Edexcel International A Level Chemistry: Organic chemistry: general principles and halogenoalkanes (YCH11)

This guide teaches Topic 10A: General principles and Topic 10B: Halogenoalkanes from the Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry specification (YCH11), Issue 1, September 2017. It covers learning outcomes 10.1 to 10.14, including Core Practical 5 and Core Practical 6. This is International AS (IAS) content, examined in Unit 2: Energetics, Group Chemistry, Halogenoalkanes and Alcohols. The specification says that related topics in Units 4 and 5 assume this material.

Course links: the Edexcel International A Level Chemistry course hub and the printable Edexcel Chemistry checklist. Once you have read this guide, summarise it with the revision notes for this topic and test yourself with the practice questions. Naming rules from Unit 1 are in the introductory organic chemistry and alkanes guide, and the alkenes made by elimination are covered in the alkenes guide. To find your weak spots quickly, use the free diagnostics.

What this topic covers

Outcome	What you need to do
10.1	Classify reactions as addition, elimination, substitution, oxidation, reduction, hydrolysis or polymerisation
10.2	Explain what a reaction mechanism is
10.3	Link heterolytic bond breaking to electrophiles and nucleophiles
10.4	Define a nucleophile
10.5	Link bond polarity to the type of mechanism
10.6	Name halogenoalkanes; draw structural, displayed and skeletal formulae
10.7	Tell primary, secondary and tertiary halogenoalkanes apart
10.8	Give the reactions with KOH(aq), ethanolic KOH, aqueous silver nitrate in ethanol, alcoholic ammonia and alcoholic KCN
10.9	Draw the substitution mechanisms of primary halogenoalkanes with aqueous KOH and with ammonia
10.10	Use observations and data to compare rates of hydrolysis
10.11	Core Practical 5: rates of hydrolysis of some halogenoalkanes
10.12	Know the reactivity trend for primary, secondary and tertiary
10.13	Explain the chloro, bromo, iodo trend using bond enthalpy
10.14	Core Practical 6: chlorination of 2-methylpropan-2-ol with concentrated hydrochloric acid

10A General principles

Classifying reactions (10.1)

• Addition: two molecules join to give one product.
• Elimination: a small molecule (HBr, H₂O) is removed and a C=C forms.
• Substitution: one atom or group is swapped for another.
• Oxidation / reduction: the organic compound gains oxygen or loses hydrogen / gains hydrogen or loses oxygen. Combustion is oxidation.
• Hydrolysis: a bond is broken by reaction with water (or hydroxide ions).
• Polymerisation: many monomers join into a long chain.

Some reactions fit two labels: hydrogenating an alkene is addition and reduction.

Worked example 1. Classify each reaction.

1. Cyclohexene + H₂ → cyclohexane (nickel catalyst, heat). This is addition (and reduction).
2. CH₃CH₂CH₂Br + KOH(aq) → CH₃CH₂CH₂OH + KBr: substitution, also called hydrolysis.
3. CH₃CHBrCH₃ + KOH (in ethanol) → CH₃CH=CH₂ + KBr + H₂O: elimination.
4. n CH₂=CHCl → poly(chloroethene): polymerisation.
5. C₂H₆ + Cl₂ → C₂H₅Cl + HCl (UV light): substitution (free-radical, from Unit 1).

Mechanisms and bond breaking (10.2, 10.3)

A reaction mechanism is the sequence of steps by which reactants become products: which bonds break, which form, and where the electrons go. A curly arrow shows an electron pair moving. It starts at a lone pair or a bond.

• Homolytic fission: each atom keeps one electron, giving two radicals.
• Heterolytic fission: one atom takes both electrons, giving a positive and a negative species.

Heterolytic fission gives the species of ionic mechanisms. An electrophile accepts an electron pair to form a covalent bond (H⁺, or the δ+ H of HBr). A nucleophile is its electron-rich partner.

Nucleophiles (10.4)

Definition: a nucleophile is a species that donates a lone pair of electrons to form a new covalent bond with an electron-deficient atom.

Examples here: OH⁻, CN⁻, NH₃ and H₂O. Ammonia and water show that a nucleophile need not be charged.

Bond polarity and mechanism type (10.5)

Bond in the organic molecule	Electron distribution	Attacked by	Typical mechanism
C–H, C–C in alkanes	Non-polar	Radicals	Free-radical substitution
C=C in alkenes	Electron-rich π bond	Electrophiles	Electrophilic addition
C–X in halogenoalkanes	Polar, C is δ+	Nucleophiles	Nucleophilic substitution

Halogens are more electronegative than carbon, so the carbon of C–X is δ+ and attracts the nucleophile’s lone pair.

10B Halogenoalkanes

Names and formulae (10.6)

Use the longest chain, the prefixes fluoro, chloro, bromo and iodo, the lowest locants, and alphabetical order.

Worked example 2.

• CH₃CH₂CH(CH₃)CH₂I: four-carbon chain numbered from the iodine end, 1-iodo-2-methylbutane. Skeletal: a four-carbon zigzag, a branch at carbon 2, “I” at the end of carbon 1.
• CH₃CHBrCH(CH₃)CH₂CH₃: locants 2,3 beat 3,4, so 2-bromo-3-methylpentane.
• (CH₃)₂CClCH₂CH₂CH₃: 2-chloro-2-methylpentane.

A displayed formula shows every atom and bond. In a skeletal formula, lines stand for the carbon chain and only the halogen is labelled.

Primary, secondary and tertiary (10.7)

Count the carbons bonded to the C–X carbon: one (or none) is primary, two is secondary, three is tertiary. In worked example 2 the three compounds are primary, secondary and tertiary in that order.

The five reactions (10.8)

For 1-bromo-2-methylpropane, (CH₃)₂CHCH₂Br:

Reagent and conditions	Role of the reagent	Organic product	Type
KOH(aq), heat under reflux	OH⁻ is a nucleophile	(CH₃)₂CHCH₂OH, 2-methylpropan-1-ol	Substitution (hydrolysis)
KOH dissolved in ethanol, heat under reflux	OH⁻ is a base	(CH₃)₂C=CH₂, 2-methylpropene	Elimination
Aqueous AgNO₃ in ethanol, warm	H₂O is a nucleophile	(CH₃)₂CHCH₂OH, plus a cream AgBr precipitate	Substitution (hydrolysis)
Ammonia in ethanol, heated in a sealed tube under pressure	NH₃ is a nucleophile	(CH₃)₂CHCH₂NH₂, 2-methylpropylamine (2-methylpropan-1-amine)	Substitution
KCN in ethanol, heat under reflux	CN⁻ is a nucleophile	(CH₃)₂CHCH₂CN, 3-methylbutanenitrile	Substitution

Two equations to learn in full: RBr + 2NH₃ → RNH₂ + NH₄Br, and RBr + H₂O → ROH + H⁺ + Br⁻ followed by Ag⁺ + Br⁻ → AgBr.

KOH gives two products. Water as solvent favours substitution; ethanol favours elimination, where OH⁻ pulls off an H⁺ from a carbon adjacent to the one carrying bromine. A secondary halogenoalkane can give more than one alkene.

The nitrile reaction lengthens the carbon chain by one. Count the CN carbon when naming: CH₃CH₂CN is propanenitrile.

Worked example 3. 5.48 g of 1-bromo-2-methylpropane is heated with excess KCN in ethanol. The yield is 75%. Find the mass of nitrile. Use H = 1.0, C = 12.0, N = 14.0, Br = 79.9.

1. M(C₄H₉Br) = 136.9 g mol⁻¹.
2. n = 5.48 ÷ 136.9 = 0.0400 mol.
3. The ratio is 1 : 1, so 0.0400 mol of C₅H₉N, M = 83.0 g mol⁻¹.
4. Theoretical mass = 0.0400 × 83.0 = 3.32 g.
5. Actual mass = 3.32 × 0.75 = 2.49 g.

Mechanisms with primary halogenoalkanes (10.9)

Hydroxide ions with 1-bromopentane.

1. Show the dipole Cδ+–Brδ−.
2. One curly arrow starts at an oxygen lone pair of OH⁻ and ends at the δ+ carbon; another moves the C–Br bonding pair onto Br.
3. Products: CH₃(CH₂)₄OH and Br⁻.

Ammonia with iodoethane.

1. Dipole Cδ+–Iδ−. The nitrogen lone pair attacks the δ+ carbon while the C–I bonding pair moves onto iodine, giving CH₃CH₂NH₃⁺ and I⁻.
2. A second NH₃ removes H⁺: arrow from its lone pair to an H on the NH₃⁺ group, and from that N–H bond to N.
3. Products: ethylamine, CH₃CH₂NH₂, and NH₄⁺.

Hence two moles of ammonia. Excess ammonia reduces further substitution of the amine. The specification says SN1 and SN2 are tested in Unit 4.

Comparing rates of hydrolysis (10.10, 10.12, 10.13)

Ethanol is a common solvent, because halogenoalkanes do not mix with water. As water substitutes the halogen, halide ions meet Ag⁺ and precipitate: AgCl white, AgBr cream, AgI yellow. Rate is taken as proportional to 1/time.

Trend with the halogen (10.13). Rate: iodoalkane > bromoalkane > chloroalkane. Use these mean bond enthalpies in kJ mol⁻¹: C–Cl 346, C–Br 290, C–I 228. The C–I bond is the weakest, so it breaks most easily. Bond polarity predicts the opposite order (C–Cl is the most polar), so bond enthalpy, not polarity, controls the rate.

Trend with structure (10.12). Rate: tertiary > secondary > primary. Know the order and show it from data; the reasons come in Unit 4.

Worked example 4. Equal amounts of three isomers, C₅H₁₁Br, were each added to silver nitrate in ethanol at 50 °C.

Isomer	Class	Time for cream precipitate / s	1/t / s⁻¹
1-bromopentane	1°	210	4.76 × 10⁻³
2-bromopentane	2°	95	1.05 × 10⁻²
2-bromo-2-methylbutane	3°	8	1.25 × 10⁻¹

The tertiary isomer reacts about 26 times faster than the primary (210 ÷ 8 = 26.25). In a second run 1-iodopentane took 35 s, so it reacts 6.0 times faster than 1-bromopentane.

Core Practical 5 (10.11)

1. Warm tubes of ethanol plus aqueous silver nitrate, and tubes of each halogenoalkane, in a water bath at a fixed temperature such as 50 °C.
2. Mix, start the timer and record the time for a precipitate to appear.

Keep temperature, volumes, silver nitrate concentration and amount of halogenoalkane the same. Ethanol is flammable, so use a water bath, not a flame.

Core Practical 6 (10.14)

(CH₃)₃COH + HCl → (CH₃)₃CCl + H₂O

The tertiary alcohol reacts at room temperature.

1. Shake the alcohol with excess concentrated hydrochloric acid in a separating funnel. Release the pressure often.
2. Allow the layers to separate. 2-chloro-2-methylpropane is less dense than water (about 0.85 g cm⁻³), so it floats on top. Drain away the aqueous layer below it.
3. Shake with sodium hydrogencarbonate solution to neutralise leftover acid, venting the CO₂, then shake with water.
4. Dry with anhydrous sodium sulfate until clear.
5. Distil, collecting the fraction that boils near 51 °C.

Worked example 5. 12.0 cm³ of 2-methylpropan-2-ol (density 0.78 g cm⁻³) gives 8.10 g of 2-chloro-2-methylpropane. Find the percentage yield. Use H = 1.0, C = 12.0, O = 16.0, Cl = 35.5.

1. Mass of alcohol = 12.0 × 0.78 = 9.36 g.
2. M(C₄H₁₀O) = 74.0 g mol⁻¹, so n = 9.36 ÷ 74.0 = 0.1265 mol.
3. M(C₄H₉Cl) = 92.5 g mol⁻¹. Theoretical mass = 0.1265 × 92.5 = 11.70 g.
4. Percentage yield = 8.10 ÷ 11.70 × 100 = 69.2%.

Further suggested practicals

To identify the halogen, warm the halogenoalkane with NaOH(aq) to release the halide ion. Acidify with dilute nitric acid to remove OH⁻, which would otherwise give brown silver oxide. Add silver nitrate and note the colour. Ammonia confirms: AgCl dissolves in dilute ammonia, AgBr dissolves in concentrated ammonia, and AgI does not dissolve.

1-bromobutane is made by heating butan-1-ol with potassium bromide and sulfuric acid, which form HBr in the flask.

Common errors

• Calling OH⁻ a nucleophile in the ethanolic KOH reaction, or writing “KOH” with no solvent.
• Starting the first curly arrow from the minus sign instead of a lone pair.
• Leaving out the second ammonia molecule and NH₄⁺ in the amine mechanism.
• Explaining the chloro, bromo, iodo trend by polarity. The specification links it to bond enthalpy.
• Naming CH₃CH₂CN as ethanenitrile. Count the CN carbon: propanenitrile.
• Placing 2-chloro-2-methylpropane in the lower layer of the separating funnel.

Where next

Condense this with the revision notes, then try the practice questions. The next part of Topic 10 is the alcohols guide. For exam technique, see preparing for the Edexcel IAL Chemistry units.

Official syllabus

Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry (XCH11/YCH11) specification, Issue 1, September 2017, Pearson Education Limited – Unit 2, Topic 10: Organic Chemistry: Halogenoalkanes, Alcohols and Spectra, parts 10A General principles and 10B Halogenoalkanes, outcomes 10.1 to 10.14.