Question

In Finkelstein reaction when acetone is replaced by water then

• A. Reaction occur in forward direction via $S_{N}1$ pathway
• B. Reaction occur in forward direction via $S_{N}2$ pathway
• C. Reaction occur in backward because $NaCl$ formed in right hand side is soluble in water and cannot precipitate out
• D. Reaction occur in forward because $NaCl$ formed in right hand side is soluble in water and can precipitate out

Answer 1

The correct option is C Reaction occur in backward because $NaCl$ formed in right hand side is soluble in water and cannot precipitate out

Finkelstein reactions occurs in the presence of dry acetone because NaI is more soluble in dry acetone whereas NaCl and NaBr are precipitated out. Thus, increasing of concentration of $I^{-}$ ion in the solution which favour the reation to proceed in forward direction. $\underset{\text{Ethyl Bromide}}{C_2{H}_{5}Br}+NaI\underset{\text{Heat}}{\overset{\text{Acetone}}{\to }}\underset{\text{Ethyl Iodide}}{C_2{H}_{5}I}+NaBr\downarrow$
If we replace acetone with water then the $NaCl$ formed in right hand side is soluble in water and due to common ion effect of $X^{-}$, it cannot precipitate out and reaction occur in backward according to Le-chatlier principle.