Halogen Compounds Revision Notes — 9701 Paper 1

Meet the halogenoalkanes

Syllabus requirement · §15.1.2

“

classify halogenoalkanes into primary, secondary and tertiary

”

What a halogenoalkane is

A halogenoalkane is an alkane in which one or more hydrogen atoms have been replaced by halogen atoms (fluoro, chloro, bromo or iodo). You made them in the AS Hydrocarbons note: free-radical substitution puts X onto an alkane chain, and electrophilic addition adds X₂ or HX across a C=C bond. With one halogen atom the general formula is $\text{C}_n\text{H}_{2n+1}\text{X}$ : an alkane's $\text{C}_n\text{H}_{2n+2}$ with one H swapped for X.

Naming uses the alkane name with a halo- prefix and a position number, like any other substituent: $\text{CH}_3\text{CH}_2\text{Br}$ is bromoethane; $\text{CH}_3\text{CH(Br)CH}_3$ is 2-bromopropane; $(\text{CH}_3)_2\text{CHCH}_2\text{Cl}$ is 1-chloro-2-methylpropane. When different substituents are present they are listed alphabetically (bromo, chloro, iodo, methyl), each with its own number: $\text{CH}_2\text{ClCHClCH}_3$ is 1,2-dichloropropane.

Why a whole topic? Because the C–X bond is polar. Chlorine (3.0) and bromine (2.8) are more electronegative than carbon (2.5), so the shared electrons are pulled towards X. (Iodine, 2.5, is about equal to carbon, but C–I is still treated as polar in this topic.) The carbon is left electron-poor, $\text{C}^{\delta+}$ , and the halogen electron-rich, $\text{X}^{\delta-}$ . That $\text{C}^{\delta+}$ is a target for nucleophiles, which is why halogenoalkanes react with reagents that alkanes ignore.

The polar C–X bond gives a δ+ carbon for nucleophiles to attack; the 1°/2°/3° class is found by counting the carbons bonded to the carbon carrying X.

Classification: count the carbons on the carbon carrying X

The class of a halogenoalkane is a count of the carbon atoms directly bonded to the carbon that carries the halogen:

primary (1°): that carbon is bonded to one other carbon — e.g. 1-bromobutane, $\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{Br}$ (the Br-carbon touches only C3)
secondary (2°): bonded to two carbons — e.g. 2-bromobutane, $\text{CH}_3\text{CHBrCH}_2\text{CH}_3$
tertiary (3°): bonded to three carbons — e.g. 2-bromo-2-methylpropane, $(\text{CH}_3)_3\text{CBr}$

Two traps. First, look only at the carbon carrying X, not at branching elsewhere: in 1-chloro-2-methylpropane, $(\text{CH}_3)_2\text{CHCH}_2\text{Cl}$ , the C–Cl carbon is a chain-end $\text{CH}_2$ touching one carbon, so it is primary even though the molecule is branched. Second, $\text{CH}_3\text{Br}$ and other methyl halides have a C–X carbon touching no other carbon; they react like primary halogenoalkanes.

The class matters because it decides the mechanism of substitution (see “SN1 and SN2: two ways to swap”).

Polarity decides where; bond strength decides how fast

Keep two properties of the C–X bond apart:

Polarity (C is $\delta+$ ) explains where nucleophiles attack: the carbon, never the halogen. It does NOT explain how fast.
Bond strength (C–F > C–Cl > C–Br > C–I) decides how fast. Polarity falls in the same direction, so if polarity controlled the rate, C–F (the most polar) would react fastest. In fact C–F hardly reacts at all.

A common multiple-choice wrong option says "the C–Cl bond is more polar, so it reacts faster". It is wrong; “Why R–I beats R–Br beats R–Cl” gives the full bond-energy argument.

Classifying a mixed set

Classify each halogenoalkane as primary, secondary or tertiary, and give its systematic name:

(i) $\text{CH}_3\text{CH}_2\text{CH(CH}_3)\text{CH}_2\text{Cl}$ (ii) $(\text{CH}_3)_2\text{CHCHBrCH}_2\text{CH}_3$ (iii) $(\text{CH}_3)_2\text{CClCH}_2\text{CH}_3$ (iv) $\text{ICH}_2\text{CH}_2\text{I}$

Show full working

1
(i) Class. The C–Cl carbon is the $\text{CH}_2$ at the right-hand end. It is bonded to one carbon → primary.
Always find the carbon carrying X first and count only its carbon neighbours. Branching further along the chain does not matter.
2
(i) Name. The longest chain runs $\text{CH}_3\text{CH}_2\text{CH}\text{–CH}_2\text{Cl}$ : four carbons, butane, with a methyl branch. Numbering from the Cl end gives Cl on C1 and methyl on C2 → 1-chloro-2-methylbutane.
Number from the end that gives the lower set of locants: {1, 2} beats {3, 4} from the other end.
3
(ii) Class. The C–Br carbon is bonded to the $(\text{CH}_3)_2\text{CH}$ – group on one side and the $\text{CH}_2\text{CH}_3$ group on the other → two carbons → secondary.
Each neighbouring group counts once, however big it is: an isopropyl group is one carbon neighbour.
4
(ii) Name. The longest chain goes from one methyl of the isopropyl group, through CH and CHBr, to the ethyl end: five carbons (pentane). Numbering from the methyl end gives methyl on C2 and Br on C3 → 3-bromo-2-methylpentane.
Only one of the two methyls of an isopropyl group can be part of the main chain; the other becomes the methyl branch. Students often count six carbons here.
5
(iii) Class. The C–Cl carbon is bonded to two $\text{CH}_3$ groups and one $\text{CH}_2\text{CH}_3$ group → three carbons → tertiary.
A tertiary C–X carbon has no hydrogen on it: three carbons plus the halogen fill its four bonds.
6
(iii) Name. The longest chain through that carbon has four carbons (butane), with Cl and a methyl both on C2 → 2-chloro-2-methylbutane.
Two substituents on the same carbon each get their own locant: 2-chloro-2-methyl, not 2-chloromethyl.
7
(iv) Each C–I carbon is a $\text{CH}_2$ bonded to one other carbon → primary at both ends. Two iodo groups on the two carbons of ethane → 1,2-diiodoethane.
With two halogens, classify each C–X carbon separately. The prefix di- shows two identical substituents, and each still needs a locant.

Answer

(i) primary, 1-chloro-2-methylbutane (ii) secondary, 3-bromo-2-methylpentane (iii) tertiary, 2-chloro-2-methylbutane (iv) primary at both ends, 1,2-diiodoethane

Your turn

19701/12 O/N 2023 Q281 mark
Which row shows the correct name and classification of the halogenoalkane shown?
$\text{CH}_3(\text{CH}_2)_2\text{CBr}(\text{CH}_3)\text{CH}_2\text{CH}_3$
Options

name classification of halogenoalkane
A 3-bromo-3-methylhexane secondary
B 3-bromo-3-methylhexane tertiary
C 3-bromo-4-methylhexane tertiary
D 4-bromo-5-methylhexane secondary
Stuck? Show hint
Count the carbons touching the C–Br carbon first; then check the numbering direction.
Show solution
1. 1
  Classify: the C–Br carbon is bonded to the propyl chain $(\text{CH}_2)_2\text{CH}_3$ on the left, the methyl group, and the ethyl group on the right — three carbons → tertiary. That eliminates A and D.
  Settle the class first: it only needs the one carbon carrying Br, and it removes two options at once.
2. 2
  Name: longest chain through C–Br = six carbons (hexane). Numbering from the ethyl end puts Br on C3 and methyl on C3; from the propyl end Br would be C4. Lowest locants → 3-bromo-3-methylhexane — option B, the published answer.
  The name depends on the whole chain and its numbering. C and D number from the wrong end or break the chain in the wrong place.
Answer
B (3-bromo-3-methylhexane, tertiary)
29701/12 F/M 2024 Q361 mark
Structural isomerism only should be considered when answering this question.

How many compounds with molecular formula $\text{C}_5\text{H}_{11}\text{Br}$ are primary halogenoalkanes?

Options

A 4 B 5 C 7 D 8
Stuck? Show hint
Draw the three pentane skeletons first, then place Br on every distinct position that leaves it on a CH₂ chain-end.
Show solution
1. 1
  Enumerate skeletons. $\text{C}_5\text{H}_{11}\text{Br}$ derivatives come from the three $\text{C}_5$ skeletons: pentane, 2-methylbutane and 2,2-dimethylpropane.
  Isomer counts start from every possible carbon skeleton. Missing one skeleton is the usual way to lose this mark.
2. 2
  Pentane. Br on an end carbon gives 1-bromopentane (both ends are the same) → 1 primary isomer.
  A primary position is an end CH₃ group that becomes CH₂Br: that carbon is bonded to one other carbon.
3. 3
  2-methylbutane. Br on a CH₃ next to the branch carbon gives 1-bromo-2-methylbutane (the two methyls on C2 are equivalent). Br on the CH₃ at the other end gives 1-bromo-3-methylbutane → 2 primary isomers.
  Equivalent positions give the same compound, so count each different position once.
4. 4
  2,2-dimethylpropane. All four methyl groups are equivalent → 1-bromo-2,2-dimethylpropane → 1 primary isomer.
  The central carbon has no hydrogen, so it cannot carry Br at all.
5. 5
  Total $1 + 2 + 1 = \mathbf{4}$ — option A, the published answer. Every other Br position is on a carbon bonded to two or three carbons (secondary or tertiary).
  Add the counts from each skeleton; a quick check of the other positions confirms none of them is primary.
Answer
A (4: one from pentane, two from 2-methylbutane, one from 2,2-dimethylpropane)

	name	classification of halogenoalkane
A	3-bromo-3-methylhexane	secondary
B	3-bromo-3-methylhexane	tertiary
C	3-bromo-4-methylhexane	tertiary
D	4-bromo-5-methylhexane	secondary

Paper 1 · MCQReal past-paper questions · Classifying halogenoalkanes Paper 2 · TheoryReal past-paper questions · Classifying halogenoalkanes

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Can you do all of these?

I can name halogenoalkanes up to C6 and classify them as primary, secondary or tertiary by counting the carbons bonded to the C–X carbon
I can recall the three routes to halogenoalkanes with conditions (UV with alkanes; X₂ or HX(g) with alkenes at room temperature; the five alcohol reagents) and write the equations for PCl₅, PCl₃ and SOCl₂
I can give reagent, solvent, conditions and product for substitution by OH⁻, CN⁻ and NH₃, write the equations (including 2NH₃ → NH₄X), and name nitriles counting the extra carbon
I can draw the SN2 mechanism (dipole, arrow from the lone pair of OH⁻ to C, arrow from the C–X bond to X, transition state) and the SN1 mechanism (arrow from the C–X bond, carbocation, arrow from the lone pair to C⁺), adding the H⁺-loss arrow for NH₃
I can say which mechanism a halogenoalkane uses from its class, explain this with steric hindrance and the inductive effect of alkyl groups, and sketch SN1 and SN2 energy profiles
I can write the elimination equation for NaOH or KOH in ethanol with heat, and find every alkene product, including cis/trans isomers and molecules with no β-hydrogen
I can describe the silver nitrate test and identify the halogen from the precipitate colour and its solubility in ammonia
I can explain the order R–I > R–Br > R–Cl using C–X bond energies (485/340/280/240 kJ mol⁻¹) and say why polarity gives the wrong order