Classification of haloalkanes
We sort haloalkanes in two ways.
- By number of halogens: mono (CH₃Cl), di (CH₂Cl₂), tri (CHCl₃), poly.
- By the carbon that holds X: count how many other carbons are joined to it. 1 → primary (1°), 2 → secondary (2°), 3 → tertiary (3°).
- By position: allylic (X on the carbon next to a C=C), benzylic (X on a carbon next to a benzene ring), vinylic (X right on a C=C carbon).
Example: CH₃CH₂Br is 1°, (CH₃)₂CHBr is 2°, (CH₃)₃CBr is 3°.
Nomenclature: naming haloalkanes
Common names say "alkyl halide": CH₃Cl is methyl chloride. IUPAC names treat the halogen as a prefix (fluoro, chloro, bromo, iodo) on the parent alkane.
- Find the longest chain that holds the halogen.
- Number it so the first substituent gets the lowest number.
- Write prefixes in alphabetical order.
(CH₃)₂CHCl → 2-chloropropane. CH₃CHBrCH₂CH₃ → 2-bromobutane. (CH₃)₃CCl → 2-chloro-2-methylpropane. CH₂=CHCH₂Cl → 3-chloroprop-1-ene.
Nature of the C–X bond
Halogens pull electrons more strongly than carbon, so carbon gets a small positive charge (δ+) and X a small negative charge (δ−). Going down the group F → Cl → Br → I, the halogen atom gets bigger, so the bond gets longer and weaker. The weaker the bond, the easier X leaves. So reactivity is R–I > R–Br > R–Cl > R–F.
Preparation of haloalkanes
- From alcohols (best route): R–OH + HCl (with ZnCl₂); R–OH + PCl₅ or PCl₃; R–OH + SOCl₂ (thionyl chloride) — the best, because the by-products SO₂ and HCl are gases and escape, leaving pure R–Cl.
- From alkanes: Cl₂ in UV light (free radical). It gives a mixture, so it is not a good lab method.
- From alkenes: add HX. With unequal alkenes, H goes to the carbon that already has more H (Markovnikov rule). With HBr and a peroxide, the opposite happens.
- Halogen exchange: Finkelstein (R–Cl + NaI in dry acetone → R–I; NaCl falls out) and Swarts (R–Br + AgF → R–F).
Physical properties
- Boiling point is higher than the parent alkane (bigger, polar molecules). It rises from RF to RI, and with chain length.
- Branching lowers the boiling point (a rounder molecule has less surface to touch neighbours).
- They hardly dissolve in water: they cannot form strong H-bonds with water. They dissolve well in organic solvents.
- Bromo, iodo and poly-chloro compounds are denser than water.
Chemical properties: SN1 and SN2
A nucleophile ("nucleus lover") is an electron-rich species such as OH⁻, CN⁻, NH₃. It attacks the δ+ carbon and pushes out X⁻. This is nucleophilic substitution.
| SN2 | SN1 | |
|---|---|---|
| Steps | One step | Two steps (carbocation first) |
| Rate depends on | [RX] and [Nu⁻] | [RX] only |
| Best for | CH₃X > 1° > 2° > 3° | 3° > 2° > 1° > CH₃X |
| Result on a chiral C | Inversion | Racemisation |
| Helped by | Aprotic solvent, strong Nu⁻ | Polar protic solvent (water, alcohol) |
Why? In SN2 the nucleophile must reach the back of the carbon, and big groups block it. In SN1 the carbocation must be stable, and more alkyl groups push electrons towards it and steady it. Benzylic and allylic halides do SN1 easily because their carbocations are shared by resonance.
Other reactions: KCN gives R–CN (cyanide), but AgCN gives R–NC (isocyanide); KNO₂ gives R–ONO, AgNO₂ gives R–NO₂. Reaction with sodium (Wurtz, dry ether) joins two R groups: 2R–X + 2Na → R–R + 2NaX. With Mg in dry ether it gives a Grignard reagent RMgX.
Optical activity and chirality
A carbon joined to four different groups is called chiral (asymmetric). Its molecule and its mirror image cannot be placed on top of each other, like your left and right hands. These two forms are enantiomers. One rotates plane-polarised light to the right (dextro, +), the other to the left (laevo, −) by the same amount.
A 50:50 mixture is a racemic mixture: the rotations cancel, so it is optically inactive. SN2 on a chiral carbon gives inversion (Walden inversion); SN1 gives mostly a racemic product.
Example: 2-bromobutane has a chiral C2 (groups H, CH₃, C₂H₅, Br). 1-bromobutane has none.
Elimination reactions
Heat a haloalkane with alcoholic KOH: the base removes an H from the β-carbon (the carbon next door) and X leaves from the α-carbon. A C=C forms. This is β-elimination (dehydrohalogenation).
Saytzeff (Zaitsev) rule: if more than one alkene can form, the main one is the alkene with more alkyl groups on the double-bond carbons. 2-bromobutane gives mainly but-2-ene.
Aqueous KOH → substitution (alcohol). Alcoholic KOH → elimination (alkene). 3° halides prefer elimination with strong bases.
Board exam focus
The unit carries about 6 marks. Common questions: name/draw isomers, arrange by SN1 or SN2 rate, explain why SOCl₂ is preferred, predict the major product (Markovnikov, Saytzeff), and "what is a racemic mixture?". Practise 2-mark "give reason" answers.
Key formulas and definitions
- R–OH + SOCl₂ → R–Cl + SO₂↑ + HCl↑
- R–Cl + NaI → R–I + NaCl (Finkelstein, dry acetone)
- R–Br + AgF → R–F + AgBr (Swarts)
- SN2 rate = k[RX][Nu⁻]; SN1 rate = k[RX]
- R–X + KOH(aq) → R–OH + KX (substitution)
- R–CH₂–CH₂–X + KOH(alc) → R–CH=CH₂ + KX + H₂O (elimination)
- Enantiomeric excess % = (major − minor) ÷ total × 100; observed rotation = ee × rotation of pure form
Worked examples
1. Classify CH₃CH₂CH₂Br, CH₃CHBrCH₃ and (CH₃)₃CBr as 1°, 2° or 3°.
Count carbons on the C that holds Br. CH₃CH₂CH₂Br: 1 → 1°. CH₃CHBrCH₃: 2 → 2°. (CH₃)₃CBr: 3 → 3°.
2. Give the IUPAC name of CH₃–CH(CH₃)–CH₂–CHCl–CH₃.
Longest chain = 5 C. Number from the right so Cl gets 2 and CH₃ gets 4: 2-chloro-4-methylpentane (chloro before methyl, alphabetical).
3. How many structural isomers does C₄H₉Br have? How many of them are chiral?
Chains: 1-bromobutane, 2-bromobutane, 1-bromo-2-methylpropane, 2-bromo-2-methylpropane = 4. Only 2-bromobutane has a carbon with four different groups (H, Br, CH₃, C₂H₅), so 1 is chiral.
4. Arrange for SN2 rate: (CH₃)₃CBr, CH₃Br, CH₃CH₂Br, (CH₃)₂CHBr.
SN2 needs a clear backside. Fewer groups = faster: CH₃Br > CH₃CH₂Br > (CH₃)₂CHBr > (CH₃)₃CBr.
5. A sample of 2-butanol-derived bromide has 80% (+) form and 20% (−) form. The pure (+) form rotates +23°. Find the observed rotation.
ee = (80 − 20) ÷ 100 × 100 = 60%. Observed rotation = 0.60 × (+23°) = +13.8°.
6. Predict the main product when 2-bromobutane is heated with alcoholic KOH, and give the reason.
β-Hs are on C1 and C3. Removing H from C3 gives but-2-ene (CH₃CH=CHCH₃, two alkyl groups on C=C); from C1 gives but-1-ene. By the Saytzeff rule the more substituted but-2-ene is the main product.
7. Why does (CH₃)₃CBr react faster than CH₃CH₂Br with water (SN1)?
SN1 speed depends on how stable the carbocation is. (CH₃)₃C⁺ has three CH₃ groups pushing electrons in (+I effect and hyperconjugation), so it is far more stable than CH₃CH₂⁺. More stable carbocation → faster SN1.
Common mistakes
- Counting hydrogens instead of carbons for 1°/2°/3°: count the carbons attached to the C–X carbon.
- Mixing up the order: SN2 is fastest for 1° (less crowding), SN1 is fastest for 3° (stable carbocation).
- Thinking aqueous KOH and alcoholic KOH do the same thing: aqueous gives an alcohol, alcoholic gives an alkene.
- Calling any compound with a halogen "chiral": the carbon must carry four different groups.