What is Finkelstein Reaction in Chemistry?
Finkelstein Reaction is essential in chemistry and helps students understand various practical and theoretical applications related to this topic. It is a key named reaction for halide exchange and is especially important for Class 12 curriculum and competitive exam preparation.
What is Finkelstein Reaction in Chemistry?
A Finkelstein reaction refers to a nucleophilic substitution (SN2 type) reaction in which a halide ion (usually iodide) replaces another halogen (chloride or bromide) atom in an alkyl halide, typically using sodium iodide in acetone.
This concept appears in chapters related to organic compound conversions, alkyl halides, and substitution reactions, making it a foundational part of your chemistry syllabus.
Molecular Formula and Composition
The Finkelstein reaction does not have a single molecular formula as it is a reaction process. It normally involves a primary or secondary alkyl halide (R–Cl or R–Br) reacting with NaI in acetone.
Here, R is the alkyl group, X is Cl or Br, and iodide acts as the nucleophile. This reaction is categorized under nucleophilic substitution reactions (SN2).
Preparation and Synthesis Methods
The Finkelstein reaction is commonly used in organic synthesis to prepare alkyl iodides from the more readily available alkyl chlorides or bromides.
In the laboratory, this synthesis is achieved by mixing the alkyl chloride/bromide with sodium iodide in dry acetone. The insolubility of NaCl/NaBr in acetone drives the reaction forward.
Physical Properties of Finkelstein Reaction
Since the Finkelstein reaction is not a pure substance, but a reaction, it does not have unique physical properties. However, the reaction occurs best in a polar aprotic solvent like acetone.
The use of dry acetone prevents reverse reaction and precipitation of the by-product (NaCl or NaBr) as a solid assists in shifting equilibrium towards the product (alkyl iodide).
Chemical Properties and Reactions
The chemical property highlighted by the Finkelstein reaction is the SN2 mechanism. It is sensitive to steric hindrance, favoring primary alkyl halides. Aromatic halides usually do not react unless catalyzed. The halide exchange can be summarized as:
R–X + NaI →(acetone) R–I + NaX
Where X = Cl or Br, R = Alkyl group. The reaction is reversible, but the precipitation of NaCl or NaBr in acetone makes it effectively go forward.
Frequent Related Errors
- Confusing Finkelstein Reaction with Swarts reaction or with SN1 mechanism.
- Using tertiary or neopentyl alkyl halides, which do not react efficiently in this method.
- Ignoring the role of acetone as a polar aprotic solvent that helps drive the reaction.
Uses of Finkelstein Reaction in Real Life
Finkelstein reaction is widely used to prepare alkyl iodides for pharmaceutical synthesis, organic compound conversion, and laboratory analysis. It serves as a test for classifying alkyl halides, and enables the production of alkyl iodides, which are valuable intermediates in many industrial syntheses.
Relation with Other Chemistry Concepts
Finkelstein reaction is closely related to SN2 mechanism and nucleophilic substitution reactions. It is also contrasted with Swarts reaction, which uses different reagents for halide exchange (often to prepare alkyl fluorides).
Step-by-Step Reaction Example
- Setup: Start with methyl bromide (CH₃Br) and sodium iodide (NaI) in dry acetone.
Write the balanced equation:
CH₃Br + NaI →(acetone)→ CH₃I + NaBr↓ - Mechanism: The I- ion attacks the methyl carbon, displacing Br- in a single concerted SN2 step.
Reaction occurs with inversion of configuration (if the carbon is chiral). NaBr precipitates due to being insoluble in acetone, driving the reaction forward.
Lab or Experimental Tips
Remember the Finkelstein reaction by the hint: "Iodide in, chloride/bromide out, acetone essential." Vedantu educators often suggest remembering that alkyl iodides are only efficiently prepared when the leaving halide is Cl or Br, not vice versa.
Try This Yourself
- Write the IUPAC name for CH₃I formed in Finkelstein reaction.
- List two reasons why secondary alkyl halides are less reactive in Finkelstein Reaction compared to primary alkyl halides.
- Give another real-life example of an alkyl halide conversion using Finkelstein method.
Final Wrap-Up
We explored Finkelstein Reaction—its definition, SN2-based mechanism, examples, comparison to related reactions, and applications. Mastering this reaction helps in understanding substitution processes essential for board exams and competitive tests.
Related Topics on Vedantu: Swarts Reaction, Substitution Reaction
FAQs on Finkelstein Reaction Explained: Mechanism, Examples & Exam Tips
1. What is the Finkelstein reaction in chemistry?
The Finkelstein reaction is a halogen exchange method used primarily to synthesise alkyl iodides. It involves treating an alkyl chloride or alkyl bromide with a solution of sodium iodide (NaI) dissolved in a polar aprotic solvent, typically dry acetone. This reaction proceeds via an SN2 mechanism.
2. Which chapter in the Class 12 NCERT syllabus covers the Finkelstein reaction?
The Finkelstein reaction is a key topic in Chapter 6: Haloalkanes and Haloarenes of the Class 12 Chemistry NCERT syllabus for the 2025-26 session. It is presented as a primary method for the preparation of haloalkanes through halogen exchange.
3. What is the mechanism of the Finkelstein reaction?
The Finkelstein reaction follows a bimolecular nucleophilic substitution (SN2) mechanism. This occurs in a single, concerted step where the iodide ion (I⁻), a strong nucleophile, attacks the carbon atom bonded to the halogen from the backside, simultaneously displacing the chloride (Cl⁻) or bromide (Br⁻) ion as the leaving group.
4. Why is dry acetone used as the solvent in the Finkelstein reaction?
Dry acetone is the preferred solvent for two main reasons based on solubility:
- It readily dissolves the reactant, sodium iodide (NaI), allowing the iodide ion to act as an effective nucleophile.
- However, it does not dissolve the by-products, sodium chloride (NaCl) or sodium bromide (NaBr). These salts precipitate out of the solution as they are formed.
5. Provide a typical example of the Finkelstein reaction.
A classic example is the conversion of ethyl bromide to ethyl iodide. The reaction is written as:
CH₃CH₂Br + NaI (in acetone) → CH₃CH₂I + NaBr (s)
Here, ethyl bromide reacts with sodium iodide in acetone. The product ethyl iodide remains dissolved, while the by-product, sodium bromide (NaBr), precipitates as a solid.
6. What is the main difference between the Finkelstein and Swarts reactions?
Both are halogen exchange reactions, but they synthesise different types of alkyl halides.
- Finkelstein Reaction: Used to prepare alkyl iodides from alkyl chlorides or bromides using sodium iodide (NaI) in acetone.
- Swarts Reaction: Used to prepare alkyl fluorides from alkyl chlorides or bromides using metallic fluorides like AgF, Hg₂F₂, or SbF₃.
7. Why does the Finkelstein reaction fail with tertiary alkyl halides?
The Finkelstein reaction fails with tertiary alkyl halides due to steric hindrance. The bulky alkyl groups surrounding the tertiary carbon atom physically block the incoming iodide nucleophile from performing the backside attack required for the SN2 mechanism. Instead of substitution, elimination reactions often become the major, unwanted pathway.
8. Can aryl halides like chlorobenzene undergo the Finkelstein reaction?
No, aryl halides like chlorobenzene do not undergo the Finkelstein reaction under normal conditions. The C-X bond in aryl halides has a partial double bond character due to resonance with the benzene ring, making it much stronger and harder to break. Furthermore, the electron-rich ring repels the incoming nucleophile, making an SN2 attack impossible.
9. What is the role of Le Chatelier's principle in the Finkelstein reaction?
Le Chatelier's principle is crucial for the success of the Finkelstein reaction. The reaction is reversible: R-X + NaI ⇌ R-I + NaX. The key is that the by-product, sodium halide (NaX, where X = Cl or Br), is insoluble in the acetone solvent and precipitates out. By continuously removing a product (NaX) from the reaction mixture, the equilibrium is forced to shift to the right, favouring the formation of more alkyl iodide (R-I) and ensuring the reaction proceeds to completion.
