Edexcel A-level Chemistry (9CH0) · Topic 6: Organic Chemistry I
Mini-Lesson
Organic Chemistry I
This mini-lesson covers Edexcel Topic 6: alkanes and free-radical substitution, alkenes and electrophilic addition (with Markovnikov's rule), halogenoalkanes and nucleophilic substitution vs elimination, and the reactions of alcohols.
Work through each screen, answer the questions as you go (some are wordy, some are calculations) and collect ⭐ stars. Every calculation is worked through for you first. Press Start when you're ready.
Foundations
Homologous series and isomerism
A homologous series is a family of compounds with the same functional group and general formula, whose members differ by CH2. They have similar chemistry and a gradual trend in physical properties.
Alkanes CnH2n+2 — saturated, only C–C and C–H single bonds
Alkenes CnH2n (one C=C) — unsaturated
Halogenoalkanes CnH2n+1X · Alcohols CnH2n+1OH
Structural isomers have the same molecular formula but different structural formulae (chain, position or functional-group isomers). E/Z isomers occur about a C=C because there is no rotation about the double bond — each carbon of the C=C must carry two different groups.
Cahn–Ingold–Prelog: compare the atomic numbers of the atoms attached to each C of the double bond. If the two higher-priority groups are on the same side it is Z; on opposite sides it is E.
Quick check
Can it show E/Z isomerism?
1Which one of these compounds can exist as a pair of E/Z isomers?
Calculate
Your turn — identify the alkane
2A gaseous alkane has Mr = 58.0. How many carbon atoms does one molecule contain? (Ar: C 12.0, H 1.0.)
C atoms
Hint: An alkane is CₙH₂ₙ₊₂, so Mₕ = 12.0n + 1.0(2n + 2) = 14n + 2. Set 14n + 2 = 58.0 and solve for n.
Alkanes
Free-radical substitution
Alkanes are unreactive (strong, non-polar C–C and C–H bonds), but in UV light they react with halogens by free-radical substitution. The bonds break homolytically — each atom keeps one electron — giving radicals (species with an unpaired electron).
CH4 + Cl2 → CH3Cl + HCloverall equation — but you must be able to write the three stages
Initiation (UV): Cl2 → 2Cl• — the only step that makes radicals from a molecule
Propagation 1: Cl• + CH4 → •CH3 + HCl
Propagation 2: •CH3 + Cl2 → CH3Cl + Cl• — the Cl• is regenerated, so the chain continues
Termination: any two radicals combine, e.g. •CH3 + Cl• → CH3Cl, or 2•CH3 → C2H6
Why it is a poor synthesis: further substitution gives CH2Cl2, CHCl3 and CCl4, and termination gives ethane — a mixture of products, so the yield of any one is low.
Quick check
Spot the propagation step
3Which of these is a propagation step in the chlorination of methane?
Alkenes
Electrophilic addition and Markovnikov's rule
The C=C double bond is a σ bond plus a π bond. The π electrons sit above and below the plane of the molecule, exposed and electron rich — so alkenes attract electrophiles (electron-pair acceptors).
Mechanism (HBr + propene):
The H of H–Br is δ+. A curly arrow goes from the C=C π bond to that H atom.
A second curly arrow goes from the H–Br bond to the Br, so Br⁻ leaves with both electrons (heterolytic fission).
A carbocation forms; the Br⁻ then attacks the positive carbon (arrow from a Br⁻ lone pair to C+).
With a non-polar molecule such as Br2, the π electrons induce a dipole in the approaching Br–Br bond — this is why bromine water is decolourised by alkenes.
Markovnikov: the H adds to the carbon that already has more H atomsbecause that route goes through the more stable carbocation: tertiary > secondary > primary
Why: alkyl groups are electron releasing (a positive inductive effect). The more alkyl groups attached to the positive carbon, the more the charge is spread out and the more stable the carbocation.
Quick check
The major product
4Propene, CH3CH=CH2, reacts with HBr. What is the major product?
Quick check
Why is a tertiary carbocation more stable?
5A tertiary carbocation is more stable than a primary one. Why?
Halogenoalkanes
Nucleophilic substitution
The C–X bond is polar (X is more electronegative), so the carbon is δ+ and is attacked by nucleophiles — species with a lone pair that donate an electron pair.
+ warm aqueous KOH (or NaOH) → an alcohol (nucleophile OH−) — called hydrolysis
+ KCN in ethanol, reflux → a nitrile, RCN (nucleophile CN−) — the chain gains a carbon
+ excess ethanolic NH3, heated in a sealed tube → a primary amine, RNH2
Two mechanisms.Primary halogenoalkanes go by SN2: the nucleophile attacks the δ+ carbon as the C–X bond breaks, through a single transition state. Tertiary halogenoalkanes go by SN1: the C–X bond breaks first to give a (relatively stable, tertiary) carbocation, which the nucleophile then attacks.
Rate of hydrolysis: R–I > R–Br > R–Clcontrolled by bond enthalpy: C–I is the weakest bond, so it breaks most easily
Classic exam trap: C–F is the most polar C–X bond, yet fluoroalkanes hydrolyse slowest. Bond enthalpy wins over bond polarity.
Quick check
Which hydrolyses fastest?
6Equal amounts of 1-chlorobutane, 1-bromobutane and 1-iodobutane are warmed with aqueous silver nitrate in ethanol. Which produces a precipitate fastest, and why?
Two routes
Substitution or elimination? The solvent decides
The same halogenoalkane and the same reagent (KOH) give two completely different products, depending on the conditions:
Warm aqueous KOH → OH− acts as a nucleophile, attacking the δ+ carbon. Product: an alcohol (substitution).
Hot ethanolic KOH (concentrated, reflux) → OH− acts as a base, removing an H+ from the carbon next to the C–X carbon. The electrons form a C=C and X− leaves. Product: an alkene (elimination).
Alcohols can be dehydrated too: heating an alcohol with concentrated H2SO4 (or passing its vapour over hot Al2O3) eliminates water and gives an alkene — the reverse of the hydration of an alkene.
Quick check
Choose the conditions
7A student wants to convert 2-bromobutane into but-2-ene. Which conditions should they use?
Alcohols
Oxidising alcohols
The oxidising agent is acidified potassium dichromate(VI), K2Cr2O7/H2SO4 (written [O] in equations). It turns from orange to green (Cr2O72− → Cr3+) when it is reduced.
Primary alcohol → aldehyde:distil the product off as it forms, so it cannot be oxidised further.
Primary alcohol → carboxylic acid: heat under reflux with excess oxidising agent.
Secondary alcohol → ketone (reflux). A ketone cannot be oxidised further.
Tertiary alcohol:no reaction — there is no H atom on the carbon bearing the OH group, so it stays orange.
CH3CH2OH + [O] → CH3CHO + H2Odistillation — ethanol to ethanal; with excess [O] under reflux you get ethanoic acid, CH3COOH
Distinguishing test: aldehydes reduce Tollens' reagent (silver mirror) and Fehling's solution (blue → brick-red); ketones do not.
Quick check
Oxidising butan-2-ol
8Butan-2-ol is heated under reflux with excess acidified potassium dichromate(VI). What is the organic product?
Calculate
Your turn — percentage yield
98.00 g of ethanol (M = 46.0 g mol⁻¹) is dehydrated over hot aluminium oxide: C2H5OH → C2H4 + H2O (M(C2H4) = 28.0) 3.36 g of ethene is collected. Calculate the percentage yield to 3 significant figures.
%
Hint: n(ethanol) = 8.00 ÷ 46.0 = 0.1739 mol. The ratio is 1 : 1, so the theoretical mass of ethene = 0.1739 × 28.0 = 4.87 g. Now do (3.36 ÷ 4.87) × 100.
Calculate
Your turn — atom economy
10For that same dehydration, C2H5OH → C2H4 + H2O, calculate the atom economy for making ethene. (M: C2H4 28.0, H2O 18.0.)
%
Hint: Atom economy = M(desired product) ÷ sum of M of all products × 100 = 28.0 ÷ (28.0 + 18.0) × 100.
Calculate
Your turn — mass of product
110.150 mol of propan-1-ol is converted into 1-bromopropane (M = 123.0 g mol⁻¹) in a 1 : 1 reaction. The yield is 62.0%. Calculate the mass of 1-bromopropane obtained, to 3 significant figures.
g
Hint: Theoretical mass = 0.150 × 123.0 = 18.45 g. Now take 62.0% of that: 18.45 × 0.620.
Sort it
Name that mechanism
Tap a reaction, then tap the mechanism it goes by.
🟫 Electrophilic addition
🟩 Nucleophilic substitution
🟪 Elimination
Quick check
Where does the first curly arrow start?
12In the mechanism for the electrophilic addition of Br2 to ethene, where does the first curly arrow begin and end?
Match it
Match the term to its meaning
Tap an item on the left, then its partner on the right.
Term
Meaning
Recap
The big ideas to know
Alkanes: free-radical substitution in UV — initiation, propagation (radical in, radical out), termination
Alkenes: electrophilic addition; the electron-rich π bond attacks the δ+ atom of the electrophile
Markovnikov: H adds to the carbon with more H atoms, because that gives the more stable carbocation
Carbocation stability: tertiary > secondary > primary (alkyl groups release electron density)
Halogenoalkanes: nucleophilic substitution with OH⁻ (alcohol), CN⁻ (nitrile) or NH3 (amine); SN2 for primary, SN1 for tertiary
Rate of hydrolysis: R–I > R–Br > R–Cl — set by bond enthalpy, not by bond polarity
Aqueous vs ethanolic KOH: aqueous = substitution (alcohol); hot ethanolic = elimination (alkene)