Organic | HSC - Wyatt's Notes

HSC chemistry study notes - Organic

flowchart TD
    A[Organic] --> B[Key Concepts]
    A --> C[Core Principles]
    A --> D[Practical Applications]
    B --> E[Fundamental definitions]
    C --> F[Design patterns]
    D --> G[Real-world usage]

1. Find the longest carbon chain containing the functional group
2. Number from the end nearest the functional group
3. Name substituents as prefixes
4. Add the appropriate suffix

Addition reactions (alkenes): Addition of H₂, H₂O, HX, Br₂

Substitution reactions (alkanes): Halogenation under UV light

Condensation reactions: Esterification, peptide bond formation

Elimination reactions: Dehydration of alcohols to form alkenes

Structural isomers: Same molecular formula, different structural formula

Stereoisomers: Same structural formula, different spatial arrangement

• Geometric (cis-trans) isomerism
• Optical isomerism

Problem: Name the following compound: CH₃CH(OH)CH₂CH₃

Solution:

Step 1: Find the longest chain containing -OH: 4 carbons (butane)

Step 2: Number from the end nearest -OH: -OH is on carbon 2

Step 3: Name: butan-2-ol

Answer: butan-2-ol

Problem: Write the equation for the reaction between ethanoic acid and ethanol. Name the product.

Solution:

Step 1: Identify the reactants:

• Ethanoic acid: CH₃COOH
• Ethanol: CH₃CH₂OH

Step 2: Write the condensation reaction: $\text{CH}_3\text{COOH} + \text{CH}_3\text{CH}_2\text{OH} \xrightarrow{\text{H}_2\text{SO}_4} \text{CH}_3\text{COOCH}_2\text{CH}_3 + \text{H}_2\text{O}$

Step 3: The product is an ester: ethyl ethanoate

Answer: $\text{CH}_3\text{COOH} + \text{CH}_3\text{CH}_2\text{OH} \rightarrow \text{CH}_3\text{COOCH}_2\text{CH}_3 + \text{H}_2\text{O}$ (ethyl ethanoate)

Problem: How many structural isomers does C₄H₁₀ have?

Solution:

Step 1: Calculate degree of unsaturation: $\frac{2(4) + 2 - 10}{2} = 0$ (saturated)

Step 2: Draw possible structures:

• Straight chain: CH₃CH₂CH₂CH₃ (butane)
• Branched: (CH₃)₃CH (2-methylpropane, isobutane)

Step 3: Count: 2 structural isomers

Answer: 2 structural isomers

1. Always identify the functional group first when naming compounds
2. Esterification is reversible; use excess reagent or remove water to drive forward
3. Cis-trans isomerism requires restricted rotation (C=C or ring) and two different groups on each carbon
4. Molecular formula alone doesn’t determine structure; always consider isomerism

1. Name: CH₃CH₂CH₂COOH
2. Write the equation for the reaction between propanoic acid and methanol
3. Draw all structural isomers of C₅H₁₂

Problem: Write the mechanism for the acid-catalysed hydration of ethene to form ethanol.

Solution:

Step 1: Protonation of ethene (electrophilic addition): $\text{CH}_2=\text{CH}_2 + \text{H}^+ \rightarrow \text{CH}_3\text{CH}_2^+$

Step 2: Nucleophilic attack by water: $\text{CH}_3\text{CH}_2^+ + \text{H}_2\text{O} \rightarrow \text{CH}_3\text{CH}_2\text{OH}_2^+$

Step 3: Deprotonation: $\text{CH}_3\text{CH}_2\text{OH}_2^+ \rightarrow \text{CH}_3\text{CH}_2\text{OH} + \text{H}^+$

Answer: The mechanism involves protonation, nucleophilic attack, and deprotonation. The acid catalyst is regenerated.

Problem: Does 2-bromobutane exhibit optical isomerism? Explain.

Solution:

Step 1: Draw the structure: CH₃CHBrCH₂CH₃

Step 2: Identify the chiral centre: Carbon 2 is bonded to four different groups: CH₃, H, Br, and CH₂CH₃

Step 3: Since carbon 2 has four different substituents, it is a chiral centre.

Step 4: The molecule exists as two enantiomers (R and 2-bromobutane), which rotate plane-polarized light in opposite directions.

Answer: Yes, 2-bromobutane exhibits optical isomerism because it has a chiral centre at carbon 2.

Problem: Predict whether the reaction of 2-bromopropan with alcoholic KOH gives mainly an alkene or an alcohol.

Solution:

Step 1: Identify the reaction conditions: alcoholic KOH (strong base in non-aqueous solvent)

Step 2: These conditions favour elimination (E2 mechanism) over substitution (SN2)

Step 3: The product is propene via dehydrohalogenation: $\text{CH}_3\text{CHBrCH}_3 + \text{KOH (alc)} \rightarrow \text{CH}_3\text{CH}=\text{CH}_2 + \text{KBr} + \text{H}_2\text{O}$

Answer: Elimination predominates, giving propene as the major product.

Organic chemistry is the basis of pharmaceuticals, plastics, food science, and biochemistry. Understanding reaction mechanisms allows chemists to design and synthesise new molecules with specific properties.

1. When naming compounds, always identify the longest chain containing the functional group
2. Degree of unsaturation = $\frac{2C + 2 - H}{2}$ helps determine if rings or double bonds are present
3. Esterification is reversible — use excess reagent or remove water to drive equilibrium
4. Cis-trans isomerism requires restricted rotation (C=C or ring) and two different groups on each carbon

Problem: Identify all functional groups in the following molecule: $\text{CH}_3\text{CH(OH)CHO}$

Solution:

Step 1: Draw the structure: $\text{CH}_3-\text{CH(OH)}-\text{CHO}$

Step 2: Identify functional groups:

• $-\text{OH}$: hydroxyl group (alcohol)
• $-\text{CHO}$: aldehyde group

Step 3: The molecule has both alcohol and aldehyde functional groups. The IUPAC name is 2-hydroxypropanal.

Answer: The molecule contains a hydroxyl group ($-\text{OH}$) and an aldehyde group ($-\text{CHO}$).

Common mistake: Confusing the aldehyde group ($-\text{CHO}$) with the alcohol group ($-\text{OH}$). The aldehyde carbon is bonded to both a hydrogen and a double-bonded oxygen.

Problem: Write the equation for the dehydration of ethanol to form ethene. What conditions are required?

Solution:

Step 1: Dehydration is an elimination reaction where water is removed from an alcohol.

Step 2: The reaction requires concentrated sulfuric acid as a catalyst and heating to approximately $170°\text{C}$: $\text{CH}_3\text{CH}_2\text{OH} \xrightarrow[\text{170°C}]{\text{conc. H}_2\text{SO}_4} \text{CH}_2=\text{CH}_2 + \text{H}_2\text{O}$

Step 3: At lower temperatures ($140°\text{C}$), the substitution product (diethyl ether) forms instead: $2\text{CH}_3\text{CH}_2\text{OH} \xrightarrow[\text{140°C}]{\text{conc. H}_2\text{SO}_4} \text{CH}_3\text{CH}_2\text{OCH}_2\text{CH}_3 + \text{H}_2\text{O}$

Answer: $\text{CH}_3\text{CH}_2\text{OH} \xrightarrow[\text{170°C}]{\text{conc. H}_2\text{SO}_4} \text{CH}_2=\text{CH}_2 + \text{H}_2\text{O}$

Common mistake: Confusing elimination (forming alkenes) with substitution (forming ethers). Temperature determines which product forms.

Problem: Draw the repeating unit of the polymer formed from the addition polymerisation of propene ($\text{CH}_3\text{CH}=\text{CH}_2$).

Solution:

Step 1: In addition polymerisation, the double bond opens up and monomers join together.

Step 2: The repeating unit is: $-\left[\text{CH}_2-\text{CH(CH}_3)\right]-$

Step 3: The polymer is polypropene (polypropylene). The methyl group ($\text{CH}_3$) is a side chain.

Step 4: The number of monomers $n$ in the polymer chain is the degree of polymerisation.

Answer: The repeating unit is $-\text{CH}_2-\text{CH(CH}_3)-$

Common mistake: Forgetting that in addition polymerisation, the double bond becomes a single bond in the polymer backbone. The side groups remain as branches.

Organic chemistry is about carbon’s unique bonding: Carbon forms four covalent bonds and can chain with other carbons indefinitely. This creates an enormous variety of molecules — from simple methane (CH₄) to complex proteins. The functional groups attached to the carbon backbone determine how the molecule behaves chemically.

Why it matters: Organic chemistry is the foundation of biochemistry, pharmaceuticals, and materials science. Understanding reaction mechanisms (SN1/SN2, addition, elimination) lets you predict products and design synthesis routes for new molecules.

The key insight: The structure of a molecule determines its reactivity — the same atoms arranged differently (isomers) can have completely different chemical properties.

Confusing SN1 and SN2 reaction mechanisms for haloalkanes. SN1 proceeds through a carbocation intermediate and gives racemisation. SN2 proceeds through backside attack and gives inversion of configuration. Students often assume all nucleophilic substitutions follow the same mechanism.

• Inorganic — Redox reactions and electrochemistry provide the foundation for understanding organic oxidation and reduction.
• Algebra — Logarithmic functions are used in rate equations for organic reaction kinetics.
• Calculus — Integration is used to derive integrated rate laws for organic reaction mechanisms.
• Reading — Close reading skills help analyse organic chemistry problems and interpret reaction mechanisms.

Forgetting Markovnikov’s rule for alkene addition reactions. When HX adds to an unsymmetrical alkene, hydrogen adds to the carbon with more hydrogens (Markovnikov product). Students sometimes add in the reverse direction, getting the anti-Markovnikov product which only forms in the presence of peroxides.

Misidentifying the functional group in organic molecules. Alcohols (-OH), ethers (-O-), aldehydes (-CHO), ketones (-CO-), and carboxylic acids (-COOH) have distinct structures and reactivities. Students often confuse esters (-COO-) with ethers (-O-) or aldehydes with ketones, leading to incorrect reaction predictions.