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MHT-CET Chemistry · Halogen Derivatives of Alkanes

Preparation of Alkyl Halides

Alkyl halides come from alcohols (HX, PCl₅, SOCl₂), from alkenes by Markovnikov addition of HX, and from other halides by exchange — Finkelstein for iodides, Swarts for fluorides; the reverse direction, Wurtz and Fittig coupling with sodium, removes the halogen and joins two carbons.

Why this matters

24 PYQs, none HARD. Of the fifteen set from 2021 to 2025, six are the two exchange reactions by name or product (Finkelstein → iodide, Swarts → fluoride, tert-butyl bromide + AgF), six are Wurtz, Fittig and Wurtz–Fittig — which is which, which halide couples to a named alkane, which alkane cannot form — and three are reagent-from-by-products or the alkene that adds HBr to a named product. Names and one Markovnikov rule.

Concept 1 of 3: From Alcohols and Alkenes

An alcohol's OH is swapped for X by HX (tertiary fastest), by PCl₅ (giving HCl and POCl₃) or by SOCl₂ (giving only gases, the cleanest). An alkene adds HX across the double bond with the halogen going to the MORE substituted carbon — Markovnikov — so 1-methylcyclohexene gives 1-bromo-1-methylcyclohexane.

Definition

  • R-OH+HX→R-X+H2O\text{R-OH} + \text{HX} \to \text{R-X} + \text{H}_2\text{O}; with HCl needs ZnCl₂ (Lucas); rate 3° > 2° > 1°.
  • R-OH+PCl5→R-Cl+HCl+POCl3\text{R-OH} + \text{PCl}_5 \to \text{R-Cl} + \text{HCl} + \text{POCl}_3 — the by-products HCl + POCl₃ identify PCl₅. 3R-OH+PCl3→3R-Cl+H3PO33\text{R-OH} + \text{PCl}_3 \to 3\text{R-Cl} + \text{H}_3\text{PO}_3. R-OH+SOCl2→R-Cl+SO2+HCl\text{R-OH} + \text{SOCl}_2 \to \text{R-Cl} + \text{SO}_2 + \text{HCl} (Darzen; gaseous by-products).
  • Alkene + HX: Markovnikov — H to the carbon with more H, X to the more substituted carbon. 1-Methylcyclohexene + HBr → 1-bromo-1-methylcyclohexane (3° bromide); propene + HBr → 2-bromopropane. Peroxide reverses it for HBr only.
  • Alkane + X₂ (light): free-radical substitution, a mixture.

Alcohol to halide

R-OH→HX / PCl5 / SOCl2R-X;alkene+HX→MarkovnikovR-X\text{R-OH} \xrightarrow{\text{HX / PCl}_5\text{ / SOCl}_2} \text{R-X};\qquad \text{alkene} + \text{HX} \xrightarrow{\text{Markovnikov}} \text{R-X}

Worked example

Name the product of 2-methylpropene with HBr, and identify Q in R-OH+Q→R-Cl+SO2+HCl\text{R-OH} + \text{Q} \to \text{R-Cl} + \text{SO}_2 + \text{HCl}.
Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 3rd May Shift 2 · Q84Easy

Example 1 · Halogen Derivatives of Alkanes · Preparation of Alkyl Halides

Identify compound Q in following reaction. R-OH+Q→R-Cl+HCl+POCl3R\text{-}OH + Q \rightarrow R\text{-}Cl + HCl + POCl_3

Adding HBr the wrong way round

Without peroxide the H goes where the hydrogens already are. On 1-methylcyclohexene that puts Br on C1 with the methyl — the tertiary product — not on C2.

Concept 2 of 3: Finkelstein and Swarts: Swapping the Halogen

Two exchanges with names. Finkelstein: chloride or bromide + NaI in dry acetone gives the iodide, pushed by NaCl or NaBr precipitating out of acetone. Swarts: chloride or bromide + a metal fluoride (AgF, Hg₂F₂, CoF₂, SbF₃) gives the fluoride. Both keep the carbon skeleton — tert-butyl bromide + AgF is 2-fluoro-2-methylpropane.

Definition

  • Finkelstein: R-Cl+NaI→dry acetoneR-I+NaCl↓\text{R-Cl} + \text{NaI} \xrightarrow{\text{dry acetone}} \text{R-I} + \text{NaCl}\downarrow — alkyl IODIDES.
  • Swarts: R-Br+AgF→R-F+AgBr\text{R-Br} + \text{AgF} \to \text{R-F} + \text{AgBr} — alkyl FLUORIDES (also Hg₂F₂, CoF₂, SbF₃).
  • tert-Butyl bromide + AgF → 2-fluoro-2-methylpropane, skeleton unchanged.
  • Fluorination of chloroform gives refrigerant-22, CHClF2\text{CHClF}_2: CHCl3→SbF3CHClF2\text{CHCl}_3 \xrightarrow{\text{SbF}_3} \text{CHClF}_2.

The two exchanges

Finkelstein: R-X+NaI→R-I;Swarts: R-X+AgF→R-F\text{Finkelstein: R-X} + \text{NaI} \to \text{R-I};\qquad \text{Swarts: R-X} + \text{AgF} \to \text{R-F}

Worked example

How would you convert 1-chloropropane into 1-iodopropane, and into 1-fluoropropane?
Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 9th May Shift 1 · Q86Easy

Example 2 · Halogen Derivatives of Alkanes · Preparation of Alkyl Halides

Which among the following reactions is used for the preparation of alkyl fluorides?

Swapping the two names

Finkelstein — Iodide (NaI). Swarts — Fluoride (AgF). The pairing is asked more often than either reaction, and always with both names as options.

Concept 3 of 3: Wurtz, Fittig and Wurtz–Fittig Coupling

Sodium in dry ether pulls the halogens off two molecules and joins the carbons. Two alkyl halides — Wurtz, an alkane with an even carbon count. Two aryl halides — Fittig, a biaryl. One of each — Wurtz–Fittig, an alkylbenzene. A MIXTURE of two alkyl halides gives three alkanes, never one with more carbons than the two combined.

Definition

  • Wurtz: 2R-X+2Na→dry etherR-R+2NaX2\text{R-X} + 2\text{Na} \xrightarrow{\text{dry ether}} \text{R-R} + 2\text{NaX}. tert-Butyl bromide → 2,2,3,3-tetramethylbutane.
  • Mixture CH₃Br + C₂H₅Br: ethane, propane, butane — NOT pentane.
  • Fittig: 2C6H5Cl+2Na→C6H5-C6H52\text{C}_6\text{H}_5\text{Cl} + 2\text{Na} \to \text{C}_6\text{H}_5\text{-C}_6\text{H}_5 (biphenyl).
  • Wurtz–Fittig: C6H5Br+CH3Br+2Na→C6H5CH3\text{C}_6\text{H}_5\text{Br} + \text{CH}_3\text{Br} + 2\text{Na} \to \text{C}_6\text{H}_5\text{CH}_3 (toluene).
  • Wurtz fails for methyl (no C₂ from one carbon... it gives ethane) and gives poor yields with tertiary halides (elimination competes).

Sodium coupling

R-X+R’-X+2Na→dry etherR-R’+2NaX\text{R-X} + \text{R'-X} + 2\text{Na} \xrightarrow{\text{dry ether}} \text{R-R'} + 2\text{NaX}

Worked example

Which halide gives 2,3-dimethylbutane by the Wurtz reaction, and what does bromobenzene with ethyl bromide and sodium give?
Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2021 · May Shift 1 · Q91Moderate

Example 3 · Halogen Derivatives of Alkanes · Preparation of Alkyl Halides

Which of the following alkyl halides is treated with sodium metal to obtain 2,2,3,3-tetramethylbutane?

Calling the aryl + alkyl coupling 'Fittig'

Fittig needs TWO aryl halides and gives a biaryl. Aryl + alkyl is Wurtz–Fittig. The 2025 papers printed the name as 'Wurtzilite'; the intended reaction is the same.

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

  • From Alcohols and Alkenes

    Alcohol to halide

    R-OH→HX / PCl5 / SOCl2R-X;alkene+HX→MarkovnikovR-X\text{R-OH} \xrightarrow{\text{HX / PCl}_5\text{ / SOCl}_2} \text{R-X};\qquad \text{alkene} + \text{HX} \xrightarrow{\text{Markovnikov}} \text{R-X}
  • Finkelstein and Swarts: Swapping the Halogen

    The two exchanges

    Finkelstein: R-X+NaI→R-I;Swarts: R-X+AgF→R-F\text{Finkelstein: R-X} + \text{NaI} \to \text{R-I};\qquad \text{Swarts: R-X} + \text{AgF} \to \text{R-F}
  • Wurtz, Fittig and Wurtz–Fittig Coupling

    Sodium coupling

    R-X+R’-X+2Na→dry etherR-R’+2NaX\text{R-X} + \text{R'-X} + 2\text{Na} \xrightarrow{\text{dry ether}} \text{R-R'} + 2\text{NaX}

Watch out for (3)

Test yourself on Halogen Derivatives of Alkanes

20 past MHT-CET questions from this chapter, timed at 18 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.