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Chapter 27: Problem 14

Identify the nucleophile, electrophile, and leaving group in each of the following substitution reactions. Predict whether equilibrium favors the reactants or products: (a) \(\mathrm{CH}_{3} \mathrm{OH}+\mathrm{I}^{-} \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}+\mathrm{NaOH}\) (b) \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCl}+\mathrm{KCN} \rightleftharpoons\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CHCN}+\mathrm{KCl}\)

Short Answer

Expert verified
For reaction (a), \( I^- \) is the nucleophile, \( CH_3OH \) is the electrophile, and \( H^+ \) is the leaving group. The equilibrium favors the products. For reaction (b), \( KCN \) is the nucleophile, \( (CH_3)_2CHCl \) is the electrophile, and \( Cl^- \) ion is the leaving group. The equilibrium favors the products.

Step by step solution

01

Understanding the Key Terms

A nucleophile is a species that donates an electron pair to an electrophile to form a chemical bond in a reaction. An electrophile is a reagent attracted to electrons that participates in a chemical reaction by accepting an electron pair in order to bond to a nucleophile. The leaving group is a molecular fragment that departs with a pair of electrons in heterolytic bond cleavage.
02

Solution for Reaction (a)

In the reaction: \( CH_3OH + I^- \rightleftharpoons CH_3O^- + HI \), the \( I^- \) ion acts as the nucleophile since it donates an electron pair. The \( CH_3OH \) acts as the electrophile as it accepts an electron pair, and the \( H^+ \) from \( CH_3OH \) becomes the leaving group. The equilibrium favors the product side due to the formation of a stronger acid, \( HI \), compared to the reactant acid \( CH_3OH \).
03

Solution for Reaction (b)

In the reaction: \( (CH_3)_2CHCl + KCN \rightleftharpoons (CH_3)_2CHCN + KCl \), the \( KCN \) acts as a nucleophile, since it donates an electron pair. The \( (CH_3)_2CHCl \) acts as the electrophile as it accepts an electron pair, and \( Cl^- \) ion becomes the leaving group. Based on the principle of Le Chatelier's, when one of the products is removed, in this case, \( KCl \) which is an insoluble salt, the equilibrium shifts to the right side, thus, favors the product side.

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