Haloalkanes and Haloarenes

This article follows the NCERT sequence chapter-for-chapter. It starts with classification (mono/di/polyhalogen compounds; sp3 C–X versus sp2 C–X; alkyl, allylic, benzylic, vinylic and aryl halides) and nomenclature (common names, IUPAC names, gem- and vic-dihalides), then the physical nature of the C–X bond (length, enthalpy, dipole moment trends). It covers the methods of preparing haloalkanes (from alcohols, from hydrocarbons, by halogen exchange) and haloarenes (electrophilic halogenation, the Sandmeyer reaction), and the physical properties (boiling point, density, solubility trends). The bulk of the chapter — and of this article — is the chemistry of haloalkanes: nucleophilic substitution (SN1 and SN2 mechanisms, ambident nucleophiles, and the stereochemistry of inversion, retention and racemization), elimination (Zaitsev's rule), and reactions with metals (Grignard reagent, Wurtz reaction). It then turns to haloarenes: why they resist nucleophilic substitution, how ortho/para nitro groups switch that resistance off, and their ordinary electrophilic substitution and metal reactions (Wurtz–Fittig, Fittig). It closes with polyhalogen compounds and their environmental footprint. Three worked derivations walk the SN2 mechanism, the SN1 mechanism, and nucleophilic aromatic substitution step by step; three worked examples cover naming, kinetics, and stereochemical calculation at board and competitive depth.

Classification: where the halogen sits decides everything

Halides are first classified by how many halogen atoms they carry — mono-, di- or polyhalogen (tri-, tetra-, and so on) compounds — and then, for monohalogen compounds, by the hybridisation of the carbon bearing the halogen. Compounds with an sp3 C–X bond include alkyl halides (R–X, the homologous series $\ce{C_nH_{2n+1}X}$, further split into primary/1°, secondary/2° and tertiary/3° according to the carbon type), allylic halides (X on an sp3 carbon adjacent to a C=C double bond) and benzylic halides (X on an sp3 carbon directly attached to an aromatic ring). Compounds with an sp2 C–X bond include vinylic halides (X on a double-bond carbon) and aryl halides (X directly on an aromatic ring carbon). This single hybridisation split is why the chapter has two very different halves: sp3 C–X (haloalkanes) undergoes fast nucleophilic substitution; sp2 C–X (haloarenes, vinylic halides) strongly resists it.

Nomenclature: common names, IUPAC names, and gem/vic-dihalides

Common names combine the alkyl or aryl group's name with "halide" (ethyl bromide, sec-butyl chloride); IUPAC names treat the halogen as a substituent prefix (bromoethane, 2-chlorobutane). For disubstituted benzenes, common names use ortho/*meta/para while IUPAC uses locants 1,2-, 1,3-, 1,4-. Dihaloalkanes carrying the same halogen atom on the same carbon are called <strong>gem-dihalides</strong> (geminal, common name: alkylidene halides), while those on adjacent* carbons are vic-dihalides (vicinal, common name: alkylene dihalides); IUPAC simply calls both dihaloalkanes with appropriate locants.

Nature of the carbon-halogen bond

Halogens are more electronegative than carbon, so the C–X bond is polarized: carbon bears $\delta+$, halogen bears $\delta-$. Going down the group, halogen atomic radius increases, so bond length increases steadily from C–F to C–I, while bond enthalpy (bond strength) falls just as steadily, since a longer bond is a weaker one. Dipole moment, however, is not monotonic — it depends on both the charge separation and the bond length ($\mu = q \times d$), and the very short C–F bond partly offsets fluorine's huge electronegativity, so CH3Cl actually carries the largest dipole moment of the four halomethanes, not CH3F.

where: the numbers are for CH3–X; bond length rises and bond enthalpy falls together down the halogen group, while dipole moment order is $\text{CH}_3\text{Cl} (1.860\ \text{D}) > \text{CH}_3\text{F} (1.847\ \text{D}) > \text{CH}_3\text{Br} (1.830\ \text{D}) > \text{CH}_3\text{I} (1.636\ \text{D})$.

Physical properties: boiling point, density and solubility

For the same alkyl group, boiling point rises RF less than RCl less than RBr less than RI, because larger, heavier halogens have stronger van der Waals attractions; branching lowers boiling point (2-bromo-2-methylpropane boils lowest among the three isomeric bromobutanes). Among isomeric dihalobenzenes, boiling points are nearly identical, but the para isomer melts noticeably higher than ortho or meta because its symmetry lets it pack more efficiently into a crystal lattice. Density rises with the number of carbon and halogen atoms and with the halogen's atomic mass — bromo, iodo and polychloro derivatives are all denser than water. Haloalkanes dissolve only slightly in water (too little energy is released forming new haloalkane–water attractions to compensate for breaking water's hydrogen bonds) but dissolve freely in organic solvents.

Preparation of haloalkanes

Alcohols are the most accessible starting material. Concentrated HX, phosphorus halides (PX3, PX5) or thionyl chloride (SOCl2) all replace –OH with halogen; SOCl2 is preferred because its byproducts, SO2 and HCl, are gases that escape, leaving a pure alkyl halide (the Darzens process). Primary and secondary alcohols need a ZnCl2 catalyst with HCl; tertiary alcohols react simply by shaking with conc. HCl at room temperature. Alkyl bromides come from constant-boiling 48% HBr; alkyl iodides come from heating the alcohol with NaI/KI in 95% orthophosphoric acid (H2SO4 is avoided because it oxidizes the HI formed to I2). Reactivity of alcohols follows $3° > 2° > 1°$. From hydrocarbons: free-radical halogenation of alkanes gives a messy isomer mixture and low yield of any single product, while alkenes react cleanly — HX addition follows Markovnikov's rule, and X2 addition (bromine water/CCl4 decolorization) gives vic-dihalides and is the classic laboratory test for unsaturation. Halogen exchange converts one halide to another: the Finkelstein reaction (R–Cl/Br + NaI in dry acetone $\rightarrow$ R–I, with NaCl/NaBr precipitating out and pulling the equilibrium forward by Le Chatelier's principle) makes iodides, while the Swarts reaction (R–Cl/Br heated with AgF, Hg2F2, CoF2 or SbF3) makes fluorides.

Preparation of haloarenes

The alcohol route does not work for aryl halides because the phenolic C–O bond has partial double-bond character and resists cleavage. Instead, aryl chlorides and bromides form by electrophilic substitution: arene + Cl2 or Br2 in the presence of a Lewis acid (Fe or FeCl3) gives ortho/para products, separable because their melting points differ substantially; iodination is reversible and needs an oxidizing agent (HNO3, HIO4) to remove the HI byproduct; fluorination is not done this way because fluorine is too reactive. The second route is the Sandmeyer reaction: a primary aromatic amine, treated with NaNO2 and cold mineral acid, forms a diazonium salt, which on mixing with Cu2Cl2 or Cu2Br2 (cuprous halide) gives the aryl chloride or bromide with loss of N2; aryl iodide needs no copper catalyst at all — simply shaking the diazonium salt with KI does the job.

Kinetics of nucleophilic substitution: SN2 versus SN1

Haloalkanes undergo nucleophilic substitution because the electrophilic, $\delta+$ carbon attracts electron-rich nucleophiles, which displace the halide leaving group. Table 6.4 of the NCERT text lists common nucleophiles and the substitution products they give: $\text{OH}^- \to$ alcohol, $\text{OR}'^- \to$ ether, $\text{I}^- \to$ alkyl iodide, $\text{NH}_3 \to$ amine, $\text{CN}^- \to$ nitrile, $\text{NO}_2^- \to$ nitroalkane, $\text{R}'\text{COO}^- \to$ ester, $\text{H}^-$ (from LiAlH4) $\to$ hydrocarbon, and $\text{R}'^-$ (a Grignard carbanion) $\to$ alkane. Two experimentally distinct mechanisms deliver these products.

where: $k$ is the rate constant, $[\text{RX}]$ the haloalkane concentration and $[\text{Nu}^-]$ the nucleophile concentration. SN2 (substitution nucleophilic bimolecular) is second-order overall — both concentrations matter because both species collide in the single rate-determining step. SN1 (substitution nucleophilic unimolecular) is first-order — only the haloalkane's slow ionization matters; the nucleophile's subsequent attack is fast and does not affect the rate. SN2 reactivity runs $1° > 2° > 3°$ (bulky groups near the reacting carbon block the nucleophile's backside approach), while SN1 reactivity runs $3° > 2° > 1°$, and allylic/benzylic halides also react fast by SN1 because their carbocations are resonance-stabilized. For a given alkyl group, both mechanisms share the leaving-group order $\text{R–I} > \text{R–Br} > \text{R–Cl} \gg \text{R–F}$, since a weaker C–X bond leaves more easily. Two full mechanisms, worked step by step with arrows, follow in the derivation blocks below.

Ambident nucleophiles: two ways to attack

Some nucleophiles have two possible atoms through which they can bond — these are ambident nucleophiles. Cyanide is a resonance hybrid, \ce{^{-}C#N <-> C=N^{-}}, and can attack through carbon (giving an alkyl cyanide, R–CN) or through nitrogen (giving an isocyanide, R–NC). With KCN, which is predominantly ionic, the more stable C–C bond forms preferentially, giving the nitrile as the main product; with AgCN, which is largely covalent, nitrogen is left free to bond, giving the isocyanide as the main product. Nitrite ion behaves the same way: O-attack gives an alkyl nitrite (R–O–N=O), N-attack gives a nitroalkane (R–NO2).

Stereochemistry of substitution: chirality, inversion and racemization

A molecule that is non-superimposable on its mirror image is chiral; such molecules rotate the plane of plane-polarized light (they are optically active) and are measured with a polarimeter — dextrorotatory ($+$, clockwise) or laevorotatory ($-$, anticlockwise). A carbon bonded to four different groups is a stereocentre; two such molecules that are non-superimposable mirror images of each other are enantiomers. Enantiomers share every ordinary physical property (melting point, boiling point, density) but rotate light in opposite directions by equal amounts; a 50:50 mixture of the two, called a racemic mixture (prefixed $(\pm)$ or dl), shows zero net rotation because the rotations cancel. In an SN2 reaction, the incoming nucleophile attacks from the side directly opposite the leaving halide, so the product's configuration at the stereocentre is inverted relative to the reactant — this is Walden inversion, exactly like an umbrella turning inside out in a gust of wind. In an SN1 reaction, the planar, sp2 carbocation intermediate can be attacked from either face with roughly equal probability, so the product is a near-racemic mixture — optically active reactant gives optically inactive (or only weakly active) product. The full step-by-step mechanisms for both are worked out in the derivation blocks that follow.

Elimination reactions: Zaitsev's rule

When a haloalkane bearing a $\beta$-hydrogen is heated with alcoholic KOH, HX is eliminated: a hydrogen leaves the $\beta$-carbon and the halogen leaves the $\alpha$-carbon, forming an alkene ($\beta$-elimination). If more than one $\beta$-hydrogen is available, more than one alkene is possible, but one usually dominates. Zaitsev's rule (Saytzeff), formulated in 1875, states that the preferred alkene is the one with the greater number of alkyl groups on the doubly bonded carbons — the more substituted, more stable alkene. So 2-bromopentane gives pent-2-ene as the major elimination product over pent-1-ene. Substitution and elimination are always in competition for a haloalkane with $\beta$-hydrogens: the outcome depends on the substrate (1° favours SN2, 3° favours SN1/elimination), the strength and bulk of the base/nucleophile (a bulky base prefers to abstract a proton rather than attack a crowded carbon), and the reaction conditions (aqueous KOH favours substitution; alcoholic KOH favours elimination).

Reactions with metals: Grignard reagents and the Wurtz reaction

Alkyl halides react with magnesium metal in dry ether to give Grignard reagents, $\text{R–MgX}$ — discovered by Victor Grignard in 1900 (Nobel Prize 1912). The C–Mg bond is covalent but strongly polarized (carbon pulls electron density from the electropositive metal), while the Mg–X bond is essentially ionic; this makes the carbon strongly nucleophilic/carbanion-like. Grignard reagents react instantly with any proton source — water, alcohols, even amines — to give the corresponding hydrocarbon, so the reaction must be run under strictly anhydrous conditions (dry ether), and this proton-quench is itself sometimes used deliberately to convert a halide to a hydrocarbon. Separately, the Wurtz reaction couples two molecules of an alkyl halide using sodium metal in dry ether, $2\text{R–X} + 2\text{Na} \rightarrow \text{R–R} + 2\text{NaX}$, giving a symmetrical hydrocarbon with double the original carbon count — useful only when a symmetrical, even-carbon product is actually wanted.

Why haloarenes resist nucleophilic substitution

Aryl halides are dramatically less reactive than alkyl halides toward nucleophilic substitution, for three physically distinct reasons that reinforce each other. First, resonance: a lone pair on the halogen conjugates into the aromatic $\pi$ system, giving the C–X bond partial double-bond character, so it is harder to break (this is the same reason chlorobenzene's dipole moment is lower than cyclohexyl chloride's — the C–Cl bond is less purely single, less purely polarized in the ionic sense). Second, hybridisation: the sp2 carbon bearing the halogen is more electronegative than an sp3 carbon (greater s-character holds bonding electrons tighter), so C–Cl measures only $169\ \text{pm}$ in a haloarene versus $177\ \text{pm}$ in a haloalkane — a shorter bond is a harder bond to break. Third, phenyl cation instability: unlike a tertiary alkyl carbocation, a phenyl cation formed by C–X ionization cannot be resonance-stabilized by the ring (the empty orbital lies in the ring plane, orthogonal to the $\pi$ system), so SN1 at an aryl carbon is essentially impossible. A fourth, softer effect is simple electrostatic repulsion between an electron-rich nucleophile and an already electron-rich aromatic ring. Chlorobenzene can be forced to phenol, but only under brutal conditions — aqueous NaOH at 623 K and 300 atmospheres.

Activation of haloarenes by electron-withdrawing groups

Strong electron-withdrawing groups at the ortho or para position dramatically accelerate nucleophilic substitution on a haloarene, but a meta substituent does essentially nothing. A single ortho- or para-nitro group already helps; several such groups (as in the classic conversion of 2,4-dinitrochlorobenzene or 2,4,6-trinitrochlorobenzene, picryl chloride) make the halide hydrolyse in cold water. The mechanistic reason is worked out fully in the third derivation block below: only ortho/*para* attack lets the developing negative charge delocalize onto the nitro-bearing carbon itself.

Electrophilic substitution and reactions with metals

Away from the C–X bond, the aromatic ring of a haloarene behaves like ordinary benzene toward electrophiles — it undergoes halogenation, nitration, sulphonation and Friedel–Crafts alkylation/acylation. The halogen substituent is ortho, para-directing because resonance pushes extra electron density into those positions, yet it is also weakly deactivating overall because of its stronger $-I$ (inductive) effect withdrawing electron density from the whole ring; the two effects do not cancel — inductive withdrawal wins for reactivity (haloarenes react slower than benzene), while resonance donation wins for orientation (substitution still lands ortho/para). Finally, aryl halides participate in two metal-coupling reactions structurally related to the Wurtz reaction: the Wurtz–Fittig reaction couples an alkyl halide and an aryl halide with sodium in dry ether to give an alkylarene, and the Fittig reaction couples two aryl halides with sodium in dry ether to give a biaryl (Ar–Ar).

Polyhalogen compounds and their environmental footprint

Compounds bearing more than one halogen have their own industrial life and their own hazards. Dichloromethane (methylene chloride) is a paint-remover and process solvent that harms the central nervous system on exposure. Chloroform (trichloromethane) dissolves fats and alkaloids and was once a surgical anaesthetic; it slowly oxidizes in air and light to the poisonous gas phosgene ($\ce{COCl2}$), which is why it must be stored in dark, completely filled bottles. Iodoform was once an antiseptic, though the antiseptic action came from liberated iodine, not the iodoform molecule itself. Carbon tetrachloride was once a common cleaning fluid and fire extinguisher but is now known to be carcinogenic and to deplete stratospheric ozone. Freons — chlorofluorocarbons of methane and ethane, made industrially by the Swarts reaction (Freon-12, $\ce{CCl2F2}$, from $\ce{CCl4}$) — are stable, non-toxic refrigerants and propellants precisely because of their unreactivity, but that same stability lets them drift unchanged into the stratosphere, where UV light triggers radical chain reactions that destroy ozone. DDT (p,p'-dichlorodiphenyltrichloroethane), the first chlorinated insecticide, was hugely effective against malaria mosquitoes and typhus lice but proved fat-soluble and resistant to metabolism, so it bioaccumulates up the food chain; it was banned in the United States in 1973 though it remains in use elsewhere.

Board level - IUPAC names and classification of the eight isomers of C5H11Br

Draw and classify (primary, secondary or tertiary) all eight structural isomers of molecular formula $\ce{C5H11Br}$, and give the IUPAC name of $\ce{(CH3)3CCH2Br}$ specifically.

Step 1. Fix the carbon skeletons available for five carbons: the straight chain (pentane), the singly branched chain (2-methylbutane), and the doubly branched chain (2,2-dimethylpropane, neopentane). Bromine can sit at each chemically distinct carbon position on each skeleton.

Step 2. Enumerate the straight-chain (pentane) isomers: 1-bromopentane (1°), 2-bromopentane (2°), 3-bromopentane (2°, note 3-bromopentane is its own mirror-symmetric position on the chain).

Step 3. Enumerate the 2-methylbutane isomers: 1-bromo-3-methylbutane (1°), 2-bromo-3-methylbutane (2°), 2-bromo-2-methylbutane (3°), 1-bromo-2-methylbutane (1°).

Step 4. Enumerate the neopentane isomer: only one position is chemically distinct on 2,2-dimethylpropane's periphery, giving 1-bromo-2,2-dimethylpropane (1°), which is exactly $\ce{(CH3)3CCH2Br}$.

Step 5. Count and name. That gives 8 distinct isomers in total (1-bromopentane, 2-bromopentane, 3-bromopentane, 1-bromo-3-methylbutane, 2-bromo-3-methylbutane, 2-bromo-2-methylbutane, 1-bromo-2-methylbutane, 1-bromo-2,2-dimethylpropane), matching NCERT's own Example 6.1 count. Answer: there are 8 structural isomers of C5H11Br, and $\ce{(CH3)3CCH2Br}$ is 1-bromo-2,2-dimethylpropane, a primary (1°) bromide.

Competitive level (JEE / NEET) - identifying SN1 versus SN2 from kinetic data and computing rate

The hydrolysis of tert-butyl bromide in aqueous acetone was studied. Doubling $[\ce{OH^-}]$ at constant $[\ce{RBr}]$ left the rate unchanged; doubling $[\ce{RBr}]$ at constant $[\ce{OH^-}]$ doubled the rate. Given $k = 1.0 \times 10^{-3}\ \text{s}^{-1}$ and $[\ce{RBr}] = 0.050\ \text{M}$, identify the mechanism, write the rate law, and calculate the rate.

Step 1. Read the mechanism off the kinetic dependence. Rate is unaffected by $[\ce{OH^-}]$ but scales linearly with $[\ce{RBr}]$ — this is first-order-overall behaviour, the signature of SN1, consistent with tert-butyl bromide's tertiary, highly stabilized carbocation intermediate (the same substrate treated in the SN1 derivation above).

Step 2. Write the correct rate law. Since only $[\ce{RBr}]$ appears, $\text{Rate} = k[\ce{RBr}]$, with units of $k$ in $\text{s}^{-1}$ (consistent with the given $k = 1.0\times10^{-3}\ \text{s}^{-1}$, a first-order rate constant).

Step 3. Substitute and compute. $\text{Rate} = (1.0\times10^{-3}\ \text{s}^{-1})(0.050\ \text{M}) = 5.0\times10^{-5}\ \text{M s}^{-1}$.

Step 4. Sanity-check against structure. tert-Butyl bromide is a 3° halide; SN1 reactivity order is $3° > 2° > 1°$, exactly the order predicted from carbocation stability in the derivation above, and it is consistent with an SN1 assignment rather than SN2 (which would demand a primary substrate and rate dependence on $[\ce{OH^-}]$ too). Answer: the mechanism is SN1, Rate $= k[\text{RBr}]$, and the rate $= 5.0\times10^{-5}\ \text{M s}^{-1}$. Timing note: under 90 seconds once the kinetic clue is recognized. The fast read is: rate independent of nucleophile concentration $\Rightarrow$ SN1 $\Rightarrow$ rate law uses only $[\text{RBr}]$. Do not be tempted to write $\text{Rate} = k[\text{RBr}][\text{OH}^-]$ out of habit — always check which concentrations the question says the rate actually depends on before writing the rate law.

Competitive level (JEE / NEET) - partial racemization in SN1 solvolysis: percentage racemization and enantiomeric excess

Pure (S)-2-bromobutane has a specific rotation of $+20°$ under the conditions of an experiment (as its solvolysis product, pure (R)-butan-2-ol, would show if formed with complete inversion and no racemization). When the actual SN1 solvolysis is carried out and the product mixture is examined, the observed net rotation is only $+15°$. Calculate the percentage optical purity (enantiomeric excess) of the product and the percentage of the product that is racemized.

Step 1. Recall why the answer is not simply "0° = fully racemic." The SN1 mechanism above predicts attack on the planar carbocation from either face with nearly equal probability, but not perfectly equal — a fleeting ion-pair intermediate can shield the front face for an instant before the leaving bromide fully departs, so real SN1 solvolysis gives net inversion with partial, not complete, racemization.

Step 2. Apply the optical-purity formula. $$\text{Optical purity (\% ee)} = \frac{\text{observed rotation}}{\text{maximum (pure enantiomer) rotation}} \times 100 = \frac{15°}{20°} \times 100 = 75\%. \qquad (5)$$

Step 3. Convert optical purity into the racemized fraction. If the product were $x\%$ racemic (an exactly 50:50 mixture contributing zero net rotation) and $(100-x)\%$ single enantiomer (contributing full rotation), the observed optical purity equals $(100-x)\%$. So $100 - x = 75 \Rightarrow x = 25\%$: a quarter of the product is present as a racemic (optically silent) pair, while three-quarters behaved as the pure, inverted enantiomer.

Step 4. Translate into actual enantiomer percentages for a full picture. Of the racemized 25%, half (12.5%) is the "wrong" (retained) enantiomer and half (12.5%) is extra inverted enantiomer; adding the pure-inversion contribution, total inverted product $= 75\% + 12.5\% = 87.5\%$ and total retained product $= 12.5\%$, using the standard relation $\%\text{major} = (100+\text{ee})/2$ and $\%\text{minor} = (100-\text{ee})/2$. Answer: the product is 75% optically pure (75% ee), 25% of it is racemized, corresponding to an 87.5 : 12.5 mixture of inverted to retained product — net inversion dominates, but SN1 is never perfectly clean racemization in practice. Timing note: about 2 minutes. The trap is assuming SN1 always means exactly 50:50 racemization (0% ee) — real ion-pair SN1 gives partial racemization with a net-inversion bias, and JEE/NEET numericals often test exactly this nuance using the optical-purity formula rather than the textbook's idealized "complete racemization" statement.

Treating "sp3 C–X" and "sp2 C–X" as a minor labelling detail. It is the single most consequential fact in the chapter — it is the entire reason haloalkanes undergo fast nucleophilic substitution while haloarenes and vinylic halides strongly resist it. Always classify a halide's hybridisation before predicting its reactivity.

Assuming SN2 always means "primary halide" and SN1 always means "tertiary halide," full stop. The correct predictors are steric bulk near the reacting carbon (for SN2) and carbocation stability (for SN1); this is exactly why allylic and benzylic halides — which can be primary by carbon count — still react readily by SN1, because their carbocations are resonance-stabilized.

Believing SN1 always gives a perfectly 50:50 racemic product. Textbook idealization aside, real SN1 solvolysis usually shows a small net inversion bias from tight ion pairing, as Worked Example 3 above computes explicitly; "racemization" in SN1 means predominant loss of optical activity, not always exactly zero.

Forgetting that dipole moment does not track electronegativity in a simple line. CH3Cl has a larger dipole moment than CH3F because dipole moment depends on bond length as well as charge separation ($\mu = qd$) — the very short C–F bond partly cancels fluorine's larger electronegativity.

Mixing up gem- and vic-dihalides. gem = both halogens on the same (geminal) carbon; vic = on adjacent (vicinal) carbons. This distinction is tested directly in nomenclature questions.

Calling chlorobenzene unreactive toward electrophiles. It is deactivated relative to benzene (slower reactions overall, due to the halogen's net $-I$ effect) but it remains strongly ortho/para-directing (due to resonance donation) — "deactivating but o,p-directing" is the precise, examinable description, not "unreactive."

Assuming AgCN and KCN must give the same product with a haloalkane. KCN (ionic, carbon-rich electron density) gives predominantly the nitrile via carbon attack; AgCN (covalent, nitrogen free to donate) gives predominantly the isocyanide via nitrogen attack — an ambident-nucleophile distinction worth memorizing exactly.

• Classify any given halide by number of halogens (mono/di/poly) and by hybridisation (sp3: alkyl, allylic, benzylic; sp2: vinylic, aryl), and identify it as primary, secondary or tertiary where relevant.
• Assign correct IUPAC names to haloalkanes and haloarenes, and distinguish gem- from vic-dihalides.
• Explain the C-X bond length, bond enthalpy and dipole moment trends across F, Cl, Br, I, including why dipole moment is not simply monotonic.
• Choose the correct preparation route for a target haloalkane or haloarene (alcohol + SOCl2/PX3/PX5/HX, alkene + HX/X2, Finkelstein, Swarts, electrophilic halogenation, Sandmeyer) and justify the choice.
• Write the SN2 mechanism with curved arrows, explain backside attack and predict Walden inversion for a given stereocentre.
• Write the SN1 mechanism with curved arrows, explain the planar carbocation intermediate, and predict (near-)racemization along with the correct reactivity order ($3° > 2° > 1°$, with allylic/benzylic accelerated).
• Predict the major elimination product using Zaitsev's rule and reason about the substitution-versus-elimination competition for a given substrate and base.
• Explain, using resonance, hybridisation and phenyl-cation instability, why haloarenes resist nucleophilic substitution, and why ortho/para (not meta) electron-withdrawing groups switch that resistance off via addition-elimination.
• Distinguish ambident nucleophiles (CN-/NC-, NO2-/ONO-) and predict which product KCN versus AgCN, or KNO2 versus AgNO2, will favour.
• Describe the industrial uses and environmental hazards of at least four polyhalogen compounds (chloroform, iodoform, CCl4, Freons, DDT).

Ask the tutor: Want to work through drawing full curved-arrow mechanisms for a specific SN1 or SN2 problem from your textbook, or practice predicting the major product for a tricky elimination-versus-substitution case?