Q.Write the structure of the major organic product in each of the following reactions:

The Finkelstein reaction converts alkyl chlorides/bromides to alkyl iodides using NaI, and it must be run in dry acetone so that NaCl/NaBr precipitates out and drives the reaction forward.

$\text{R-X (X = Cl, Br)} + \text{NaI} \xrightarrow{\text{dry acetone}} \text{R-I} + \text{NaX}\downarrow$

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The Finkelstein reaction converts alkyl chlorides/bromides to alkyl iodides using NaI in dry acetone.

R-Cl/R-Br + NaI $\xrightarrow{\text{dry acetone}}$ R-I + NaCl/NaBr↓

The reaction works because NaI is soluble in dry acetone while NaCl and NaBr are not — the precipitation of NaCl/NaBr drives the equilibrium (Le Chatelier's principle) toward the iodide product. This distinguishes it from Sandmeyer's reaction (diazonium salt → ary …

The Finkelstein reaction is the specific name for halide exchange using NaI in acetone, driven by NaCl's insolubility in acetone.

CH3CH2Cl + NaI --(acetone)--> CH3CH2I + NaCl

This is the Finkelstein reaction: NaI is soluble in dry acetone while the NaCl (or NaBr) byproduct is not, so it precipitates out, driving the equilibrium (Le Chatelier) towards the iodide product. (Wurtz reaction couples two alkyl halides with Na metal to form a larger alkane; Sandmeyer reaction …

The exchange of Br in an alkyl bromide for I using NaI in dry acetone is the Finkelstein reaction, a method for preparing alkyl iodides.

CH3–CH2–Br + NaI --(dry acetone)--> CH3–CH2–I + NaBr

This halogen-exchange reaction is called the Finkelstein reaction. It is used to prepare alkyl iodides from the corresponding alkyl chlorides or bromides. The reaction works because NaI is soluble in dry acetone while the NaCl/NaBr formed is not — it precipitates out, driving the equilibrium (by Le Chatelier's principle) towards …

Halogen exchange of an alkyl chloride/bromide with NaI in dry acetone to give the corresponding alkyl iodide is the textbook definition of the Finkelstein reaction — option (a).

The reaction

$CH_3-CH_2-Br + NaI \xrightarrow{\text{dry acetone}} CH_3-CH_2-I + NaBr\downarrow$

This is an $S_N2$ halide-exchange reaction. It is driven forward because $NaBr$ (and $NaCl$) are insoluble in dry acetone and precipitate out, removing them from the equilibrium (Le Chatelier's principle), while $NaI$ is soluble in acetone — the selective precipitation pulls the reaction to completion.

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Finkelstein reaction: $CH_3CH_2Br + NaI \xrightarrow{\text{dry acetone}} CH_3CH_2I + NaBr$. The missing product is ethyl iodide.

The Finkelstein reaction converts an alkyl chloride/bromide into an alkyl iodide by treatment with sodium iodide in dry acetone:

$CH_3CH_2Br + NaI \xrightarrow{\text{dry acetone}} CH_3CH_2I + NaBr$

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Two molecules of an alkyl halide react with sodium in dry ether to form a symmetrical alkane with double the number of carbons: $2R{-}X + 2Na \to R{-}R + 2NaX$.

Concept: In the Wurtz reaction, sodium abstracts the halogen from the alkyl halide, and the two alkyl groups couple to form a new C–C bond, lengthening the carbon chain.

General equation: …

Heating an alkyl halide with NaI in dry acetone replaces the halogen with iodine, precipitating NaCl (or NaBr) out of the acetone solution - this halide-exchange reaction is called the Finkelstein reaction.

CH3-CH2-Cl + NaI --(dry acetone, heat)--> CH3-CH2-I + NaCl.

This works because NaI is soluble in acetone while NaCl (and NaBr) are not, so the precipitation of NaCl continuously drives the equilibrium (Le Chatelier's principle) towards the iodide product. This general method - converting an alkyl chloride or bromide into the corresponding iodide using NaI/dry acetone - is known as the Finkelstein reaction, and is a standard laborat …