★ India’s Student Guidance Platform
Ask Doubt

Haloalkanes and Haloarenes — Class 12 Chemistry Notes & Practice

September 11, 2026
Haloalkanes and Haloarenes — Class 12 Chemistry Notes & Practice

Take a breath. If someone has told you that Haloalkanes and Haloarenes is the chapter where organic chemistry stops making sense, they were wrong — they just met it in the wrong order. This chapter is really one small story told over and over: a carbon atom is holding a halogen slightly too loosely, and the entire syllabus is about who takes that halogen away, how, and what the carbon looks like afterwards. Once you see that single thread, every reaction here becomes a variation, not a new burden.

We are going to build the chapter from absolute zero. No prior comfort with mechanisms is assumed. You will get plain explanations, a code-drawn decision chart, more than a dozen worked examples, an original memory device for the tricky calls, and a practice worksheet with answers you can check yourself. Along the way we cover exactly what CBSE asks: the SN1 and SN2 mechanism with examples, the full haloalkanes and haloarenes name reactions list, and the kind of haloalkanes and haloarenes class 12 important questions that actually turn up in the paper. Nothing outside the 2026–27 syllabus, nothing padded.

One promise before we start: you do not need to memorise sixty equations. You need about eight ideas and the confidence to apply them. Let’s go and get them.

Meet Your Tutor

Haloalkanes and Haloarenes is easier when you stop treating every reaction as a separate memory test. I will help you read the substrate, reagent, solvent and leaving group first, then predict the pathway and product. Keep a small reaction map beside you and update it as each mechanism becomes clear.

What You’ll Learn

Jump straight to any topic

Your Game Plan

Work through the chapter in this order. It is deliberately built so that each step makes the next one easier, and so that you are never asked to memorise something you have not first understood.

  1. Day 1 — Vocabulary. Classification and IUPAC naming. Nothing else. If you can name and classify a structure on sight, half the exam’s short questions are already yours.
  2. Day 2 — The bond. Understand why the C–X bond is polar and why it gets weaker down the group. Every reaction in the chapter is downstream of this one fact.
  3. Day 3 — Preparations. Learn the four doorways to R–X (alcohol, alkane, alkene, halogen exchange) plus the two routes to haloarenes. Use the hub diagram, not a list.
  4. Day 4 — Substitution. SN1 versus SN2, with the stereochemistry attached. This is the highest-scoring topic in the chapter; give it a full sitting.
  5. Day 5 — Elimination and metals. Saytzeff’s rule, Grignard reagents, Wurtz. Short, satisfying, very examinable.
  6. Day 6 — Haloarenes. Why they refuse nucleophiles, how they behave with electrophiles, and the Fittig pair.
  7. Day 7 — Polyhalogen compounds and revision. Then do the worksheet at the end of this page under timed conditions, closed book.
Key Idea — The one sentence that runs through the whole chapter
Carbon is less electronegative than every halogen it bonds to, so in R–X the carbon carries a partial positive charge (δ+) and the halogen a partial negative charge (δ−). Electron-rich species (nucleophiles) are drawn to that δ+ carbon, and the halogen is happy to walk away with the bonding pair as X−. Every preparation makes that bond; every reaction breaks it.

Study Notes

Classification of Haloalkanes and Haloarenes

Before you can react with something, you have to be able to sort it. Chemists classify these compounds in two independent ways, and the exam expects both.

First sort: how many halogens? One halogen makes a monohaloalkane, two make a dihalo compound, and three or more make a polyhalogen compound. Dihalides split further: if both halogens sit on the same carbon they are gem-dihalides (from the Latin geminus, twin — think of twins sharing one seat), and if they sit on adjacent carbons they are vic-dihalides (vicinal, meaning neighbouring — think of next-door neighbours).

Second sort: what kind of carbon is holding the halogen? This is the sort that actually predicts reactivity, so it matters more.

ClassCarbon holding XEveryday exampleWhy it matters
Alkyl halide (1°, 2°, 3°)sp3, attached to 1, 2 or 3 other carbonsCH3CH2Cl (1°), (CH3)2CHBr (2°), (CH3)3CCl (3°)Decides whether substitution goes SN1 or SN2
Allylic halidesp3 carbon next to a C=CCH2=CH–CH2ClVery reactive — its carbocation is resonance-stabilised
Benzylic halidesp3 carbon attached to a benzene ringC6H5CH2ClAlso very reactive, same resonance reason
Vinylic halidesp2 carbon of a C=CCH2=CH–ClExtremely unreactive to substitution
Aryl halide (haloarene)sp2 carbon of a benzene ringC6H5ClExtremely unreactive — the whole second half of this chapter

Why it works: notice that the two unreactive classes are exactly the two where the halogen sits on an sp2 carbon. That is not a coincidence, and we will unpack it properly in the section on the C–X bond. Keep it as a flag for now: sp2–X means stubborn.

Example 1 — Sorting five structures at a glance
Classify: (a) CH3CHClCH3 (b) C6H5CH2Br (c) CH2=CHBr (d) CH3CHClCH2Cl (e) (CH3)3CI.
Working: (a) the Cl-bearing carbon touches two other carbons → 2° alkyl halide. (b) the Br sits on a CH2 attached to a ring → benzylic halide. (c) the Br is on a doubly bonded (sp2) carbon → vinylic halide. (d) two chlorines on neighbouring carbons → vic-dichloride. (e) the I-bearing carbon touches three carbons → 3° alkyl halide.
Common Mistake — Benzylic is not aryl
C6H5CH2Cl is benzylic — the chlorine is on a side-chain carbon, not on the ring. C6H5Cl is aryl, with the chlorine bolted directly onto the ring. They behave in opposite ways: the benzylic one is one of the most reactive halides you will meet, the aryl one is one of the least. Students lose easy marks by treating them as the same thing.

↑ Back to top

IUPAC Nomenclature Without the Panic

Naming is a mechanical skill, and mechanical skills are the cheapest marks in chemistry. Four rules cover everything CBSE asks.

  1. Find the longest carbon chain that contains the maximum number of carbon atoms. That chain gives the parent name (ethane, propane, butane…).
  2. Treat the halogen as a prefix: fluoro, chloro, bromo, iodo. It is never a suffix, so it never gets priority over the chain in naming — but see rule 3 for numbering.
  3. Number the chain to give the lowest set of locants to all substituents taken together. If two numberings tie, give the lower number to whichever substituent comes first alphabetically.
  4. List prefixes alphabetically, ignoring multiplying prefixes (di, tri) when alphabetising, and separate numbers from words with hyphens.
Example 2 — Naming CH3–CHCl–CH2–CH2Br
Step 1: longest chain = 4 carbons = butane.
Step 2: substituents are bromo and chloro.
Step 3: number from the CH2Br end → Br at C-1, Cl at C-3, locant set {1, 3}. Number from the other end → Cl at C-2, Br at C-4, locant set {2, 4}. Compare at the first point of difference: 1 < 2, so {1, 3} wins.
Step 4: alphabetise — bromo before chloro.
Answer: 1-bromo-3-chlorobutane.
Example 3 — When the locants tie: CH2Br–CH2Cl
Numbering either way gives {1, 2}, a dead tie. The tie-break rule says the substituent cited first alphabetically gets the lower number, and bromo beats chloro. So Br is C-1.
Answer: 1-bromo-2-chloroethane.
Example 4 — A branched one: (CH3)2CH–CH2–CH2Br
Longest chain is CH3–CH(CH3)–CH2–CH2Br, i.e. four carbons = butane, with a methyl branch. Numbering from the Br end gives bromo at C-1 and methyl at C-3, set {1, 3}; from the other end {2, 4}. First set wins.
Answer: 1-bromo-3-methylbutane.
StructureCommon nameIUPAC name
CH3CH2CH2Cln-propyl chloride1-chloropropane
(CH3)2CHBrisopropyl bromide2-bromopropane
(CH3)3CCltert-butyl chloride2-chloro-2-methylpropane
CH2Cl2methylene chloridedichloromethane
CHCl3chloroformtrichloromethane
CHI3iodoformtriiodomethane
CCl4carbon tetrachloridetetrachloromethane
C6H5Clchlorobenzenechlorobenzene
C6H5CH2Clbenzyl chloride(chloromethyl)benzene
Exam Tip — For rings, ortho / meta / para still earn full marks
In disubstituted benzenes CBSE accepts both systems. 1,2-dichlorobenzene is ortho-dichlorobenzene, 1,3- is meta, 1,4- is para. Write the IUPAC name first and put the common name in brackets — you look thorough and you cannot be marked down.
Common Mistake — Alphabetise the letters, not the multipliers
In 1,1-dibromo-2-chloroethane you alphabetise using b (from bromo), not d (from dibromo). Getting this backwards is one of the most common one-mark losses in the whole paper.

↑ Back to top

Nature of the Carbon–Halogen Bond

C–X bond enthalpy in CH3–X (kJ mol−1)C–F452kJ mol−1C–Cl351kJ mol−1C–Br293kJ mol−1C–I234kJ mol−1reactivity risesShorter bar = weaker bond = snaps first, so R–I > R–Br > R–Cl > R–F in substitution
Bond enthalpies for the methyl halides — the numbers behind the reactivity order R–I > R–Br > R–Cl > R–F.

Here is the physics behind everything else. Every halogen is more electronegative than carbon, so the shared pair in C–X sits closer to the halogen. The result is a permanent dipole: the carbon is electron-poor (δ+) and the halogen is electron-rich (δ−). Think of it as a slightly stretched handshake — the two are still joined, but one of them is already leaning towards the door.

Now go down the group. Fluorine is tiny; iodine is large. A large halogen puts its bonding orbital further from the carbon nucleus, so the bond gets longer and weaker.

BondBond length / pmBond enthalpy in CH3–X / kJ mol−1Dipole moment of CH3X / D
C–F1394521.847
C–Cl1783511.860
C–Br1932931.830
C–I2142341.636
Key Idea — Weaker bond, faster reaction
Because bond enthalpy falls steadily from 452 kJ mol−1 (C–F) to 234 kJ mol−1 (C–I), the ease of breaking C–X runs C–I > C–Br > C–Cl > C–F. That single ordering explains why, for the same alkyl group, reactivity in nucleophilic substitution is R–I > R–Br > R–Cl > R–F. Iodide is also the best leaving group of the four, so both halves of the argument point the same way.

The dipole moment column contains a trap. You would predict CH3F to be the most polar, because fluorine is the most electronegative element there is. But dipole moment is charge separation multiplied by distance, and the C–F bond is so short that the small distance cancels part of the large charge separation. The measured order is therefore CH3Cl > CH3F > CH3Br > CH3I, with chloromethane just edging ahead.

Example 5 — Which C–X bond breaks first?
Question: Arrange CH3F, CH3Cl, CH3Br and CH3I in increasing order of reactivity towards aqueous KOH, and justify with numbers.
Working: the slow step is C–X cleavage. Using the enthalpies above, the energy needed falls 452 → 351 → 293 → 234 kJ mol−1. Less energy needed means a lower activation barrier and a faster reaction.
Answer: CH3F < CH3Cl < CH3Br < CH3I. Quoting even two of those numbers turns a guess into a justified answer.
Common Mistake — Do not confuse polarity with reactivity
C–F is the most polar bond by electronegativity difference, yet fluoroalkanes are the least reactive. Polarity tells you where a nucleophile will aim; bond enthalpy tells you how hard it is to finish the job. In this chapter, bond strength wins the argument almost every time.

↑ Back to top

Preparation of Haloalkanes From Alcohols

AlcoholR–OHHX + anhyd. ZnCl2 · SOCl2PCl5 · PCl3 · red P + Br2/I2AlkaneR–HX2 with UV light or heatfree-radical substitution (mixtures!)AlkeneC=CHX — Markovnikov (or peroxide → anti)or X2 adds right across the C=CAnother halideR–Cl or R–BrNaI in dry acetone — FinkelsteinAgF, Hg2F2 or SbF3 — SwartsR–Xthe haloalkaneyou were asked for
Four doorways into the same room: every syllabus route to a haloalkane, with its reagent.

Alcohols are the workhorse starting material, because the –OH group is already sitting on the carbon you want to halogenate. The whole job is to persuade a rather unwilling OH− to leave. Four families of reagent do it, and each has a personality worth knowing. If you want to revise how alcohols themselves are made and named first, the Alcohols, Phenols and Ethers chapter pairs naturally with this section.

Route 1 — hydrogen halides. R–OH + HX → R–X + H2O. Tertiary alcohols react with concentrated HCl at room temperature just by shaking. Primary and secondary alcohols are lazier and need anhydrous ZnCl2 alongside concentrated HCl — that mixture is Lucas reagent. Reactivity of the acid runs HI > HBr > HCl, and reactivity of the alcohol runs 3° > 2° > 1°.

Route 2 — phosphorus halides. These are clean and reliable:

  • R–OH + PCl5 → R–Cl + POCl3 + HCl
  • 3 R–OH + PCl3 → 3 R–Cl + H3PO3
  • 3 R–OH + PBr3 → 3 R–Br + H3PO3  (PBr3 generated in the flask from red phosphorus + Br2)
  • 3 R–OH + PI3 → 3 R–I + H3PO3  (PI3 generated from red phosphorus + I2)

Why generate PBr3 and PI3 in the flask rather than bottling them? Because they are unstable and awkward to store. Making them on the spot from red phosphorus and the halogen is simply easier chemistry housekeeping.

Route 3 — thionyl chloride, the tidy one. R–OH + SOCl2 → R–Cl + SO2↑ + HCl↑. This is the preferred laboratory method for chlorides, and the reason is beautiful: both by-products are gases. They bubble straight out of the flask, so the alkyl chloride is left behind essentially pure and the equilibrium is dragged forward at the same time. No filtration, no distillation to separate a solid.

Key Idea — Why thionyl chloride is the examiner’s favourite
SOCl2 is preferred because the only other products, SO2 and HCl, escape as gases. Purity is high and separation is trivial. If a question asks you to choose a reagent ‘for a pure product’, SOCl2 is almost always the intended answer.
Example 6 — Choosing between PCl5 and SOCl2
Question: You need a very pure sample of 1-chloropropane from propan-1-ol. Which reagent, and why?
Working: PCl5 would work but leaves POCl3, a liquid, mixed with your product — you would have to distil them apart. SOCl2 leaves only SO2 and HCl, both gases at room temperature.
Answer: SOCl2, because the by-products are gaseous and simply leave the reaction mixture, giving a purer product with no separation step.
Example 7 — A mass calculation you can be asked for
Question: 7.4 g of butan-1-ol is treated with excess SOCl2. If the yield is 80%, what mass of 1-chlorobutane forms?
Working: M(C4H9OH) = 4(12.011) + 10(1.008) + 15.999 = 74.12 g mol−1. Moles = 7.4 ÷ 74.12 = 0.0998 mol. The equation is 1 : 1, so theoretical moles of C4H9Cl = 0.0998 mol. M(C4H9Cl) = 48.044 + 9.072 + 35.45 = 92.57 g mol−1. Theoretical mass = 0.0998 × 92.57 = 9.24 g. At 80% yield: 9.24 × 0.80.
Answer: 7.39 g, i.e. about 7.4 g of 1-chlorobutane.
Common Mistake — This route does not work on phenol
Phenol + HCl will not give chlorobenzene. In phenol the C–O bond has partial double bond character because the oxygen lone pair is delocalised into the ring, so it is far too strong to break this way. Aryl halides need an entirely different strategy — see the haloarenes section below.

↑ Back to top

Preparation From Hydrocarbons: Free Radical Halogenation and Addition to Alkenes

From alkanes — free radical halogenation. Shine ultraviolet light on a mixture of an alkane and a halogen and hydrogen atoms get swapped for halogen atoms:

CH4 + Cl2  —hv→  CH3Cl → CH2Cl2 → CHCl3 → CCl4

Why it works, and why it is a bad lab method: the light splits Cl2 into chlorine radicals, which are indiscriminate. They attack whichever C–H bond they happen to collide with, and they keep going after the first substitution. You end up with a mixture of mono-, di-, tri- and tetra-substituted products that is expensive to separate. Industrially, where the whole mixture is saleable, this is fine. In an exam, the correct comment is always ‘gives a mixture of products, so it is of limited synthetic use’.

From alkenes — adding HX. The π bond is a pool of loose electrons, so it grabs the H+ of an acid readily:

Key Idea — Markovnikov’s rule, stated so it is actually usable
When HX adds to an unsymmetrical alkene, the hydrogen goes to the carbon that already has more hydrogens, and the halogen goes to the other one. The real reason is that this pathway passes through the more stable (more substituted) carbocation.

So propene + HBr gives 2-bromopropane, not 1-bromopropane.

The peroxide effect (Kharasch effect). Add an organic peroxide and the mechanism switches from ionic to free-radical, and the orientation flips. Propene + HBr in the presence of benzoyl peroxide gives 1-bromopropane — anti-Markovnikov addition.

Exam Tip — The peroxide effect is HBr only
Peroxides reverse the orientation for HBr and for nothing else. With HCl the H–Cl bond is too strong to be broken by the radical chain, and with HI the C–I bond formed is too weak, so the chain cannot propagate. Every year students write ‘anti-Markovnikov with HCl/peroxide’ and lose the mark. HBr only.

From alkenes — adding a halogen. CH2=CH2 + Br2 (in CCl4) → BrCH2CH2Br, a vic-dibromide. The reddish-brown colour of the bromine solution disappears as it reacts, which is exactly why bromine water is the classic bench test for unsaturation.

Example 8 — Predicting the product two ways
Question: What is the major product when but-1-ene reacts with HBr (a) in the absence of peroxide, (b) in the presence of peroxide?
Working: but-1-ene is CH2=CH–CH2CH3. (a) Ionic path, Markovnikov: H joins C-1 (which already carries two hydrogens), Br joins C-2, and the intermediate is a secondary carbocation — more stable than the primary alternative. (b) Radical path: the bromine radical adds first, and it adds to the terminal carbon because that leaves the more stable secondary radical.
Answer: (a) 2-bromobutane; (b) 1-bromobutane.

↑ Back to top

Halogen Exchange: Finkelstein and Swarts Reactions

Sometimes the halide you want is hard to make directly, but a different halide is easy. Halogen exchange lets you swap.

Finkelstein reaction — getting to iodides.

R–X + NaI  —(dry acetone)→  R–I + NaX   (X = Cl, Br)

Key Idea — Why dry acetone is not optional
NaI dissolves in dry acetone, but NaCl and NaBr do not. As soon as the swap happens the sodium chloride or bromide crashes out as a solid, removing a product from the solution. By Le Chatelier’s principle the equilibrium keeps shifting to the right until the exchange is essentially complete. The solvent is doing the work of driving the reaction, which is why the exam always wants the words ‘dry acetone’ written in.

Swarts reaction — getting to fluorides. Alkyl fluorides are made by heating an alkyl chloride or bromide with a metallic fluoride such as AgF, Hg2F2, CoF2 or SbF3:

CH3–Br + AgF → CH3–F + AgBr

Keep the board definition narrow: the Swarts reaction converts an alkyl chloride or alkyl bromide into the corresponding alkyl fluoride. Polyhalogen compounds such as freons are discussed separately later in this chapter.

Example 9 — Two-step synthesis using Finkelstein
Question: Suggest a route from propan-1-ol to 1-iodopropane.
Working: going straight to the iodide with HI is possible but messy. A cleaner two-step route: (1) propan-1-ol + SOCl2 → 1-chloropropane + SO2 + HCl; (2) 1-chloropropane + NaI in dry acetone → 1-iodopropane + NaCl↓.
Answer: SOCl2 first, then a Finkelstein exchange with NaI in dry acetone. The precipitated NaCl drives step 2 to completion.
Common Mistake — Finkelstein runs the wrong way in water
In an aqueous solvent all the sodium halides stay dissolved, nothing precipitates, and the equilibrium sits in the middle giving a useless mixture. ‘Acetone’ alone is not enough either — it must be dry, since water dissolves the salt you are relying on to fall out.

↑ Back to top

Preparation of Haloarenes: Electrophilic Substitution and Sandmeyer

Route 1 — direct halogenation of benzene. Benzene will not react with chlorine on its own; the halogen molecule is not electrophilic enough. A Lewis acid ‘halogen carrier’ fixes that by polarising the Cl–Cl bond:

C6H6 + Cl2  —(anhydrous FeCl3)→  C6H5Cl + HCl

Bromination works the same way with anhydrous FeBr3 or iron filings. The catalysts here are transition-metal Lewis acids, and if you want the underlying reason d-block ions are so good at this job, the d- and f-Block Elements notes on transition metal catalysis covers it.

Iodination is different. The reaction with iodine is reversible — the HI produced simply reduces the aryl iodide straight back to benzene. The fix is to add an oxidising agent such as HNO3 or HIO4 to destroy the HI as it forms. Aryl fluorides cannot be made this way at all; direct fluorination is far too violent to control.

Route 2 — from diazonium salts (the reliable one). Start from an aromatic amine. Treat it with nitrous acid at 273–278 K (that is 0–5 °C, made in the flask from NaNO2 and HCl) to make a diazonium salt:

C6H5NH2  —(NaNO2 + HCl, 273–278 K)→  C6H5N2+Cl−

The –N2+ group is the best leaving group in organic chemistry, because it departs as harmless nitrogen gas. Now you can install almost any halogen:

Reagent on the diazonium saltProductName of the reaction
Cu2Cl2 / HClC6H5Cl + N2Sandmeyer reaction
Cu2Br2 / HBrC6H5Br + N2Sandmeyer reaction
Copper powder / HCl or HBrC6H5Cl or C6H5BrGattermann reaction
KI (no copper needed)C6H5I + N2direct iodination of the diazonium salt
HBF4, then warm the dry saltC6H5F + N2 + BF3Balz–Schiemann reaction
Key Idea — Diazonium salts are the master key
Direct halogenation gives you only chloro- and bromobenzene, and it gives you an ortho/para mixture whenever the ring already carries a group. The diazonium route puts the halogen exactly where the amino group used to be — no isomers to separate — and it is the only sensible way to reach aryl iodides and aryl fluorides.
Example 10 — Why copper(I) is in the flask
Question: Why does the Sandmeyer reaction need Cu2Cl2 rather than just HCl?
Working: the diazonium ion needs a one-electron push to lose N2 and generate an aryl radical; the copper(I) ion supplies that electron and is regenerated afterwards, so it behaves as a catalyst. Plain HCl provides chloride but no electron transfer, so the diazonium salt just sits there or decomposes to phenol.
Answer: Cu(I) acts as the electron-transfer catalyst. Copper’s ability to shuttle between +1 and +2 is exactly the redox flexibility discussed in the Coordination Compounds chapter.
Example 11 — Building 4-bromotoluene cleanly
Question: How would you obtain a clean sample of 4-bromoaniline-derived bromobenzene ring system — specifically, convert 4-methylaniline into 4-bromotoluene?
Working: diazotise first, then swap. (1) 4-methylaniline + NaNO2/HCl at 273–278 K → 4-methylbenzenediazonium chloride. (2) Warm with Cu2Br2/HBr → 4-bromotoluene + N2.
Answer: a Sandmeyer reaction. Direct bromination of toluene would give a mixture of the 2- and 4-isomers; here the position is fixed in advance by where the amino group sat.
Common Mistake — Watch the temperature
Diazotisation must be done at 273–278 K. Warm the solution and the diazonium salt hydrolyses to phenol instead, and your synthesis is finished before it starts. If a question gives you an ice bath, that detail is worth a mark.

↑ Back to top

Physical Properties: Boiling Point, Density and Solubility

Pure haloalkanes are colourless liquids, though bromides and iodides slowly darken in the light as traces of halogen are released. Many are sweet-smelling. Beyond that, three properties get examined, and all three come from the same two forces: the permanent dipole of C–X, and London dispersion forces that grow with molecular size.

Boiling point. A haloalkane always boils higher than the alkane it came from, for two reasons stacked together: the molecule is now polar so there are dipole–dipole attractions, and it is now much heavier with a bigger, more polarisable electron cloud so the dispersion forces are stronger too.

  • Same alkyl group, different halogen: R–I > R–Br > R–Cl > R–F. Iodine has the largest, softest electron cloud, so dispersion forces are strongest.
  • Same halogen, longer chain: boiling point rises steadily, because a longer molecule has more surface in contact with its neighbours.
  • Branching lowers the boiling point. A branched molecule is more spherical, so less of its surface can touch a neighbour. Among the C4H9Br isomers, n-butyl bromide boils highest (about 375 K) and tert-butyl bromide lowest (about 346 K).
Exam Tip — The dihalobenzene trap
For the isomeric dichlorobenzenes the boiling points are almost identical, but the melting point of the para isomer (about 323 K) is far above ortho (about 256 K) and meta (about 249 K). The reason is not intermolecular force at all — it is symmetry. The para isomer is the most symmetrical, so it packs into the crystal lattice most efficiently and more energy is needed to pull that lattice apart. Whenever a question compares melting points of ortho/meta/para isomers, the word the examiner wants is symmetry.

Density. Density rises with the number of carbons, the number of halogen atoms, and the atomic mass of the halogen. So for the same alkyl group, R–I > R–Br > R–Cl. Bromo, iodo and polychloro compounds are denser than water — which is why a layer of chloroform sits at the bottom of a separating funnel, not the top.

Solubility. Haloalkanes are only very slightly soluble in water. To dissolve, a haloalkane would have to break the hydrogen bonds holding water molecules to each other, and break its own dipole–dipole attractions. The new haloalkane–water attractions released less energy than that costs, so the process is not favourable. They dissolve happily in organic solvents, where the forces being broken and made are of a similar kind.

Example 12 — Ordering four boiling points
Question: Arrange in increasing boiling point: 1-bromobutane, 1-bromopropane, 2-bromo-2-methylpropane, 1-bromo-2-methylpropane.
Working: 1-bromopropane is the shortest chain, so it is lowest. The remaining three are all C4H9Br isomers, so molecular mass is identical and only shape decides. 2-bromo-2-methylpropane is the most branched (nearly spherical) → lowest of the three; 1-bromo-2-methylpropane has one branch; 1-bromobutane is a straight chain → highest.
Answer: 1-bromopropane < 2-bromo-2-methylpropane < 1-bromo-2-methylpropane < 1-bromobutane.
Common Mistake — ‘Polar, so it must dissolve in water’
Being polar is necessary but nowhere near sufficient. Water is held together by hydrogen bonds, which are much stronger than the dipole–dipole attraction a haloalkane can offer in return. A haloalkane cannot donate a hydrogen bond at all, since it has no O–H or N–H. Compare it with an alcohol of similar size, which does dissolve — the Alcohols, Phenols and Ethers notes makes that contrast very clearly.

↑ Back to top

Nucleophilic Substitution: The SN1 and SN2 Mechanisms

Alkyl halide R–XAsk first: how crowded is the carbon holding X?Methyl or 1° carbonroomy, easy to reachSN2 pathnucleophile attacksthe back — one stepRate = k[R–X][Nu]bimolecular slow stepInversionthe umbrella flipsinside out2° carbonthe borderline caseEither pathsolvent and nucleophilecast the deciding voteBlend of both rate lawsPart inverted,part racemised3° carbonthree groups in the waySN1 pathX leaves first →flat carbocationRate = k[R–X]unimolecular slow stepRacemisationattack from either face
The whole SN1-versus-SN2 decision on one board: read it top to bottom, column by column.

This is the heart of the chapter. A nucleophile is any species with a lone pair or a negative charge that is looking for a positive centre — OH−, CN−, NH3, RO−, and so on. In R–X the carbon is already δ+, so the nucleophile aims there, the halogen leaves with the bonding pair, and you have swapped one group for another.

Here is the entire product list CBSE expects. Learn it as a table, not as sentences.

ReagentProductClass of product
aqueous KOH (or moist Ag2O)R–OHalcohol
NaOR′R–O–R′ether (Williamson synthesis)
NaI in dry acetoneR–Ialkyl iodide (Finkelstein)
KCNR–C≡Nalkyl cyanide (nitrile)
AgCNR–N≡Calkyl isocyanide
KNO2R–O–N=Oalkyl nitrite
AgNO2R–NO2nitroalkane
NH3 (excess)R–NH2primary amine
R′COOAgR′COORester
LiAlH4R–Halkane
Key Idea — Ambident nucleophiles — the KCN / AgCN pair
The cyanide ion can attack through carbon or through nitrogen; a nucleophile with two possible attacking atoms is called ambident. KCN is an ionic salt, so free CN− is released and attacks through the carbon (the better nucleophilic site), giving the cyanide R–CN. AgCN is largely covalent, so the cyanide is locked to silver through carbon and only the nitrogen lone pair is available, giving the isocyanide R–NC. Exactly the same logic explains KNO2 → nitrite and AgNO2 → nitroalkane.

Now the mechanisms. There are only two, and they differ in one thing: whether the halogen leaves before or during the nucleophile’s attack.

SN2 — substitution, nucleophilic, bimolecular. One single step. The nucleophile comes in from the side directly opposite the halogen (the ‘back door’, because the halogen’s own electron cloud blocks the front) and pushes the halogen out as it arrives. For a moment the carbon is bonded to five things at once in a transition state; then the old bond is fully gone and the new one fully formed.

  • Rate = k[R–X][Nu−] — second order, because both partners appear in the one and only step.
  • No intermediate is ever isolated; there is a transition state, not a carbocation.
  • Reactivity: CH3X > 1° > 2° > 3°. Bulky groups physically block the back door.
  • Favoured by a strong, concentrated nucleophile and a polar aprotic solvent such as acetone.

SN1 — substitution, nucleophilic, unimolecular. Two steps. First the C–X bond breaks all by itself (slow, rate-determining), leaving a flat, positively charged carbocation. Then the nucleophile attacks that carbocation (fast).

  • Rate = k[R–X] — first order. The nucleophile does not appear in the rate law at all, because it is not involved until after the slow step. If the rate-law language feels shaky, the Chemical Kinetics chapter on rate laws and order of reaction is the place to firm it up.
  • Reactivity: 3° > 2° > 1° > CH3X, exactly the reverse of SN2, because more alkyl groups stabilise the carbocation.
  • Allylic and benzylic halides react fastest of all, because their carbocations are stabilised by resonance with the neighbouring π system.
  • Favoured by a polar protic solvent such as water or ethanol, which surrounds and stabilises the ions as they separate — the same solvation idea you met when ionic conductance was explained in the Electrochemistry chapter.
Key Idea — the memory device: “Lonely ONE, Team TWO”
Lonely ONE leaves first and flips a coin. SN1: one molecule in the slow step, the halide leaves on its own (lonely), a flat carbocation forms, and the nucleophile can land on either face — a coin flip — so you get racemisation. Crowded 3° carbons prefer this, because crowding is what makes the cation stable.

Team TWO pushes through the back door and turns the umbrella inside out. SN2: two molecules must meet in the slow step, the nucleophile attacks from behind, and the three remaining groups get swept backwards like an umbrella in a gale — inversion. Roomy methyl and 1° carbons prefer this.

One line to carry into the hall: “Crowded is lonely; roomy works as a team.”
Example 13 — Reading a rate law
Question: Hydrolysis of a bromoalkane is found to have rate = k[RBr], independent of [OH−]. What does this tell you about the substrate?
Working: the nucleophile is absent from the rate law, so it plays no part in the slow step. That is the signature of SN1, where the slow step is the unaided ionisation of C–Br. SN1 is preferred by substrates that give a stable carbocation.
Answer: the reaction is SN1 and the substrate is very likely tertiary (or allylic/benzylic).
Example 14 — Predicting the mechanism from the structure
Question: Which mechanism dominates for (a) CH3CH2Br + CN− in acetone, (b) (CH3)3CBr + H2O?
Working: (a) primary substrate (roomy, back door open) + strong nucleophile + polar aprotic solvent — every factor points one way. (b) tertiary substrate (back door blocked, carbocation very stable) + weak nucleophile + polar protic solvent — again all factors agree.
Answer: (a) SN2, giving propanenitrile; (b) SN1, giving 2-methylpropan-2-ol.
Example 15 — Ranking SN2 reactivity
Question: Arrange in decreasing rate of SN2 reaction: 2-bromo-2-methylpropane, 1-bromobutane, 2-bromobutane, bromomethane.
Working: SN2 depends on how easily the nucleophile reaches the back of the carbon, so less substitution means faster. Count the alkyl groups on the C–Br carbon: bromomethane 0, 1-bromobutane 1, 2-bromobutane 2, 2-bromo-2-methylpropane 3.
Answer: bromomethane > 1-bromobutane > 2-bromobutane > 2-bromo-2-methylpropane.
Common Mistake — The two orders are opposites — never quote one for both
3° > 2° > 1° is the SN1 order (carbocation stability). CH3 > 1° > 2° > 3° is the SN2 order (steric access). Writing the wrong one is the single most expensive error in this chapter. Anchor it with the mnemonic: crowding helps the lonely path and hurts the team path.

Picture the SN2 transition state: the nucleophile approaches from the side opposite the leaving group. Both the incoming nucleophile–carbon bond and the carbon–leaving-group bond are only partly formed or broken at that instant. As one bond forms and the other breaks, the three unchanged groups turn through to give inversion of configuration.

↑ Back to top

Stereochemistry: Retention, Inversion and Racemisation

A carbon carrying four different groups has no plane of symmetry. Its mirror image cannot be superimposed on it, exactly like your left and right hands — hence the word chiral, from the Greek for hand. The two non-superimposable forms are enantiomers.

Enantiomers are chemically identical in every ordinary test, but they rotate the plane of plane-polarised light by equal angles in opposite directions. The one that rotates it clockwise is dextrorotatory, written (+) or d; the anticlockwise one is laevorotatory, written (−) or l. A 50:50 mixture of the two is a racemic mixture: the rotations cancel exactly, so the sample is optically inactive. That cancellation is the whole point of the word.

Key Idea — Three outcomes, three names
Retention — the arrangement in space around the chiral carbon is unchanged.
Inversion (Walden inversion) — the arrangement is turned inside out because the nucleophile arrives opposite the departing halogen. This is the SN2 signature.
Racemisation — both configurations form because a planar intermediate can be attacked from either face. An exactly equal mixture is racemic and optically inactive; in practice, shielding by the departing group can make inversion slightly greater than retention. Formation of both configurations is the SN1 signature.
Example 16 — Retention — when nothing at the chiral centre is touched
Question: (−)-2-methylbutan-1-ol is treated with concentrated HCl and anhydrous ZnCl2. The product is (+)-1-chloro-2-methylbutane. Has the configuration changed?
Working: the chiral centre is C-2. The reaction replaces the OH on C-1, so no bond to C-2 is ever broken. The spatial arrangement at C-2 therefore survives untouched.
Answer: no — this is retention of configuration. The change of sign from (−) to (+) is a red herring: the direction of rotation is a measured property of the whole molecule, not a label for the arrangement in space.
Example 17 — Inversion — the umbrella turning inside out
Question: (−)-2-bromooctane is hydrolysed with aqueous NaOH under SN2 conditions. Predict the stereochemistry of the octan-2-ol formed.
Working: SN2 forces backside attack, so every single molecule that reacts is inverted. There is no way for a molecule to react with retention on this pathway.
Answer: (+)-octan-2-ol, formed with complete inversion of configuration. The product is optically active, and essentially 100% of one enantiomer.
Example 18 — Racemisation through a planar intermediate
Question: Optically active 2-bromobutane is hydrolysed under SN1 conditions. What stereochemical change should you expect?
Working: the slow step gives a planar carbocation. Water can attack from either face, so products of both retention and inversion form. The simplified board model shows complete racemisation; in a real ion pair, the departing bromide can partly shield one face, so inversion may be slightly greater than retention.
Answer: expect racemisation, often accompanied by some excess inversion. A perfectly racemic 50:50 mixture is optically inactive because the equal and opposite rotations cancel.
Example 19 — Putting a number on partial racemisation
Question: The pure (+) enantiomer of an alcohol has specific rotation +13.6°. A sample made by hydrolysing a chiral halide shows +3.4°. What fraction of the sample is racemic?
Working: optical purity (enantiomeric excess) = observed ÷ pure = 3.4 ÷ 13.6 = 0.25, i.e. 25%. That 25% is pure single enantiomer; the rest of the sample must be the 50:50 racemic part.
Answer: the sample is 25% optically pure, so 75% of it is a racemic mixture — the reaction went mostly, but not entirely, through a planar carbocation.
Exam Tip — Say ‘so it is optically inactive’ out loud
When a question asks about the optical activity of an SN1 product, the mark is usually split: one for ‘racemic mixture’ and one for ‘therefore optically inactive, because the rotations of the two enantiomers cancel’. Write both halves, every time.
Common Mistake — (+) and (−) do not mean ‘same’ and ‘opposite’
The sign of rotation is measured in a polarimeter; the configuration is a fact about geometry. A reaction can proceed with complete retention and still flip the sign, as in the example above. Never argue ‘the sign changed, so inversion happened’ — argue from whether a bond to the chiral carbon was broken.

↑ Back to top

Elimination Reactions and the Saytzeff Rule

Substitution is not the only thing a nucleophile can do. A base can instead pull off a hydrogen from the carbon next door to the one holding the halogen. The electrons left behind swing in to form a π bond, and the halide departs. Because the hydrogen comes from the β-carbon, this is called β-elimination or dehydrohalogenation.

CH3–CH2–CH2–Br + KOH (alcoholic) → CH3–CH=CH2 + KBr + H2O

Key Idea — Aqueous swaps, alcoholic strips
Aqueous KOH supplies plenty of free, well-solvated OH−, which behaves as a nucleophile — you get substitution to the alcohol. Alcoholic KOH supplies alkoxide ions, which are bulkier and more strongly basic, so they grab a β-hydrogen instead — you get elimination to the alkene. Two words in the question, two completely different products. Remember it as: water swaps, alcohol strips.

Which alkene forms when there is a choice? If the halide has β-hydrogens on more than one side, two different alkenes are possible, and Saytzeff’s rule tells you which dominates.

Key Idea — Saytzeff’s rule, and how to remember it
In dehydrohalogenation, the hydrogen is removed from the β-carbon that has the fewest hydrogen atoms. The result is the alkene carrying the greatest number of alkyl groups on its doubly bonded carbons — and more alkyl groups means a more stable alkene, through hyperconjugation.

Memory device: “S for Saytzeff, S for Stuffed.” The major alkene is the stuffed one, the one with the most alkyl groups packed around the C=C. Or, if you prefer it as a slogan: the double bond goes where the crowd already is.
Example 20 — Applying Saytzeff to 2-bromobutane
Question: Give the major and minor products of 2-bromobutane with alcoholic KOH.
Working: CH3–CHBr–CH2–CH3. There are β-hydrogens on C-1 (a CH3 with three H) and on C-3 (a CH2 with two H). Removing from C-3, the carbon with fewer hydrogens, gives but-2-ene, whose C=C carries two alkyl groups. Removing from C-1 gives but-1-ene, whose C=C carries only one.
Answer: but-2-ene is the major product and but-1-ene the minor one.
Example 21 — A harder Saytzeff call
Question: What is the major alkene from 2-bromo-2-methylbutane with alcoholic KOH?
Working: the structure is (CH3)2CBr–CH2–CH3. β-hydrogens are available on the two methyl groups and on the CH2. Taking a hydrogen from the CH2 gives 2-methylbut-2-ene, with three alkyl groups on the double bond. Taking one from a methyl gives 2-methylbut-1-ene, with only two.
Answer: 2-methylbut-2-ene is the major product — more substituted, therefore more stable.
Exam Tip — Substitution and elimination always compete
They are rivals for the same substrate. Primary halides with a good nucleophile lean towards substitution; tertiary halides with a strong bulky base and heat lean heavily towards elimination. If a question gives you ‘alcoholic KOH’ and ‘heat’, it is signposting elimination as loudly as it can.

↑ Back to top

Reaction With Metals: Grignard Reagents and the Wurtz Reaction

Grignard reagents. Magnesium metal inserts itself into the C–X bond in dry ether:

R–X + Mg  —(dry ether)→  R–Mg–X   (an alkyl magnesium halide)

Something remarkable has happened to the polarity. Carbon is more electronegative than magnesium, so the C–Mg bond is polarised the other way round — the carbon is now δ−. A carbon atom that carries negative charge behaves almost like a carbanion, which makes the Grignard reagent one of the most useful reactive species in all of organic chemistry.

Key Idea — Why the ether must be bone dry
The δ− carbon of R–MgX will snatch a proton from anything even slightly acidic — water, alcohols, even the N–H of an amine. With water: R–MgX + H2O → R–H + Mg(OH)X. Your expensive reagent is destroyed and you get the plain alkane back. Every glass item is dried and the ether is distilled over sodium before the reaction is set up.
Example 22 — What ruins a Grignard
Question: A student prepares ethylmagnesium bromide but forgets to dry the apparatus. On working up, the only organic product is ethane. Explain.
Working: traces of water on the glass react with C2H5MgBr as fast as it forms. The strongly basic δ− carbon takes a proton from water: C2H5MgBr + H2O → C2H6 + Mg(OH)Br.
Answer: the moisture protonated the Grignard reagent, giving ethane. This is why the reaction is always run in dry ether under anhydrous conditions.

Wurtz reaction. Two molecules of an alkyl halide are stitched together by sodium in dry ether:

2 R–X + 2 Na  —(dry ether)→  R–R + 2 NaX

This doubles the carbon chain, so it always produces an alkane with an even number of carbon atoms when you start from a single halide. Two ethyl bromides give butane; two methyl iodides give ethane.

Example 23 — The limitation of Wurtz
Question: Can propane be made by the Wurtz reaction? Can it be made by mixing CH3I and C2H5I with sodium?
Working: propane has three carbons, an odd number, so it cannot come from coupling two identical halides. Mixing two different halides does give some CH3–C2H5 (propane), but it also gives CH3–CH3 (ethane) and C2H5–C2H5 (butane), because the couplings are random.
Answer: not usefully. The Wurtz reaction is only practical for symmetrical alkanes; with a mixture of halides you get three products of similar boiling point that are painful to separate.
Common Mistake — Wurtz cannot make odd-numbered alkanes from one halide
‘Prepare propane by the Wurtz reaction from a single alkyl halide’ is an impossible instruction, and questions do test whether you notice. Say clearly that a single halide can only give an even-numbered chain.

↑ Back to top

Why Haloarenes Resist Nucleophilic Substitution

Chlorobenzene and chloroethane look like cousins on paper. In the flask they behave nothing alike. Boil chloroethane with aqueous NaOH and you get ethanol in minutes. Boil chlorobenzene with aqueous NaOH and… nothing happens. To get phenol out of chlorobenzene industrially you need 623 K and 300 atmospheres. Something is holding that chlorine on very tightly. There are four reasons, and a good answer names at least three.

Key Idea — remember it as R–S–P
R for Resonance. A lone pair on the halogen delocalises into the ring, giving the C–X bond partial double bond character. A partial double bond is shorter and stronger, and far harder to break.
S for sp2. The carbon holding X in a haloarene is sp2, with 33% s-character, against sp3 and 25% in a haloalkane. More s-character means the electrons sit closer to the nucleus, so the bond is shorter and stronger again. Measured C–Cl lengths: 169 pm in chlorobenzene against 177 pm in chloromethane.
P for Phenyl cation. An SN1 route would need a phenyl cation, and that cation gets no resonance stabilisation at all — so the SN1 door is bolted shut.
And a fourth for good measure: the π electron cloud above and below the ring repels an approaching electron-rich nucleophile.

Say it as “rock–scissors–paper: Resonance, sp2, Phenyl cation” and you will not forget the three-mark answer.

Read the resonance argument as a sequence: a lone pair on chlorine overlaps with the benzene-ring π system; electron density is delocalised to the two ortho positions and then the para position; returning the electrons restores the neutral chlorobenzene form. These contributing structures explain the partial double-bond character of the C–Cl bond.

So how do you ever do it? Two ways.

(a) Brute force. Chlorobenzene + NaOH at 623 K and 300 atm gives sodium phenoxide, and acidifying that gives phenol. This is the Dow process. The phenol chemistry that follows is covered in the Alcohols, Phenols and Ethers chapter.

(b) Persuasion — put electron-withdrawing groups on the ring. A nitro group at the ortho or para position transforms the situation. Once the nucleophile has added, a negative charge sits on the ring; a nitro group at ortho or para can pull that charge onto its own oxygen atoms by resonance, stabilising the intermediate enormously.

CompoundConditions needed to give the phenolComment
Chlorobenzene623 K, 300 atm NaOHthe halogen is extremely reluctant to go
4-nitrochlorobenzenewarm dilute NaOH, 443 Kone activating group already helps a great deal
2,4-dinitrochlorobenzenewarm aqueous NaOH, about 368 Ktwo groups, far milder conditions
2,4,6-trinitrochlorobenzenejust warm watergives picric acid; the ring is now hugely activated
Example 24 — The meta trap
Question: Why does 3-nitrochlorobenzene react with NaOH scarcely faster than chlorobenzene itself, while 4-nitrochlorobenzene reacts far faster?
Working: when the nucleophile adds, the negative charge is spread over the carbons ortho and para to the point of attack. A nitro group at the ortho or para position sits exactly on one of those carbons, so it can drain the charge away by resonance. A nitro group at the meta position does not sit on a charge-bearing carbon, so it can only offer a weak inductive pull.
Answer: only ortho and para nitro groups can stabilise the negatively charged intermediate by resonance; a meta nitro group cannot, so it barely activates the ring.
Example 25 — Ranking four aryl halides
Question: Arrange in increasing reactivity towards aqueous NaOH: chlorobenzene, 2,4-dinitrochlorobenzene, 4-nitrochlorobenzene, 2,4,6-trinitrochlorobenzene.
Working: count the electron-withdrawing groups sitting at ortho or para positions relative to the C–Cl carbon: 0, 2, 1, 3 respectively. More such groups means a more stabilised intermediate and a faster reaction.
Answer: chlorobenzene < 4-nitrochlorobenzene < 2,4-dinitrochlorobenzene < 2,4,6-trinitrochlorobenzene.
Common Mistake — Do not credit ‘the ring is aromatic’ as the reason
Aromaticity by itself is not the explanation, and it earns no marks on its own. The examiner wants the specific reasons: resonance giving partial double bond character, sp2 hybridisation shortening the bond, and the instability of the phenyl cation. Give the 169 pm versus 177 pm figures if you can — a number turns a vague answer into a confident one.

↑ Back to top

Electrophilic Substitution in Haloarenes

Nucleophiles get nowhere with haloarenes, but electrophiles do react — the ring is still a pool of π electrons. The halogen does two apparently contradictory things at once, and the exam loves this contradiction.

Key Idea — deactivating and ortho/para directing
A halogen is strongly electronegative, so its −I effect drains electron density out of the whole ring. That makes the ring poorer in electrons than benzene, so it reacts more slowly: the halogen is deactivating.

But the halogen also has lone pairs, and its +R (resonance) effect pushes electron density back into the ring — and resonance can only deliver that density to the ortho and para positions, never to meta. So although every position is worse off than in benzene, ortho and para are less badly off than meta. The incoming electrophile therefore still goes ortho and para.

One line: the halogen slows the whole ring down, but it decides where the traffic that does get through will park.

Because the halogen is bulky, the para product usually dominates over the ortho — there is simply less crowding at the far end of the ring.

ReactionReagent and conditionsProducts from chlorobenzene
HalogenationCl2, anhydrous FeCl31,2-dichlorobenzene + 1,4-dichlorobenzene (para major)
Nitrationconc. HNO3 + conc. H2SO42-nitrochlorobenzene + 4-nitrochlorobenzene (para major)
Sulphonationconc. H2SO42- and 4-chlorobenzenesulphonic acid (para major)
Friedel–Crafts alkylationCH3Cl, anhydrous AlCl32- and 4-chlorotoluene (para major)
Friedel–Crafts acylationCH3COCl, anhydrous AlCl32- and 4-chloroacetophenone (para major)
Example 26 — Explaining the apparent contradiction
Question: Chlorine withdraws electrons, so why is chlorobenzene not meta directing like a nitro group?
Working: the nitro group has no lone pair to donate; it is purely electron-withdrawing, so all three positions are deactivated and meta — the least deactivated — wins. Chlorine has three lone pairs, and its resonance donation is aimed specifically at ortho and para. The −I effect lowers the whole ring; the +R effect then lifts ortho and para back up relative to meta.
Answer: the two effects act on different things. The −I effect controls the rate (slower than benzene, hence deactivating), while the +R effect controls the position (ortho and para).
Example 27 — A two-step prediction
Question: Benzene is treated with Cl2/anhydrous FeCl3 to give A, and A is treated with the same reagents again to give B (major). Identify A and B.
Working: the first electrophilic substitution on benzene gives chlorobenzene, A. The chlorine already present is ortho/para directing, and the para position is less hindered, so the second chlorine goes mainly para.
Answer: A = chlorobenzene; B = 1,4-dichlorobenzene as the major product, with 1,2-dichlorobenzene as a minor product.
Exam Tip — Stick to monosubstituted starting materials
The 2026–27 syllabus asks for the directive influence of halogen in monosubstituted compounds only. You will never be asked to predict where a third group goes on a ring that already carries two different substituents, so do not burn revision time on it.

↑ Back to top

Fittig and Wurtz–Fittig Reactions

Sodium in dry ether joins carbon chains, and it will do the same job on aromatic rings. Two closely related name reactions come out of this, and telling them apart is worth easy marks.

Fittig reaction — two aryl halides couple to give a biaryl:

2 C6H5Cl + 2 Na  —(dry ether)→  C6H5–C6H5 (biphenyl) + 2 NaCl

Wurtz–Fittig reaction — an aryl halide and an alkyl halide couple to give an alkylbenzene:

C6H5Cl + CH3Cl + 2 Na  —(dry ether)→  C6H5–CH3 (toluene) + 2 NaCl

Key Idea — Three coupling reactions, one sentence each
Wurtz: alkyl + alkyl → alkane. Fittig: aryl + aryl → biaryl. Wurtz–Fittig: aryl + alkyl → alkylbenzene — the hyphenated name literally tells you it is one of each. All three need sodium in dry ether.
Example 28 — Naming the right reaction
Question: Name the reaction and give the product: bromobenzene + bromoethane + sodium in dry ether.
Working: one partner is aromatic and one is aliphatic, so this is the hyphenated case.
Answer: the Wurtz–Fittig reaction, giving ethylbenzene (C6H5–CH2CH3) together with sodium bromide.
Common Mistake — Wurtz–Fittig is never a clean reaction
Alongside the wanted ethylbenzene you will also get some biphenyl (aryl–aryl) and some butane (alkyl–alkyl), because the couplings happen at random. If a question asks for a comment on the yield, that is the point to make.

↑ Back to top

Polyhalogen Compounds and Their Environmental Story

Compounds with more than one halogen have been enormously useful and, in several cases, enormously damaging. This section is short on mechanism and long on consequence — and it is very commonly examined, because it is where chemistry meets everyday life.

CompoundMain usesHazards and environmental effects
Dichloromethane, CH2Cl2
(methylene chloride)		Paint remover, aerosol propellant, process solvent in the manufacture of medicines, metal cleaning and finishingHigh concentrations in air impair vision and hearing; direct skin contact causes intense burning; toxic in large doses
Trichloromethane, CHCl3
(chloroform)		Solvent for fats, waxes, resins and rubber; once used as an inhalation anaesthetic; today mostly used to manufacture the refrigerant R-22Vapour depresses the central nervous system — dizziness, headache, unconsciousness; damages the liver and kidneys on long exposure
Triiodomethane, CHI3
(iodoform)		Formerly a wound antiseptic, because it slowly liberates free iodineIts objectionable smell led to it being replaced by other iodine formulations
Tetrachloromethane, CCl4Solvent, dry-cleaning and spot-removing fluid, feedstock for refrigerants and propellants, once used in fire extinguishersDamages the liver, causes nausea and dizziness; released into the air it depletes the ozone layer, increasing ultraviolet exposure, skin cancer and eye disease
Freons (chlorofluorocarbons), e.g. Freon-12, CCl2F2Refrigerant in refrigerators and air conditioners, aerosol propellant — chosen because they are unreactive, non-corrosive and easily liquefiedChemically inert, so they drift unchanged into the stratosphere, where ultraviolet light snaps C–Cl bonds and releases chlorine radicals that destroy ozone catalytically
DDT, C14H9Cl5The first chlorinated organic insecticide; used against malaria-carrying mosquitoes and typhus-carrying liceInsects developed resistance; being highly fat-soluble it is not metabolised quickly and accumulates in fatty tissue up the food chain; toxic to fish
Key Idea — The chloroform storage question, answered in full
Chloroform is kept in dark-coloured bottles filled right up to the brim. In light and in the presence of air it is slowly oxidised to phosgene (carbonyl chloride, COCl2), which is extremely poisonous:
2 CHCl3 + O2  —(light)→  2 COCl2 + 2 HCl
The dark glass keeps out the light and filling to the brim leaves no air space, removing both requirements at once. About 1% ethanol is added as well, which converts any phosgene that does form into harmless ethyl carbonate.
Example 29 — Why the bottle is full to the brim
Question: A laboratory bottle of chloroform is half empty and stored on a sunny bench. What is the danger, and give the equation.
Working: the air gap supplies oxygen and the sunlight supplies the energy, so the two conditions for photo-oxidation are both met.
Answer: chloroform is oxidised to the highly poisonous gas phosgene: 2 CHCl3 + O2 → 2 COCl2 + 2 HCl. The bottle should be dark-coloured, filled completely, and contain about 1% ethanol as a preservative.
Example 30 — The freon problem, in three steps
Question: Explain how Freon-12 damages the ozone layer, given that it is famously unreactive.
Working: (1) because CCl2F2 is chemically inert, nothing in the lower atmosphere destroys it, so it survives long enough to diffuse up into the stratosphere. (2) There, high-energy ultraviolet light breaks a C–Cl bond and releases a chlorine radical. (3) That radical destroys ozone and is regenerated afterwards, so a single chlorine atom goes on to destroy many ozone molecules.
Answer: its unreactivity is precisely the problem — it lets the molecule reach the stratosphere intact, where ultraviolet light turns it into a catalytic ozone-destroying radical.
Example 31 — Why DDT is banned in most uses
Question: DDT kills insects very effectively. Give two chemical reasons it was withdrawn from general use.
Working: reason one is biological — populations of insects that survived exposure passed on their resistance, so the same dose stopped working. Reason two is a direct consequence of its structure: with five chlorine atoms and two aromatic rings it is extremely fat-soluble and very poorly water-soluble, so animals cannot excrete it. It is stored in fatty tissue and concentrates further at each step up the food chain.
Answer: insect resistance, and bioaccumulation in fatty tissue arising from its very high fat solubility, which also makes it toxic to fish and to animals higher up the chain.
Exam Tip — Know one fact per compound
Questions here are usually one mark. Bank one distinctive fact each: dichloromethane → paint remover; chloroform → phosgene and dark bottles; iodoform → antiseptic that liberates iodine; carbon tetrachloride → ozone depletion; freons → refrigerants that release chlorine radicals; DDT → the first chlorinated insecticide, now a bioaccumulation problem.

↑ Back to top

Haloalkanes and Haloarenes Name Reactions List for Quick Revision

Print this table, stick it above your desk, and test yourself on it every second day. Almost every ‘name the reaction’ question in the paper comes from these rows.

Name reactionWhat it doesReagents and conditions
FinkelsteinR–Cl or R–Br → R–INaI in dry acetone
SwartsR–Cl or R–Br → R–FAgF, Hg2F2, CoF2 or SbF3, heat
SandmeyerArN2+X− → Ar–Cl or Ar–BrCu2Cl2/HCl or Cu2Br2/HBr
GattermannArN2+X− → Ar–Cl or Ar–BrCopper powder with HCl or HBr
Balz–SchiemannArN2+ → Ar–FHBF4, then warm the dry diazonium fluoroborate
Wurtz2 R–X → R–RSodium in dry ether
Fittig2 Ar–X → Ar–ArSodium in dry ether
Wurtz–FittigAr–X + R–X → Ar–RSodium in dry ether
Williamson synthesisR–X → etherSodium alkoxide, NaOR′
Markovnikov additionAlkene + HX → the more substituted halideHX, no peroxide
Peroxide (Kharasch) effectAlkene + HBr → the less substituted bromideHBr with an organic peroxide — HBr only
Saytzeff rulePredicts the major alkene in eliminationAlcoholic KOH, heat
Walden inversionThe configuration flip in every SN2 reactionStrong nucleophile, polar aprotic solvent
Grignard formationR–X → R–MgXMagnesium turnings in dry ether
Dow processChlorobenzene → phenolNaOH at 623 K and 300 atm, then acidify

↑ Back to top

Haloalkanes and Haloarenes Class 12 Important Questions With Solved Examples

Two full-length answers written the way you should write them in the hall — stating the rule, applying it, and finishing with a clear conclusion. Notice how short they are. Marks come from precision, not from length.

Example 32 — A three-mark comparison question
Question: Explain why chlorobenzene does not undergo nucleophilic substitution under conditions where chloroethane reacts readily. (3 marks)
Model answer: (1) In chlorobenzene a lone pair on chlorine is delocalised into the ring, so the C–Cl bond acquires partial double bond character and becomes shorter and stronger — 169 pm against 177 pm in chloromethane. (2) The carbon bearing chlorine is sp2 hybridised, with greater s-character than the sp3 carbon of chloroethane, which shortens and strengthens the bond further. (3) An SN1 route is unavailable because the phenyl cation it would require is not resonance stabilised. Together these make the C–Cl bond in chlorobenzene very difficult to break.
Example 33 — A mechanism-and-stereochemistry question
Question: An optically active alkyl halide is hydrolysed and the product is found to be optically inactive. Name the mechanism, explain the observation, and state the rate law. (3 marks)
Model answer: The mechanism is SN1. In the slow, rate-determining step the C–X bond ionises to give a planar carbocation, so the original three-dimensional arrangement is lost. Water can attack that planar carbocation from either face, producing both configurations. In this sample their rotations cancel, so the product is racemic and optically inactive. The rate law is rate = k[R–X], first order overall, since the nucleophile takes no part in the slow step.
Exam Tip — Answer in the examiner’s currency
A three-mark question wants three distinct chemical points, not one point explained three times. Number them in your head as you write. If you can attach a figure — a bond length, a bond enthalpy, a temperature — do it; it converts a plausible answer into an unarguable one.

↑ Back to top

Practice Worksheet

Ten questions, written fresh for this page. Attempt each one on paper before you open the answer — the reveal is only useful after you have committed to something. Give yourself 40 minutes.

Q1. Give the IUPAC name of (CH3)2CH–CH2–CH2Br.
Show Answer
The longest chain is four carbons (butane) with a methyl branch. Numbering from the bromine end puts bromo at C-1 and methyl at C-3, locant set {1, 3}; numbering the other way gives {2, 4}, which is higher at the first point of difference. Alphabetically bromo precedes methyl.
Answer: 1-bromo-3-methylbutane.
Q2. Arrange in increasing order of reactivity towards an SN1 reaction: bromobenzene, 1-bromobutane, 2-bromobutane, 2-bromo-2-methylpropane. Justify your order.
Show Answer
SN1 rate follows the stability of the carbocation formed in the slow step. Bromobenzene cannot form a carbocation at all — a phenyl cation is not resonance stabilised — so it is effectively unreactive. Among the rest, stability runs 3° > 2° > 1°.
Answer: bromobenzene < 1-bromobutane < 2-bromobutane < 2-bromo-2-methylpropane.
Q3. Bromoethane is treated separately with (a) KCN and (b) AgCN. Name both products and explain the difference.
Show Answer
Cyanide is an ambident nucleophile, able to attack through carbon or through nitrogen. KCN is ionic, releasing free CN− which attacks through the more nucleophilic carbon atom. AgCN is largely covalent, so the carbon end is tied up with silver and only the nitrogen lone pair is available.
Answer: (a) KCN gives propanenitrile, CH3CH2–C≡N (ethyl cyanide). (b) AgCN gives ethyl isocyanide, CH3CH2–N≡C.
Q4. 2-bromopentane is heated with alcoholic KOH. Give the major and minor alkenes and name the rule you used.
Show Answer
The structure is CH3–CHBr–CH2CH2CH3. β-hydrogens are available on C-1 (a CH3, three hydrogens) and on C-3 (a CH2, two hydrogens). By Saytzeff’s rule the hydrogen is removed from the β-carbon with fewer hydrogens, which is C-3, giving the more substituted and more stable alkene.
Answer: pent-2-ene is the major product (disubstituted C=C) and pent-1-ene the minor product (monosubstituted C=C). The rule is Saytzeff’s rule.
Q5. Give the organic product when 2-bromopropane reacts with (a) aqueous KOH and (b) alcoholic KOH, and explain why they differ.
Show Answer
Aqueous KOH provides abundant free, well-solvated hydroxide ions that behave as nucleophiles and attack the δ+ carbon, so substitution wins. Alcoholic KOH provides alkoxide ions, which are bulkier and more strongly basic, so they abstract a β-hydrogen and elimination wins.
Answer: (a) propan-2-ol, CH3CH(OH)CH3, by nucleophilic substitution. (b) propene, CH3CH=CH2, by β-elimination.
Q6. The C–Cl bond length is 169 pm in chlorobenzene but 177 pm in chloromethane. Give two reasons.
Show Answer
Reason 1 — resonance. In chlorobenzene a lone pair on chlorine is delocalised into the benzene ring, so the C–Cl bond acquires partial double bond character; a partial double bond is shorter than a pure single bond.
Reason 2 — hybridisation. The carbon bearing chlorine is sp2 in chlorobenzene (about 33% s-character) but sp3 in chloromethane (25% s-character). Greater s-character holds the bonding electrons closer to the nucleus, shortening the bond.
Consequence: the shorter, stronger bond is a large part of why haloarenes resist nucleophilic substitution.
Q7. 7.4 g of butan-1-ol is treated with excess thionyl chloride. Assuming an 80% yield, calculate the mass of 1-chlorobutane obtained. Write the equation.
Show Answer
Equation: C4H9OH + SOCl2 → C4H9Cl + SO2↑ + HCl↑
Working: M(butan-1-ol) = 4(12.011) + 10(1.008) + 15.999 = 74.12 g mol−1. Moles = 7.4 ÷ 74.12 = 0.0998 mol. The stoichiometry is 1 : 1, so theoretical moles of 1-chlorobutane = 0.0998 mol. M(C4H9Cl) = 48.044 + 9.072 + 35.45 = 92.57 g mol−1. Theoretical mass = 0.0998 × 92.57 = 9.24 g. Actual = 9.24 × 0.80.
Answer: 7.39 g, i.e. about 7.4 g.
Q8. The pure (+) enantiomer of a chiral alcohol has a specific rotation of +13.6°. A sample of the same alcohol prepared by hydrolysing a chiral bromide shows +3.4°. Calculate the optical purity and state what fraction of the sample is racemic. What does this tell you about the mechanism?
Show Answer
Working: optical purity (enantiomeric excess) = observed rotation ÷ rotation of the pure enantiomer = 3.4 ÷ 13.6 = 0.25 = 25%. Only that 25% is present as a single enantiomer; the remaining 75% must be a 50:50 racemic mixture, since a racemic mixture contributes zero rotation.
Interpretation: substantial racemisation means most molecules passed through a planar carbocation, so the reaction is largely SN1. The 25% excess of one enantiomer shows a competing SN2 pathway (or attack before the leaving group had fully departed) was also operating.
Q9. Benzene is treated with Cl2 and anhydrous FeCl3 to give A. A is treated with the same reagents to give B as the major product. Identify A and B, and explain why B is the major isomer.
Show Answer
A = chlorobenzene, from electrophilic substitution on benzene; anhydrous FeCl3 acts as a Lewis acid halogen carrier that polarises Cl–Cl.
B = 1,4-dichlorobenzene (para), with 1,2-dichlorobenzene as a minor product.
Why: the chlorine already on the ring is deactivating (its −I effect drains the ring) but ortho/para directing, because its +R effect can only push electron density to the ortho and para positions. Between those two, the bulky chlorine sterically hinders the ortho positions, so the para product dominates.
Q10. (a) Name the reaction of bromobenzene with bromoethane and sodium in dry ether, and give the main product. (b) Explain why propane cannot be prepared in good yield by the Wurtz reaction.
Show Answer
(a) One aromatic halide plus one alkyl halide with sodium in dry ether is the Wurtz–Fittig reaction. The main product is ethylbenzene, C6H5–CH2CH3, along with sodium bromide. Biphenyl and butane are formed as by-products because the couplings are random.
(b) The Wurtz reaction joins two alkyl groups, so from a single halide it can only give an alkane with an even number of carbon atoms; propane has three. Using a mixture of iodomethane and iodoethane does produce some propane, but also ethane and butane in comparable amounts, and the three have similar boiling points, so the yield of pure propane is poor.
Exam Tip — Mark yourself honestly
Award the mark only if you wrote the reason, not just the product. In this chapter almost every question carries a ‘why’ that is worth as much as the answer itself. A product with no justification is usually half a mark.

↑ Back to top

One Last Thing

You will not master this chapter in one sitting, and you are not supposed to. Kaizen — steady, unglamorous improvement — beats heroic all-nighters every single time. Tomorrow, aim for exactly one more correct answer than you managed today. One extra name reaction recalled without looking. One more Saytzeff call made confidently. One more mechanism written without hesitating over which order goes with which.

Do that for a fortnight and you will not have to revise this chapter before the exam — you will simply know it. Come back to this page whenever a doubt surfaces, redo the worksheet cold, and keep the running total going in your own favour. That is the whole method.

Build the chapter sequence: revise Solutions, strengthen redox and conductance ideas with Electrochemistry, and practise rate-law reasoning in Chemical Kinetics.

Written & reviewed by Team Principal Saab — Meet the team →