Sodium Iodide (Finkelstein) Test A solution of sodium iodide in acetone is a test for some alkyl chlorides and bromides. The mechanism is largely SN2, so primary alkyl halides react faster than secondary alkyl halides, and tertiary alkyl halides generally give no reaction.

1. Iodide ion is a good nucleophile and sodium iodide is quite soluble in acetone. On the other hand, sodium chloride and sodium bromide have low solubilities in acetone. As a result, the reaction of alkyl bromides and alkyl chlorides with NaI/acetone can serve as a simple test reaction as indicated below.

Confirmation of your product, 1-bromobutane can also be performed by reacting the product with a solution of sodium iodide in acetone. The primary alkyl halide will react by means of an SN2 mechanism with the sodium iodide to form an insoluble precipitate.

In the presence of sodium iodide in acetone, alkyl halides that are not sterically hindered (i.e., primary or secodary) react with the iodide nucleophile through an SN2 mechanism. The halogen of the alkyl halide serves as leaving group at the reactive sp3 hybridized carbon.

For SN2 reactions, NaI in acetone is chosen as iodide ion is a good nucleophile and acetone is a polar aprotic solvent, favoring a SN2 mechanism.

Thus,in 1-bromobutane, the carbon containing Bromine is 1° and in 2-bromobutane the carbon containing Bromine is 2°. So, as per the reactivity order, 2-bromobutane will react faster than 1-bromobutane in SN1 Reaction.

In 1-chlorobutane and 1-bromobutane, the leaving group was attached to a primary carbon, or primary electrophile.

All primary alkyl halides 1-chlorobutane, 1-bromobutane undergo SN2 reaction. Secondary halides: 2-chlorobutane undergoes both SN1 and SN2 reactions. Because in secondary alkyl halides, the steric hindrance is less so it favours SN2 reaction.

Sodium Iodide (Finkelstein) Test The reaction is driven by the precipitation of the NaCl or NaBr in the acetone solvent. Therefore, a positive test result is the appearance of a white cloudiness (NaX solid). Procedure: In a small test tube (13 x 100mm), add 2mL of 15%NaI in acetone solution.

In the presence of ethanolic silver nitrate, alkyl halides that can form “stable” carbocations react, through an SN1 mechanism to provide a substitution product, in this case, an ether derived from the alkyl halide and the ethanol nucleophile.

Characterization: Since 1-Chlorobutane did not react strongly it is not a Sn2 reaction even though NaI is a strong nucleophile. 1-Chlorobutane is also 1 degree in reactivity of structure so it should have a strong reaction in Sn2 but 1-Chlorbutane is not a strong leaving group. There was no reaction. Thereof, why is sn2 favored by NAI in acetone?

SN2, Sodium Iodide In acetone (NAI in Acetone. SN1, Silver Nitrate in Ethanol (AgNO3 in EtOH) With These: 2-bromobutane,= NO SN2 Reaction, but SN1 in 3 sec. 1-bromobutane= SN2 56 sec.

SN2, Sodium Iodide In Acetone (NAI In AcetoneSN1, Silver Nitrate In Ethanol (AgNO3 In EtOH)With

Finkelstein reaction is the substitution reaction between alkyl halides and NaI in acetone as solvent. The driving force for this reaction is that one of the products (NaBr or NaCl) is not soluble in acetone.