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Conversion questions in haloalkanes often hinge on one reagent: NaI in acetone, or AgF. Students who know both halogen exchange reactions pick up easy marks; students who mix them up lose them. This post explains why each reaction works, works two yield calculations and lists the mistakes to avoid.
Where is this in the syllabus?
The NCERT Class 12 chapter Haloalkanes and Haloarenes (2026-27 reprint) covers both in section 6.4.3, Halogen Exchange. The CBSE 2026-27 Chemistry curriculum lists methods of preparation of haloalkanes. The JEE (Main) 2026 syllabus and the NEET (UG) 2026 syllabus list general methods of preparation and the mechanisms of substitution reactions for organic compounds containing halogens. The JEE (Advanced) 2026 brochure lists nucleophilic substitution reactions of alkyl halides. Check the current year's documents.
Why does the Finkelstein reaction go forward?
NCERT describes it this way: alkyl chlorides or bromides react with NaI in dry acetone to give alkyl iodides.
R–Cl + NaI → R–I + NaCl
Iodide is the nucleophile and the halide ion leaves, so the exchange could run either way. The trick is the solvent. NaI dissolves in dry acetone, but, as NCERT notes, the NaCl or NaBr formed precipitates, which pushes the reaction forward by Le Chatelier's principle. If water were present the salt would dissolve and the advantage would be lost, which is why "dry" matters.
The reaction is named after Hans Finkelstein, whose 1910 paper described preparing organic iodides from the corresponding bromides and chlorides. Because the exchange is a nucleophilic substitution with an SN2 mechanism, NCERT's SN2 reactivity order applies: primary halide > secondary halide > tertiary halide. Our post on SN1 and SN2 in haloalkanes explains why.
Worked example 1: a Finkelstein yield
Sample numbers: 13.7 g of 1-bromobutane is stirred with 1.5 equivalents of NaI in dry acetone.
Moles of 1-bromobutane (M = 137.0 g/mol): 13.7 ÷ 137.0 = 0.100 mol.
NaI needed (M = 149.9 g/mol): 1.5 × 0.100 × 149.9 ≈ 22.5 g. The excess also helps push the equilibrium.
Theoretical 1-iodobutane (M = 184.0 g/mol): 0.100 × 184.0 = 18.4 g.
Most NaBr that can precipitate (M = 102.9 g/mol): about 10.3 g.
If 15.6 g of 1-iodobutane is collected, the percentage yield is 15.6 ÷ 18.4 × 100 ≈ 84.8%.
How is the Swarts reaction different?
To make an alkyl fluoride, NCERT says to heat an alkyl chloride or bromide with a metallic fluoride such as AgF, Hg₂F₂, CoF₂ or SbF₃. The fluoride comes from the metal salt, not from NaF in acetone. Frédéric Swarts studied organic fluorine compounds; his 1901 memoir is titled Contribution to the study of organic fluorine compounds.
Worked example 2: two Swarts calculations
Bromoethane with silver fluoride: CH₃CH₂Br + AgF → CH₃CH₂F + AgBr.
10.9 g of bromoethane (M = 109.0 g/mol) is 0.100 mol.
It needs 0.100 × 126.9 ≈ 12.7 g of AgF.
It gives at most 0.100 × 48.06 ≈ 4.81 g of fluoroethane.
Freon 12. NCERT says CCl₂F₂ is manufactured from tetrachloromethane by the Swarts reaction. An idealised equation with SbF₃ is:
3CCl₄ + 2SbF₃ → 3CCl₂F₂ + 2SbCl₃
Check the balance: 12 Cl, 6 F and 2 Sb on each side. For 100 g of CCl₂F₂ (M = 120.9 g/mol), that is 0.827 mol, needing 0.827 × 2/3 = 0.551 mol of SbF₃ (about 98.6 g) and 0.827 mol of CCl₄ (about 127 g). NCERT's same chapter explains why freons that reach the stratosphere damage the ozone layer.
Which mistakes cost marks?
Writing NaI in water. The driving force depends on NaCl or NaBr being insoluble in dry acetone.
Using Finkelstein on a tertiary halide. SN2 is slowest there.
Using either reaction on chlorobenzene. NCERT notes that aryl halides are extremely less reactive towards nucleophilic substitution.
Inventing reagents. Stick to the fluorides NCERT names.
Leaving out the inorganic product (NaCl, NaBr, AgBr, SbCl₃), which you need for balancing and for the Le Chatelier argument.
A practice routine
Answer NCERT Exercise 6.14(i): the product of CH₃CH₂CH₂Cl with NaI.
Write a two-step route from propan-1-ol to 1-iodopropane (an alcohol to chloride step, then Finkelstein).
Redo worked example 1 for 9.25 g of 1-chlorobutane (M = 92.6 g/mol). You should get a theoretical 18.4 g of 1-iodobutane again, since both start at 0.100 mol.
Add both reactions to a reagent card; our list of named organic reactions shows a format.
When a conversion chain goes wrong and you cannot see which step failed, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer.
Today, write both equations from memory with their inorganic products, then check them against NCERT section 6.4.3.
