produced from renewable biological sources through fermentation of plant materials.

Production: Fermentation of glucose from crops (sugar cane, corn, wheat)

\mathrm{C_6H_{12}O_6} \xrightarrow{\text{yeast}} 2\mathrm{C_2H_5OH} + 2\mathrm{CO_2}

Combustion:

\mathrm{C_2H_5OH} + 3\mathrm{O_2} \rightarrow 2\mathrm{CO_2} + 3\mathrm{H_2O}

Advantages:

Renewable resource (crops regrow yearly)
Carbon neutral or nearly carbon neutral (see below)
Burns cleaner than petrol (fewer pollutants like sulfur compounds)
Can be blended with petrol (E10 = 10% ethanol, 90% petrol)
Supports agricultural economies

Disadvantages:

Lower energy density than petrol (~66% of petrol's energy per liter)
Requires large land areas for crop production
May compete with food production for agricultural land
Energy required for farming, harvesting, and fermentation
Corrosive to some engine components
More expensive than fossil fuels currently

Carbon Neutrality

Definition: A fuel is carbon neutral if the amount of CO₂ released during combustion equals the amount of CO₂ absorbed from the atmosphere during production.

Why bioethanol is considered carbon neutral:

Step 1 - Plant growth (CO₂ absorption):

Plants absorb CO₂ from the atmosphere during photosynthesis to produce glucose:

6\mathrm{CO_2} + 6\mathrm{H_2O} \xrightarrow{\text{sunlight}} \mathrm{C_6H_{12}O_6} + 6\mathrm{O_2}

Step 2 - Fermentation (no net CO₂ to atmosphere):

\mathrm{C_6H_{12}O_6} \rightarrow 2\mathrm{C_2H_5OH} + 2\mathrm{CO_2}

This CO₂ was just absorbed from atmosphere, so it's being returned

Step 3 - Combustion (CO₂ released):

\mathrm{C_2H_5OH} + 3\mathrm{O_2} \rightarrow 2\mathrm{CO_2} + 3\mathrm{H_2O}

Overall carbon cycle:

CO₂ absorbed (photosynthesis) = CO₂ released (fermentation + combustion)

Net change in atmospheric CO₂ = zero

Why fossil fuels are NOT carbon neutral:

Fossil fuels release carbon that was locked underground for millions of years. This carbon was removed from the atmosphere long ago, so burning fossil fuels adds NEW carbon to the current atmosphere, increasing total atmospheric CO₂.

Reality check: Bioethanol is not perfectly carbon neutral because:

Tractors and farm equipment use fossil fuels
Fertilizer production requires energy (often from fossil fuels)
Transportation of crops and ethanol uses fossil fuels
Fermentation and distillation require energy

However, bioethanol is still much closer to carbon neutral than fossil fuels, with ~70-90% reduction in net CO₂ emissions.

Energy Density Comparison

Energy density: The amount of energy released per unit mass or volume of fuel.

Approximate values:

Petrol (octane): ~48 MJ/kg or ~35 MJ/L
Diesel: ~45 MJ/kg or ~38 MJ/L
Ethanol: ~30 MJ/kg or ~24 MJ/L

Implication: Ethanol has about 66% of petrol's energy density (by volume). This means:

Vehicles using ethanol need larger fuel tanks for same range
Fuel consumption (L/100km) is higher with ethanol
More frequent refueling required

Why the difference? Ethanol already contains oxygen (C₂H₅OH), so it releases less energy when combusted compared to pure hydrocarbons which must gain all their oxygen from air during combustion.

Worked Example

Explain why bioethanol is considered carbon neutral while petrol is not. Use equations to support your explanation.

Bioethanol - Carbon neutral cycle:

1. Plant growth (CO₂ absorbed from atmosphere):

$6\mathrm{CO_2} + 6\mathrm{H_2O} \rightarrow \mathrm{C_6H_{12}O_6} + 6\mathrm{O_2}$

2. Fermentation (some CO₂ released back):

$\mathrm{C_6H_{12}O_6} \rightarrow 2\mathrm{C_2H_5OH} + 2\mathrm{CO_2}$

3. Combustion (remaining carbon released as CO₂):

$\mathrm{C_2H_5OH} + 3\mathrm{O_2} \rightarrow 2\mathrm{CO_2} + 3\mathrm{H_2O}$

Net result: All the carbon in bioethanol came from atmospheric CO₂ absorbed during recent plant growth (this growing season). When bioethanol burns, it returns this CO₂ to the atmosphere. Since the CO₂ released equals the CO₂ absorbed (within the same year or few years), there is no net increase in atmospheric CO₂. The carbon simply cycles: atmosphere → plant → ethanol → atmosphere.

Petrol - NOT carbon neutral:

Combustion of octane (petrol component):

$2\mathrm{C_8H_{18}} + 25\mathrm{O_2} \rightarrow 16\mathrm{CO_2} + 18\mathrm{H_2O}$

Carbon source: The carbon in petrol was locked underground as fossil remains for millions of years. This carbon was removed from the atmosphere millions of years ago when ancient organisms photosynthesized and then died.

Net result: Burning petrol takes carbon that has been out of circulation for millions of years and puts it back into the current atmosphere. This represents a NET ADDITION of CO₂ to the present atmosphere. The carbon is not being recycled from recent photosynthesis—it's ancient carbon being added to the modern carbon cycle, causing atmospheric CO₂ to increase.

Key difference: Bioethanol cycles carbon over months/years (carbon neutral), while fossil fuels add ancient stored carbon to today's atmosphere (not carbon neutral).

Practice Question

Compare the advantages and disadvantages of bioethanol versus petrol as vehicle fuels. In your answer, discuss energy density and environmental impact.

Show Answer

Advantages of bioethanol over petrol:

1. Carbon neutrality: Bioethanol is renewable and approximately carbon neutral because the CO₂ released during combustion was recently absorbed from the atmosphere during plant photosynthesis. This creates a closed carbon cycle with no net addition of CO₂ to the atmosphere. In contrast, petrol adds ancient carbon to the atmosphere, contributing to climate change.

2. Cleaner combustion: Bioethanol produces fewer harmful pollutants such as sulfur dioxide (SO₂) and particulates because plant-derived ethanol doesn't contain sulfur compounds that are present in petroleum. This reduces acid rain and air pollution.

3. Renewable resource: Crops for bioethanol can be grown annually, making it a renewable fuel source, whereas petroleum reserves are finite and non-renewable.

Disadvantages of bioethanol compared to petrol:

1. Lower energy density: Bioethanol has an energy density of approximately 24 MJ/L compared to petrol's 35 MJ/L—only about 66% of petrol's energy content by volume. This means:

Vehicles need to carry more fuel for the same driving range
Larger or more frequent refueling required
Lower fuel efficiency (more liters consumed per 100 km)

2. Land use concerns: Large areas of agricultural land are needed to grow crops for bioethanol production. This can:

Compete with food production, potentially affecting food prices
Lead to deforestation if land is cleared for fuel crops
Reduce biodiversity if monoculture crops replace diverse ecosystems

3. Production energy requirements: Growing crops requires fertilizers, farm machinery, harvesting, transportation, and fermentation—all of which require energy input. While bioethanol is much better than petrol, it's not perfectly carbon neutral when these factors are considered.

4. Cost: Currently, bioethanol is more expensive to produce than petrol from crude oil, though this may change as petroleum becomes scarcer and renewable technology improves.

Conclusion:

Bioethanol offers significant environmental benefits, particularly in reducing net CO₂ emissions and air pollutants. However, its lower energy density and land use requirements present practical challenges. The ideal solution may involve a combination of improved biofuel technology, increased engine efficiency, and a transition to multiple renewable energy sources rather than relying solely on either fuel type.

Esters

Esters are organic compounds with the functional group -COO-. They are formed by the reaction between carboxylic acids and alcohols, and are known for their pleasant, fruity aromas.

Structure and Properties

General structure: R-COO-R' where R and R' are alkyl groups

Functional group: The ester linkage (-COO-) consists of a carbonyl group (C=O) bonded to an oxygen atom, which is bonded to another carbon.

Properties:

Pleasant, fruity odors (used in perfumes and flavorings)
Lower boiling points than corresponding carboxylic acids (no hydrogen bonding between ester molecules)
Slightly soluble in water (small esters)
Good solvents for non-polar compounds

Examples:

Ethyl ethanoate (ethyl acetate): CH₃-COO-CH₂CH₃ - nail polish remover smell

Methyl butanoate: CH₃CH₂CH₂-COO-CH₃ - apple/pineapple aroma

Pentyl ethanoate: CH₃-COO-C₅H₁₁ - banana aroma

Esterification Reaction

Definition: The condensation reaction between a carboxylic acid and an alcohol to form an ester and water.

General equation:

\text{Carboxylic acid} + \text{Alcohol} \rightleftharpoons \text{Ester} + \text{H}_2\text{O}

Or more explicitly:

\text{R-COOH} + \text{R'-OH} \rightleftharpoons \text{R-COO-R'} + \text{H}_2\text{O}

Example: Formation of ethyl ethanoate

\mathrm{CH_3COOH} + \mathrm{C_2H_5OH} \rightleftharpoons \mathrm{CH_3COOC_2H_5} + \mathrm{H_2O}

Ethanoic acid + Ethanol ⇌ Ethyl ethanoate + Water

Conditions for esterification:

Catalyst: Concentrated sulfuric acid (H₂SO₄) - acts as both catalyst and dehydrating agent
Reflux: Heat under reflux to prevent loss of volatile reactants/products
Remove water: Shift equilibrium to the right (Le Chatelier's principle)

Why reflux? Esterification is slow and reversible. Heating speeds up the reaction, but alcohols and esters have low boiling points and would evaporate. Reflux allows heating while preventing loss of materials - vapors condense and return to the reaction flask.

Why remove water? Esterification is an equilibrium reaction. Removing water (one of the products) shifts equilibrium to the right, increasing ester yield.

Naming Esters (Band 6 Critical Habit)

Rule: The alcohol part becomes the start (-yl suffix), and the acid part becomes the end (-oate suffix).

Pattern: [Alcohol-yl] + [Acid-oate]

Step-by-step naming:

1. Identify the alcohol side (R'-O-) - name it with -yl ending

2. Identify the acid side (R-COO-) - name it with -oate ending

3. Combine: [alcohol name-yl] + [acid name-oate]

Examples:

Example 1: CH₃-COO-CH₂CH₃

Alcohol side: -CH₂CH₃ (from ethanol) → ethyl

Acid side: CH₃-COO- (from ethanoic acid) → ethanoate

Name: Ethyl ethanoate

Example 2: CH₃CH₂-COO-CH₃

Alcohol side: -CH₃ (from methanol) → methyl

Acid side: CH₃CH₂-COO- (from propanoic acid) → propanoate

Name: Methyl propanoate

Example 3: CH₃-COO-CH₂CH₂CH₃

Alcohol side: -CH₂CH₂CH₃ (from propan-1-ol) → propyl

Acid side: CH₃-COO- (from ethanoic acid) → ethanoate

Name: Propyl ethanoate

Common mistake: Naming it backwards (acid-yl alcohol-oate) - always check that the alcohol part comes first!

Reflux Diagrams (Band 6 Critical Habit)

Critical safety feature: The reflux apparatus must have an open top condenser - never sealed, or pressure builds up and the apparatus explodes!

Key components of a reflux setup:

Round-bottom flask containing reactants
Heat source (heating mantle or water bath, NOT direct flame for flammable liquids)
Vertical condenser attached to flask
Water jacket around condenser
Water enters at the BOTTOM and exits at the TOP (critical!)
Open top (never sealed - allows pressure relief)

Why water enters at bottom: Ensures the entire condenser is filled with cold water for maximum cooling efficiency. If water entered at top, only the top would cool properly.

How it works:

1. Mixture is heated and vaporizes

2. Vapor rises into the condenser

3. Cold water jacket cools the vapor, condensing it back to liquid

4. Liquid drips back into the flask

5. Volatile components don't escape, but pressure can escape through open top

When drawing reflux apparatus:

Show arrows indicating water flow (in at bottom, out at top)
Show the open top clearly
Show heat source below flask
Label all components

Worked Example

(a) Write a balanced equation for the formation of propyl methanoate from the appropriate carboxylic acid and alcohol. (b) Name the conditions required. (c) Explain why reflux is used rather than simple heating.

(a) Equation:

Step 1: Identify the reactants from the ester name

Propyl methanoate = propyl (from propan-1-ol) + methanoate (from methanoic acid)

Step 2: Write the structures

Methanoic acid: H-COOH

Propan-1-ol: CH₃CH₂CH₂OH

Propyl methanoate: H-COO-CH₂CH₂CH₃

Step 3: Write the balanced equation

$\mathrm{HCOOH} + \mathrm{C_3H_7OH} \rightleftharpoons \mathrm{HCOOC_3H_7} + \mathrm{H_2O}$

$\text{Methanoic acid} + \text{Propan-1-ol} \rightleftharpoons \text{Propyl methanoate} + \text{Water}$

(b) Conditions:

1. Catalyst: Concentrated sulfuric acid (conc. H₂SO₄)

2. Reflux (heat under reflux conditions)

3. Remove water to shift equilibrium toward products

Problem with simple heating: Both the alcohol (propan-1-ol, bp 97°C) and the ester product (propyl methanoate, bp ~80°C) have relatively low boiling points. If we simply heat the mixture in an open container, these volatile components would evaporate and escape, reducing the yield dramatically.

Why reflux solves this: The reflux apparatus has a vertical condenser above the reaction flask. When volatile components vaporize and rise, they encounter the cold condenser walls and condense back to liquid, dripping back into the flask. This allows us to heat the reaction mixture to speed up the rate while preventing loss of reactants and products.

Additional benefit: The reflux setup has an open top, which prevents dangerous pressure buildup while still containing the volatile components through condensation.

Practice Question

A student wants to prepare methyl butanoate. (a) Name the carboxylic acid and alcohol required. (b) Explain why concentrated sulfuric acid is added to the reaction mixture. (c) Describe how to maximize the yield of ester.

Show Answer

(a) Reactants required:

Breaking down the ester name: "Methyl butanoate"

- "Methyl" comes from the alcohol → Methanol (CH₃OH)

- "Butanoate" comes from the carboxylic acid → Butanoic acid (CH₃CH₂CH₂COOH)

Reactants: Butanoic acid and methanol

(b) Role of concentrated sulfuric acid:

Concentrated sulfuric acid serves two important roles:

1. Catalyst: H₂SO₄ acts as an acid catalyst that speeds up the esterification reaction. Esterification is naturally very slow at room temperature. The acid catalyst protonates the carbonyl oxygen of the carboxylic acid, making the carbon more susceptible to nucleophilic attack by the alcohol, thus accelerating the reaction.

2. Dehydrating agent: Concentrated H₂SO₄ is hygroscopic (absorbs water). Since esterification is a reversible reaction that produces water as a byproduct, removing the water shifts the equilibrium to the right (toward products) according to Le Chatelier's principle, increasing ester yield.

1. Use reflux conditions: Heat the reaction mixture under reflux to speed up the reaction while preventing loss of volatile reactants and products. The condenser must have water entering at the bottom and exiting at the top, with an open top to prevent pressure buildup.

2. Remove water continuously: Since esterification is an equilibrium reaction, removing the water product as it forms shifts equilibrium to the right (Le Chatelier's principle), increasing the yield of ester. This can be done using:

A dehydrating agent like concentrated H₂SO₄
Distillation if the water can be selectively removed
Molecular sieves or drying agents

3. Use excess reactant: Using an excess of either the alcohol or the carboxylic acid will shift equilibrium toward products. Since methanol is cheaper and more readily available than butanoic acid, using excess methanol is economically sensible.

4. Allow sufficient reaction time: Esterification can be slow even with heating and catalyst, so refluxing for an extended period (several hours) allows the reaction to approach equilibrium.

Soaps & Detergents

Soaps and detergents are surfactants that clean by emulsifying grease and oils in water. Understanding their structure and mechanism of action explains both how they work and why detergents perform better than soaps in hard water.

Structure of Soaps and Detergents

Key structural feature: Both soaps and detergents have a dual nature - one end is hydrophilic (water-loving) and the other end is hydrophobic (water-fearing).

Soap structure:

Hydrophobic tail: Long hydrocarbon chain (12-18 carbons), non-polar

Drawn as a zig-zag line or wavy line

Hydrophilic head: Carboxylate ion (-COO⁻), ionic, polar

Drawn as a circle with negative charge

Example: Sodium stearate CH₃(CH₂)₁₆COO⁻Na⁺

Detergent structure:

Hydrophobic tail: Long hydrocarbon chain, similar to soap

Hydrophilic head: Sulfonate group (-SO₃⁻) or sulfate group (-OSO₃⁻), ionic

Example: Sodium dodecylbenzenesulfonate (LAS)

Why this dual nature matters:

The hydrophobic tail dissolves in grease/oil

The hydrophilic head dissolves in water

This allows soap/detergent molecules to bridge the grease-water interface

Emulsification and Micelle Formation (Band 6 Critical Habit)

Emulsification: The process of dispersing grease/oil droplets throughout water using soap or detergent.

How cleaning works - step by step:

Step 1: Soap molecules surround grease

When soap/detergent is added to water containing grease, the hydrophobic tails are attracted to the grease while the hydrophilic heads remain in the water.

Step 2: Micelle formation

Soap/detergent molecules arrange themselves around the grease droplet with:

- Hydrophobic tails pointing INWARD (into the grease droplet)

- Hydrophilic heads pointing OUTWARD (toward the water)

This spherical structure is called a micelle

Step 3: Grease disperses in water

The micelles have charged surfaces (from the hydrophilic heads), so they repel each other and remain suspended in water rather than clumping together. The grease is now emulsified - dispersed as tiny droplets throughout the water.

Step 4: Rinse away

The water (containing suspended micelles with grease trapped inside) can be rinsed away, removing the grease from the surface being cleaned.

CRITICAL for Band 6: Drawing micelles

When drawing a micelle diagram:

Draw a grease droplet in the center
Draw hydrophobic tails (zig-zag lines) pointing INTO the grease
Draw hydrophilic heads (circles, often with - charge) pointing OUT toward water
Label: "hydrophobic tails inside grease", "hydrophilic heads in water"
Show water molecules around the outside

Common mistake: Drawing tails pointing outward - this is wrong! The tails must be INSIDE the grease because they are hydrophobic (grease-loving, water-hating).

Hard Water Problem

Hard water: Water containing dissolved calcium (Ca²⁺) and magnesium (Mg²⁺) ions, typically from limestone regions.

How soaps react with hard water:

Soap molecules (carboxylate ions) react with Ca²⁺ and Mg²⁺ ions to form insoluble precipitates:

2\text{C}_{17}\text{H}_{35}\text{COO}^- + \text{Ca}^{2+} \rightarrow (\text{C}_{17}\text{H}_{35}\text{COO})_2\text{Ca}

Stearate ion + Calcium ion → Calcium stearate (solid)

Problems caused:

Scum formation: The insoluble calcium/magnesium soap forms a gray scum that floats on water and sticks to surfaces (bathtubs, sinks, hair)
Waste of soap: Soap is consumed forming scum rather than cleaning
Reduced cleaning efficiency: Less soap available to form micelles
Hard to rinse: Scum is sticky and difficult to remove

Why detergents don't form scum:

Detergents have sulfonate (-SO₃⁻) or sulfate (-OSO₃⁻) heads instead of carboxylate (-COO⁻) heads. The calcium and magnesium salts of these groups are SOLUBLE in water, so no precipitate forms.

\text{R-SO}_3^- + \text{Ca}^{2+} \rightarrow \text{(R-SO}_3\text{)}_2\text{Ca}

Detergent + Calcium ion → Soluble calcium salt (no scum!)

Advantage of detergents: Work effectively in both soft and hard water, no scum formation, more efficient cleaning in hard water areas.

Worked Example

Explain how soap removes grease from a dirty plate. In your answer, describe the structure of soap molecules and the formation of micelles. Include a labeled diagram.

Structure of soap molecules:

Soap molecules have a dual structure:

1. Hydrophobic tail: A long hydrocarbon chain (typically 12-18 carbons) that is non-polar and water-repelling. This tail is attracted to grease and oils because they are also non-polar ("like dissolves like").

2. Hydrophilic head: An ionic carboxylate group (-COO⁻) that is polar and water-attracting. This head dissolves readily in water.

This dual nature allows soap to interact with both grease and water.

How soap removes grease:

Step 1: When soapy water contacts grease on the plate, soap molecules are attracted to the grease. The hydrophobic tails dissolve into the grease while the hydrophilic heads remain in the water.

Step 2: As more soap molecules accumulate, they arrange themselves around the grease droplet, forming a structure called a micelle. In a micelle:

- Hydrophobic tails point INWARD, buried inside the grease droplet

- Hydrophilic heads point OUTWARD, facing the surrounding water

Step 3: The grease is now surrounded by a layer of soap molecules with charged heads on the outside. This makes the grease droplet (with its soap coating) soluble in water.

Step 4: The micelles repel each other due to their charged surfaces, keeping the grease droplets dispersed throughout the water rather than clumping together. This process is called emulsification.

Step 5: When you rinse the plate, the water carries away the micelles (with grease trapped inside), leaving the plate clean.

Labeled diagram of a micelle:

[In an exam, you would draw:]

- A circle representing the grease droplet in the center

- Zig-zag lines (hydrophobic tails) pointing INWARD into the grease

- Circles with - charges (hydrophilic heads) on the outer edge

- Water molecules (H₂O) surrounding the micelle

Labels:

- "Grease droplet" (center)

- "Hydrophobic tails inside grease"

- "Hydrophilic heads in water"

- "Water molecules"

Practice Question

Explain why soap forms scum in hard water but detergents do not. Include equations in your answer.

Show Answer

Why soap forms scum in hard water:

Hard water contains dissolved calcium (Ca²⁺) and magnesium (Mg²⁺) ions. These ions are present in water that has passed through limestone or chalk regions, where minerals dissolve into the water.

Soap molecules have a carboxylate head (-COO⁻). When soap is added to hard water, the carboxylate ions react with Ca²⁺ and Mg²⁺ ions to form insoluble salts:

$2\text{C}_{17}\text{H}_{35}\text{COO}^- + \text{Ca}^{2+} \rightarrow (\text{C}_{17}\text{H}_{35}\text{COO})_2\text{Ca} \downarrow$

(Stearate ion + Calcium ion → Calcium stearate precipitate)

Similarly with magnesium:

$2\text{C}_{17}\text{H}_{35}\text{COO}^- + \text{Mg}^{2+} \rightarrow (\text{C}_{17}\text{H}_{35}\text{COO})_2\text{Mg} \downarrow$

The calcium and magnesium salts of soap are insoluble in water. They precipitate out as a gray, sticky solid called scum. This scum:

Floats on the water surface
Sticks to bathtubs, sinks, and hair
Wastes soap (consumed forming scum instead of cleaning)
Reduces cleaning efficiency

Why detergents do not form scum:

Detergents have a different hydrophilic head group - typically sulfonate (-SO₃⁻) or sulfate (-OSO₃⁻) instead of carboxylate (-COO⁻).

When detergents react with Ca²⁺ or Mg²⁺ ions, they also form salts:

$2\text{R-SO}_3^- + \text{Ca}^{2+} \rightarrow (\text{R-SO}_3)_2\text{Ca}$

(Detergent + Calcium ion → Calcium detergent salt)

However, the crucial difference is that calcium and magnesium salts of sulfonate/sulfate groups are soluble in water. They remain dissolved and do not precipitate, so no scum forms.

This means detergents:

Work effectively in both soft and hard water
Don't form scum deposits
Don't leave residue on surfaces
Are more efficient in hard water areas

Conclusion:

The difference in scum formation comes down to solubility. Soap forms insoluble Ca²⁺/Mg²⁺ salts (scum), while detergents form soluble Ca²⁺/Mg²⁺ salts (no scum). This is why detergents have largely replaced soaps for washing clothes and dishes, especially in areas with hard water.

Pillar 3 of 3

Polymers & Materials

Master addition polymerization to form polyethylene, PVC, polystyrene, and Teflon, then explore condensation polymerization to create polyesters and polyamides including nylon, identifying the small molecules released during synthesis.

Addition Polymers

Addition polymers are formed when alkene monomers undergo repeated addition reactions to form long chains. No small molecules are eliminated during the reaction—all atoms from the monomers are incorporated into the polymer.

General Mechanism

Process: Addition polymerization involves breaking the C=C double bond in alkene monomers and forming new C-C single bonds between monomers to create a long chain.

General equation:

n\text{CH}_2\text{=CHR} \xrightarrow{\text{catalyst, heat, pressure}} \text{(CH}_2\text{-CHR)}_n

Where n = degree of polymerization (number of monomer units)

R = substituent group (H, CH₃, Cl, C₆H₅, etc.)

Key features:

Monomers must contain C=C double bonds
The double bond opens up to form single bonds between monomers
No atoms are lost (no small molecule eliminated)
All atoms from monomers end up in the polymer
Requires catalyst, heat, and/or pressure
Forms very long chains (n can be thousands or millions)

Conditions:

Catalyst (e.g., Ziegler-Natta catalyst, peroxides)
High temperature and/or pressure
Free radical or ionic initiators

Drawing polymer structures:

1. Remove the double bond from the monomer

2. Add bonds extending from each end carbon (shown as lines through brackets)

3. Enclose the repeating unit in brackets with subscript n

4. Example: CH₂=CH₂ becomes -(CH₂-CH₂)_n- or more commonly shown as [-CH₂-CH₂-]_n

Polyethylene (PE)

Monomer: Ethene (ethylene), CH₂=CH₂

Polymerization reaction:

n\text{CH}_2\text{=CH}_2 \rightarrow \text{(CH}_2\text{-CH}_2\text{)}_n

or shown as: [-CH₂-CH₂-]_n

Two types of polyethylene:

1. LDPE (Low-Density Polyethylene):

Highly branched chains
Lower density (~0.92 g/cm³)
More flexible, softer
Lower melting point (~110°C)
Weaker intermolecular forces (branches prevent close packing)
Uses: plastic bags, squeeze bottles, packaging film, cling wrap

2. HDPE (High-Density Polyethylene):

Linear chains with minimal branching
Higher density (~0.95 g/cm³)
More rigid, harder
Higher melting point (~130°C)
Stronger intermolecular forces (chains pack closely)
Uses: milk bottles, detergent bottles, pipes, toys, cutting boards

Why the difference? Different polymerization conditions (temperature, pressure, catalyst) produce different degrees of branching, which affects how closely the polymer chains can pack together.

PVC (Polyvinyl Chloride)

Monomer: Chloroethene (vinyl chloride), CH₂=CHCl

Polymerization reaction:

n\text{CH}_2\text{=CHCl} \rightarrow \text{(CH}_2\text{-CHCl)}_n

or shown as: [-CH₂-CHCl-]_n

Structure: Similar to polyethylene but every second carbon has a chlorine atom attached.

Properties:

Rigid and strong (in unplasticized form)
Can be made flexible by adding plasticizers
Good chemical resistance
Fire resistant (due to chlorine)
Weathers well outdoors

Uses:

Rigid PVC: pipes, window frames, siding, credit cards
Flexible PVC: electrical cable insulation, inflatable products, clothing, upholstery

Environmental concern: When burned, PVC releases hydrogen chloride (HCl) gas and can produce toxic dioxins, making disposal problematic.

Polystyrene (PS)

Monomer: Styrene (phenylethene), CH₂=CH-C₆H₅

Polymerization reaction:

n\text{CH}_2\text{=CH-C}_6\text{H}_5 \rightarrow \text{(CH}_2\text{-CH(C}_6\text{H}_5\text{))}_n

or shown as: [-CH₂-CH(C₆H₅)-]_n

Structure: Similar to polyethylene but every second carbon has a benzene ring (phenyl group) attached.

Two forms:

1. Solid polystyrene:

Hard, rigid, brittle
Transparent or can be colored
Uses: CD cases, disposable cutlery, plastic models, laboratory equipment

2. Expanded polystyrene (foam, Styrofoam™):

Contains gas bubbles (~95% air)
Very light and excellent insulator
Uses: packaging material, disposable cups, insulation, flotation devices

Advantages: Low cost, good insulation, lightweight

Disadvantages: Non-biodegradable, takes up large volume in landfills, releases styrene if burned

Teflon (PTFE - Polytetrafluoroethylene)

Monomer: Tetrafluoroethene, CF₂=CF₂

Polymerization reaction:

n\text{CF}_2\text{=CF}_2 \rightarrow \text{(CF}_2\text{-CF}_2\text{)}_n

or shown as: [-CF₂-CF₂-]_n

Structure: Similar to polyethylene but ALL hydrogen atoms are replaced by fluorine atoms.

Properties:

Extremely non-stick surface (very low coefficient of friction)
Chemically inert (resistant to almost all chemicals)
High heat resistance (can withstand ~260°C)
Low friction
Non-reactive and non-toxic
Electrical insulator

Uses:

Non-stick cookware coating
Chemical-resistant containers and tubing
Electrical wire insulation
Bearings and seals (low friction)
Medical implants (biocompatible)
Fabric waterproofing (Gore-Tex™)

Why it's non-stick: The C-F bonds are very strong and the fluorine atoms create a surface that has very weak intermolecular forces with other substances, so nothing sticks to it.

Worked Example

(a) Draw the structure of the monomer used to make polystyrene. (b) Write an equation for the polymerization reaction. (c) Explain why expanded polystyrene is used for packaging rather than solid polystyrene.

(a) Monomer structure:

Name: Styrene (also called phenylethene or vinylbenzene)

Structure: CH₂=CH-C₆H₅

or more explicitly:

CH₂=CH- (with a benzene ring attached to the second carbon)

The monomer contains a C=C double bond (ethene part) with a benzene ring (phenyl group) attached to one of the carbons.

(b) Polymerization equation:

$n\text{CH}_2\text{=CH-C}_6\text{H}_5 \xrightarrow{\text{catalyst, heat}} \text{[-CH}_2\text{-CH(C}_6\text{H}_5\text{)-]}_n$

Where:

- n = a very large number (thousands to millions)

- The C=C double bond breaks

- Monomers link through new C-C single bonds

- The benzene ring remains unchanged as a side group

Structure of expanded polystyrene: Contains millions of tiny gas bubbles trapped within the polymer structure, making it approximately 95% air.

Advantages for packaging:

1. Lightweight: The high air content makes expanded polystyrene extremely light, reducing shipping costs and making packages easier to handle. Solid polystyrene would be much heavier and more expensive to transport.

2. Shock absorption: The air bubbles compress when impacted, absorbing shocks and protecting the packaged items from damage during transport. Solid polystyrene is rigid and brittle—it would transfer impacts directly to the packaged goods and could shatter.

3. Thermal insulation: Air is an excellent insulator. Expanded polystyrene keeps items at stable temperatures, which is important for food packaging and temperature-sensitive products. Solid polystyrene has no insulating air gaps.

4. Cost-effective: Expanded polystyrene uses less plastic material (more air, less polymer), making it cheaper to produce than solid polystyrene of the same volume.

5. Cushioning: The soft, compressible nature provides cushioning around fragile items, while solid polystyrene would provide no cushioning effect.

Practice Question

Compare LDPE and HDPE in terms of structure, properties, and uses. Explain the relationship between molecular structure and physical properties.

Show Answer

Structure differences:

LDPE (Low-Density Polyethylene): Has highly branched polymer chains. During polymerization under high pressure and temperature, side chains form off the main carbon backbone, creating a branched, tree-like structure.

HDPE (High-Density Polyethylene): Has linear polymer chains with minimal or no branching. Produced using special catalysts (Ziegler-Natta) at lower temperatures and pressures, resulting in straight-chain polymers.

How structure affects properties:

Chain packing: Linear HDPE chains can pack closely together in an ordered, crystalline arrangement, similar to how straight logs stack neatly. Branched LDPE chains cannot pack as closely because the branches create gaps and disorder, like trying to stack tree branches—there are many empty spaces.

Intermolecular forces: When polymer chains are close together (HDPE), the dispersion forces between chains are stronger because there's more surface contact. When chains are far apart (LDPE), intermolecular forces are weaker.

Property comparison:

Density:

LDPE: ~0.92 g/cm³ (lower density due to gaps between branches)
HDPE: ~0.95 g/cm³ (higher density from tight chain packing)

Flexibility:

LDPE: More flexible and softer (chains can slide past each other easily due to weaker forces)
HDPE: More rigid and harder (chains held firmly together by stronger forces)

Melting point:

LDPE: ~110°C (lower due to weaker intermolecular forces)
HDPE: ~130°C (higher because stronger forces require more energy to overcome)

Tensile strength:

LDPE: Lower strength (chains can be pulled apart more easily)
HDPE: Higher strength (chains held together more strongly)

Uses based on properties:

LDPE uses: Applications requiring flexibility and transparency:

Plastic shopping bags (need to be flexible and tear-resistant)
Cling wrap (must conform to food shape)
Squeeze bottles (need to be soft enough to squeeze)
Packaging film

HDPE uses: Applications requiring rigidity and strength:

Milk bottles (need to hold shape and support weight)
Detergent bottles (require strength and rigidity)
Water pipes (must be strong and not deform)
Cutting boards (need to be rigid)
Toys (require durability)

Structure-property relationship summary:

The key principle is that molecular structure determines how closely polymer chains can pack, which determines the strength of intermolecular forces, which determines physical properties. Branching in LDPE prevents tight packing → weaker forces → softer, more flexible, lower melting point. Linear structure in HDPE allows tight packing → stronger forces → harder, more rigid, higher melting point. This demonstrates how microscopic structure directly determines macroscopic properties.

Condensation Polymers

Condensation polymers are formed when monomers join together with the elimination of a small molecule, usually water (H₂O). Unlike addition polymers, not all atoms from the monomers end up in the polymer chain.

General Principles

Key features of condensation polymerization:

Monomers have TWO functional groups (one at each end)
Functional groups react to form a link between monomers
A small molecule (usually H₂O) is eliminated with each link formed
The polymer contains a different functional group than the monomers
Can involve one type of monomer (self-condensation) or two different monomers

Difference from addition polymerization:

Addition: All atoms from monomer → polymer (no elimination)

Condensation: Monomer atoms → polymer + small molecule (H₂O)

Common types:

Polyesters: Formed from carboxylic acids + alcohols (eliminates H₂O)
Polyamides: Formed from carboxylic acids + amines (eliminates H₂O)

General pattern:

Monomer A + Monomer B → Polymer + H₂O (or other small molecule)

Polyesters

Formation: Condensation reaction between dicarboxylic acids and diols (dialcohols).

Functional groups involved:

Carboxylic acid: -COOH

Alcohol: -OH

Ester link formed: -COO-

General reaction:

\text{HOOC-R-COOH} + \text{HO-R'-OH} \rightarrow \text{[-OC-R-CO-O-R'-O-]}_n + n\text{H}_2\text{O}

Dicarboxylic acid + Diol → Polyester + Water

Example: PET (Polyethylene Terephthalate)

Monomers:

1. Terephthalic acid (benzene-1,4-dicarboxylic acid): HOOC-C₆H₄-COOH

2. Ethylene glycol (ethane-1,2-diol): HO-CH₂-CH₂-OH

How water is eliminated:

The -OH from the acid and an H from the alcohol combine to form H₂O

The remaining parts join to form an ester link (-COO-)

Properties of PET:

Strong and lightweight
Transparent or can be colored
Good barrier to moisture and gases
Recyclable (recycling code #1)
Resistant to impact

Uses:

Soft drink and water bottles
Food containers
Synthetic fibers (polyester clothing, Dacron™)
Film and videotape
Fleece fabric (from recycled bottles)

Why it's called polyethylene terephthalate: The name comes from the two monomers—ethylene glycol and terephthalic acid.

Polyamides (Nylon)

Formation: Condensation reaction between dicarboxylic acids and diamines.

Functional groups involved:

Carboxylic acid: -COOH

Amine: -NH₂

Amide link formed: -CONH-

General reaction:

\text{HOOC-R-COOH} + \text{H}_2\text{N-R'-NH}_2 \rightarrow \text{[-OC-R-CONH-R'-NH-]}_n + n\text{H}_2\text{O}

Dicarboxylic acid + Diamine → Polyamide + Water

Example: Nylon-6,6

Monomers:

1. Adipic acid (hexanedioic acid): HOOC-(CH₂)₄-COOH

(6 carbons total, hence "6" in the name)

2. Hexamethylenediamine (1,6-diaminohexane): H₂N-(CH₂)₆-NH₂

(6 carbons total, hence the other "6" in the name)

Name explanation: "Nylon-6,6" indicates that both monomers contain 6 carbon atoms.

How water is eliminated:

The -OH from the acid and one H from the amine combine to form H₂O

The remaining parts join to form an amide link (-CO-NH-)

One H₂O molecule is eliminated for each amide link formed

Properties of Nylon:

Strong and durable
Elastic and flexible
Resistant to abrasion and chemicals
Can be drawn into fibers
Low friction coefficient
Can absorb small amounts of water (unlike polyester)

Uses:

Clothing (stockings, activewear, swimwear)
Ropes and fishing line
Carpets and upholstery
Toothbrush bristles
Zip fasteners
Bearings and gears (low friction)
Parachutes

Other nylon types:

Nylon-6: Made from a single monomer (caprolactam) through self-condensation

Nylon-6,10: Made from a 6-carbon diamine and a 10-carbon diacid

Identifying the Small Molecule Released (Band 6 Critical Habit)

Critical skill: Being able to identify which small molecule is eliminated during condensation polymerization.

For polyesters (acid + alcohol):

Look at what's removed from each monomer:

- From acid: -OH

- From alcohol: -H

- Combine: -OH + H = H₂O (water)

For polyamides (acid + amine):

Look at what's removed from each monomer:

- From acid: -OH

- From amine: -H

- Combine: -OH + H = H₂O (water)

General rule for condensation polymers:

In most condensation polymerizations you'll encounter in HSC Chemistry, the small molecule eliminated is water (H₂O).

How to show this in equations:

When writing condensation polymerization equations, ALWAYS include "+ nH₂O" on the product side, where n is the degree of polymerization.

Example for polyester:

n\text{HOOC-R-COOH} + n\text{HO-R'-OH} \rightarrow \text{[-OC-R-CO-O-R'-O-]}_n + 2n\text{H}_2\text{O}

Note: 2nH₂O because each diacid has two -COOH groups, so two water molecules form per repeat unit

Example for polyamide:

n\text{HOOC-R-COOH} + n\text{H}_2\text{N-R'-NH}_2 \rightarrow \text{[-OC-R-CO-NH-R'-NH-]}_n + 2n\text{H}_2\text{O}

Why this matters: Exam questions often ask you to identify the small molecule released or to write balanced equations. Missing the "+ nH₂O" loses marks!

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Carbons

Prefix

Alkane

meth-

methane

eth-

ethane

prop-

propane

but-

butane

pent-

pentane

hex-

hexane

hept-

heptane

oct-

octane