IB Chemistry SL — Organic Chemistry

Complete Study Guide

Topics Covered

1. Functional Groups & Homologous Series
2. IUPAC Naming
3. Structural Isomers
4. Combustion Reactions
5. Addition Reactions (Alkenes)
6. Oxidation of Alcohols
7. Nucleophilic Substitution (SN1 & SN2) HL
8. Free Radical Substitution
9. Polymerization
10. Properties of Organic Compounds

1. Functional Groups & Homologous Series

Organic chemistry is the study of carbon-containing compounds. Carbon is unique because it can form four bonds and chain together almost indefinitely, creating an enormous variety of molecules. To make sense of this variety, chemists group organic molecules into homologous series — families of molecules that share the same functional group. The functional group is the reactive part of the molecule — it determines how the compound behaves chemically. For example, all alcohols contain -OH and react similarly, regardless of how long their carbon chain is. Understanding functional groups is the foundation of everything else in organic chemistry: once you know what functional group is present, you can predict the reactions, the products, and even the physical properties.

Key Definitions

Term	Definition / Notes
Functional group	An atom or group of atoms that gives an organic molecule its characteristic chemical properties
Homologous series	A family of compounds with the same functional group, differing by $\text{CH}_2$ each time. Same general formula, similar chemical properties, gradually changing physical properties
Homologue	A member of a homologous series (e.g. methane, ethane, propane are all alkane homologues)

The Main Functional Groups You Must Know

Functional Group	Description & Example
Alkane	Single C-C bonds only. General formula: $\text{C}_n\text{H}_{2n+2}$. e.g. $\text{CH}_3\text{CH}_3$ (ethane)
Alkene	Contains a C=C double bond. General formula: $\text{C}_n\text{H}_{2n}$. e.g. $\text{CH}_2\text{=CH}_2$ (ethene)
Alcohol (-OH)	Contains an -OH group. e.g. $\text{CH}_3\text{CH}_2\text{OH}$ (ethanol). Suffix: -ol
Aldehyde (-CHO)	Contains -CHO at the END of the chain. e.g. $\text{CH}_3\text{CHO}$ (ethanal). Suffix: -al
Ketone (C=O)	Contains C=O in the MIDDLE of the chain. e.g. $\text{CH}_3\text{COCH}_3$ (propanone). Suffix: -one
Carboxylic acid (-COOH)	Contains -COOH at the end. e.g. $\text{CH}_3\text{COOH}$ (ethanoic acid). Suffix: -oic acid
Halogenoalkane	Contains a halogen (F, Cl, Br, I) bonded to carbon. e.g. $\text{CH}_3\text{CH}_2\text{Br}$ (bromoethane)
Ester (-COO-)	Contains a -COO- group. Formed from carboxylic acid + alcohol. e.g. $\text{CH}_3\text{COOCH}_2\text{CH}_3$ (ethyl ethanoate). Sweet/fruity smell. Suffix: -yl -anoate
Amine (-NH₂)	Contains an $\text{-NH}_2$ group bonded to carbon. e.g. $\text{CH}_3\text{NH}_2$ (methylamine). Suffix: -amine

Complete Homologous Series Reference

Series	General Formula	Functional Group	Suffix	Example
Alkanes	$\text{C}_n\text{H}_{2n+2}$	C-C single bonds only	-ane	$\text{CH}_4$ (methane), $\text{C}_2\text{H}_6$ (ethane)
Alkenes	$\text{C}_n\text{H}_{2n}$	C=C double bond	-ene	$\text{C}_2\text{H}_4$ (ethene), $\text{C}_3\text{H}_6$ (propene)
Alcohols	$\text{C}_n\text{H}_{2n+1}\text{OH}$	-OH (hydroxyl)	-ol	$\text{CH}_3\text{OH}$ (methanol), $\text{C}_2\text{H}_5\text{OH}$ (ethanol)
Aldehydes	$\text{C}_n\text{H}_{2n}\text{O}$	-CHO (at chain end)	-al	$\text{HCHO}$ (methanal), $\text{CH}_3\text{CHO}$ (ethanal)
Ketones	$\text{C}_n\text{H}_{2n}\text{O}$	C=O (in chain middle)	-one	$\text{CH}_3\text{COCH}_3$ (propanone)
Carboxylic acids	$\text{C}_n\text{H}_{2n}\text{O}_2$	-COOH	-oic acid	$\text{HCOOH}$ (methanoic acid), $\text{CH}_3\text{COOH}$ (ethanoic acid)
Esters	$\text{C}_n\text{H}_{2n}\text{O}_2$	-COO-	-yl -anoate	$\text{CH}_3\text{COOCH}_3$ (methyl ethanoate)
Halogenoalkanes	$\text{C}_n\text{H}_{2n+1}\text{X}$	C-X (X = F, Cl, Br, I)	prefix: fluoro/chloro/bromo/iodo	$\text{CH}_3\text{Cl}$ (chloromethane)
Amines	$\text{C}_n\text{H}_{2n+1}\text{NH}_2$	$\text{-NH}_2$	-amine	$\text{CH}_3\text{NH}_2$ (methylamine)

Practice Questions

MCQ (inline answers — students see answer immediately):

Written Questions (answers at end of guide):

W1. Define the term “homologous series” and explain why members of a homologous series have similar chemical properties but gradually changing physical properties. [3 marks]

Watch: Organic Chemistry — Functional Groups and Naming

Crash Course · 10 min · Hydrocarbons, functional groups, and why carbon is the backbone of organic chemistry

Crash Course · 11 min · IUPAC nomenclature — systematic naming of organic compounds from scratch

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2. IUPAC Naming

With millions of organic compounds known to science, a universal naming system is essential. The IUPAC (International Union of Pure and Applied Chemistry) system gives every compound a unique, systematic name that tells you exactly what the molecule looks like — the length of the carbon chain, the location and type of any branches, and the functional group present. The name is like an address: it pinpoints the structure precisely. The key insight is that IUPAC names are built in layers: first identify the longest chain (the parent), then number it to give branches and functional groups the lowest possible position numbers, then add prefixes and suffixes. Once you master this logic, you can name — or draw — any organic molecule from its name.

Prefix = Number of Carbon Atoms

Prefix	# of Carbons
meth-	1 carbon
eth-	2 carbons
prop-	3 carbons
but-	4 carbons
pent-	5 carbons
hex-	6 carbons

Naming Rules — Step by Step

• Step 1: Find the longest carbon chain — this gives the parent name (e.g. ‘pentane’ for 5 carbons)
• Step 2: Number the chain from the end closest to the functional group or branch
• Step 3: Name any branches (methyl, ethyl) with their position number (e.g. 2-methyl)
• Step 4: Add the suffix for the functional group (-ol, -al, -one, -oic acid)

Common Examples

Name	Explanation
2-methylbutane	Butane (4C) with a methyl branch on carbon 2. NOT 3-methylbutane (number from closest end)
2-methylpentane	Pentane (5C) with methyl on carbon 2
propan-1-ol	3-carbon chain, OH group on carbon 1 (primary alcohol)
propan-2-ol	3-carbon chain, OH group on carbon 2 (secondary alcohol)
1,2-dibromoethane	2-carbon chain with Br on carbons 1 and 2
2-methylpropan-2-ol	4C chain, OH on C2, methyl on C2 (tertiary alcohol)

Practice Questions

MCQ (inline answers — students see answer immediately):

Written Questions (answers at end of guide):

W1. Draw the structural formula and give the IUPAC name for the following compound: a 5-carbon alcohol with the hydroxyl group on carbon 3 and a methyl branch on carbon 2. [3 marks]

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3. Structural Isomers & Primary/Secondary/Tertiary

Because carbon can bond in so many ways, two molecules can share the same molecular formula (the same atoms) yet have completely different structures — and therefore different names, properties, and reactions. These are called structural isomers. Think of it like having the same Lego bricks but building different models. For example, $\text{C}_4\text{H}_{10}$ can be a straight chain (butane) or a branched chain (2-methylpropane) — same formula, different structure, different boiling point. Within alcohols and halogenoalkanes, a further important classification is whether the carbon bearing the functional group is bonded to one, two, or three other carbons — giving us primary, secondary, and tertiary compounds. This classification is critical because it completely determines which reactions are possible (e.g. tertiary alcohols cannot be oxidised; tertiary halogenoalkanes react by SN1 not SN2 HL).

Structural Isomers

Structural isomers have the SAME molecular formula but DIFFERENT structural arrangements.

Example: $\text{C}_4\text{H}_{10}$ has 2 structural isomers:

• $\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_3$ — butane (straight chain)
• $(\text{CH}_3)_2\text{CHCH}_3$ — 2-methylpropane (branched)

Primary, Secondary & Tertiary

This classification depends on how many carbon atoms are bonded to the carbon that carries the functional group:

Type	Description & Example
Primary (1°)	The carbon with the -OH or -X is bonded to only 1 other carbon. e.g. propan-1-ol, 1-bromopropane
Secondary (2°)	The carbon with the -OH or -X is bonded to 2 other carbons. e.g. propan-2-ol, 2-bromopropane
Tertiary (3°)	The carbon with the -OH or -X is bonded to 3 other carbons. e.g. 2-methylpropan-2-ol, $(\text{CH}_3)_3\text{CBr}$

Practice Questions

MCQ (inline answers — students see answer immediately):

Written Questions (answers at end of guide):

W1. Draw all the structural isomers of $\text{C}_3\text{H}_7\text{Br}$ and classify each as primary, secondary, or tertiary. Predict which isomer would react faster with aqueous $\text{NaOH}$ and explain why. [4 marks]

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4. Combustion Reactions

Combustion is the reaction of an organic compound with oxygen, releasing energy as heat and light — it is the basis of fuels, from petrol engines to gas cookers. All hydrocarbons (and most organic compounds) combust, but the products depend on how much oxygen is available. With plenty of oxygen, combustion is complete and clean — every carbon becomes $\text{CO}_2$ and every hydrogen becomes $\text{H}_2\text{O}$. With limited oxygen (such as in a poorly ventilated engine), combustion is incomplete, and more dangerous products form: carbon monoxide ($\text{CO}$), which is a colourless, odourless, highly toxic gas that binds to haemoglobin and prevents oxygen transport in the blood, and carbon soot ($\text{C}$). This is why car engines, boilers, and fires in enclosed spaces can be lethal. Being able to write balanced equations for both types — and know which products form under which conditions — is an essential IB skill.

Complete Combustion

Occurs with excess oxygen. Products are ONLY $\text{CO}_2$ and $\text{H}_2\text{O}$.

Complete Combustion General Equation:

$\text{C}_x\text{H}_y + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O}$

Example: $\text{C}_3\text{H}_8 + 5\text{O}_2 \rightarrow 3\text{CO}_2 + 4\text{H}_2\text{O}$ (propane)

Incomplete Combustion

Occurs with LIMITED oxygen. Produces some or all of: carbon monoxide ($\text{CO}$), carbon (soot/$\text{C}$), and water.

Incomplete Combustion (limited $\text{O}_2$):

$\text{C}_x\text{H}_y + \text{O}_2 \rightarrow \text{CO} + \text{C} + \text{H}_2\text{O}$

The IB may ask: ‘which products are formed?’ — answer includes CO and/or C (soot), but NOT $\text{CO}_2$ if very limited $\text{O}_2$

Concept	Details
Products of complete combustion	$\text{CO}_2$ and $\text{H}_2\text{O}$ only
Products of incomplete combustion	$\text{CO}$ (carbon monoxide), $\text{C}$ (carbon/soot), $\text{H}_2\text{O}$ — but NOT $\text{CO}_2$ in very limited $\text{O}_2$
Why incomplete combustion is dangerous	$\text{CO}$ is colourless, odourless, and toxic (binds to haemoglobin)

Practice Questions

MCQ (inline answers — students see answer immediately):

Written Questions (answers at end of guide):

W1. Explain why incomplete combustion of hydrocarbons in enclosed spaces is dangerous. In your answer, identify the toxic product formed and describe how it affects the human body. [3 marks]

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5. Addition Reactions of Alkenes

Alkenes are significantly more reactive than alkanes, and the reason comes down to one structural feature: the C=C double bond. A double bond consists of a strong sigma ($\sigma$) bond and a weaker pi ($\pi$) bond. It is this pi bond that is electron-rich and relatively easy to break, making alkenes attractive targets for molecules looking to gain electrons. In an addition reaction, the double bond opens up and two new groups are added across the two carbons — no atoms are lost. This is fundamentally different from substitution (where one atom replaces another) or elimination (where atoms are removed). Addition reactions are the key to understanding how alkenes are converted into alcohols, alkanes, halogenoalkanes, and ultimately polymers — making them one of the most commercially important reaction types in chemistry.

Key Addition Reactions

Reaction Type	Equation & Notes
Hydrogenation (+ $\text{H}_2$)	Alkene + $\text{H}_2$ $\rightarrow$ Alkane. Catalyst: Ni, heat. $\text{CH}_2\text{=CH}_2 + \text{H}_2 \rightarrow \text{CH}_3\text{CH}_3$
Halogenation (+ $\text{Br}_2$)	Alkene + $\text{Br}_2$ $\rightarrow$ dibromoalkane. No catalyst needed. $\text{CH}_2\text{=CH}_2 + \text{Br}_2 \rightarrow \text{CH}_2\text{BrCH}_2\text{Br}$. Test: bromine water turns from orange/brown to colourless
Hydrohalogenation (+ HBr)	Alkene + HBr $\rightarrow$ bromoalkane. $\text{CH}_2\text{=CH}_2 + \text{HBr} \rightarrow \text{CH}_3\text{CH}_2\text{Br}$
Hydration (+ $\text{H}_2\text{O}$)	Alkene + $\text{H}_2\text{O}$ $\rightarrow$ Alcohol. Catalyst: $\text{H}_3\text{PO}_4$, high temp & pressure. $\text{CH}_2\text{=CH}_2 + \text{H}_2\text{O} \rightarrow \text{CH}_3\text{CH}_2\text{OH}$ (important!)

Reaction Type = Addition

When an alkene reacts, the reaction type is ALWAYS called addition (or electrophilic addition at HL). At SL, just say ‘addition reaction.‘

Terminal vs Internal Alkenes — Alkene 1 and Alkene 2

IB questions often refer to ‘compound A’ and ‘compound B’ being different alkenes. The key distinction is whether the C=C double bond is at the end of the chain (terminal alkene, e.g. but-1-ene) or in the middle (internal alkene, e.g. but-2-ene). The number in the IUPAC name tells you the position of the double bond.

Type	Explanation
Terminal alkene (e.g. but-1-ene)	C=C double bond starts at C1 — at the END of the chain. Structure: $\text{CH}_2\text{=CH-CH}_2\text{-CH}_3$. Adding HBr can give two possible products (different carbons get H and Br).
Internal alkene (e.g. but-2-ene)	C=C double bond starts at C2 — in the MIDDLE of the chain. Structure: $\text{CH}_3\text{-CH=CH-CH}_3$. but-2-ene is symmetrical, so adding HBr gives only ONE product.
How to identify which is which	The number in the name = position of the LOWER carbon of the C=C. but-1-ene: double bond at C1. but-2-ene: double bond at C2. Both are $\text{C}_4\text{H}_8$ — they are structural isomers.
Why it matters in IB questions	In multi-step questions (e.g. Q33, Q35), ‘compound A’ is often one alkene and ‘compound B’ another. You must draw out the full structural formula for each and show different products.

Example — but-1-ene vs but-2-ene:

but-1-ene: $\text{CH}_2\text{=CHCH}_2\text{CH}_3$ (double bond at C1, terminal)

but-2-ene: $\text{CH}_3\text{CH=CHCH}_3$ (double bond at C2, internal, symmetrical)

Adding HBr to but-2-ene:

$\text{CH}_3\text{CH=CHCH}_3 + \text{HBr} \rightarrow \text{CH}_3\text{CH}_2\text{CHBrCH}_3 \text{ (one product, symmetrical)}$

Adding HBr to but-1-ene:

$\text{CH}_2\text{=CHCH}_2\text{CH}_3 + \text{HBr} \rightarrow \text{1-bromobutane OR 2-bromobutane (two possible products)}$

Both are $\text{C}_4\text{H}_8$ — same molecular formula, different position of the C=C. They are structural isomers.

Electrophilic Addition — Why It Is Called That HL

At SL you call it ‘addition reaction’. The full name is electrophilic addition — because the reagent that attacks the alkene is an electrophile (electron-seeking species). The C=C $\pi$ bond is electron-rich, attracting electrophiles. Understanding this helps you answer ‘explain’ questions.

Term	Explanation
Electrophile	Electron-deficient species — ACCEPTS electrons. Has $\delta$+ or full + charge. Examples: $\text{H}^+$ (in HBr), $\text{Br}_2$, carbocations
Nucleophile	Electron-rich species — DONATES electrons. Has lone pairs or negative charge. Examples: $\text{OH}^-$, $\text{CN}^-$, $\text{NH}_3$, $\text{Br}^-$
Why C=C is attacked	The $\pi$ bond electrons sit above/below the bond axis — this negative cloud attracts electrophiles.
What ‘addition’ means	Electrophile adds ACROSS the double bond. Both carbons of C=C get a new bond. No atoms lost.

Mechanism: Electrophilic Addition of HBr to Ethene

Step	Detail
Step 1	HBr approaches. Br is electronegative, pulling electron density from H, making H $\delta$+. H is the electrophile.
Step 2	$\pi$ electrons attack H. C-H bond forms on one carbon. Other carbon becomes carbocation ($\text{C}^+$). $\text{Br}^-$ released.
Step 3	$\text{Br}^-$ (nucleophile) attacks the carbocation. C-Br bond forms. Product: bromoethane.
Overall	$\text{CH}_2\text{=CH}_2 + \text{HBr} \rightarrow \text{CH}_3\text{CH}_2\text{Br}$. Type: addition (electrophilic addition at HL, just ‘addition’ at SL).

Practice Questions

MCQ (inline answers — students see answer immediately):

Written Questions (answers at end of guide):

W1. Describe a chemical test you could use to distinguish between ethane and ethene. State the reagent, the procedure, and the expected observations for each compound. [3 marks]

W2. But-1-ene and but-2-ene are structural isomers with the molecular formula $\text{C}_4\text{H}_8$. When each reacts with HBr, a different number of organic products is possible. Draw the structural formula of but-1-ene and but-2-ene, show the products formed with HBr in each case, and explain why the number of products differs. [4 marks]

Watch: Reactions of Organic Compounds — Addition, Combustion, and Derivatives

Crash Course · 10 min · Alkenes, alkynes, and addition reactions — double bond chemistry explained

Crash Course · 11 min · Hydrocarbon derivatives — alcohols, aldehydes, ketones, carboxylic acids, and their reactions

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6. Oxidation of Alcohols

Oxidation in organic chemistry generally means the addition of oxygen (or removal of hydrogen) to a molecule. Alcohols are particularly interesting because how far they can be oxidised depends entirely on their structure. The -OH group in an alcohol sits on a carbon atom. If that carbon still has a hydrogen attached (as in primary and secondary alcohols), oxidation can remove that hydrogen and form a new C=O bond. Primary alcohols can be oxidised in two stages: first to an aldehyde (still has that C-H bond next to the C=O), and then all the way to a carboxylic acid if excess oxidising agent is used. Secondary alcohols can only be oxidised to a ketone, because ketones have no adjacent C-H to lose. Tertiary alcohols have no such hydrogen at all — so they simply cannot be oxidised under normal conditions.

The reagent used in IB is acidified potassium dichromate(VI), written as $\text{K}_2\text{Cr}_2\text{O}_7$ / $\text{H}_2\text{SO}_4$. The chromium in this reagent starts as $\text{Cr}^{6+}$ (orange dichromate ion) and is reduced to $\text{Cr}^{3+}$ (green chromium ion) as it oxidises the alcohol. This colour change — orange to green — is your experimental proof that oxidation has occurred. If the solution stays orange, no oxidation has taken place (this is what happens with tertiary alcohols).

What Gets Oxidised to What? — Full Breakdown

Alcohol Type	Products, Steps & Colour
Primary alcohol (1°) e.g. ethanol, propan-1-ol	STEP 1 — Mild oxidation (limited $\text{K}_2\text{Cr}_2\text{O}_7$, distil off product immediately): $\rightarrow$ Aldehyde (still has a C-H on the carbonyl carbon). STEP 2 — Further oxidation (excess $\text{K}_2\text{Cr}_2\text{O}_7$, reflux): $\rightarrow$ Carboxylic acid (no more C-H to remove). Colour change: orange $\rightarrow$ green at BOTH steps
Secondary alcohol (2°) e.g. propan-2-ol, butan-2-ol	ONE STEP ONLY: $\rightarrow$ Ketone (C=O in middle of chain, NO adjacent H). Ketones CANNOT be oxidised further at SL conditions. Colour change: orange $\rightarrow$ green
Tertiary alcohol (3°) e.g. 2-methylpropan-2-ol	NO REACTION — the central carbon has NO hydrogen to remove. Solution STAYS ORANGE — this is the negative result. No colour change = proof it is tertiary

Full Worked Examples — Equations

Primary alcohol $\rightarrow$ Aldehyde (mild oxidation, distillation):

$\text{CH}_3\text{CH}_2\text{OH} + [\text{O}] \rightarrow \text{CH}_3\text{CHO} + \text{H}_2\text{O}$

ethanol $\rightarrow$ ethanal

$\text{CH}_3\text{CH}_2\text{CH}_2\text{OH} + [\text{O}] \rightarrow \text{CH}_3\text{CH}_2\text{CHO} + \text{H}_2\text{O}$

propan-1-ol $\rightarrow$ propanal

Use [O] to represent the oxidising agent in equations. Distil off the aldehyde immediately to prevent further oxidation.

Primary alcohol $\rightarrow$ Carboxylic acid (excess oxidising agent, reflux):

$\text{CH}_3\text{CH}_2\text{OH} + 2[\text{O}] \rightarrow \text{CH}_3\text{COOH} + \text{H}_2\text{O}$

ethanol $\rightarrow$ ethanoic acid

$\text{CH}_3\text{CH}_2\text{CH}_2\text{OH} + 2[\text{O}] \rightarrow \text{CH}_3\text{CH}_2\text{COOH} + \text{H}_2\text{O}$

propan-1-ol $\rightarrow$ propanoic acid

Use excess $\text{K}_2\text{Cr}_2\text{O}_7$/$\text{H}_2\text{SO}_4$ under reflux conditions. The aldehyde intermediate is oxidised again before it can escape.

Secondary alcohol $\rightarrow$ Ketone:

$\text{CH}_3\text{CH(OH)CH}_3 + [\text{O}] \rightarrow \text{CH}_3\text{COCH}_3 + \text{H}_2\text{O}$

propan-2-ol $\rightarrow$ propanone

$\text{CH}_3\text{CH(OH)CH}_2\text{CH}_3 + [\text{O}] \rightarrow \text{CH}_3\text{COCH}_2\text{CH}_3 + \text{H}_2\text{O}$

butan-2-ol $\rightarrow$ butanone

Colour change: orange $\rightarrow$ green. But if you add MORE $\text{K}_2\text{Cr}_2\text{O}_7$, nothing further happens — solution stays green.

Tertiary alcohol — NO reaction:

$(\text{CH}_3)_3\text{COH} + \text{K}_2\text{Cr}_2\text{O}_7/\text{H}_2\text{SO}_4 \rightarrow \text{NO REACTION}$

2-methylpropan-2-ol (solution stays orange)

The carbon bonded to -OH already has 3 carbon groups attached and NO hydrogen. There is nothing for the oxidising agent to remove.

How to Control Whether You Get Aldehyde or Carboxylic Acid

Goal	Conditions & Method
To get the ALDEHYDE (stop at first stage)	Use LIMITED $\text{K}_2\text{Cr}_2\text{O}_7$/$\text{H}_2\text{SO}_4$. Heat gently and DISTIL the product off immediately as it forms. The aldehyde has a lower boiling point than the alcohol — it escapes before it can be oxidised further. Practical setup: distillation apparatus, gentle heat
To get the CARBOXYLIC ACID (go all the way)	Use EXCESS $\text{K}_2\text{Cr}_2\text{O}_7$/$\text{H}_2\text{SO}_4$. Heat under REFLUX (the condenser returns vapours back to the flask). The aldehyde is trapped in the flask and oxidised again. Practical setup: reflux condenser, prolonged heating

How Do You PROVE Oxidation Has Occurred? — Experimental Tests

This is a key IB skill: not just knowing what happens, but being able to describe a test and its expected result that proves a reaction has occurred. For oxidation of alcohols you have several options:

Test	How to Do It & What It Proves
Test 1: Acidified $\text{K}_2\text{Cr}_2\text{O}_7$ (potassium dichromate)	Reagent: orange $\text{K}_2\text{Cr}_2\text{O}_7$ solution + dilute $\text{H}_2\text{SO}_4$. Positive result (oxidation occurred): solution turns GREEN. Negative result (no oxidation): solution stays ORANGE. Proves: the alcohol has been oxidised (e.g. primary or secondary alcohol present). Does NOT prove: which product formed (aldehyde or carboxylic acid)
Test 2: Tollens’ Reagent (silver mirror test)	Reagent: ammoniacal silver nitrate solution ($\text{Ag(NH}_3)_2^+$). Positive result: silver mirror forms on inside of test tube. Negative result: no change, solution stays colourless. Proves: an ALDEHYDE is present (distinguishes aldehyde from ketone). Why: aldehydes reduce $\text{Ag}^+$ to Ag metal. Ketones cannot do this.
Test 3: Fehling’s / Benedict’s Solution	Reagent: blue copper(II) solution ($\text{Cu}^{2+}$). Positive result: blue $\rightarrow$ brick-red precipitate ($\text{Cu}_2\text{O}$ formed). Negative result: stays blue. Proves: an ALDEHYDE is present. Why: aldehydes reduce $\text{Cu}^{2+}$ to $\text{Cu}^+$ (copper(I) oxide). Ketones cannot.
Test 4: pH / litmus test	Reagent: universal indicator or pH paper. Positive result: low pH (acidic, pH 1-4), indicator turns red. Negative result: neutral or slightly acidic. Proves: a CARBOXYLIC ACID has formed (from full oxidation of primary alcohol). Cross-check: also check for sharp, vinegar-like smell
Test 5: Smell / physical observation	Not an official IB test, but useful to describe: Aldehydes: often sweet, fruity smell. Carboxylic acids: sharp, vinegar-like smell (ethanoic acid = vinegar). Alcohols: distinctive alcoholic smell. Note: never rely on smell alone in an IB answer — always use a chemical test

Distinguishing Aldehydes from Ketones — Summary

Test	Aldehyde vs Ketone Result
Tollens’ reagent (ammoniacal $\text{AgNO}_3$)	Aldehyde: silver mirror on glass. Ketone: no reaction, stays colourless. Proves: aldehyde present vs ketone
Fehling’s / Benedict’s (blue $\text{Cu}^{2+}$ solution)	Aldehyde: blue $\rightarrow$ brick-red precipitate ($\text{Cu}_2\text{O}$). Ketone: stays blue. Proves: aldehyde present vs ketone
Acidified $\text{K}_2\text{Cr}_2\text{O}_7$ (orange)	Aldehyde: orange $\rightarrow$ green (can be further oxidised to carboxylic acid). Ketone: NO colour change — stays orange. Proves: whether further oxidation is possible
$\text{Na}_2\text{CO}_3$ solution	Carboxylic acid: fizzing / $\text{CO}_2$ bubbles produced. Aldehyde / ketone: no bubbles. Proves: carboxylic acid present (not just any carbonyl compound)

Reduction of Carbonyl Compounds — Back to Alcohol

Oxidation and reduction are reversible: just as alcohols are oxidised to carbonyls, carbonyl compounds can be reduced back to alcohols by adding hydrogen. The reducing agent used in IB is $\text{NaBH}_4$ (sodium tetrahydridoborate). In equations, write [H] to represent the reducing agent. Reduction converts C=O to C-OH.

Transformation	Product, Reagent & Example
Aldehyde + [H] $\rightarrow$ Primary alcohol	C=O at the END of the chain is reduced to C-OH. Reagent: $\text{NaBH}_4$ (or [H]) in water. Example: $\text{CH}_3\text{CHO} + 2[\text{H}] \rightarrow \text{CH}_3\text{CH}_2\text{OH}$ (ethanal $\rightarrow$ ethanol)
Ketone + [H] $\rightarrow$ Secondary alcohol	C=O in the MIDDLE of the chain is reduced to C-OH. Reagent: $\text{NaBH}_4$ (or [H]) in water. Example: $\text{CH}_3\text{COCH}_3 + 2[\text{H}] \rightarrow \text{CH}_3\text{CH(OH)CH}_3$ (propanone $\rightarrow$ propan-2-ol)
Carboxylic acid + [H] $\rightarrow$ Primary alcohol	Requires stronger reducing agent ($\text{LiAlH}_4$, more common at HL). Example: $\text{CH}_3\text{COOH} + 4[\text{H}] \rightarrow \text{CH}_3\text{CH}_2\text{OH} + \text{H}_2\text{O}$

Reduction Worked Examples

1. Aldehyde $\rightarrow$ primary alcohol:

$\text{CH}_3\text{CHO} + 2[\text{H}] \rightarrow \text{CH}_3\text{CH}_2\text{OH}$

ethanal $\rightarrow$ ethanol

$\text{CH}_3\text{CH}_2\text{CH}_2\text{CHO} + 2[\text{H}] \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{OH}$

butanal $\rightarrow$ butan-1-ol

2. Ketone $\rightarrow$ secondary alcohol:

$\text{CH}_3\text{COCH}_3 + 2[\text{H}] \rightarrow \text{CH}_3\text{CH(OH)CH}_3$

propanone $\rightarrow$ propan-2-ol

$\text{CH}_3\text{COCH}_2\text{CH}_3 + 2[\text{H}] \rightarrow \text{CH}_3\text{CH(OH)CH}_2\text{CH}_3$

butanone $\rightarrow$ butan-2-ol

Reduction = add [H]. Oxidation = add [O] or remove [H]. They are exact reverse operations.

Complete Oxidation-Reduction Pathway for a Primary Alcohol

Primary alcohol $\xrightarrow{\text{K}_2\text{Cr}_2\text{O}_7/\text{H}_2\text{SO}_4,\ \text{distil}}$ Aldehyde $\xrightarrow{\text{K}_2\text{Cr}_2\text{O}_7/\text{H}_2\text{SO}_4,\ \text{reflux}}$ Carboxylic acid

Carboxylic acid $\xleftarrow{\text{LiAlH}_4}$ Aldehyde $\xleftarrow{\text{NaBH}_4}$ Primary alcohol

Oxidation goes UP (more oxygen). Reduction goes DOWN (more hydrogen).

How to Prove Reduction Has Occurred

Stage	Test & Expected Result
Before reduction: test for aldehyde/ketone	Aldehyde: Tollens’ $\rightarrow$ silver mirror. Fehling’s $\rightarrow$ red precipitate. Ketone: $\text{K}_2\text{Cr}_2\text{O}_7$ stays orange. Tollens’ gives no mirror.
After reduction: test for alcohol	$\text{K}_2\text{Cr}_2\text{O}_7$/$\text{H}_2\text{SO}_4$: if primary or secondary alcohol formed $\rightarrow$ orange $\rightarrow$ green. This confirms the carbonyl has been converted to an alcohol.
Confirm aldehyde is gone	After reduction: Tollens’ gives NO silver mirror. Fehling’s stays blue. This proves the aldehyde has been consumed.

Practice Questions

MCQ (inline answers — students see answer immediately):

Written Questions (answers at end of guide):

W1. Explain why tertiary alcohols cannot be oxidised by acidified potassium dichromate. [3 marks]

W2. A student has two unlabelled bottles, one containing ethanal ($\text{CH}_3\text{CHO}$) and one containing propanone ($\text{CH}_3\text{COCH}_3$). Describe two different chemical tests the student could use to identify which bottle contains the aldehyde. For each test, state the reagent, the expected positive result, and the expected negative result. [4 marks]

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6A. Reduction of Carbonyl Compounds — The Reverse of Oxidation

Just as alcohols can be oxidised to carbonyl compounds (aldehydes, ketones, carboxylic acids), the reverse is also possible: carbonyl compounds can be reduced back to alcohols. Reduction in organic chemistry means addition of hydrogen (or removal of oxygen). The C=O double bond in a carbonyl is reduced to a C-OH group. This is the reverse of the oxidation pathway — and understanding both directions together (oxidation up, reduction down) is a powerful tool for synthesis questions.

The Reducing Agent

The standard reducing agent in IB organic chemistry is $\text{NaBH}_4$ (sodium tetrahydridoborate / sodium borohydride), used in aqueous or alcoholic solution. It delivers a hydride ion ($\text{H}^-$) to the carbonyl carbon, converting C=O to C-OH. Another reagent you may see is $\text{LiAlH}_4$ (lithium aluminium hydride) — a stronger reducing agent, used in dry ether solvent (more common at HL). At SL, the key is to know that [H] represents the reducing agent in equations, and to know what each carbonyl compound reduces to.

Reduction Reactions — What Reduces to What

Transformation	Product, Reagent & Example
Aldehyde $\rightarrow$ Primary alcohol	The C=O at the end of the chain gains H, becoming CH-OH. Reagent: $\text{NaBH}_4$ (or [H] in equations). e.g. $\text{CH}_3\text{CHO} + 2[\text{H}] \rightarrow \text{CH}_3\text{CH}_2\text{OH}$ (ethanal $\rightarrow$ ethanol)
Ketone $\rightarrow$ Secondary alcohol	The C=O in the middle of the chain gains H, becoming CH-OH. Reagent: $\text{NaBH}_4$ (or [H] in equations). e.g. $\text{CH}_3\text{COCH}_3 + 2[\text{H}] \rightarrow \text{CH}_3\text{CH(OH)CH}_3$ (propanone $\rightarrow$ propan-2-ol)
Carboxylic acid $\rightarrow$ Primary alcohol	Requires a stronger reducing agent ($\text{LiAlH}_4$). Not common at SL but worth knowing. e.g. $\text{CH}_3\text{COOH} + 4[\text{H}] \rightarrow \text{CH}_3\text{CH}_2\text{OH} + \text{H}_2\text{O}$ (ethanoic acid $\rightarrow$ ethanol)
Ester $\rightarrow$ Two alcohols	Full reduction of an ester gives two alcohols (HL topic mostly). e.g. $\text{CH}_3\text{COOCH}_2\text{CH}_3 + 4[\text{H}] \rightarrow \text{CH}_3\text{CH}_2\text{OH} + \text{CH}_3\text{CH}_2\text{OH}$

Full Worked Examples — Reduction Equations

The Full Oxidation-Reduction Pathway (Primary Alcohol)

Complete reversible pathway — primary alcohol family:

Primary alcohol $\xrightarrow{\text{oxidation, K}_2\text{Cr}_2\text{O}_7/\text{H}_2\text{SO}_4,\ \text{distil}}$ Aldehyde

Aldehyde $\xrightarrow{\text{oxidation, K}_2\text{Cr}_2\text{O}_7/\text{H}_2\text{SO}_4,\ \text{reflux}}$ Carboxylic acid

Carboxylic acid $\xrightarrow{\text{reduction, LiAlH}_4}$ Primary alcohol

Aldehyde $\xrightarrow{\text{reduction, NaBH}_4}$ Primary alcohol

Oxidation goes UP (more oxygen, less hydrogen). Reduction goes DOWN (more hydrogen, less oxygen).

How to Prove Reduction Has Occurred

Stage	Test & Expected Result
Before reduction: test for aldehyde or ketone	Aldehyde: Tollens’ reagent gives silver mirror. Fehling’s gives red precipitate. Ketone: $\text{K}_2\text{Cr}_2\text{O}_7$ stays orange. Tollens’ gives no mirror.
After reduction: test for alcohol	Test with $\text{K}_2\text{Cr}_2\text{O}_7$/$\text{H}_2\text{SO}_4$: if the product is a PRIMARY or SECONDARY alcohol, solution turns orange $\rightarrow$ green. If TERTIARY alcohol formed (impossible from reduction), no colour change.
Confirm it is no longer an aldehyde	After reduction, Tollens’ reagent should give NO silver mirror (aldehyde has been converted). Fehling’s should stay blue (no aldehyde remaining).
Additional confirmation	IR spectroscopy (HL): the C=O stretch at ~1700 cm$^{-1}$ disappears and broad O-H stretch appears at ~3300 cm$^{-1}$. At SL: smell changes (aldehydes are sweet/fruity; alcohols smell ‘alcoholic’).

Practice Questions

MCQ (inline answers — students see answer immediately):

Written Questions (answers at end of guide):

W1. Propanal ($\text{CH}_3\text{CH}_2\text{CHO}$) can be reduced to an alcohol. Write the equation for this reaction using [H], name the product, classify it as primary/secondary/tertiary, and state the reagent used. Describe how you could confirm that the reduction has occurred using a chemical test. [4 marks]

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6C. Esterification — Making Esters

Esters are organic compounds with a distinctive sweet, fruity smell — they are responsible for the aromas of many fruits (banana, pineapple, pear) and are widely used in perfumes, food flavourings, and solvents. An ester is formed when a carboxylic acid reacts with an alcohol in the presence of an acid catalyst (concentrated $\text{H}_2\text{SO}_4$). This reaction is called esterification (or condensation), because a small molecule — water — is released as a by-product. The reaction is reversible and reaches an equilibrium, which is why an acid catalyst and reflux conditions are used to push the reaction forward and increase yield.

The Esterification Reaction

$\text{Carboxylic acid} + \text{Alcohol} \xrightarrow{\text{conc. } \text{H}_2\text{SO}_4, \text{ reflux}} \text{Ester} + \text{H}_2\text{O}$

How Ester Names Work

Ester names have two parts: the alcohol part comes first (as an -yl group), and the acid part comes second (as an -anoate group).

Acid	Alcohol	Ester Produced	Ester Name
$\text{CH}_3\text{COOH}$ (ethanoic acid)	$\text{CH}_3\text{OH}$ (methanol)	$\text{CH}_3\text{COOCH}_3$	Methyl ethanoate
$\text{CH}_3\text{COOH}$ (ethanoic acid)	$\text{CH}_3\text{CH}_2\text{OH}$ (ethanol)	$\text{CH}_3\text{COOCH}_2\text{CH}_3$	Ethyl ethanoate
$\text{HCOOH}$ (methanoic acid)	$\text{CH}_3\text{OH}$ (methanol)	$\text{HCOOCH}_3$	Methyl methanoate
$\text{CH}_3\text{CH}_2\text{COOH}$ (propanoic acid)	$\text{CH}_3\text{OH}$ (methanol)	$\text{CH}_3\text{CH}_2\text{COOCH}_3$	Methyl propanoate

Worked Examples

Hydrolysis of Esters — The Reverse Reaction

Esters can be broken back down into the original carboxylic acid and alcohol by hydrolysis (reaction with water). This can be done under:

Condition	Details
Acid hydrolysis	Ester + $\text{H}_2\text{O}$ with dilute $\text{H}_2\text{SO}_4$ or $\text{HCl}$, heat under reflux → carboxylic acid + alcohol. Reversible.
Base hydrolysis (saponification)	Ester + $\text{NaOH}$(aq), heat under reflux → sodium salt of carboxylic acid + alcohol. Irreversible. Used in soap-making.

$\text{CH}_3\text{COOCH}_2\text{CH}_3 + \text{NaOH} \rightarrow \text{CH}_3\text{COONa} + \text{CH}_3\text{CH}_2\text{OH}$

Key Facts for IB

Point	Detail
Reaction type	Condensation (esterification) — small molecule ($\text{H}_2\text{O}$) released
Catalyst	Concentrated $\text{H}_2\text{SO}_4$
Conditions	Heat under reflux
Reversibility	Reversible — reaches equilibrium
How to detect an ester	Sweet, fruity smell. Does NOT turn litmus red (unlike carboxylic acid).
Hydrolysis	Reverse of esterification — ester + water → acid + alcohol
Uses of esters	Perfumes, food flavourings, solvents, plasticisers

Practice Questions

MCQ (inline answers — students see answer immediately):

Written Questions (answers at end of guide):

W1. The ester propyl methanoate ($\text{HCOOCH}_2\text{CH}_2\text{CH}_3$) is hydrolysed using aqueous $\text{NaOH}$. Write the equation for this reaction, name both organic products, and state the conditions required. Explain why base hydrolysis is described as irreversible, unlike acid hydrolysis. [4 marks]

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6B. Identifying Reactions & Experimental Tests

One of the most common IB question types gives you a reaction or a structural formula and asks you to identify the reaction type, reagents, conditions, or products. This section gives you worked examples of every major reaction type so you can recognise them on sight. The key is to look at what changes between reactant and product — then ask: what type of change is this?

The Reaction Identification Checklist

• Addition: Two reactants combine into ONE product. A double bond disappears. No atoms lost.
• Substitution: One atom or group is REPLACED by another. Two products always formed.
• Elimination: Atoms are REMOVED from adjacent carbons to form a double bond. HX or $\text{H}_2\text{O}$ lost.
• Oxidation: Oxygen added or hydrogen removed. Use [O]. Colour change orange $\rightarrow$ green with $\text{K}_2\text{Cr}_2\text{O}_7$.
• Hydrolysis: Bond broken by water (or $\text{OH}^-$). Often seen in esters and halogenoalkanes.
• Polymerization: Many identical monomers join into a long chain. n appears in the equation.

Example Set 1 — Identify the Reaction Type

Reaction Equation	Reaction Type & Key Clues
$\text{CH}_2\text{=CH}_2 + \text{H}_2 \rightarrow \text{CH}_3\text{CH}_3$	ADDITION (hydrogenation). The C=C double bond opens and $\text{H}_2$ adds across it. One product formed. Catalyst: Ni, heat.
$\text{CH}_2\text{=CH}_2 + \text{Br}_2 \rightarrow \text{CH}_2\text{BrCH}_2\text{Br}$	ADDITION (halogenation). Bromine adds across the double bond. Bromine water decolourises (orange $\rightarrow$ colourless). One product.
$\text{CH}_2\text{=CH}_2 + \text{HBr} \rightarrow \text{CH}_3\text{CH}_2\text{Br}$	ADDITION (hydrohalogenation). H and Br both add across the C=C.
$\text{CH}_2\text{=CH}_2 + \text{H}_2\text{O} \rightarrow \text{CH}_3\text{CH}_2\text{OH}$	ADDITION (hydration). Water adds across the double bond to form an alcohol. Catalyst: $\text{H}_3\text{PO}_4$, high temperature and pressure.
$\text{CH}_4 + \text{Cl}_2 \xrightarrow{\text{UV}} \text{CH}_3\text{Cl} + \text{HCl}$	FREE RADICAL SUBSTITUTION. Cl replaces H. UV light required. Two products. Alkane + halogen.
$\text{CH}_3\text{CH}_2\text{Br} + \text{NaOH(aq)} \rightarrow \text{CH}_3\text{CH}_2\text{OH} + \text{NaBr}$	NUCLEOPHILIC SUBSTITUTION (SN2) HL. -OH replaces -Br. Primary halogenoalkane. Two products.
$(\text{CH}_3)_3\text{CBr} + \text{NaOH(aq)} \rightarrow (\text{CH}_3)_3\text{COH} + \text{NaBr}$	NUCLEOPHILIC SUBSTITUTION (SN1) HL. Tertiary halogenoalkane. Via carbocation intermediate.
$\text{CH}_3\text{CH}_2\text{OH} + [\text{O}] \rightarrow \text{CH}_3\text{CHO} + \text{H}_2\text{O}$	OXIDATION. Primary alcohol $\rightarrow$ aldehyde. $\text{K}_2\text{Cr}_2\text{O}_7$/$\text{H}_2\text{SO}_4$, orange $\rightarrow$ green. Distil off product.
$\text{CH}_3\text{CHO} + [\text{O}] \rightarrow \text{CH}_3\text{COOH}$	OXIDATION. Aldehyde $\rightarrow$ carboxylic acid. $\text{K}_2\text{Cr}_2\text{O}_7$/$\text{H}_2\text{SO}_4$, orange $\rightarrow$ green. Reflux conditions.
$\text{CH}_3\text{CH(OH)CH}_3 + [\text{O}] \rightarrow \text{CH}_3\text{COCH}_3 + \text{H}_2\text{O}$	OXIDATION. Secondary alcohol (propan-2-ol) $\rightarrow$ ketone (propanone). Orange $\rightarrow$ green. Cannot oxidise further.
$n\ \text{CH}_2\text{=CH}_2 \rightarrow [-\text{CH}_2\text{-CH}_2-]_n$	ADDITION POLYMERIZATION. Many ethene monomers join. Double bond opens. No by-product.

Example Set 2 — Identify the Product

Starting Materials	Product & Reaction Type
Propene + $\text{Br}_2$ $\rightarrow$ ?	$\text{CH}_3\text{CHBrCH}_2\text{Br}$ (1,2-dibromopropane). Type: addition. Both Br atoms add across the C=C of propene.
But-2-ene + HBr $\rightarrow$ ?	$\text{CH}_3\text{CH}_2\text{CHBrCH}_3$ (2-bromobutane). Type: addition. H adds to one end of C=C, Br to the other.
Butan-1-ol + excess $\text{K}_2\text{Cr}_2\text{O}_7$/$\text{H}_2\text{SO}_4$ $\rightarrow$ ?	$\text{CH}_3\text{CH}_2\text{CH}_2\text{COOH}$ (butanoic acid). Type: oxidation (full). Primary alcohol + excess oxidant under reflux $\rightarrow$ carboxylic acid.
Butan-2-ol + $\text{K}_2\text{Cr}_2\text{O}_7$/$\text{H}_2\text{SO}_4$ $\rightarrow$ ?	$\text{CH}_3\text{COCH}_2\text{CH}_3$ (butanone). Type: oxidation. Secondary alcohol $\rightarrow$ ketone. Cannot go further.
2-methylpropan-2-ol + $\text{K}_2\text{Cr}_2\text{O}_7$/$\text{H}_2\text{SO}_4$ $\rightarrow$ ?	NO REACTION. Tertiary alcohol. Solution stays orange.
1-bromopropane + NaOH(aq) $\rightarrow$ ?	$\text{CH}_3\text{CH}_2\text{CH}_2\text{OH}$ + NaBr (propan-1-ol). Type: SN2 nucleophilic substitution HL. -OH replaces -Br.
2-methylpropane + $\text{Cl}_2$ $\xrightarrow{\text{UV}}$ ?	2-chloro-2-methylpropane (major) + HCl. Type: free radical substitution. Multiple products possible.
Ethene + $\text{H}_2\text{O}$ $\rightarrow$ ?	$\text{CH}_3\text{CH}_2\text{OH}$ (ethanol). Type: addition (hydration). Catalyst: $\text{H}_3\text{PO}_4$, 300°C, high pressure.

Example Set 3 — Identify the Reagent or Condition

Transformation	Reagent, Condition & Type
$\text{CH}_2\text{=CH}_2 \rightarrow \text{CH}_3\text{CH}_3$ (What reagent and condition?)	Reagent: $\text{H}_2$ (hydrogen gas). Condition: Ni catalyst, heat (about 150°C). Type: hydrogenation (addition)
$\text{CH}_3\text{CH}_2\text{Br} \rightarrow \text{CH}_3\text{CH}_2\text{OH}$ (What reagent and condition?)	Reagent: NaOH(aq) or KOH(aq) — aqueous sodium/potassium hydroxide. Condition: warm. Type: nucleophilic substitution (SN2) HL
$\text{CH}_3\text{CH}_2\text{OH} \rightarrow \text{CH}_3\text{CHO}$ (What reagent and condition?)	Reagent: acidified $\text{K}_2\text{Cr}_2\text{O}_7$ ($\text{K}_2\text{Cr}_2\text{O}_7$ + $\text{H}_2\text{SO}_4$). Condition: gentle heat, DISTIL product off immediately. Type: oxidation (limited)
$\text{CH}_3\text{CH}_2\text{OH} \rightarrow \text{CH}_3\text{COOH}$ (What reagent and condition?)	Reagent: excess acidified $\text{K}_2\text{Cr}_2\text{O}_7$. Condition: heat under REFLUX. Type: oxidation (full)
$\text{CH}_4 \rightarrow \text{CH}_3\text{Cl}$ (What reagent and condition?)	Reagent: $\text{Cl}_2$ (chlorine gas). Condition: UV light (sunlight). Type: free radical substitution
$\text{CH}_2\text{=CH}_2 \rightarrow [-\text{CH}_2\text{CH}_2-]_n$ (What reagent and condition?)	Reagent: none (self-reaction of monomers). Condition: high pressure, catalyst (Ziegler-Natta). Type: addition polymerization

Example Set 4 — Given a Colour Change, What Does It Tell You?

These are ‘interpret the observation’ questions — very common in IB Paper 2.

Observation	What It Tells You
$\text{K}_2\text{Cr}_2\text{O}_7$/$\text{H}_2\text{SO}_4$ turns orange $\rightarrow$ green	Oxidation has occurred. The substance is either a primary or secondary alcohol, OR an aldehyde. Something has been oxidised.
$\text{K}_2\text{Cr}_2\text{O}_7$/$\text{H}_2\text{SO}_4$ stays orange (no change)	No oxidation has occurred. The substance is either a tertiary alcohol, a ketone, or an alkane — none of these can be oxidised under these conditions.
Bromine water (orange/brown) $\rightarrow$ colourless	An addition reaction has occurred. A C=C double bond was present (alkene). The compound is UNSATURATED.
Bromine water stays orange/brown	No addition reaction. No C=C present. The compound is saturated (e.g. alkane, alcohol, ketone).
Tollens’ reagent $\rightarrow$ silver mirror	An aldehyde is present. The aldehyde reduced $\text{Ag}^+$ to Ag metal.
Fehling’s solution (blue) $\rightarrow$ brick-red precipitate	An aldehyde is present. $\text{Cu}^{2+}$ was reduced to $\text{Cu}_2\text{O}$. NOT a ketone.
Both solutions stay unchanged (Tollens’ + Fehling’s)	A KETONE is present (or a carboxylic acid, or an alcohol). NOT an aldehyde.
$\text{Na}_2\text{CO}_3$ solution produces $\text{CO}_2$ bubbles	A CARBOXYLIC ACID is present. Acid + carbonate $\rightarrow$ salt + water + $\text{CO}_2$.
pH paper shows strongly acidic (pH 1-3)	A carboxylic acid is likely present. Confirms acid functional group.

7. Nucleophilic Substitution (SN1 & SN2) HL

Halogenoalkanes contain a polar C-X bond (where X is a halogen). Because halogens are highly electronegative, they pull electron density away from the carbon they are bonded to, leaving that carbon slightly electron-deficient ($\delta$+). This makes it a target for nucleophiles — electron-rich species such as $\text{OH}^-$ that are attracted to positive (or partially positive) centres. In a nucleophilic substitution reaction, the nucleophile attacks the $\delta$+ carbon and the halogen leaves as a halide ion ($\text{X}^-$). The key question for the IB is: does this happen in one step or two? The answer depends on whether the halogenoalkane is primary, secondary, or tertiary. Primary halogenoalkanes react by SN2 (one concerted step, back-side attack). Tertiary halogenoalkanes react by SN1 (two steps, via a stable carbocation intermediate). Understanding these two mechanisms — and being able to draw them with curly arrows — is a major IB assessment skill.

SN2 — Primary Halogenoalkanes

Feature	Detail
Substrate	Primary halogenoalkane (1 carbon attached to the C-X carbon)
Mechanism	One step — $\text{OH}^-$ attacks the carbon from the BACK while $\text{X}^-$ leaves from the front simultaneously
Transition state	Carbon is bonded to 5 atoms momentarily — pentavalent carbon (draw with dashes)
Stereochemistry	Inversion of configuration (Walden inversion) — not required at SL
Rate	Depends on both [RX] and [$\text{OH}^-$] — bimolecular
Example	$\text{CH}_3\text{CH}_2\text{CH}_2\text{Br} + \text{NaOH(aq)} \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{OH} + \text{NaBr}$

SN1 — Tertiary Halogenoalkanes

Feature	Detail
Substrate	Tertiary halogenoalkane (3 carbons attached to the C-X carbon)
Mechanism	Two steps: Step 1: halogen leaves $\rightarrow$ carbocation intermediate. Step 2: $\text{OH}^-$ attacks carbocation
Intermediate	Carbocation (positively charged carbon, planar). Draw this in the mechanism.
Rate	Depends only on [RX] — unimolecular. Rate-determining step is step 1.
Why tertiary?	Tertiary carbocations are more stable (3 alkyl groups donate electron density)
Example	$(\text{CH}_3)_3\text{CBr} + \text{NaOH(aq)} \rightarrow (\text{CH}_3)_3\text{COH} + \text{NaBr}$

Curly Arrow Rules (Very Important for Marks)

• Curly arrows show movement of ELECTRON PAIRS (not atoms)
• Arrow starts from lone pair or bond, ends at atom or between atoms
• For SN2: one arrow from $\text{OH}^-$ lone pair to C, one arrow from C-X bond to X
• For SN1 Step 1: one arrow from C-X bond to X (heterolytic fission)
• For SN1 Step 2: one arrow from $\text{OH}^-$ lone pair to carbocation

Worked Examples — Nucleophilic Substitution Equations

Conditions: aqueous NaOH (or KOH), warm. $\text{OH}^-$ replaces the halogen. Always two products: the alcohol + sodium halide salt.

Addition vs Substitution — How to Tell Apart

Feature	Addition vs Substitution
Number of products	Addition: ONE product. Substitution: TWO products (organic + small molecule like HCl, NaBr).
Starting material	Addition: always an ALKENE (has C=C). Substitution: an ALKANE (+ UV light) or a HALOGENOALKANE (+ NaOH).
What happens to double bond	Addition: C=C disappears — becomes C-C. Substitution: no double bond involved — C-X single bond is replaced.
Bromine water test	Addition: bromine water decolourises (C=C reacted). Substitution: bromine water stays orange (no C=C).
Example comparison	Addition: $\text{CH}_2\text{=CH}_2 + \text{HBr} \rightarrow \text{CH}_3\text{CH}_2\text{Br}$ (1 product). Substitution: $\text{CH}_3\text{CH}_2\text{Br} + \text{NaOH} \rightarrow \text{CH}_3\text{CH}_2\text{OH} + \text{NaBr}$ (2 products)

Practice Questions

MCQ (inline answers — students see answer immediately):

Written Questions (answers at end of guide):

W1. 1-Bromobutane ($\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{Br}$) reacts with aqueous $\text{NaOH}$. Write the equation for the reaction, name the organic product, state the mechanism (SN1 or SN2), and explain why this mechanism is favoured for this substrate. [4 marks]

W2. Compare the mechanisms of SN1 and SN2 nucleophilic substitution. In your answer, state the number of steps for each, explain the role of the carbocation in SN1, and describe how the type of halogenoalkane (primary vs tertiary) determines the mechanism. [4 marks]

Watch: Nucleophilic Substitution — SN1 and SN2 Mechanisms

Crash Course · 13 min · Introduction to substitution reactions — what nucleophiles do and why leaving groups leave

Khan Academy · 9 min · SN2 stereochemistry — backside attack, inversion of configuration, and steric effects

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8. Free Radical Substitution

Alkanes are actually quite unreactive under normal conditions — they have no polar bonds, no double bonds, and no lone pairs, so they don’t attract nucleophiles or electrophiles. However, they will react with halogens in the presence of UV light via a mechanism involving free radicals — highly reactive species with an unpaired electron. UV light provides the energy to break the Cl-Cl (or Br-Br) bond homolytically, meaning each atom takes one electron. This produces two Cl radicals that kick off a self-sustaining chain reaction. The reaction proceeds through three distinct stages — initiation, propagation, and termination — and the IB expects you to write equations for all three and explain what happens in each. A key limitation of this reaction is that it can produce a mixture of products (e.g. $\text{CH}_3\text{Cl}$, $\text{CH}_2\text{Cl}_2$, $\text{CHCl}_3$, $\text{CCl}_4$), making it non-selective.

Overall Reaction (e.g. methane + chlorine):

$\text{CH}_4 + \text{Cl}_2 \xrightarrow{\text{UV light}} \text{CH}_3\text{Cl} + \text{HCl}$

Reaction type: free radical substitution

The Three Steps

Step	What Happens
1. Initiation	UV light breaks the Cl-Cl bond by HOMOLYTIC FISSION. $\text{Cl}_2 \rightarrow 2\ \text{Cl}\cdot$ (Each Cl atom gets one electron — shown as $\text{Cl}\cdot$, a free radical)
2. Propagation	Two steps that repeat in a chain reaction: (a) $\text{Cl}\cdot + \text{CH}_4 \rightarrow \text{HCl} + \text{CH}_3\cdot$ (b) $\text{CH}_3\cdot + \text{Cl}_2 \rightarrow \text{CH}_3\text{Cl} + \text{Cl}\cdot$ The $\text{Cl}\cdot$ produced in (b) re-enters step (a)
3. Termination	Two radicals combine — the chain ends: $\text{Cl}\cdot + \text{Cl}\cdot \rightarrow \text{Cl}_2$ / $\text{CH}_3\cdot + \text{Cl}\cdot \rightarrow \text{CH}_3\text{Cl}$ / $\text{CH}_3\cdot + \text{CH}_3\cdot \rightarrow \text{C}_2\text{H}_6$

What is a Free Radical?

Property	Detail
Definition	A species with an UNPAIRED electron
Charge	NEUTRAL (no charge)
Formation	By HOMOLYTIC fission (bond splits equally, each atom gets one electron)
Notation	Written with a dot: $\text{Cl}\cdot$, $\text{CH}_3\cdot$
Reactivity	Very reactive because the unpaired electron makes it unstable

Homolytic vs Heterolytic Bond Fission

When a covalent bond breaks, the two bonding electrons must go somewhere. Free radical substitution uses homolytic fission; ionic mechanisms (SN1, SN2 HL, electrophilic addition) use heterolytic fission. The IB mark scheme requires you to use these exact words.

Type	Definition, Result & Example
Homolytic fission (‘homo’ = equal)	Bond breaks EQUALLY — each atom gets ONE electron. Result: two FREE RADICALS (neutral, one unpaired electron each). Shown as: $\text{Cl-Cl} \xrightarrow{\text{UV}} \text{Cl}\cdot + \text{Cl}\cdot$. When: FREE RADICAL reactions (alkane + halogen, UV light). Memory: HOME run = EQUAL split
Heterolytic fission (‘hetero’ = unequal)	Bond breaks UNEQUALLY — one atom takes BOTH electrons. Result: two IONS (one cation +, one anion -). Shown as: $\text{C-Br} \rightarrow \text{C}^+ + \text{Br}^-$. When: IONIC mechanisms (SN1, SN2 HL, electrophilic addition). Memory: HETERO = DIFFERENT = unequal split

Analogy: Two people sharing 2 sweets and then splitting up:

• Homolytic: Each person takes 1 sweet — both leave with something (radicals, neutral, reactive)
• Heterolytic: One person takes both sweets — one leaves with 2 (anion), one with 0 (cation)

Reaction Type	Fission Type
Free radical substitution (this section)	Homolytic — UV light splits Cl-Cl equally $\rightarrow$ $\text{Cl}\cdot + \text{Cl}\cdot$
SN1 mechanism (Section 7) HL	Heterolytic — C-X breaks: $\text{C}^+$ (carbocation) + $\text{X}^-$ (halide ion)
SN2 mechanism (Section 7) HL	Heterolytic — C-X breaks as $\text{OH}^-$ attacks; $\text{Br}^-$ leaves as ion
Electrophilic addition (Section 5)	Heterolytic — H-Br breaks: $\text{H}^+$ adds to alkene carbon, $\text{Br}^-$ released

Homolytic fission — free radical substitution:

$\text{Cl:Cl} \xrightarrow{\text{UV}} \text{Cl}\cdot + \text{Cl}\cdot$

(1 electron each — both neutral radicals)

Required IB wording: ‘UV light causes homolytic fission of the Cl-Cl bond to form two Cl radicals.’

Heterolytic fission — SN1 tertiary halogenoalkane: HL

$\text{C-Br} \rightarrow \text{C}^+ + \text{Br}^-$

(both electrons go to Br — forms ions)

The $\text{C}^+$ is the carbocation intermediate in the SN1 mechanism.

Practice Questions

MCQ (inline answers — students see answer immediately):

Written Questions (answers at end of guide):

W1. Write equations for the initiation, two propagation steps, and one termination step for the free radical substitution of ethane ($\text{C}_2\text{H}_6$) with bromine ($\text{Br}_2$) in the presence of UV light. State the overall equation and explain why a mixture of products is obtained. [5 marks]

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9. Polymerization

Polymers are giant molecules made by linking thousands of small repeating units called monomers together. They are the basis of all plastics, synthetic fibres, and many natural materials (DNA, proteins, and cellulose are all biological polymers). In IB SL, you need to understand two types of polymerization. Addition polymerization occurs when alkene monomers (containing C=C) simply join end-to-end as the double bond opens up — nothing is lost or produced as a by-product. Condensation polymerization involves monomers with two functional groups reacting to form a polymer chain while releasing a small molecule (usually water) each time a bond forms. The real-world scale of polymers explains why they behave so differently from their monomers: ethene is a gas, but polyethene is a tough solid plastic used in everything from food packaging to bulletproof vests.

Addition Polymerization

Small molecules (monomers) containing C=C double bonds join together. The double bond opens up and monomers link in a chain. No other products are formed.

Addition polymerization of ethene $\rightarrow$ polyethene:

$n\ \text{CH}_2\text{=CH}_2 \rightarrow [-\text{CH}_2\text{-CH}_2-]_n$

The monomer must contain a C=C double bond. The polymer has no double bonds.

Monomer	Polymer & Uses
Chloroethene ($\text{CH}_2\text{=CHCl}$)	$\rightarrow$ poly(chloroethene), PVC. Used in pipes, window frames
Ethene ($\text{CH}_2\text{=CH}_2$)	$\rightarrow$ poly(ethene)/polyethene. Plastic bags, bottles
Propene ($\text{CH}_2\text{=CHCH}_3$)	$\rightarrow$ poly(propene). Ropes, carpets
Tetrafluoroethene ($\text{CF}_2\text{=CF}_2$)	$\rightarrow$ PTFE (Teflon). Non-stick coatings

How to Draw the Repeating Unit

• Remove the double bond from the monomer
• Draw the single bond chain, connecting to the next unit with bonds extending outside the brackets
• Add a subscript ‘n’ outside the brackets

Example for chloroethene: monomer is $\text{CH}_2\text{=CHCl}$ $\rightarrow$ repeating unit is $[-\text{CH}_2\text{-CHCl}-]_n$

Condensation Polymerization

Monomers with TWO functional groups react together, releasing a small molecule (usually water or HCl) with each bond formed.

Type	Details
Polyester	Diol + dicarboxylic acid $\rightarrow$ polyester + water. e.g. PET (drinks bottles, clothing)
Polyamide (nylon)	Diamine + dicarboxylic acid $\rightarrow$ polyamide + water. e.g. Nylon-6,6, Kevlar
Small molecule produced	$\text{H}_2\text{O}$ (in polyester and polyamide from carboxylic acid groups)

Why Are Monomers Gases/Liquids but Polymers Are Solids?

Monomers have small molecular masses $\rightarrow$ low intermolecular forces $\rightarrow$ low boiling points $\rightarrow$ gases or volatile liquids.

Polymers have very large molecular masses (thousands of monomer units) $\rightarrow$ very strong intermolecular forces $\rightarrow$ high melting points $\rightarrow$ solids at room temperature.

Practice Questions

MCQ (inline answers — students see answer immediately):

Written Questions (answers at end of guide):

W1. Propene ($\text{CH}_2\text{=CHCH}_3$) undergoes addition polymerization. Draw the repeating unit of the polymer formed, name the polymer, and explain why the monomer is a gas at room temperature but the polymer is a solid. [3 marks]

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10. Properties of Organic Compounds

The physical properties of organic compounds — particularly boiling point and solubility — are largely determined by the intermolecular forces between molecules. The stronger the intermolecular forces, the more energy is needed to separate molecules from each other, and the higher the boiling point. All organic molecules have London dispersion forces (van der Waals), which increase with molecular size and surface area. Branching reduces surface area and weakens these forces, lowering the boiling point. Compounds with -OH groups (alcohols, carboxylic acids) can form hydrogen bonds — the strongest intermolecular force at SL — dramatically raising their boiling points compared to similarly sized alkanes. Organic chemistry also has enormous industrial importance: crude oil is the world’s primary source of fuels, solvents, and feedstocks for making plastics, medicines, and fertilisers. Understanding where these products come from — and the environmental consequences of their use — is part of the IB curriculum.

Boiling Points

Factor	Effect on Boiling Point
Molecular mass increases	Boiling point increases (stronger London/van der Waals forces)
Branching increases	Boiling point decreases (less surface area, weaker dispersion forces). e.g. pentane > 2-methylbutane > 2,2-dimethylpropane
Hydrogen bonding	Alcohols and carboxylic acids have H-bonding $\rightarrow$ much higher boiling points than similar alkanes
Polar functional groups	Aldehydes/ketones have dipole-dipole forces $\rightarrow$ higher bp than similar alkanes

Crude Oil & Petroleum

Product	Source & Notes
Fractional distillation	Separates crude oil by boiling point into fractions. Shorter chains = lower bp = collected at top
Plastics	From alkenes via polymerization (e.g. ethene $\rightarrow$ polyethene)
Margarine	From vegetable oils + $\text{H}_2$ (hydrogenation of C=C double bonds)
Motor fuel	Mainly alkanes (petrol = C5-C10 chain lengths)

Crude Oil Concerns

• Non-renewable resource — will run out
• Burning releases $\text{CO}_2$ $\rightarrow$ greenhouse effect / climate change
• Incomplete combustion releases $\text{CO}$ $\rightarrow$ toxic
• Combustion of S impurities $\rightarrow$ $\text{SO}_2$ $\rightarrow$ acid rain

Quick Reaction Summary Table

Reaction	Type & Product
Alkane + halogen (UV)	Free radical substitution $\rightarrow$ halogenoalkane + HX
Alkene + $\text{Br}_2$	Addition $\rightarrow$ dibromoalkane (decolourises bromine water)
Alkene + $\text{H}_2$ (Ni, heat)	Addition (hydrogenation) $\rightarrow$ alkane
Alkene + HBr	Addition (hydrohalogenation) $\rightarrow$ bromoalkane
Alkene + $\text{H}_2\text{O}$ ($\text{H}_3\text{PO}_4$)	Addition (hydration) $\rightarrow$ alcohol
Primary alcohol + $\text{K}_2\text{Cr}_2\text{O}_7$	Oxidation $\rightarrow$ aldehyde $\rightarrow$ carboxylic acid
Secondary alcohol + $\text{K}_2\text{Cr}_2\text{O}_7$	Oxidation $\rightarrow$ ketone
Tertiary alcohol + $\text{K}_2\text{Cr}_2\text{O}_7$	No reaction (stays orange)
Halogenoalkane (1°) + NaOH HL	SN2 nucleophilic substitution $\rightarrow$ alcohol
Halogenoalkane (3°) + NaOH HL	SN1 nucleophilic substitution $\rightarrow$ alcohol (via carbocation)
Alkene + n (polymerize)	Addition polymerization $\rightarrow$ polymer
Carboxylic acid + Alcohol (conc. $\text{H}_2\text{SO}_4$)	Esterification (condensation) $\rightarrow$ ester + $\text{H}_2\text{O}$
Ester + NaOH(aq)	Base hydrolysis (saponification) $\rightarrow$ sodium salt + alcohol

Practice Questions

MCQ (inline answers — students see answer immediately):

Written Questions (answers at end of guide):

W1. Arrange the following compounds in order of increasing boiling point: pentane, pentan-1-ol, 2,2-dimethylpropane. Justify your answer by identifying the types of intermolecular forces present in each compound and explaining how molecular structure affects boiling point. [4 marks]

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11. What You MUST Memorise — IB Exam Checklist

This section is your final revision checklist. These are the facts, colour changes, reagents, conditions, and definitions that the IB repeatedly tests. If you can answer every item below from memory, you are well prepared for the organic chemistry section of your exam.

1. Colour Changes — Know These Cold

2. Reagents & Conditions — Exact Answers Required

3. Definitions — Write These Out Word for Word

4. General Formulas — Must Know

5. Key Reaction Sequences — The Big Picture

6. Things Students Always Get Wrong — Final Warnings

Common Mistake	Correct Understanding
Aldehyde vs Ketone position	Aldehyde: C=O is at the END of the chain (always on C1). Has a C-H next to C=O. Ketone: C=O is in the MIDDLE of the chain. No C-H on the carbonyl carbon.
Tertiary alcohol oxidation	Tertiary alcohols DO NOT react with $\text{K}_2\text{Cr}_2\text{O}_7$. Solution stays orange. This is NOT a failure of the test — it is the expected result. Say: ‘no oxidation occurs because the carbon bearing -OH has no hydrogen.‘
Distillation vs reflux	Distil = remove product early = only get aldehyde. Reflux = keep product in = get carboxylic acid. Never swap these — it loses marks every time.
Naming the mechanism	Say ‘free radical substitution’ not just ‘substitution’ for alkane + halogen/UV. Say ‘nucleophilic substitution’ not just ‘substitution’ for halogenoalkane + NaOH. HL The IB mark scheme requires the full name.
Two products in substitution	Substitution ALWAYS gives two products. If you write only one, you will lose a mark. e.g. $\text{CH}_3\text{Br} + \text{NaOH} \rightarrow \text{CH}_3\text{OH}$ is INCOMPLETE — must add + NaBr.
Homolytic fission in initiation	Do not say ‘the bond breaks’. Say: ‘UV light causes homolytic fission of the Cl-Cl bond to form two Cl radicals.’ The word homolytic is required.
Addition adds across C=C, not C-C	Addition only works if there is a C=C double bond. Alkanes have no double bonds, so they cannot undergo addition — only free radical substitution with UV.
Reduction uses [H], not $\text{H}^+$	[H] represents a hydride ion from $\text{NaBH}_4$. Do not write $\text{H}^+$ (that is an acid, not a reductant). The IB accepts [H] or $\text{NaBH}_4$ in equations.

IB Chemistry SL — Organic Chemistry Study Guide | Good luck on your exam!

12. Written Question Answers

Section 1, W1:

A homologous series is a family of compounds that share the same functional group and the same general formula, with each successive member differing by $\text{CH}_2$ [1 mark]. Members have similar chemical properties because they all contain the same functional group, which determines how the molecule reacts [1 mark]. Physical properties (such as boiling point) change gradually because as the carbon chain lengthens, the molecular mass increases, leading to stronger London dispersion forces between molecules, which require more energy to overcome [1 mark].

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Section 2, W1:

The compound is 2-methylpentan-3-ol [1 mark]. It has a 5-carbon longest chain (pentane) with a hydroxyl group on carbon 3 and a methyl branch on carbon 2 [1 mark]. Structural formula: $\text{CH}_3\text{CH(CH}_3\text{)CH(OH)CH}_2\text{CH}_3$ [1 mark].

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Section 3, W1:

There are two structural isomers of $\text{C}_3\text{H}_7\text{Br}$:

• 1-Bromopropane ($\text{CH}_3\text{CH}_2\text{CH}_2\text{Br}$) — this is a primary halogenoalkane (the carbon bonded to Br is attached to only 1 other carbon) [1 mark].
• 2-Bromopropane ($\text{CH}_3\text{CHBrCH}_3$) — this is a secondary halogenoalkane (the carbon bonded to Br is attached to 2 other carbons) [1 mark].

2-Bromopropane would react faster with aqueous $\text{NaOH}$ [1 mark] because the C-Br bond is more accessible in the secondary isomer. At SL level, the key point is that the type of halogenoalkane affects the rate of substitution [1 mark].

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Section 4, W1:

Incomplete combustion produces carbon monoxide ($\text{CO}$) [1 mark]. Carbon monoxide is a colourless and odourless gas, making it undetectable without specialised equipment [1 mark]. It is toxic because it binds irreversibly to haemoglobin in red blood cells (forming carboxyhaemoglobin), preventing oxygen from being transported around the body, which can lead to oxygen deprivation and death [1 mark].

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Section 5, W1:

Add bromine water ($\text{Br}_2\text{(aq)}$, which is orange/brown) to separate samples of each compound [1 mark]. With ethene: the bromine water decolourises (turns from orange to colourless) because $\text{Br}_2$ undergoes an addition reaction across the C=C double bond, forming 1,2-dibromoethane [1 mark]. With ethane: the bromine water remains orange/brown because ethane is saturated (no C=C) and cannot undergo addition reactions under these conditions [1 mark].

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Section 5, W2:

But-1-ene: $\text{CH}_2\text{=CHCH}_2\text{CH}_3$ (terminal alkene, C=C at carbon 1) [1 mark]. But-2-ene: $\text{CH}_3\text{CH=CHCH}_3$ (internal alkene, C=C at carbon 2) [1 mark].

When but-1-ene reacts with HBr, two products are possible: 1-bromobutane ($\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{Br}$) and 2-bromobutane ($\text{CH}_3\text{CHBrCH}_2\text{CH}_3$), because the C=C is asymmetric — H and Br can add in two different orientations [1 mark].

When but-2-ene reacts with HBr, only one product is formed: 2-bromobutane ($\text{CH}_3\text{CHBrCH}_2\text{CH}_3$), because the C=C is symmetric — adding H to either carbon of the double bond gives the same product [1 mark].

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Section 6, W1:

In a tertiary alcohol, the carbon atom bonded to the -OH group is also bonded to three other carbon atoms [1 mark]. This means there is no hydrogen atom on the carbon bearing the -OH group [1 mark]. Oxidation of alcohols requires the removal of a hydrogen from the C-OH carbon to form a C=O bond. Since there is no hydrogen available on this carbon in a tertiary alcohol, oxidation cannot occur, and the acidified $\text{K}_2\text{Cr}_2\text{O}_7$ solution remains orange [1 mark].

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Section 6, W2:

Test 1 — Tollens’ reagent (silver mirror test): Add ammoniacal silver nitrate solution ($\text{Ag(NH}_3)_2^+$) to each sample and warm gently. With ethanal (aldehyde): a silver mirror forms on the inside of the test tube ($\text{Ag}^+$ is reduced to Ag metal). With propanone (ketone): no change — solution remains colourless [2 marks].

Test 2 — Fehling’s (or Benedict’s) solution: Add blue Fehling’s solution ($\text{Cu}^{2+}$) to each sample and warm. With ethanal (aldehyde): the blue solution changes to a brick-red precipitate ($\text{Cu}_2\text{O}$ formed as $\text{Cu}^{2+}$ is reduced to $\text{Cu}^+$). With propanone (ketone): the solution remains blue — no reaction occurs [2 marks].

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Section 6A, W1:

Equation: $\text{CH}_3\text{CH}_2\text{CHO} + 2[\text{H}] \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{OH}$ [1 mark]. The product is propan-1-ol, a primary alcohol (the -OH is on a carbon bonded to only one other carbon) [1 mark]. Reagent: $\text{NaBH}_4$ (sodium tetrahydridoborate / sodium borohydride) in water [1 mark]. To confirm reduction has occurred: test the product with acidified $\text{K}_2\text{Cr}_2\text{O}_7$ — if propan-1-ol is present, the solution will change from orange to green (the alcohol can be oxidised back). Additionally, test the product with Tollens’ reagent — if no silver mirror forms, the aldehyde has been fully consumed [1 mark].

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Section 6C, W1:

Equation: $\text{HCOOCH}_2\text{CH}_2\text{CH}_3 + \text{NaOH} \rightarrow \text{HCOONa} + \text{CH}_3\text{CH}_2\text{CH}_2\text{OH}$ [1 mark]. Products: sodium methanoate ($\text{HCOONa}$, the sodium salt of methanoic acid) and propan-1-ol ($\text{CH}_3\text{CH}_2\text{CH}_2\text{OH}$) [1 mark]. Conditions: heat under reflux with aqueous $\text{NaOH}$ [1 mark]. Base hydrolysis is irreversible because the carboxylic acid product reacts with $\text{NaOH}$ to form the sodium salt (carboxylate ion), which cannot react back with the alcohol to reform the ester. In acid hydrolysis, the free carboxylic acid can recombine with the alcohol, making it reversible [1 mark].

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Section 7, W1: HL

Equation: $\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{Br} + \text{NaOH(aq)} \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{OH} + \text{NaBr}$ [1 mark]. The organic product is butan-1-ol [1 mark]. The mechanism is SN2 (bimolecular nucleophilic substitution) [1 mark]. SN2 is favoured because 1-bromobutane is a primary halogenoalkane — the carbon bonded to Br has only one other carbon attached, leaving relatively little steric hindrance. This allows the nucleophile ($\text{OH}^-$) to attack the $\delta$+ carbon directly from the back in a single concerted step, simultaneously displacing $\text{Br}^-$ [1 mark].

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Section 7, W2: HL

SN2 mechanism: Occurs in one step. The nucleophile ($\text{OH}^-$) attacks the $\delta$+ carbon from the opposite side to the halogen, forming a new C-OH bond while the C-X bond breaks simultaneously. There is no intermediate — a single transition state is formed [1 mark]. SN2 is favoured for primary halogenoalkanes because the carbon bonded to the halogen is relatively unhindered (only one other carbon attached), allowing direct back-side attack by the nucleophile [1 mark].

SN1 mechanism: Occurs in two steps. Step 1 (slow, rate-determining): the C-X bond breaks by heterolytic fission, and the halogen leaves as $\text{X}^-$, forming a carbocation intermediate ($\text{C}^+$). Step 2 (fast): the nucleophile ($\text{OH}^-$) attacks the positively charged carbocation [1 mark]. SN1 is favoured for tertiary halogenoalkanes because tertiary carbocations are stabilised by the electron-donating effect (inductive effect) of three alkyl groups, making the formation of the carbocation energetically favourable. The steric crowding around the carbon also prevents direct SN2 attack [1 mark].

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Section 8, W1:

Initiation: $\text{Br}_2 \xrightarrow{\text{UV}} 2\text{Br}\cdot$ (UV light causes homolytic fission of the Br-Br bond to form two bromine radicals) [1 mark].

Propagation step 1: $\text{Br}\cdot + \text{C}_2\text{H}_6 \rightarrow \text{HBr} + \text{C}_2\text{H}_5\cdot$ (a bromine radical abstracts a hydrogen atom from ethane, forming HBr and an ethyl radical) [1 mark].

Propagation step 2: $\text{C}_2\text{H}_5\cdot + \text{Br}_2 \rightarrow \text{C}_2\text{H}_5\text{Br} + \text{Br}\cdot$ (the ethyl radical reacts with a bromine molecule, forming bromoethane and regenerating a bromine radical) [1 mark].

Termination (one example): $\text{C}_2\text{H}_5\cdot + \text{Br}\cdot \rightarrow \text{C}_2\text{H}_5\text{Br}$ (two radicals combine, ending the chain) [1 mark].

Overall equation: $\text{C}_2\text{H}_6 + \text{Br}_2 \xrightarrow{\text{UV}} \text{C}_2\text{H}_5\text{Br} + \text{HBr}$. A mixture of products is obtained because the organic product (bromoethane) can undergo further substitution — a bromine radical can abstract another hydrogen, leading to dibromoethane and other poly-substituted products. The chain reaction is non-selective [1 mark].

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Section 9, W1:

Repeating unit of poly(propene): $[-\text{CH}_2\text{-CH(CH}_3\text{)}-]_n$ — the C=C double bond opens, the methyl group ($\text{CH}_3$) remains as a substituent, and bonds extend outside the brackets [1 mark]. The polymer is called poly(propene) [1 mark]. The monomer (propene) is a gas at room temperature because it has a small molecular mass and only weak London dispersion forces between molecules, requiring little energy to separate them. The polymer is a solid because it consists of thousands of monomer units joined into a very long chain with a very high molecular mass. The extensive surface contact between polymer chains creates much stronger London dispersion forces overall, requiring far more energy to separate the chains [1 mark].

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Section 10, W1:

Order of increasing boiling point: 2,2-dimethylpropane < pentane < pentan-1-ol [1 mark].

2,2-dimethylpropane and pentane are both alkanes with the molecular formula $\text{C}_5\text{H}_{12}$ (structural isomers). They have only London dispersion forces between molecules. However, 2,2-dimethylpropane is highly branched (spherical shape), giving it a smaller surface area and weaker London dispersion forces than the straight-chain pentane. Therefore 2,2-dimethylpropane has the lowest boiling point [1 mark].

Pentane has a longer, more extended chain shape, giving it a greater surface area for intermolecular contact and thus stronger London dispersion forces than 2,2-dimethylpropane [1 mark].

Pentan-1-ol ($\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{CH}_2\text{OH}$) has the highest boiling point because, in addition to London dispersion forces, the -OH group enables hydrogen bonding between molecules. Hydrogen bonds are significantly stronger than London dispersion forces alone, requiring considerably more energy to break [1 mark].